3.177 \(\int \frac{1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=326 \[ \frac{\left (\frac{25}{16}+\frac{21 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{\left (\frac{25}{16}+\frac{21 i}{16}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 d^{3/2} f}+\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{25}{8 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f (a+i a \tan (e+f x))^2 \sqrt{d \tan (e+f x)}} \]

[Out]

((25/16 + (21*I)/16)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(3/2)*f) - ((25/16 + (
21*I)/16)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(3/2)*f) - ((25/32 - (21*I)/32)*L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(3/2)*f) + ((25/32 - (21*I)/
32)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(3/2)*f) - 25/(8*a^2*d*
f*Sqrt[d*Tan[e + f*x]]) + 7/(8*a^2*d*f*(1 + I*Tan[e + f*x])*Sqrt[d*Tan[e + f*x]]) + 1/(4*d*f*Sqrt[d*Tan[e + f*
x]]*(a + I*a*Tan[e + f*x])^2)

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Rubi [A]  time = 0.520725, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3559, 3596, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\frac{25}{16}+\frac{21 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{\left (\frac{25}{16}+\frac{21 i}{16}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 d^{3/2} f}+\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{25}{8 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f (a+i a \tan (e+f x))^2 \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^2),x]

[Out]

((25/16 + (21*I)/16)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(3/2)*f) - ((25/16 + (
21*I)/16)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*d^(3/2)*f) - ((25/32 - (21*I)/32)*L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(3/2)*f) + ((25/32 - (21*I)/
32)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*d^(3/2)*f) - 25/(8*a^2*d*
f*Sqrt[d*Tan[e + f*x]]) + 7/(8*a^2*d*f*(1 + I*Tan[e + f*x])*Sqrt[d*Tan[e + f*x]]) + 1/(4*d*f*Sqrt[d*Tan[e + f*
x]]*(a + I*a*Tan[e + f*x])^2)

Rule 3559

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

Rule 3596

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

Rule 3529

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

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

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

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{1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx &=\frac{1}{4 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac{\int \frac{\frac{9 a d}{2}-\frac{5}{2} i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx}{4 a^2 d}\\ &=\frac{7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac{\int \frac{\frac{25 a^2 d^2}{2}-\frac{21}{2} i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{8 a^4 d^2}\\ &=-\frac{25}{8 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac{\int \frac{-\frac{21}{2} i a^2 d^3-\frac{25}{2} a^2 d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{8 a^4 d^4}\\ &=-\frac{25}{8 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{21}{2} i a^2 d^4-\frac{25}{2} a^2 d^3 x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{4 a^4 d^4 f}\\ &=-\frac{25}{8 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2}+-\frac{\left (\frac{25}{16}+\frac{21 i}{16}\right ) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 d f}+\frac{\left (\frac{25}{16}-\frac{21 i}{16}\right ) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 d f}\\ &=-\frac{25}{8 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2}+-\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}+2 x}{-d-\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{3/2} f}+-\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}-2 x}{-d+\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{3/2} f}+-\frac{\left (\frac{25}{32}+\frac{21 i}{32}\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 d f}+-\frac{\left (\frac{25}{32}+\frac{21 i}{32}\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 d f}\\ &=-\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{3/2} f}+\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{25}{8 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2}+-\frac{\left (\frac{25}{16}+\frac{21 i}{16}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{3/2} f}+\frac{\left (\frac{25}{16}+\frac{21 i}{16}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{3/2} f}\\ &=\frac{\left (\frac{25}{16}+\frac{21 i}{16}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{\left (\frac{25}{16}+\frac{21 i}{16}\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{3/2} f}+\frac{\left (\frac{25}{32}-\frac{21 i}{32}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a^2 d^{3/2} f}-\frac{25}{8 a^2 d f \sqrt{d \tan (e+f x)}}+\frac{7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt{d \tan (e+f x)}}+\frac{1}{4 d f \sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 1.07567, size = 231, normalized size = 0.71 \[ \frac{\sec ^3(e+f x) \left (43 i \sin (e+f x)+43 i \sin (3 (e+f x))+23 \cos (e+f x)+41 \cos (3 (e+f x))+(21-25 i) \sqrt{\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\sin (2 (e+f x))-i \cos (2 (e+f x)))+(-21-25 i) \sin ^{\frac{3}{2}}(2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )-(25-21 i) \sqrt{\sin (2 (e+f x))} \cos (2 (e+f x)) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{32 a^2 d f (\tan (e+f x)-i)^2 \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^2),x]

