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

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

[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

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Rubi [A]  time = 0.505928, antiderivative size = 453, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3498, 3501, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac{7 i a^{3/2} e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{8 \sqrt{2} d}-\frac{7 i a^{3/2} e^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{8 \sqrt{2} d}-\frac{7 i a^{3/2} e^{5/2} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{16 \sqrt{2} d}+\frac{7 i a^{3/2} e^{5/2} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{16 \sqrt{2} d}+\frac{7 i a^2 (e \sec (c+d x))^{5/2}}{12 d \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a e^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}{8 d}+\frac{i a \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}}{3 d} \]

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 3498

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 3501

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 3495

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 297

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 1162

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 617

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 204

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

Rule 1165

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 628

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]

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 3.33488, size = 358, normalized size = 0.79 \[ \frac{(\cos (c)-i \sin (c)) \cos ^4(c+d x) (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{5/2} \left (\frac{42 \sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \tan ^{-1}\left (\frac{\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )}{\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}-\frac{42 \sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )}{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}+\sec ^3(c+d x) (14 \sin (2 (c+d x))-7 i \cos (2 (c+d x))+9 i)\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^4*(e*Sec[c + d*x])^(5/2)*(Cos[c] - I*Sin[c])*(Cos[d*x] - I*Sin[d*x])*(Sec[c + d*x]^3*(9*I - (7*I
)*Cos[2*(c + d*x)] + 14*Sin[2*(c + d*x)]) - (42*ArcTan[(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[I + Tan[(d*x)/2]])/(Sq
rt[-1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]]) + (42*ArcTan[(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[I + Tan[(d*x)
/2]])/(Sqrt[-1 - I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]]))*(a + I*a*Tan[c + d*x])^(3/2))/(48*d)

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Maple [A]  time = 0.325, size = 414, normalized size = 0.9 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

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

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Maxima [B]  time = 2.98898, size = 4070, normalized size = 8.98 \begin{align*} \text{result too large to display} \end{align*}

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]

