\(\int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [423]

Optimal result

Integrand size = 30, antiderivative size = 529 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i e^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{\sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {3 i e^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right ) \sec (c+d x)}{\sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {3 i e^{7/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i e^{7/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-I*e^2*(e*sec(d*x+c))^(3/2)/a/d/(a+I*a*tan(d*x+c))^(1/2)-3/2*I*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*2^(1/2)/a^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+
c))^(1/2)+3/2*I*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*2^(1/2)/a^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+3/4*I*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*2^(1/2)/a^(1/2
)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-3/4*I*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*2^(1/2)/a^(1/2)/(a-I*a*tan(d*x+c))^
(1/2)/(a+I*a*tan(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3581, 3579, 3580, 3576, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {3 i e^{7/2} \sec (c+d x) \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {3 i e^{7/2} \sec (c+d x) \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {3 i e^{7/2} \sec (c+d x) \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i e^{7/2} \sec (c+d x) \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Int[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-I)*e^2*(e*Sec[c + d*x])^(3/2))/(a*d*Sqrt[a + I*a*Tan[c + d*x]]) - ((3*I)*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]*Sqrt[a]*d*Sqrt[a - I*a*Ta
n[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + ((3*I)*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]*Sqrt[a]*d*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Ta
n[c + d*x]]) + (((3*I)/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]*Sqrt[a]*d*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[
a + I*a*Tan[c + d*x]]) - (((3*I)/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]*Sqrt[a]*d*Sqrt[a - I*a*Tan[c + d
*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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 3581

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

Rubi steps \begin{align*} \text {integral}& = -\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 e^2\right ) \int (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx}{a^2} \\ & = -\frac {i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 e^2\right ) \int \frac {(e \sec (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{2 a} \\ & = -\frac {i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 e^3 \sec (c+d x)\right ) \int \sqrt {e \sec (c+d x)} \sqrt {a-i a \tan (c+d x)} \, dx}{2 a \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = -\frac {i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (6 i 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 )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = -\frac {i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}}-\frac {\left (3 i 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 )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i 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 )}{d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = -\frac {i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i 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 )}{2 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i 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 )}{2 d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i 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 )}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i 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 )}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = -\frac {i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}}+\frac {3 i 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)}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i 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)}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (3 i 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 )}{\sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (3 i 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 )}{\sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = -\frac {i e^2 (e \sec (c+d x))^{3/2}}{a d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i e^{7/2} \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)}{\sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {3 i e^{7/2} \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)}{\sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {3 i 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)}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {3 i 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)}{2 \sqrt {2} \sqrt {a} d \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.66 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.64 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {e (e \sec (c+d x))^{5/2} \left (-i \cos (c+d x)+\sin (c+d x)+\frac {3 \cos (c+d x) (\cos (c)+i \sin (c)) \left (\text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)}-\text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)}\right ) (\cos (d x)+i \sin (d x))^2 \sqrt {i+\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}\right )}{d (a+i a \tan (c+d x))^{3/2}} \]

[In]

Integrate[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(e*(e*Sec[c + d*x])^(5/2)*((-I)*Cos[c + d*x] + Sin[c + d*x] + (3*Cos[c + d*x]*(Cos[c] + I*Sin[c])*(ArcTanh[(Sq
rt[1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[
-1 - I*Cos[c] - Sin[c]]*Sqrt[1 + I*Cos[c] - Sin[c]] - ArcTanh[(Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/
2]])/(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt[-1 + I*Cos[c] + S
in[c]])*(Cos[d*x] + I*Sin[d*x])^2*Sqrt[I + Tan[(d*x)/2]])/(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[-1 + I*Cos[c] + S
in[c]]*Sqrt[I - Tan[(d*x)/2]])))/(d*(a + I*a*Tan[c + d*x])^(3/2))

Maple [A] (verified)