[Out]

(Sec[e + f*x]^3*(23*Cos[e + f*x] + 41*Cos[3*(e + f*x)] + (43*I)*Sin[e + f*x] - (25 - 21*I)*Cos[2*(e + f*x)]*Lo
g[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] - (21 + 25*I)*Log[Cos[e + f*x]
+ Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sin[2*(e + f*x)]^(3/2) + (21 - 25*I)*ArcSin[Cos[e + f*x] - Sin[e + f*
x]]*Sqrt[Sin[2*(e + f*x)]]*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]) + (43*I)*Sin[3*(e + f*x)]))/(32*a^2*d*f*
Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x])^2)

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Maple [A]  time = 0.057, size = 163, normalized size = 0.5 \begin{align*} -{\frac{9}{8\,f{a}^{2}d \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{11\,i}{8}}}{f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{23}{8\,f{a}^{2}d}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}-2\,{\frac{1}{f{a}^{2}d\sqrt{d\tan \left ( fx+e \right ) }}}-{\frac{1}{4\,f{a}^{2}d}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id}}}} \right ){\frac{1}{\sqrt{id}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x)

[Out]

-9/8/f/a^2/d/(-I*d+d*tan(f*x+e))^2*(d*tan(f*x+e))^(3/2)+11/8*I/f/a^2/(-I*d+d*tan(f*x+e))^2*(d*tan(f*x+e))^(1/2
)-23/8/f/a^2/d/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))-2/a^2/d/f/(d*tan(f*x+e))^(1/2)-1/4/f/a^2
/d/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.65044, size = 1863, normalized size = 5.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/16*(4*(a^2*d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sqrt(1/16*I/(a^4*d^3*f^2))*log(((8*I*
a^2*d^2*f*e^(2*I*f*x + 2*I*e) + 8*I*a^2*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)
)*sqrt(1/16*I/(a^4*d^3*f^2)) - 2*I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 4*(a^2*d^2*f*e^(6*I*f*x + 6*
I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sqrt(1/16*I/(a^4*d^3*f^2))*log(((-8*I*a^2*d^2*f*e^(2*I*f*x + 2*I*e) - 8*
I*a^2*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/16*I/(a^4*d^3*f^2)) - 2*I
*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 4*(a^2*d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*
e))*sqrt(-529/64*I/(a^4*d^3*f^2))*log(1/8*(8*(a^2*d*f*e^(2*I*f*x + 2*I*e) + a^2*d*f)*sqrt((-I*d*e^(2*I*f*x + 2
*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-529/64*I/(a^4*d^3*f^2)) + 23)*e^(-2*I*f*x - 2*I*e)/(a^2*d*f)) +
4*(a^2*d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sqrt(-529/64*I/(a^4*d^3*f^2))*log(-1/8*(8*(a
^2*d*f*e^(2*I*f*x + 2*I*e) + a^2*d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-5
29/64*I/(a^4*d^3*f^2)) - 23)*e^(-2*I*f*x - 2*I*e)/(a^2*d*f)) - sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f
*x + 2*I*e) + 1))*(-42*I*e^(6*I*f*x + 6*I*e) - 33*I*e^(4*I*f*x + 4*I*e) + 10*I*e^(2*I*f*x + 2*I*e) + I))/(a^2*
d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**2,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.22615, size = 306, normalized size = 0.94 \begin{align*} -\frac{1}{8} \, d^{3}{\left (\frac{2 \, \sqrt{2} \arctan \left (\frac{16 \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{a^{2} d^{\frac{9}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{23 i \, \sqrt{2} \arctan \left (-\frac{16 i \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{a^{2} d^{\frac{9}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{16}{\sqrt{d \tan \left (f x + e\right )} a^{2} d^{4} f} + \frac{9 \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) - 11 i \, \sqrt{d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} d^{4} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-1/8*d^3*(2*sqrt(2)*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d
)))/(a^2*d^(9/2)*f*(I*d/sqrt(d^2) + 1)) - 23*I*sqrt(2)*arctan(-16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2
)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^2*d^(9/2)*f*(I*d/sqrt(d^2) + 1)) + 16/(sqrt(d*tan(f*x + e))*a^2*d
^4*f) + (9*sqrt(d*tan(f*x + e))*d*tan(f*x + e) - 11*I*sqrt(d*tan(f*x + e))*d)/((d*tan(f*x + e) - I*d)^2*a^2*d^
4*f))