3.4.100 \(\int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{3/2} \, dx\) [400]

Optimal. Leaf size=453 \[ \frac {7 i a^{3/2} e^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d}-\frac {7 i a^{3/2} e^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{8 \sqrt {2} d}-\frac {7 i a^{3/2} e^{5/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{16 \sqrt {2} d}+\frac {7 i a^{3/2} e^{5/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{16 \sqrt {2} d}+\frac {7 i a^2 (e \sec (c+d x))^{5/2}}{12 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {i a (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[Out]

7/16*I*a^(3/2)*e^(5/2)*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))/d*2^(1/
2)-7/16*I*a^(3/2)*e^(5/2)*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))/d*2^
(1/2)-7/32*I*a^(3/2)*e^(5/2)*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)))/d*2^(1/2)+7/32*I*a^(3/2)*e^(5/2)*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)))/d*2^(1/2)+7/12*I*a^2*(e*sec(d*x+c))^(5/2)/d/(a+I*a*tan(d
*x+c))^(1/2)+1/3*I*a*(e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2)/d-7/8*I*a*e^2*(e*sec(d*x+c))^(1/2)*(a+I*a*t
an(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.36, antiderivative size = 453, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3579, 3582, 3576, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {7 i a^{3/2} e^{5/2} \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 )}{8 \sqrt {2} d}-\frac {7 i a^{3/2} e^{5/2} \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 )}{8 \sqrt {2} d}-\frac {7 i a^{3/2} e^{5/2} \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}+\frac {7 i a^{3/2} e^{5/2} \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}+\frac {7 i a^2 (e \sec (c+d x))^{5/2}}{12 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{8 d}+\frac {i a \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]
time = 4.57, size = 376, normalized size = 0.83 \begin {gather*} -\frac {a (e \sec (c+d x))^{5/2} \left (2 i \sqrt {1+\cos (2 c)+i \sin (2 c)} (-9+7 \cos (2 c+2 d x)+14 i \sin (2 c+2 d x)) \sqrt {i-\tan \left (\frac {d x}{2}\right )}+84 \tanh ^{-1}\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 ) \cos ^3(c+d x) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}-84 \tanh ^{-1}\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 ) \cos ^3(c+d x) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}\right ) \sqrt {a+i a \tan (c+d x)}}{96 d \sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

-1/96*(a*(e*Sec[c + d*x])^(5/2)*((2*I)*Sqrt[1 + Cos[2*c] + I*Sin[2*c]]*(-9 + 7*Cos[2*c + 2*d*x] + (14*I)*Sin[2
*c + 2*d*x])*Sqrt[I - Tan[(d*x)/2]] + 84*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]])]*Cos[c + d*x]^3*Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[1 + I*Cos[c] -
Sin[c]]*Sqrt[I + Tan[(d*x)/2]] - 84*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]])]*Cos[c + d*x]^3*Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt[-1 + I*Cos[c] + Sin
[c]]*Sqrt[I + Tan[(d*x)/2]])*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[1 + Cos[2*c] + I*Sin[2*c]]*Sqrt[I - Tan[(d*x)
/2]])

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Maple [A]
time = 0.97, size = 414, normalized size = 0.91

method	result	size
default	\(\frac {\left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{3} \left (21 i \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right ) \left (\cos ^{3}\left (d x +c \right )\right )-21 i \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right ) \left (\cos ^{3}\left (d x +c \right )\right )-42 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-28 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-42 \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right )+21 \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right ) \left (\cos ^{3}\left (d x +c \right )\right )+21 \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right ) \left (\cos ^{3}\left (d x +c \right )\right )+16 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-14 \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+44 \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}+16 \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\right ) a}{48 d \sin \left (d x +c \right )^{5} \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \left (\frac {1}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}\)	\(414\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/48/d*(e/cos(d*x+c))^(5/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^3*(21*I*arctanh(1/2
*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)^3-21*I*arctanh(1/2*(1/(1+cos(d*x+c)))^(1/2)*(c
os(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)^3-42*I*(1/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-28*I*(1/(1+cos(d*x
+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)-42*(1/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3+21*arctanh(1/2*(1/(1+cos(d*x+c)))^(
1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)^3+21*arctanh(1/2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))
)*cos(d*x+c)^3+16*I*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-14*(1/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+44*cos(d*x+c)
*(1/(1+cos(d*x+c)))^(1/2)+16*(1/(1+cos(d*x+c)))^(1/2))/sin(d*x+c)^5/(I*sin(d*x+c)+cos(d*x+c)-1)/(1/(1+cos(d*x+
c)))^(5/2)*a

