3.1098 \(\int \frac{1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=273 \[ \frac{d \left (c^2-8 i c d-3 d^2\right )}{2 a f (c-i d)^2 (c+i d)^3 (c+d \tan (e+f x))}+\frac{2 d^2 \left (3 c^2-2 i c d-d^2\right ) \log (c \cos (e+f x)+d \sin (e+f x))}{a f (c+i d)^4 (d+i c)^3}+\frac{x \left (6 c^2 d^2+4 i c^3 d+c^4-12 i c d^3-3 d^4\right )}{2 a (c-i d)^3 (c+i d)^4}+\frac{d (c-2 i d)}{2 a f (c-i d) (c+i d)^2 (c+d \tan (e+f x))^2}-\frac{1}{2 f (-d+i c) (a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \]

[Out]

((c^4 + (4*I)*c^3*d + 6*c^2*d^2 - (12*I)*c*d^3 - 3*d^4)*x)/(2*a*(c - I*d)^3*(c + I*d)^4) + (2*d^2*(3*c^2 - (2*
I)*c*d - d^2)*Log[c*Cos[e + f*x] + d*Sin[e + f*x]])/(a*(c + I*d)^4*(I*c + d)^3*f) + ((c - (2*I)*d)*d)/(2*a*(c
- I*d)*(c + I*d)^2*f*(c + d*Tan[e + f*x])^2) - 1/(2*(I*c - d)*f*(a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^2)
+ (d*(c^2 - (8*I)*c*d - 3*d^2))/(2*a*(c - I*d)^2*(c + I*d)^3*f*(c + d*Tan[e + f*x]))

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Rubi [A]  time = 0.494064, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3552, 3529, 3531, 3530} \[ \frac{d \left (c^2-8 i c d-3 d^2\right )}{2 a f (c-i d)^2 (c+i d)^3 (c+d \tan (e+f x))}+\frac{2 d^2 \left (3 c^2-2 i c d-d^2\right ) \log (c \cos (e+f x)+d \sin (e+f x))}{a f (c+i d)^4 (d+i c)^3}+\frac{x \left (6 c^2 d^2+4 i c^3 d+c^4-12 i c d^3-3 d^4\right )}{2 a (c-i d)^3 (c+i d)^4}+\frac{d (c-2 i d)}{2 a f (c-i d) (c+i d)^2 (c+d \tan (e+f x))^2}-\frac{1}{2 f (-d+i c) (a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c^4 + (4*I)*c^3*d + 6*c^2*d^2 - (12*I)*c*d^3 - 3*d^4)*x)/(2*a*(c - I*d)^3*(c + I*d)^4) + (2*d^2*(3*c^2 - (2*
I)*c*d - d^2)*Log[c*Cos[e + f*x] + d*Sin[e + f*x]])/(a*(c + I*d)^4*(I*c + d)^3*f) + ((c - (2*I)*d)*d)/(2*a*(c
- I*d)*(c + I*d)^2*f*(c + d*Tan[e + f*x])^2) - 1/(2*(I*c - d)*f*(a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^2)
+ (d*(c^2 - (8*I)*c*d - 3*d^2))/(2*a*(c - I*d)^2*(c + I*d)^3*f*(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 3531

