3.7.80 \(\int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx\) [680]

Optimal. Leaf size=719 \[ \frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a \cos ^2(c+d x)}{8 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{3/2} e^{7/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{8 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{3/2} e^{7/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{8 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{3/2} e^{7/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{16 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{3/2} e^{7/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{16 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}} \]

[Out]

1/3*I*a/d/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(1/2)+5/8*I*a*cos(d*x+c)^2/d/(e*cos(d*x+c))^(7/2)/(a+I*a*tan
(d*x+c))^(1/2)-5/16*I*a^(3/2)*e^(7/2)*arctan(1-2^(1/2)*e^(1/2)*(a-I*a*tan(d*x+c))^(1/2)/a^(1/2)/(e*sec(d*x+c))
^(1/2))*sec(d*x+c)/d/(e*cos(d*x+c))^(7/2)/(e*sec(d*x+c))^(7/2)*2^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x
+c))^(1/2)+5/16*I*a^(3/2)*e^(7/2)*arctan(1+2^(1/2)*e^(1/2)*(a-I*a*tan(d*x+c))^(1/2)/a^(1/2)/(e*sec(d*x+c))^(1/
2))*sec(d*x+c)/d/(e*cos(d*x+c))^(7/2)/(e*sec(d*x+c))^(7/2)*2^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))
^(1/2)+5/32*I*a^(3/2)*e^(7/2)*ln(a-2^(1/2)*a^(1/2)*e^(1/2)*(a-I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(1/2)+cos(d
*x+c)*(a-I*a*tan(d*x+c)))*sec(d*x+c)/d/(e*cos(d*x+c))^(7/2)/(e*sec(d*x+c))^(7/2)*2^(1/2)/(a-I*a*tan(d*x+c))^(1
/2)/(a+I*a*tan(d*x+c))^(1/2)-5/32*I*a^(3/2)*e^(7/2)*ln(a+2^(1/2)*a^(1/2)*e^(1/2)*(a-I*a*tan(d*x+c))^(1/2)/(e*s
ec(d*x+c))^(1/2)+cos(d*x+c)*(a-I*a*tan(d*x+c)))*sec(d*x+c)/d/(e*cos(d*x+c))^(7/2)/(e*sec(d*x+c))^(7/2)*2^(1/2)
/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-5/12*I*cos(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/d/(e*cos(d*x+c
))^(7/2)

________________________________________________________________________________________

Rubi [A]
time = 0.58, antiderivative size = 719, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {3596, 3579, 3582, 3580, 3576, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {5 i a^{3/2} e^{7/2} \sec (c+d x) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}+\frac {5 i a^{3/2} e^{7/2} \sec (c+d x) \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}+\frac {5 i a^{3/2} e^{7/2} \sec (c+d x) \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}-\frac {5 i a^{3/2} e^{7/2} \sec (c+d x) \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{16 \sqrt {2} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}}+\frac {5 i a \cos ^2(c+d x)}{8 d \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2}}+\frac {i a}{3 d \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Tan[c + d*x]]/(e*Cos[c + d*x])^(7/2),x]

[Out]

((I/3)*a)/(d*(e*Cos[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((5*I)/8)*a*Cos[c + d*x]^2)/(d*(e*Cos[c + d
*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/8)*a^(3/2)*e^(7/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*T
an[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(Sqrt[2]*d*(e*Cos[c + d*x])^(7/2)*(e*Sec[c + d*x])
^(7/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (((5*I)/8)*a^(3/2)*e^(7/2)*ArcTan[1 + (Sqrt[2]
*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(Sqrt[2]*d*(e*Cos[c + d*x])
^(7/2)*(e*Sec[c + d*x])^(7/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (((5*I)/16)*a^(3/2)*e^(
7/2)*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a
*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d*(e*Cos[c + d*x])^(7/2)*(e*Sec[c + d*x])^(7/2)*Sqrt[a - I*a*Tan[c + d*
x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/16)*a^(3/2)*e^(7/2)*Log[a + (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[a - I*a*Ta
n[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d*(e*Cos[c + d
*x])^(7/2)*(e*Sec[c + d*x])^(7/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/12)*Cos[c +
d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*(e*Cos[c + d*x])^(7/2))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3576