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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. 2734 vs. \(2 (304) = 608\).
time = 0.83, size = 2734, normalized size = 6.04 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-192*(42*(sqrt(2)*a*cos(6*d*x + 6*c) + 3*sqrt(2)*a*cos(4*d*x + 4*c) + 3*sqrt(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)
*a*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a*sin(4*d*x + 4*c) + 3*I*sqrt(2)*a*sin(2*d*x + 2*c) + sqrt(2)*a)*arctan2(sqr
t(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) + 42*(sqrt(2)*a*cos(6*d*x + 6*c) + 3*sqrt(2)*a*cos(4*d*x + 4*c) + 3*sqrt(2)*a*cos(2*d*x + 2
*c) + I*sqrt(2)*a*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a*sin(4*d*x + 4*c) + 3*I*sqrt(2)*a*sin(2*d*x + 2*c) + sqrt(2)
*a)*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) + 42*(sqrt(2)*a*cos(6*d*x + 6*c) + 3*sqrt(2)*a*cos(4*d*x + 4*c) + 3*sqrt(2)
*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a*sin(4*d*x + 4*c) + 3*I*sqrt(2)*a*sin(2*d*x
+ 2*c) + sqrt(2)*a)*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) + 42*(sqrt(2)*a*cos(6*d*x + 6*c) + 3*sqrt(2)*a*cos(4*d*x + 4
*c) + 3*sqrt(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a*sin(4*d*x + 4*c) + 3*I*sqrt(
2)*a*sin(2*d*x + 2*c) + sqrt(2)*a)*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) + 42*(-I*sqrt(2)*a*cos(6*d*x + 6*c) - 3*I*sq
rt(2)*a*cos(4*d*x + 4*c) - 3*I*sqrt(2)*a*cos(2*d*x + 2*c) + sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x
+ 4*c) + 3*sqrt(2)*a*sin(2*d*x + 2*c) - I*sqrt(2)*a)*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) + 42*(I*sqrt(2)*a*cos(6*d*x +
6*c) + 3*I*sqrt(2)*a*cos(4*d*x + 4*c) + 3*I*sqrt(2)*a*cos(2*d*x + 2*c) - sqrt(2)*a*sin(6*d*x + 6*c) - 3*sqrt(2
)*a*sin(4*d*x + 4*c) - 3*sqrt(2)*a*sin(2*d*x + 2*c) + I*sqrt(2)*a)*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(si
n(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) + 336*a*cos(11/
4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 288*a*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) -
112*a*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 21*(sqrt(2)*a*cos(6*d*x + 6*c) + 3*sqrt(2)*a*cos(
4*d*x + 4*c) + 3*sqrt(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a*sin(4*d*x + 4*c) +
3*I*sqrt(2)*a*sin(2*d*x + 2*c) + sqrt(2)*a)*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*(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(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(si
n(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 21*(sqrt(2)*a*cos(6*d*x + 6*c) + 3*sqrt(2)*a*cos(4*d*x + 4*c) + 3*sq
rt(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(6*d*x + 6*c) + 3*I*sqrt(2)*a*sin(4*d*x + 4*c) + 3*I*sqrt(2)*a*sin(2
*d*x + 2*c) + sqrt(2)*a)*log(-2*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(s
in(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(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), co
s(2*d*x + 2*c))) + 1) + 21*(I*sqrt(2)*a*cos(6*d*x + 6*c) + 3*I*sqrt(2)*a*cos(4*d*x + 4*c) + 3*I*sqrt(2)*a*cos(
2*d*x + 2*c) - sqrt(2)*a*sin(6*d*x + 6*c) - 3*sqrt(2)*a*sin(4*d*x + 4*c) - 3*sqrt(2)*a*sin(2*d*x + 2*c) + I*sq
rt(2)*a)*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), co
s(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*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 21*(-I*sqrt(2)*a*cos(6*d*x + 6*c) - 3*I*sqrt(2)*a*cos(4*d*x + 4*
c) - 3*I*sqrt(2)*a*cos(2*d*x + 2*c) + sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 4*c) + 3*sqrt(2)*a*
sin(2*d*x + 2*c) - I*sqrt(2)*a)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arcta
n2(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) + 21*(I*sqrt(2)*a*cos(6*d*x + 6*c) + 3*I*sq
rt(2)*a*cos(4*d*x + 4*c) + 3*I*sqrt(2)*a*cos(2*...

