3.1128 \(\int \frac{1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=205 \[ \frac{d (c-5 i d)}{2 a f (c-i d) (c+i d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{2 a f (c-i d)^{3/2}}+\frac{(-4 d+i c) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{2 a f (c+i d)^{5/2}} \]

[Out]

((-I/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a*(c - I*d)^(3/2)*f) + ((I*c - 4*d)*ArcTanh[Sqrt[c +
d*Tan[e + f*x]]/Sqrt[c + I*d]])/(2*a*(c + I*d)^(5/2)*f) + ((c - (5*I)*d)*d)/(2*a*(c - I*d)*(c + I*d)^2*f*Sqrt
[c + d*Tan[e + f*x]]) - 1/(2*(I*c - d)*f*(a + I*a*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])

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Rubi [A]  time = 0.468487, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3552, 3529, 3539, 3537, 63, 208} \[ \frac{d (c-5 i d)}{2 a f (c-i d) (c+i d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{1}{2 f (-d+i c) (a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{2 a f (c-i d)^{3/2}}+\frac{(-4 d+i c) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{2 a f (c+i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a*(c - I*d)^(3/2)*f) + ((I*c - 4*d)*ArcTanh[Sqrt[c +
d*Tan[e + f*x]]/Sqrt[c + I*d]])/(2*a*(c + I*d)^(5/2)*f) + ((c - (5*I)*d)*d)/(2*a*(c - I*d)*(c + I*d)^2*f*Sqrt
[c + d*Tan[e + f*x]]) - 1/(2*(I*c - d)*f*(a + I*a*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])

Rule 3552

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(a
*(c + d*Tan[e + f*x])^(n + 1))/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c +
d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*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] &&  !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 3539

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

Rule 3537

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{\frac{1}{2} a (2 i c-5 d)+\frac{3}{2} i a d \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx}{2 a^2 (i c-d)}\\ &=\frac{(c-5 i d) d}{2 a (c-i d) (c+i d)^2 f \sqrt{c+d \tan (e+f x)}}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{\frac{1}{2} a (c+3 i d) (2 i c+d)+\frac{1}{2} a d (i c+5 d) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 a^2 (i c-d) \left (c^2+d^2\right )}\\ &=\frac{(c-5 i d) d}{2 a (c-i d) (c+i d)^2 f \sqrt{c+d \tan (e+f x)}}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{4 a (c-i d)}+\frac{(c+4 i d) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{4 a (c+i d)^2}\\ &=\frac{(c-5 i d) d}{2 a (c-i d) (c+i d)^2 f \sqrt{c+d \tan (e+f x)}}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{(i c-4 d) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{4 a (c+i d)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{4 a (i c+d) f}\\ &=\frac{(c-5 i d) d}{2 a (c-i d) (c+i d)^2 f \sqrt{c+d \tan (e+f x)}}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{2 a (c-i d) d f}+\frac{(i (i c-4 d)) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{2 a (c+i d)^2 d f}\\ &=-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{2 a (c-i d)^{3/2} f}+\frac{(i c-4 d) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{2 a (c+i d)^{5/2} f}+\frac{(c-5 i d) d}{2 a (c-i d) (c+i d)^2 f \sqrt{c+d \tan (e+f x)}}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.84985, size = 297, normalized size = 1.45 \[ \frac{\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (\frac{2 \cos (e+f x) (\sin (f x)+i \cos (f x)) \sqrt{c+d \tan (e+f x)} \left (\left (c^2-i c d-4 d^2\right ) \cos (e+f x)+d (c-5 i d) \sin (e+f x)\right )}{(c-i d) (c+i d)^2 (c \cos (e+f x)+d \sin (e+f x))}-\frac{2 (\cos (e)+i \sin (e)) \left (i (-c-i d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c+i d}}\right )-i \sqrt{-c+i d} \left (c^2+3 i c d+4 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c-i d}}\right )\right )}{(-c-i d)^{5/2} (-c+i d)^{3/2}}\right )}{4 f (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*((-2*((-I)*Sqrt[-c + I*d]*(c^2 + (3*I)*c*d + 4*d^2)*ArcTan[Sqrt[c + d*Ta
n[e + f*x]]/Sqrt[-c - I*d]] + I*(-c - I*d)^(5/2)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c + I*d]])*(Cos[e] + I*
Sin[e]))/((-c - I*d)^(5/2)*(-c + I*d)^(3/2)) + (2*Cos[e + f*x]*(I*Cos[f*x] + Sin[f*x])*((c^2 - I*c*d - 4*d^2)*
Cos[e + f*x] + (c - (5*I)*d)*d*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/((c - I*d)*(c + I*d)^2*(c*Cos[e + f*x]
+ d*Sin[e + f*x]))))/(4*f*(a + I*a*Tan[e + f*x]))