Int[Sqrt[(d_.)*sec[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-4*b*(d^
2/f), Subst[Int[x^2/(a^2 + d^2*x^4), x], x, Sqrt[a + b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]]], x] /; FreeQ[{a, b,
d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rule 3579

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3580

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[d*(Sec
[e + f*x]/(Sqrt[a - b*Tan[e + f*x]]*Sqrt[a + b*Tan[e + f*x]])), Int[Sqrt[d*Sec[e + f*x]]*Sqrt[a - b*Tan[e + f*
x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rule 3582

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d^2*(
d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Dist[d^2*((m - 2)/(a*(m + n - 1
))), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2
+ b^2, 0] && LtQ[n, 0] && GtQ[m, 1] &&  !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx &=\frac {\int (e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)} \, dx}{(e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=\frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {(5 a) \int \frac {(e \sec (c+d x))^{7/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{6 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=\frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}}+\frac {\left (5 e^2\right ) \int (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx}{8 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=\frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a \cos ^2(c+d x)}{8 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}}+\frac {\left (5 a e^2\right ) \int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{16 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=\frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a \cos ^2(c+d x)}{8 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}}+\frac {\left (5 a e^3 \sec (c+d x)\right ) \int \sqrt {e \sec (c+d x)} \sqrt {a-i a \tan (c+d x)} \, dx}{16 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a \cos ^2(c+d x)}{8 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}}+\frac {\left (5 i a^2 e^5 \sec (c+d x)\right ) \text {Subst}\left (\int \frac {x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{4 d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a \cos ^2(c+d x)}{8 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}}-\frac {\left (5 i a^2 e^4 \sec (c+d x)\right ) \text {Subst}\left (\int \frac {a-e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{8 d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (5 i a^2 e^4 \sec (c+d x)\right ) \text {Subst}\left (\int \frac {a+e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{8 d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a \cos ^2(c+d x)}{8 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}}+\frac {\left (5 i a^2 e^3 \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{16 d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (5 i a^2 e^3 \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{16 d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (5 i a^{3/2} e^{7/2} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}+2 x}{-\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{16 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (5 i a^{3/2} e^{7/2} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}-2 x}{-\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{16 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a \cos ^2(c+d x)}{8 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{3/2} e^{7/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{16 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{3/2} e^{7/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{16 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}}+\frac {\left (5 i a^{3/2} e^{7/2} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (5 i a^{3/2} e^{7/2} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=\frac {i a}{3 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a \cos ^2(c+d x)}{8 d (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{3/2} e^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{8 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{3/2} e^{7/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{8 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {5 i a^{3/2} e^{7/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{16 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^{3/2} e^{7/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{16 \sqrt {2} d (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d (e \cos (c+d x))^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 2.96, size = 305, normalized size = 0.42 \begin {gather*} \frac {\sqrt {\cos (c+d x)} \left (-\frac {40}{3} i \cos ^{\frac {3}{2}}(c+d x)+\frac {5}{8} i e^{-\frac {7}{2} i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^3 \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (2 \text {ArcTan}\left (1-\sqrt {2} e^{\frac {1}{2} i (c+d x)}\right )-2 \text {ArcTan}\left (1+\sqrt {2} e^{\frac {1}{2} i (c+d x)}\right )+\log \left (1-\sqrt {2} e^{\frac {1}{2} i (c+d x)}+e^{i (c+d x)}\right )-\log \left (1+\sqrt {2} e^{\frac {1}{2} i (c+d x)}+e^{i (c+d x)}\right )\right )+\frac {32}{3} \sqrt {\cos (c+d x)} (i \cos (c+d x)+\sin (c+d x))+20 \cos ^{\frac {5}{2}}(c+d x) (i \cos (c+d x)+\sin (c+d x))\right ) \sqrt {a+i a \tan (c+d x)}}{32 d (e \cos (c+d x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*a*Tan[c + d*x]]/(e*Cos[c + d*x])^(7/2),x]

