3.431 \(\int \frac{(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=527 \[ \frac{i \sqrt{2} e^{7/2} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{a^{3/2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i \sqrt{2} e^{7/2} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{a^{3/2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i e^{7/2} \sec (c+d x) \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt{2} a^{3/2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i e^{7/2} \sec (c+d x) \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt{2} a^{3/2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{3 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

(((4*I)/3)*e^2*(e*Sec[c + d*x])^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (I*Sqrt[2]*e^(7/2)*ArcTan[1 - (Sqr
t[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(a^(3/2)*d*Sqrt[a - I*a
*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (I*Sqrt[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])/(a^(3/2)*d*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan
[c + d*x]]) - (I*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]] + C
os[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*a^(3/2)*d*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*T
an[c + d*x]]) + (I*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]*a^(3/2)*d*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a
*Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.45357, antiderivative size = 527, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3500, 3499, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac{i \sqrt{2} e^{7/2} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{a^{3/2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i \sqrt{2} e^{7/2} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{a^{3/2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i e^{7/2} \sec (c+d x) \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt{2} a^{3/2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i e^{7/2} \sec (c+d x) \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt{2} a^{3/2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{3 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((4*I)/3)*e^2*(e*Sec[c + d*x])^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (I*Sqrt[2]*e^(7/2)*ArcTan[1 - (Sqr
t[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(a^(3/2)*d*Sqrt[a - I*a
*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (I*Sqrt[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])/(a^(3/2)*d*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan
[c + d*x]]) - (I*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]] + C
os[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*a^(3/2)*d*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*T
an[c + d*x]]) + (I*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]*a^(3/2)*d*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a
*Tan[c + d*x]])

Rule 3500

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]

Rule 3499

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

Mathematica [A]  time = 11.6297, size = 422, normalized size = 0.8 \[ -\frac{2 e^2 (e \sec (c+d x))^{3/2} (\cos (2 (c+d x))+i \sin (2 (c+d x))) \left (\sqrt{-\sin (c)+i \cos (c)+1} \left (2 \sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} (\tan (c+d x)+i)-3 \sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} (\tan (c+d x)-i) \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )\right )-3 \sqrt{-\sin (c)-i \cos (c)+1} \sqrt{-\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} (\tan (c+d x)-i) \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)+1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)+1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )\right )}{3 a^2 d \sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\sin (c)+i \cos (c)+1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(-2*e^2*(e*Sec[c + d*x])^(3/2)*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*(-3*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]]*Sqr
t[-1 + I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]]*(-I + Tan[c + d*x]) + Sqrt[1 + I*Cos[c] - Sin[c]]*(-3*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]])]*S
qrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]]*(-I + Tan[c + d*x]) + 2*Sqrt[-1 + I*Cos[c] - Sin[c]]*Sqrt[I
- Tan[(d*x)/2]]*(I + Tan[c + d*x]))))/(3*a^2*d*Sqrt[-1 + I*Cos[c] - Sin[c]]*Sqrt[1 + I*Cos[c] - Sin[c]]*Sqrt[
I - Tan[(d*x)/2]]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.303, size = 1324, normalized size = 2.5 \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))^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/6/d/a^3*(cos(d*x+c)-1)^3*(-3*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))+3*arctanh(1/2*(
1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*sin(d*x+c)-3*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+
c)+1+sin(d*x+c)))*sin(d*x+c)-12*cos(d*x+c)^2*sin(d*x+c)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin
(d*x+c)))+12*cos(d*x+c)^2*sin(d*x+c)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+6*cos(d*x
+c)*sin(d*x+c)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))-6*cos(d*x+c)*sin(d*x+c)*arctanh
(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))-12*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)))-12*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)))-8*c
os(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)+9*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x
+c)+9*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*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)))+8*(1/(cos(d*x+c)+1))^(1/2)+3*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(
d*x+c)+1-sin(d*x+c)))*sin(d*x+c)+3*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*sin(d*x+c
)+12*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)^3-12*I*arctanh(1/2*(1/(cos(d
*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)^3-6*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1
-sin(d*x+c)))*cos(d*x+c)^2+6*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)^2-9*
I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)+9*I*arctanh(1/2*(1/(cos(d*x+c)+1)
)^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)+3*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c
)))-3*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+6*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)
*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)^2+6*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos
(d*x+c)^2-8*I*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d*x+c)-12*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*
x+c)+1-sin(d*x+c)))*cos(d*x+c)^2*sin(d*x+c)-12*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)
))*cos(d*x+c)^2*sin(d*x+c)+6*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)*sin(
d*x+c)+6*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)*sin(d*x+c))*(a*(I*sin(d*
x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^4*(e/cos(d*x+c))^(7/2)/(4*I*sin(d*x+c)*cos(d*x+c)^2+4*cos(d*x+c)
^3-I*sin(d*x+c)-3*cos(d*x+c))/(1/(cos(d*x+c)+1))^(7/2)/sin(d*x+c)^7