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Maple [B]  time = 0.068, size = 580, normalized size = 2.8 \begin{align*}{\frac{-{\frac{i}{2}}{c}^{2}}{af \left ( c+id \right ) ^{2}}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id-c}}}} \right ) \left ( id-c \right ) ^{-{\frac{3}{2}}}}+{\frac{{\frac{i}{2}}{d}^{2}}{af \left ( c+id \right ) ^{2}}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id-c}}}} \right ) \left ( id-c \right ) ^{-{\frac{3}{2}}}}+{\frac{cd}{af \left ( c+id \right ) ^{2}}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id-c}}}} \right ) \left ( id-c \right ) ^{-{\frac{3}{2}}}}-{\frac{{c}^{2}d}{2\,af \left ( id-c \right ) \left ( c+id \right ) ^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) }\sqrt{c+d\tan \left ( fx+e \right ) }}-{\frac{{d}^{3}}{2\,af \left ( id-c \right ) \left ( c+id \right ) ^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) }\sqrt{c+d\tan \left ( fx+e \right ) }}+{\frac{{\frac{i}{2}}{c}^{3}}{af \left ( id-c \right ) \left ( c+id \right ) ^{3}}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}}+{\frac{{\frac{i}{2}}{d}^{2}c}{af \left ( id-c \right ) \left ( c+id \right ) ^{3}}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}}-2\,{\frac{{c}^{2}d}{af \left ( id-c \right ) \left ( c+id \right ) ^{3}\sqrt{-id-c}}\arctan \left ({\frac{\sqrt{c+d\tan \left ( fx+e \right ) }}{\sqrt{-id-c}}} \right ) }-2\,{\frac{{d}^{3}}{af \left ( id-c \right ) \left ( c+id \right ) ^{3}\sqrt{-id-c}}\arctan \left ({\frac{\sqrt{c+d\tan \left ( fx+e \right ) }}{\sqrt{-id-c}}} \right ) }+{\frac{2\,i{d}^{2}}{af \left ( ic+d \right ) \left ( c+id \right ) \left ( ic-d \right ) }{\frac{1}{\sqrt{c+d\tan \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 5.36425, size = 3787, normalized size = 18.47 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(4*I*f*x + 4*I*e) + (a*c^4 + 2*I*a*c^3*d + 2*I*a*c*d^3 - a*d^4)*f*e^(2*I*f
*x + 2*I*e))*sqrt(I/((-4*I*a^2*c^3 - 12*a^2*c^2*d + 12*I*a^2*c*d^2 + 4*a^2*d^3)*f^2))*log((((4*I*a*c^2 + 8*a*c
*d - 4*I*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (4*I*a*c^2 + 8*a*c*d - 4*I*a*d^2)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*
e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(I/((-4*I*a^2*c^3 - 12*a^2*c^2*d + 12*I*a^2*c*d^2 + 4*a^2*d^3)*f^
2)) + 2*(c - I*d)*e^(2*I*f*x + 2*I*e) + 2*c)*e^(-2*I*f*x - 2*I*e)) - ((a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(4*I*f
*x + 4*I*e) + (a*c^4 + 2*I*a*c^3*d + 2*I*a*c*d^3 - a*d^4)*f*e^(2*I*f*x + 2*I*e))*sqrt(I/((-4*I*a^2*c^3 - 12*a^
2*c^2*d + 12*I*a^2*c*d^2 + 4*a^2*d^3)*f^2))*log((((-4*I*a*c^2 - 8*a*c*d + 4*I*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (
-4*I*a*c^2 - 8*a*c*d + 4*I*a*d^2)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))
*sqrt(I/((-4*I*a^2*c^3 - 12*a^2*c^2*d + 12*I*a^2*c*d^2 + 4*a^2*d^3)*f^2)) + 2*(c - I*d)*e^(2*I*f*x + 2*I*e) +
2*c)*e^(-2*I*f*x - 2*I*e)) + ((a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(4*I*f*x + 4*I*e) + (a*c^4 + 2*I*a*c^3*d + 2*I
*a*c*d^3 - a*d^4)*f*e^(2*I*f*x + 2*I*e))*sqrt((I*c^2 - 8*c*d - 16*I*d^2)/((-4*I*a^2*c^5 + 20*a^2*c^4*d + 40*I*
a^2*c^3*d^2 - 40*a^2*c^2*d^3 - 20*I*a^2*c*d^4 + 4*a^2*d^5)*f^2))*log(-(c^2 + 5*I*c*d - 4*d^2 - ((2*I*a*c^3 - 6
*a*c^2*d - 6*I*a*c*d^2 + 2*a*d^3)*f*e^(2*I*f*x + 2*I*e) + (2*I*a*c^3 - 6*a*c^2*d - 6*I*a*c*d^2 + 2*a*d^3)*f)*s
qrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt((I*c^2 - 8*c*d - 16*I*d^2)/((-4*
I*a^2*c^5 + 20*a^2*c^4*d + 40*I*a^2*c^3*d^2 - 40*a^2*c^2*d^3 - 20*I*a^2*c*d^4 + 4*a^2*d^5)*f^2)) + (c^2 + 4*I*