[Out]

(Sqrt[Cos[c + d*x]]*(((-40*I)/3)*Cos[c + d*x]^(3/2) + (((5*I)/8)*(1 + E^((2*I)*(c + d*x)))^3*Sqrt[(1 + E^((2*I
)*(c + d*x)))/E^(I*(c + d*x))]*(2*ArcTan[1 - Sqrt[2]*E^((I/2)*(c + d*x))] - 2*ArcTan[1 + Sqrt[2]*E^((I/2)*(c +
d*x))] + Log[1 - Sqrt[2]*E^((I/2)*(c + d*x)) + E^(I*(c + d*x))] - Log[1 + Sqrt[2]*E^((I/2)*(c + d*x)) + E^(I*
(c + d*x))]))/E^(((7*I)/2)*(c + d*x)) + (32*Sqrt[Cos[c + d*x]]*(I*Cos[c + d*x] + Sin[c + d*x]))/3 + 20*Cos[c +
d*x]^(5/2)*(I*Cos[c + d*x] + Sin[c + d*x]))*Sqrt[a + I*a*Tan[c + d*x]])/(32*d*(e*Cos[c + d*x])^(7/2))

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Maple [A]
time = 1.10, size = 417, normalized size = 0.58

method	result	size
default	\(\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right )^{4} \left (30 i \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-15 i \left (\cos ^{3}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right )+15 i \left (\cos ^{3}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )+20 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-30 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}+15 \left (\cos ^{3}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right )+15 \left (\cos ^{3}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )+16 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-10 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}+4 \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )-16 \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\right )}{48 d \sin \left (d x +c \right )^{7} \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \left (\frac {1}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \left (e \cos \left (d x +c \right )\right )^{\frac {7}{2}}}\)	\(417\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/48/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^4*(30*I*sin(d*x+c)*cos(d*x+c)
^2*(1/(1+cos(d*x+c)))^(1/2)-15*I*cos(d*x+c)^3*arctanh(1/2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))+
15*I*cos(d*x+c)^3*arctanh(1/2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+20*I*sin(d*x+c)*cos(d*x+c)*(
1/(1+cos(d*x+c)))^(1/2)-30*cos(d*x+c)^3*(1/(1+cos(d*x+c)))^(1/2)+15*cos(d*x+c)^3*arctanh(1/2*(1/(1+cos(d*x+c))
)^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))+15*cos(d*x+c)^3*arctanh(1/2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+
c)))+16*I*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)-10*cos(d*x+c)^2*(1/(1+cos(d*x+c)))^(1/2)+4*(1/(1+cos(d*x+c)))^(1
/2)*cos(d*x+c)-16*(1/(1+cos(d*x+c)))^(1/2))/sin(d*x+c)^7/(I*sin(d*x+c)+cos(d*x+c)-1)/(1/(1+cos(d*x+c)))^(7/2)/
(e*cos(d*x+c))^(7/2)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2638 vs. \(2 (469) = 938\).
time = 0.85, size = 2638, normalized size = 3.67 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-192*(30*(sqrt(2)*cos(6*d*x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)*sin(6
*d*x + 6*c) + 3*I*sqrt(2)*sin(4*d*x + 4*c) + 3*I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*arctan2(sqrt(2)*cos(1/4*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))
+ 1) + 30*(sqrt(2)*cos(6*d*x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)*sin(