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Fricas [A]
time = 0.42, size = 594, normalized size = 1.31 \begin {gather*} \frac {6 \, \sqrt {\frac {49 i \, a^{3} e^{5}}{64 \, d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (\frac {7 \, {\left (a e^{\frac {5}{2}} + a e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} \sqrt {\frac {a}{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 {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 8 \, \sqrt {\frac {49 i \, a^{3} e^{5}}{64 \, d^{2}}} d\right )} e^{\left (-\frac {5}{2}\right )}}{7 \, a}\right ) - 6 \, \sqrt {\frac {49 i \, a^{3} e^{5}}{64 \, d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (\frac {7 \, {\left (a e^{\frac {5}{2}} + a e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} \sqrt {\frac {a}{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 {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - 8 \, \sqrt {\frac {49 i \, a^{3} e^{5}}{64 \, d^{2}}} d\right )} e^{\left (-\frac {5}{2}\right )}}{7 \, a}\right ) - 6 \, \sqrt {-\frac {49 i \, a^{3} e^{5}}{64 \, d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (\frac {7 \, {\left (a e^{\frac {5}{2}} + a e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} \sqrt {\frac {a}{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 {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 8 \, \sqrt {-\frac {49 i \, a^{3} e^{5}}{64 \, d^{2}}} d\right )} e^{\left (-\frac {5}{2}\right )}}{7 \, a}\right ) + 6 \, \sqrt {-\frac {49 i \, a^{3} e^{5}}{64 \, d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (\frac {7 \, {\left (a e^{\frac {5}{2}} + a e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} \sqrt {\frac {a}{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 {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - 8 \, \sqrt {-\frac {49 i \, a^{3} e^{5}}{64 \, d^{2}}} d\right )} e^{\left (-\frac {5}{2}\right )}}{7 \, a}\right ) + \frac {{\left (-21 i \, a e^{\left (5 i \, d x + 5 i \, c + \frac {5}{2}\right )} + 18 i \, a e^{\left (3 i \, d x + 3 i \, c + \frac {5}{2}\right )} + 7 i \, a e^{\left (i \, d x + i \, c + \frac {5}{2}\right )}\right )} \sqrt {\frac {a}{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 {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/12*(6*sqrt(49/64*I*a^3*e^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2/7*(7*(a*e^(5/2)
+ a*e^(2*I*d*x + 2*I*c + 5/2))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I
*c) + 1) + 8*sqrt(49/64*I*a^3*e^5/d^2)*d)*e^(-5/2)/a) - 6*sqrt(49/64*I*a^3*e^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2
*d*e^(2*I*d*x + 2*I*c) + d)*log(2/7*(7*(a*e^(5/2) + a*e^(2*I*d*x + 2*I*c + 5/2))*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1) - 8*sqrt(49/64*I*a^3*e^5/d^2)*d)*e^(-5/2)/a) - 6*sq
rt(-49/64*I*a^3*e^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2/7*(7*(a*e^(5/2) + a*e^(2*
I*d*x + 2*I*c + 5/2))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1)
+ 8*sqrt(-49/64*I*a^3*e^5/d^2)*d)*e^(-5/2)/a) + 6*sqrt(-49/64*I*a^3*e^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2
*I*d*x + 2*I*c) + d)*log(2/7*(7*(a*e^(5/2) + a*e^(2*I*d*x + 2*I*c + 5/2))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^
(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1) - 8*sqrt(-49/64*I*a^3*e^5/d^2)*d)*e^(-5/2)/a) + (-21*I*a*e
^(5*I*d*x + 5*I*c + 5/2) + 18*I*a*e^(3*I*d*x + 3*I*c + 5/2) + 7*I*a*e^(I*d*x + I*c + 5/2))*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x
+ 2*I*c) + d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(5/2)*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8569 deep

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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((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*e^(5/2)*sec(d*x + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^(3/2), x)

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