c*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/((2*I*a*c^3 - 6*a*c^2*d - 6*I*a*c*d^2 + 2*a*d^3)*f)) - ((a*c^4
+ 2*a*c^2*d^2 + a*d^4)*f*e^(4*I*f*x + 4*I*e) + (a*c^4 + 2*I*a*c^3*d + 2*I*a*c*d^3 - a*d^4)*f*e^(2*I*f*x + 2*I*
e))*sqrt((I*c^2 - 8*c*d - 16*I*d^2)/((-4*I*a^2*c^5 + 20*a^2*c^4*d + 40*I*a^2*c^3*d^2 - 40*a^2*c^2*d^3 - 20*I*a
^2*c*d^4 + 4*a^2*d^5)*f^2))*log(-(c^2 + 5*I*c*d - 4*d^2 - ((-2*I*a*c^3 + 6*a*c^2*d + 6*I*a*c*d^2 - 2*a*d^3)*f*
e^(2*I*f*x + 2*I*e) + (-2*I*a*c^3 + 6*a*c^2*d + 6*I*a*c*d^2 - 2*a*d^3)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e)
+ c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt((I*c^2 - 8*c*d - 16*I*d^2)/((-4*I*a^2*c^5 + 20*a^2*c^4*d + 40*I*a^2
*c^3*d^2 - 40*a^2*c^2*d^3 - 20*I*a^2*c*d^4 + 4*a^2*d^5)*f^2)) + (c^2 + 4*I*c*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f
*x - 2*I*e)/((2*I*a*c^3 - 6*a*c^2*d - 6*I*a*c*d^2 + 2*a*d^3)*f)) + (I*c^2 + I*d^2 + (I*c^2 + 2*c*d - 9*I*d^2)*
e^(4*I*f*x + 4*I*e) + (2*I*c^2 + 2*c*d - 8*I*d^2)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c
+ I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(4*(a*c^4 + 2*a*c^2*d^2 + a*d^4)*f*e^(4*I*f*x + 4*I*e) + (4*a*c^4 + 8*I*a*
c^3*d + 8*I*a*c*d^3 - 4*a*d^4)*f*e^(2*I*f*x + 2*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/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.547, size = 666, normalized size = 3.25 \begin{align*} 2 \, d^{2}{\left (\frac{2 \,{\left (i \, c - 4 \, d\right )} \arctan \left (-\frac{4 \,{\left (\sqrt{d \tan \left (f x + e\right ) + c} c - \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} + i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{{\left (2 \, a c^{2} d^{2} f + 4 i \, a c d^{3} f - 2 \, a d^{4} f\right )} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}{\left (\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} - \frac{{\left (-i \, d \tan \left (f x + e\right ) - i \, c\right )} c - 5 \,{\left (d \tan \left (f x + e\right ) + c\right )} d + 4 \, c d + 4 i \, d^{2}}{{\left (4 \, a c^{3} d f + 4 i \, a c^{2} d^{2} f + 4 \, a c d^{3} f + 4 i \, a d^{4} f\right )}{\left (i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} - i \, \sqrt{d \tan \left (f x + e\right ) + c} c + \sqrt{d \tan \left (f x + e\right ) + c} d\right )}} + \frac{i \, \arctan \left (\frac{4 \,{\left (\sqrt{d \tan \left (f x + e\right ) + c} c - \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} - i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{{\left (a c d^{2} f - i \, a d^{3} f\right )} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}{\left (-\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d^2*(2*(I*c - 4*d)*arctan(-4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt
(-8*c + 8*sqrt(c^2 + d^2)) + I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-8*c + 8*sqrt(c^2 + d^2
))))/((2*a*c^2*d^2*f + 4*I*a*c*d^3*f - 2*a*d^4*f)*sqrt(-8*c + 8*sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) +
1)) - ((-I*d*tan(f*x + e) - I*c)*c - 5*(d*tan(f*x + e) + c)*d + 4*c*d + 4*I*d^2)/((4*a*c^3*d*f + 4*I*a*c^2*d^2
*f + 4*a*c*d^3*f + 4*I*a*d^4*f)*(I*(d*tan(f*x + e) + c)^(3/2) - I*sqrt(d*tan(f*x + e) + c)*c + sqrt(d*tan(f*x
+ e) + c)*d)) + I*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-8*
c + 8*sqrt(c^2 + d^2)) - I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-8*c + 8*sqrt(c^2 + d^2))))
/((a*c*d^2*f - I*a*d^3*f)*sqrt(-8*c + 8*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))