6*d*x + 6*c) + 3*I*sqrt(2)*sin(4*d*x + 4*c) + 3*I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*arctan2(sqrt(2)*cos(1/4*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))
) + 1) + 30*(sqrt(2)*cos(6*d*x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)*si
n(6*d*x + 6*c) + 3*I*sqrt(2)*sin(4*d*x + 4*c) + 3*I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*arctan2(sqrt(2)*cos(1/
4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
)) + 1) + 30*(sqrt(2)*cos(6*d*x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)*s
in(6*d*x + 6*c) + 3*I*sqrt(2)*sin(4*d*x + 4*c) + 3*I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*arctan2(sqrt(2)*cos(1
/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c))) + 1) - 30*(-I*sqrt(2)*cos(6*d*x + 6*c) - 3*I*sqrt(2)*cos(4*d*x + 4*c) - 3*I*sqrt(2)*cos(2*d*x + 2*c) + sq
rt(2)*sin(6*d*x + 6*c) + 3*sqrt(2)*sin(4*d*x + 4*c) + 3*sqrt(2)*sin(2*d*x + 2*c) - I*sqrt(2))*arctan2(sqrt(2)*
sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))), s
qrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c))) + 1) - 30*(I*sqrt(2)*cos(6*d*x + 6*c) + 3*I*sqrt(2)*cos(4*d*x + 4*c) + 3*I*sqrt(2)*cos(2*d*x + 2*c) - sqr
t(2)*sin(6*d*x + 6*c) - 3*sqrt(2)*sin(4*d*x + 4*c) - 3*sqrt(2)*sin(2*d*x + 2*c) + I*sqrt(2))*arctan2(-sqrt(2)*
sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))), -
sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c))) + 1) + 15*(sqrt(2)*cos(6*d*x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2
)*sin(6*d*x + 6*c) + 3*I*sqrt(2)*sin(4*d*x + 4*c) + 3*I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*log(2*sqrt(2)*sin(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*(sqr
t(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 15*(sqrt(2)*cos(6*d*
x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)*sin(6*d*x + 6*c) + 3*I*sqrt(2)*
sin(4*d*x + 4*c) + 3*I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*log(-2*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), co
s(2*d*x + 2*c)))*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2*(sqrt(2)*cos(1/4*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) - 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c)))^2 + 2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*co
s(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 15*(-I*sqrt(2)*cos(6*d*x + 6*c) - 3*I*sqrt(2)*cos(4*
d*x + 4*c) - 3*I*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2)*sin(6*d*x + 6*c) + 3*sqrt(2)*sin(4*d*x + 4*c) + 3*sqrt(2)*
sin(2*d*x + 2*c) - I*sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*s
qrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - 15*(I*sqrt(2)*cos(6*d*x + 6*c) + 3*I*sqrt(2
)*cos(4*d*x + 4*c) + 3*I*sqrt(2)*cos(2*d*x + 2*c) - sqrt(2)*sin(6*d*x + 6*c) - 3*sqrt(2)*sin(4*d*x + 4*c) - 3*
sqrt(2)*sin(2*d*x + 2*c) + I*sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
)) - 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - 15*(-I*sqrt(2)*cos(6*d*x + 6*c) - 3
*I*sqrt(2)*cos(4*d*x + 4*c) - 3*I*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2)*sin(6*d*x + 6*c) + 3*sqrt(2)*sin(4*d*x +
4*c) + 3*sqrt(2)*sin(2*d*x + 2*c) - I*sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 +
2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d
*x + 2*c))) + 2*sqrt(2)*sin(1/4*arctan2(sin(2*d...