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Maxima [B]  time = 2.26944, size = 1979, normalized size = 3.76 \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))^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/12*(6*I*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1, sqrt(2
)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 6*I*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/3*
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1, -sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3
/2*d*x + 3/2*c))) + 1) + 6*I*sqrt(2)*e^3*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3
/2*c))) - 1, sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 6*I*sqrt(2)*e^3*arcta
n2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 1, -sqrt(2)*sin(1/3*arctan2(sin(3/2*
d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 6*sqrt(2)*e^3*arctan2(sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c)
, cos(3/2*d*x + 3/2*c))) + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))), sqrt(2)*cos(1/3*arcta
n2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))
+ 1) + 6*sqrt(2)*e^3*arctan2(-sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sin(2/3*
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))), -sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d
*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 3*I*sqrt(2)*e^3*log(2*sqrt
(2)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2
*d*x + 3/2*c))) + 2*(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(2/3*arctan2
(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2
+ 2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), co
s(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 3*I*sqrt(2)*e^3*log(-2*sqrt(2)*sin(2/3*arctan2(sin
(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*(sq
rt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 1)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*cos(1/3*arctan2(si
n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2
+ 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3
/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 3*sqrt(2)*e^3*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3
/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3
/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)
)) + 2) + 3*sqrt(2)*e^3*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arcta
n2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*
x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 3*sqrt(2)*e^3*log
(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), co
s(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*si
n(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 3*sqrt(2)*e^3*log(2*cos(1/3*arctan2(sin(3/2*
d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*
sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x +
3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 16*I*e^3*cos(3/2*d*x + 3/2*c) + 16*e^3*sin(3/2*d*x + 3/2*c))*sqrt(e)/(a
^(5/2)*d)

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Fricas [A]  time = 2.40564, size = 1663, normalized size = 3.16 \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))^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/6*(3*a^3*d*sqrt(4*I*e^7/(a^5*d^2))*e^(3*I*d*x + 3*I*c)*log((a^3*d*sqrt(4*I*e^7/(a^5*d^2))*e^(2*I*d*x + 2*I*c
) + 2*(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^(3
/2*I*d*x + 3/2*I*c))*e^(-2*I*d*x - 2*I*c)/e^3) - 3*a^3*d*sqrt(4*I*e^7/(a^5*d^2))*e^(3*I*d*x + 3*I*c)*log(-(a^3
*d*sqrt(4*I*e^7/(a^5*d^2))*e^(2*I*d*x + 2*I*c) - 2*(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^(3/2*I*d*x + 3/2*I*c))*e^(-2*I*d*x - 2*I*c)/e^3) - 3*a^3*d*sqrt(-4*
I*e^7/(a^5*d^2))*e^(3*I*d*x + 3*I*c)*log((a^3*d*sqrt(-4*I*e^7/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + 2*(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^(3/2*I*d*x + 3/2*I*c)
)*e^(-2*I*d*x - 2*I*c)/e^3) + 3*a^3*d*sqrt(-4*I*e^7/(a^5*d^2))*e^(3*I*d*x + 3*I*c)*log(-(a^3*d*sqrt(-4*I*e^7/(
a^5*d^2))*e^(2*I*d*x + 2*I*c) - 2*(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^(3/2*I*d*x + 3/2*I*c))*e^(-2*I*d*x - 2*I*c)/e^3) + 2*(4*I*e^3*e^(2*I*d*x + 2*I*c) +
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^(3/2*I*d*x + 3/2*I*c))*e^(-3*I*
d*x - 3*I*c)/(a^3*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))**(7/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(7/2)/(I*a*tan(d*x + c) + a)^(5/2), x)