Time = 16.85 (sec) , antiderivative size = 712, normalized size of antiderivative = 1.35

method	result	size
default	\(\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \sqrt {e \sec \left (d x +c \right )}\, e^{3} \left (6 i \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+2 i \tan \left (d x +c \right ) \sec \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-3 i \sec \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 i \tan \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 i \sec \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-6 \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-6 \,\operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-3 i \tan \left (d x +c \right ) \operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-6 i \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-2 \tan \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+3 i \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-3 \tan \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-3 \,\operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-2 \tan \left (d x +c \right ) \sec \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 \sec \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+3 \sec \left (d x +c \right ) \operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{d \left (-\tan \left (d x +c \right )+i\right ) a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right )}\)	\(712\)

[In]

int((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

(1/4-1/4*I)/d*(e*sec(d*x+c))^(1/2)*e^3/(-tan(d*x+c)+I)/a/(a*(1+I*tan(d*x+c)))^(1/2)/(1/(cos(d*x+c)+1))^(1/2)/(
cos(d*x+c)+1)*(6*I*cos(d*x+c)*arctanh(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))+2
*I*tan(d*x+c)*sec(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)-3*I*sec(d*x+c)*arctanh(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*
x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))+2*I*tan(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)+2*I*(1/(cos(d*x+c)+1))^(1/2)+2*I*sec
(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)-6*sin(d*x+c)*arctanh(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(1/(cos(d*x
+c)+1))^(1/2))-6*arctanh(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-3*
I*tan(d*x+c)*arctanh(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))-6*I*sin(d*x+c)*ar
ctanh(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))-2*tan(d*x+c)*(1/(cos(d*x+c)+1))^
(1/2)+2*(1/(cos(d*x+c)+1))^(1/2)+3*I*arctanh(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(
1/2))-3*tan(d*x+c)*arctanh(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))-3*arctanh(1/
2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))-2*tan(d*x+c)*sec(d*x+c)*(1/(cos(d*x+c)+1
))^(1/2)+2*sec(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)+3*sec(d*x+c)*arctanh(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)
+1)/(1/(cos(d*x+c)+1))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 483, normalized size of antiderivative = 0.91 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {-4 i \, e^{3} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{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 )} - \sqrt {\frac {9 i \, e^{7}}{a^{3} d^{2}}} a^{2} d \log \left (-\frac {2 \, {\left (i \, \sqrt {\frac {9 i \, e^{7}}{a^{3} d^{2}}} a^{2} d - 3 \, {\left (e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + e^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{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 )}\right )}}{3 \, e^{3}}\right ) + \sqrt {\frac {9 i \, e^{7}}{a^{3} d^{2}}} a^{2} d \log \left (-\frac {2 \, {\left (-i \, \sqrt {\frac {9 i \, e^{7}}{a^{3} d^{2}}} a^{2} d - 3 \, {\left (e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + e^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{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 )}\right )}}{3 \, e^{3}}\right ) - \sqrt {-\frac {9 i \, e^{7}}{a^{3} d^{2}}} a^{2} d \log \left (-\frac {2 \, {\left (i \, \sqrt {-\frac {9 i \, e^{7}}{a^{3} d^{2}}} a^{2} d - 3 \, {\left (e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + e^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{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 )}\right )}}{3 \, e^{3}}\right ) + \sqrt {-\frac {9 i \, e^{7}}{a^{3} d^{2}}} a^{2} d \log \left (-\frac {2 \, {\left (-i \, \sqrt {-\frac {9 i \, e^{7}}{a^{3} d^{2}}} a^{2} d - 3 \, {\left (e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + e^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{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 )}\right )}}{3 \, e^{3}}\right )}{2 \, a^{2} d} \]

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/2*(-4*I*e^3*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) - sq
rt(9*I*e^7/(a^3*d^2))*a^2*d*log(-2/3*(I*sqrt(9*I*e^7/(a^3*d^2))*a^2*d - 3*(e^3*e^(2*I*d*x + 2*I*c) + e^3)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/e^3) + sqrt(9*I*e^7/(
a^3*d^2))*a^2*d*log(-2/3*(-I*sqrt(9*I*e^7/(a^3*d^2))*a^2*d - 3*(e^3*e^(2*I*d*x + 2*I*c) + e^3)*sqrt(a/(e^(2*I*
d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/e^3) - sqrt(-9*I*e^7/(a^3*d^2))*
a^2*d*log(-2/3*(I*sqrt(-9*I*e^7/(a^3*d^2))*a^2*d - 3*(e^3*e^(2*I*d*x + 2*I*c) + e^3)*sqrt(a/(e^(2*I*d*x + 2*I*
c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/e^3) + sqrt(-9*I*e^7/(a^3*d^2))*a^2*d*log(
-2/3*(-I*sqrt(-9*I*e^7/(a^3*d^2))*a^2*d - 3*(e^3*e^(2*I*d*x + 2*I*c) + e^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*
sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/e^3))/(a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1817 vs. \(2 (401) = 802\).