-(64512*a*e^2*cos(11/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 55296*a*e^2*cos(7/4*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) - 21504*a*e^2*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 64512*I*a*e^2*si
n(11/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 55296*I*a*e^2*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*
x + 2*c))) - 21504*I*a*e^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (8064*sqrt(2)*a*e^2*cos(6*d*
x + 6*c) + 24192*sqrt(2)*a*e^2*cos(4*d*x + 4*c) + 24192*sqrt(2)*a*e^2*cos(2*d*x + 2*c) + 8064*I*sqrt(2)*a*e^2*
sin(6*d*x + 6*c) + 24192*I*sqrt(2)*a*e^2*sin(4*d*x + 4*c) + 24192*I*sqrt(2)*a*e^2*sin(2*d*x + 2*c) + 8064*sqrt
(2)*a*e^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(s
in(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (8064*sqrt(2)*a*e^2*cos(6*d*x + 6*c) + 24192*sqrt(2)*a*e^2*cos(4*d*
x + 4*c) + 24192*sqrt(2)*a*e^2*cos(2*d*x + 2*c) + 8064*I*sqrt(2)*a*e^2*sin(6*d*x + 6*c) + 24192*I*sqrt(2)*a*e^
2*sin(4*d*x + 4*c) + 24192*I*sqrt(2)*a*e^2*sin(2*d*x + 2*c) + 8064*sqrt(2)*a*e^2)*arctan2(sqrt(2)*cos(1/4*arct
an2(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) + (8064*sqrt(2)*a*e^2*cos(6*d*x + 6*c) + 24192*sqrt(2)*a*e^2*cos(4*d*x + 4*c) + 24192*sqrt(2)*a*e^2*cos(2*d
*x + 2*c) + 8064*I*sqrt(2)*a*e^2*sin(6*d*x + 6*c) + 24192*I*sqrt(2)*a*e^2*sin(4*d*x + 4*c) + 24192*I*sqrt(2)*a
*e^2*sin(2*d*x + 2*c) + 8064*sqrt(2)*a*e^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) + (8064*sqrt(2)*a*e^2*cos(6*d*x + 6*
c) + 24192*sqrt(2)*a*e^2*cos(4*d*x + 4*c) + 24192*sqrt(2)*a*e^2*cos(2*d*x + 2*c) + 8064*I*sqrt(2)*a*e^2*sin(6*
d*x + 6*c) + 24192*I*sqrt(2)*a*e^2*sin(4*d*x + 4*c) + 24192*I*sqrt(2)*a*e^2*sin(2*d*x + 2*c) + 8064*sqrt(2)*a*
e^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) - (8064*I*sqrt(2)*a*e^2*cos(6*d*x + 6*c) + 24192*I*sqrt(2)*a*e^2*cos(4*d*x
+ 4*c) + 24192*I*sqrt(2)*a*e^2*cos(2*d*x + 2*c) - 8064*sqrt(2)*a*e^2*sin(6*d*x + 6*c) - 24192*sqrt(2)*a*e^2*s
in(4*d*x + 4*c) - 24192*sqrt(2)*a*e^2*sin(2*d*x + 2*c) + 8064*I*sqrt(2)*a*e^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) - (-8
064*I*sqrt(2)*a*e^2*cos(6*d*x + 6*c) - 24192*I*sqrt(2)*a*e^2*cos(4*d*x + 4*c) - 24192*I*sqrt(2)*a*e^2*cos(2*d*
x + 2*c) + 8064*sqrt(2)*a*e^2*sin(6*d*x + 6*c) + 24192*sqrt(2)*a*e^2*sin(4*d*x + 4*c) + 24192*sqrt(2)*a*e^2*si
n(2*d*x + 2*c) - 8064*I*sqrt(2)*a*e^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) - (4032*sqrt(2)*a*e^2*cos(6*d*x + 6*c) + 12
096*sqrt(2)*a*e^2*cos(4*d*x + 4*c) + 12096*sqrt(2)*a*e^2*cos(2*d*x + 2*c) + 4032*I*sqrt(2)*a*e^2*sin(6*d*x + 6
*c) + 12096*I*sqrt(2)*a*e^2*sin(4*d*x + 4*c) + 12096*I*sqrt(2)*a*e^2*sin(2*d*x + 2*c) + 4032*sqrt(2)*a*e^2)*lo
g(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(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (403
2*sqrt(2)*a*e^2*cos(6*d*x + 6*c) + 12096*sqrt(2)*a*e^2*cos(4*d*x + 4*c) + 12096*sqrt(2)*a*e^2*cos(2*d*x + 2*c)
+ 4032*I*sqrt(2)*a*e^2*sin(6*d*x + 6*c) + 12096*I*sqrt(2)*a*e^2*sin(4*d*x + 4*c) + 12096*I*sqrt(2)*a*e^2*sin(
2*d*x + 2*c) + 4032*sqrt(2)*a*e^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*(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), cos(2*d*x + 2*c))) + 1) - (-4032*I*sqrt(2)*a*e^2*cos(6*d*x + 6*c) - 12096*I*sqrt(2)*a*e^2*cos(4*d*x +
4*c) - 12096*I*sqrt(2)*a*e^2*cos(2*d*x + 2*c) + 4032*sqrt(2)*a*e^2*sin(6*d*x + 6*c) + 12096*sqrt(2)*a*e^2*sin(
4*d*x + 4*c) + 12096*sqrt(2)*a*e^2*sin(2*d*x + 2*c) - 4032*I*sqrt(2)*a*e^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*arct
an2(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)
- (4032*I*sqrt(2)*a*e^2*cos(6*d*x + 6*c) + 12096*I*sqrt(2)*a*e^2*cos(4*d*x + 4*c) + 12096*I*sqrt(2)*a*e^2*cos
(2*d*x + 2*c) - 4032*sqrt(2)*a*e^2*sin(6*d*x + 6*c) - 12096*sqrt(2)*a*e^2*sin(4*d*x + 4*c) - 12096*sqrt(2)*a*e
^2*sin(2*d*x + 2*c) + 4032*I*sqrt(2)*a*e^2)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*s
in(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) - (-4032*I*sqrt(2)*a*e^2*cos(6*
d*x + 6*c) - 12096*I*sqrt(2)*a*e^2*cos(4*d*x + 4*c) - 12096*I*sqrt(2)*a*e^2*cos(2*d*x + 2*c) + 4032*sqrt(2)*a*
e^2*sin(6*d*x + 6*c) + 12096*sqrt(2)*a*e^2*sin(4*d*x + 4*c) + 12096*sqrt(2)*a*e^2*sin(2*d*x + 2*c) - 4032*I*sq
rt(2)*a*e^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*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - (4032*I*sqrt(2)*a*e^2*cos(6*d*x + 6*c) + 12096*I*sqrt(2)*a*e
^2*cos(4*d*x + 4*c) + 12096*I*sqrt(2)*a*e^2*cos(2*d*x + 2*c) - 4032*sqrt(2)*a*e^2*sin(6*d*x + 6*c) - 12096*sqr
t(2)*a*e^2*sin(4*d*x + 4*c) - 12096*sqrt(2)*a*e^2*sin(2*d*x + 2*c) + 4032*I*sqrt(2)*a*e^2)*log(2*cos(1/4*arcta
n2(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))*sqrt(a)*sqrt(e)/(d*(-36864*I*cos(6*d*x + 6*c) - 110592*I*cos(4*d*x + 4*c) - 110592*I*cos(2*d*
x + 2*c) + 36864*sin(6*d*x + 6*c) + 110592*sin(4*d*x + 4*c) + 110592*sin(2*d*x + 2*c) - 36864*I))

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Fricas [B]  time = 2.27307, size = 1989, normalized size = 4.39 \begin{align*} \text{result too large to display} \end{align*}

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*((-21*I*a*e^2*e^(4*I*d*x + 4*I*c) + 18*I*a*e^2*e^(2*I*d*x + 2*I*c) + 7*I*a*e^2)*sqrt(a/(e^(2*I*d*x + 2*I*
c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x + 3/2*I*c) - 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(1/7*(14*(a*e^2*e^(2*I*d*x + 2*I*c) + a*e^2)*sqrt(a/(e^(2*I*d*x +
2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x + 3/2*I*c) + 16*I*sqrt(-49/64*I*a^3*e^5/d^2)*d*e^
(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(a*e^2)) + 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(1/7*(14*(a*e^2*e^(2*I*d*x + 2*I*c) + a*e^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt
(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x + 3/2*I*c) - 16*I*sqrt(-49/64*I*a^3*e^5/d^2)*d*e^(2*I*d*x + 2*I*c))
*e^(-2*I*d*x - 2*I*c)/(a*e^2)) - 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^2*e^(2*I*d*x + 2*I*c) + a*e^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I
*c) + 1))*e^(3/2*I*d*x + 3/2*I*c) + 8*I*sqrt(49/64*I*a^3*e^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/
(a*e^2)) + 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^2
*e^(2*I*d*x + 2*I*c) + a*e^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x
+ 3/2*I*c) - 8*I*sqrt(49/64*I*a^3*e^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(a*e^2)))/(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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

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