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Fricas [A]
time = 0.34, size = 600, normalized size = 0.83 \begin {gather*} \frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-5 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 42 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 15 i \, e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - 6 \, {\left (d e^{\frac {7}{2}} + d e^{\left (6 i \, d x + 6 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {7}{2}\right )}\right )} \sqrt {\frac {25 i \, a e^{\left (-7\right )}}{64 \, d^{2}}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + \frac {8}{5} \, d \sqrt {\frac {25 i \, a e^{\left (-7\right )}}{64 \, d^{2}}} e^{\frac {7}{2}}\right ) + 6 \, {\left (d e^{\frac {7}{2}} + d e^{\left (6 i \, d x + 6 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {7}{2}\right )}\right )} \sqrt {\frac {25 i \, a e^{\left (-7\right )}}{64 \, d^{2}}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - \frac {8}{5} \, d \sqrt {\frac {25 i \, a e^{\left (-7\right )}}{64 \, d^{2}}} e^{\frac {7}{2}}\right ) + 6 \, {\left (d e^{\frac {7}{2}} + d e^{\left (6 i \, d x + 6 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {7}{2}\right )}\right )} \sqrt {-\frac {25 i \, a e^{\left (-7\right )}}{64 \, d^{2}}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + \frac {8}{5} \, d \sqrt {-\frac {25 i \, a e^{\left (-7\right )}}{64 \, d^{2}}} e^{\frac {7}{2}}\right ) - 6 \, {\left (d e^{\frac {7}{2}} + d e^{\left (6 i \, d x + 6 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {7}{2}\right )}\right )} \sqrt {-\frac {25 i \, a e^{\left (-7\right )}}{64 \, d^{2}}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - \frac {8}{5} \, d \sqrt {-\frac {25 i \, a e^{\left (-7\right )}}{64 \, d^{2}}} e^{\frac {7}{2}}\right )}{12 \, {\left (d e^{\frac {7}{2}} + d e^{\left (6 i \, d x + 6 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c + \frac {7}{2}\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {7}{2}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/12*(sqrt(2)*sqrt(1/2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-5*I*e^(5*I*d*x + 5*I*c) + 42*I*e^(3*I*d*x + 3*I*c)
+ 15*I*e^(I*d*x + I*c))*sqrt(e^(2*I*d*x + 2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c) - 6*(d*e^(7/2) + d*e^(6*I*d*x
+ 6*I*c + 7/2) + 3*d*e^(4*I*d*x + 4*I*c + 7/2) + 3*d*e^(2*I*d*x + 2*I*c + 7/2))*sqrt(25/64*I*a*e^(-7)/d^2)*log
(sqrt(2)*sqrt(1/2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e^(2*I*d*x + 2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c) +
8/5*d*sqrt(25/64*I*a*e^(-7)/d^2)*e^(7/2)) + 6*(d*e^(7/2) + d*e^(6*I*d*x + 6*I*c + 7/2) + 3*d*e^(4*I*d*x + 4*I*
c + 7/2) + 3*d*e^(2*I*d*x + 2*I*c + 7/2))*sqrt(25/64*I*a*e^(-7)/d^2)*log(sqrt(2)*sqrt(1/2)*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1))*sqrt(e^(2*I*d*x + 2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c) - 8/5*d*sqrt(25/64*I*a*e^(-7)/d^2)*e^(7/
2)) + 6*(d*e^(7/2) + d*e^(6*I*d*x + 6*I*c + 7/2) + 3*d*e^(4*I*d*x + 4*I*c + 7/2) + 3*d*e^(2*I*d*x + 2*I*c + 7/
2))*sqrt(-25/64*I*a*e^(-7)/d^2)*log(sqrt(2)*sqrt(1/2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e^(2*I*d*x + 2*I*
c) + 1)*e^(-1/2*I*d*x - 1/2*I*c) + 8/5*d*sqrt(-25/64*I*a*e^(-7)/d^2)*e^(7/2)) - 6*(d*e^(7/2) + d*e^(6*I*d*x +
6*I*c + 7/2) + 3*d*e^(4*I*d*x + 4*I*c + 7/2) + 3*d*e^(2*I*d*x + 2*I*c + 7/2))*sqrt(-25/64*I*a*e^(-7)/d^2)*log(
sqrt(2)*sqrt(1/2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e^(2*I*d*x + 2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c) - 8
/5*d*sqrt(-25/64*I*a*e^(-7)/d^2)*e^(7/2)))/(d*e^(7/2) + d*e^(6*I*d*x + 6*I*c + 7/2) + 3*d*e^(4*I*d*x + 4*I*c +
7/2) + 3*d*e^(2*I*d*x + 2*I*c + 7/2))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(1/2)/(e*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*e^(-7/2)/cos(d*x + c)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(1/2)/(e*cos(c + d*x))^(7/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^(1/2)/(e*cos(c + d*x))^(7/2), x)

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