Time = 0.51 (sec) , antiderivative size = 1817, normalized size of antiderivative = 3.43 \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-8*(6*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 6*sqrt(2)*e^3*
arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 6*sqrt(2)*e^3*arctan2(sqrt(2)*c
os(1/2*d*x + 1/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 6*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c
) - 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 3*I*sqrt(2)*e^3*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2
*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*I*sqrt(2)*e^3*log(2*cos(1/2*d
*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2
) + 3*I*sqrt(2)*e^3*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) +
2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*I*sqrt(2)*e^3*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2
- 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 16*e^3*cos(1/2*d*x + 1/2*c) + 16*I*e
^3*sin(1/2*d*x + 1/2*c) - 6*(-I*sqrt(2)*e^3*cos(2*d*x + 2*c) + sqrt(2)*e^3*sin(2*d*x + 2*c) - I*sqrt(2)*e^3)*a
rctan2(sqrt(2)*sin(1/2*d*x + 1/2*c) + sin(d*x + c), sqrt(2)*cos(1/2*d*x + 1/2*c) + cos(d*x + c) + 1) - 6*(I*sq
rt(2)*e^3*cos(2*d*x + 2*c) - sqrt(2)*e^3*sin(2*d*x + 2*c) + I*sqrt(2)*e^3)*arctan2(-sqrt(2)*sin(1/2*d*x + 1/2*
c) + sin(d*x + c), -sqrt(2)*cos(1/2*d*x + 1/2*c) + cos(d*x + c) + 1) + 3*(2*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/
2*d*x + 1/2*c) + 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1
, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1, sqrt(2)*sin(1/2
*d*x + 1/2*c) + 1) + 2*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1
) + I*sqrt(2)*e^3*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2
*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - I*sqrt(2)*e^3*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2
*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + I*sqrt(2)*e^3*log(2*cos(1/2*d*x + 1/2*c)
^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - I*sqrt(
2)*e^3*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*si
n(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c) + 3*(sqrt(2)*e^3*cos(2*d*x + 2*c) + I*sqrt(2)*e^3*sin(2*d*x + 2*c) +
sqrt(2)*e^3)*log(2*sqrt(2)*sin(d*x + c)*sin(1/2*d*x + 1/2*c) + 2*(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1)*cos(d*x +
c) + cos(d*x + c)^2 + 2*cos(1/2*d*x + 1/2*c)^2 + sin(d*x + c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/
2*d*x + 1/2*c) + 1) - 3*(sqrt(2)*e^3*cos(2*d*x + 2*c) + I*sqrt(2)*e^3*sin(2*d*x + 2*c) + sqrt(2)*e^3)*log(-2*s
qrt(2)*sin(d*x + c)*sin(1/2*d*x + 1/2*c) - 2*(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1)*cos(d*x + c) + cos(d*x + c)^2
+ 2*cos(1/2*d*x + 1/2*c)^2 + sin(d*x + c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 1) -
3*(-2*I*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 2*I*sqrt(2)
*e^3*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 2*I*sqrt(2)*e^3*arctan2(sq
rt(2)*cos(1/2*d*x + 1/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 2*I*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/2*d*
x + 1/2*c) - 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + sqrt(2)*e^3*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x
+ 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*e^3*log(2*cos(1/2*
d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) +
2) + sqrt(2)*e^3*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*
sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*e^3*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sq
rt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(2*d*x + 2*c))*sqrt(a)*sqrt(e)/((-64*I*a^
2*cos(2*d*x + 2*c) + 64*a^2*sin(2*d*x + 2*c) - 64*I*a^2)*d)

Giac [F]

\[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(7/2)/(I*a*tan(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

[In]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^(3/2), x)