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

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^3} \, dx &=-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^2}+\frac{\int \frac{a (i c-4 d)+3 i a d \tan (e+f x)}{(c+d \tan (e+f x))^3} \, dx}{2 a^2 (i c-d)}\\ &=\frac{(c-2 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^2}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^2}+\frac{\int \frac{-a \left (4 c d-i \left (c^2+3 d^2\right )\right )+2 a d (i c+2 d) \tan (e+f x)}{(c+d \tan (e+f x))^2} \, dx}{2 a^2 (i c-d) \left (c^2+d^2\right )}\\ &=\frac{(c-2 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^2}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^2}+\frac{d \left (c^2-8 i c d-3 d^2\right )}{2 a (c-i d)^2 (c+i d)^3 f (c+d \tan (e+f x))}-\frac{\int \frac{a \left (i c^3-4 c^2 d+5 i c d^2+4 d^3\right )+a d \left (i c^2+8 c d-3 i d^2\right ) \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{2 a^2 (i c-d)^3 (c-i d)^2}\\ &=\frac{\left (c^4+4 i c^3 d+6 c^2 d^2-12 i c d^3-3 d^4\right ) x}{2 a (c-i d)^3 (c+i d)^4}+\frac{(c-2 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^2}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^2}+\frac{d \left (c^2-8 i c d-3 d^2\right )}{2 a (c-i d)^2 (c+i d)^3 f (c+d \tan (e+f x))}+\frac{\left (2 d^2 \left (3 c^2-2 i c d-d^2\right )\right ) \int \frac{d-c \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{a (c+i d)^4 (i c+d)^3}\\ &=\frac{\left (c^4+4 i c^3 d+6 c^2 d^2-12 i c d^3-3 d^4\right ) x}{2 a (c-i d)^3 (c+i d)^4}+\frac{2 d^2 \left (3 c^2-2 i c d-d^2\right ) \log (c \cos (e+f x)+d \sin (e+f x))}{a (c+i d)^4 (i c+d)^3 f}+\frac{(c-2 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^2}-\frac{1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^2}+\frac{d \left (c^2-8 i c d-3 d^2\right )}{2 a (c-i d)^2 (c+i d)^3 f (c+d \tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 6.62863, size = 474, normalized size = 1.74 \[ \frac{\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (\frac{4 d^2 \left (3 i c^2+2 c d-i d^2\right ) \left (\cos \left (\frac{e}{2}\right )+i \sin \left (\frac{e}{2}\right )\right )^2 \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )}{f (c-i d)^3}+\frac{8 d^2 \left (-3 c^2+2 i c d+d^2\right ) \left (\cos \left (\frac{e}{2}\right )+i \sin \left (\frac{e}{2}\right )\right )^2 \tan ^{-1}\left (\frac{c \sin (f x)+d \cos (f x)}{d \sin (f x)-c \cos (f x)}\right )}{f (c-i d)^3}+\frac{8 d^2 x \left (-3 c^2+2 i c d+d^2\right ) (\cos (e)+i \sin (e))}{(c-i d)^3}+\frac{2 x \left (6 c^2 d^2+4 i c^3 d+c^4-12 i c d^3-3 d^4\right ) (\cos (e)+i \sin (e))}{(c-i d)^3}+\frac{2 d^4 (c+i d) (\sin (e)-i \cos (e))}{f (c-i d)^2 (c \cos (e+f x)+d \sin (e+f x))^2}+\frac{4 d^3 (c+i d) (d+4 i c) (\cos (e)+i \sin (e)) \sin (f x)}{f (c-i d)^2 (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac{(c+i d) (\sin (e)+i \cos (e)) \cos (2 f x)}{f}+\frac{(c+i d) (\cos (e)-i \sin (e)) \sin (2 f x)}{f}\right )}{4 (c+i d)^4 (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*((8*d^2*(-3*c^2 + (2*I)*c*d + d^2)*ArcTan[(d*Cos[f*x] + c*Sin[f*x])/(-(c
*Cos[f*x]) + d*Sin[f*x])]*(Cos[e/2] + I*Sin[e/2])^2)/((c - I*d)^3*f) + (4*d^2*((3*I)*c^2 + 2*c*d - I*d^2)*Log[
(c*Cos[e + f*x] + d*Sin[e + f*x])^2]*(Cos[e/2] + I*Sin[e/2])^2)/((c - I*d)^3*f) + (8*d^2*(-3*c^2 + (2*I)*c*d +
d^2)*x*(Cos[e] + I*Sin[e]))/(c - I*d)^3 + (2*(c^4 + (4*I)*c^3*d + 6*c^2*d^2 - (12*I)*c*d^3 - 3*d^4)*x*(Cos[e]
+ I*Sin[e]))/(c - I*d)^3 + ((c + I*d)*Cos[2*f*x]*(I*Cos[e] + Sin[e]))/f + ((c + I*d)*(Cos[e] - I*Sin[e])*Sin[
2*f*x])/f + (2*(c + I*d)*d^4*((-I)*Cos[e] + Sin[e]))/((c - I*d)^2*f*(c*Cos[e + f*x] + d*Sin[e + f*x])^2) + (4*
(c + I*d)*d^3*((4*I)*c + d)*(Cos[e] + I*Sin[e])*Sin[f*x])/((c - I*d)^2*f*(c*Cos[e] + d*Sin[e])*(c*Cos[e + f*x]
+ d*Sin[e + f*x]))))/(4*(c + I*d)^4*(a + I*a*Tan[e + f*x]))

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Maple [B]  time = 0.067, size = 542, normalized size = 2. \begin{align*}{\frac{-{\frac{i}{4}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) c}{af \left ( c+id \right ) ^{4}}}+{\frac{7\,\ln \left ( \tan \left ( fx+e \right ) -i \right ) d}{4\,af \left ( c+id \right ) ^{4}}}+{\frac{1}{2\,af \left ( c+id \right ) ^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{4}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{af \left ( id-c \right ) ^{3}}}+{\frac{{\frac{i}{2}}{d}^{2}{c}^{4}}{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{2}}}+{\frac{i{d}^{4}{c}^{2}}{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{2}}}+{\frac{{\frac{i}{2}}{d}^{6}}{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{2}}}+{\frac{3\,i{d}^{2}{c}^{3}}{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4} \left ( c+d\tan \left ( fx+e \right ) \right ) }}+{\frac{3\,i{d}^{4}c}{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4} \left ( c+d\tan \left ( fx+e \right ) \right ) }}+{\frac{{d}^{3}{c}^{2}}{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4} \left ( c+d\tan \left ( fx+e \right ) \right ) }}+{\frac{{d}^{5}}{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4} \left ( c+d\tan \left ( fx+e \right ) \right ) }}-{\frac{6\,i{d}^{2}\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{2}}{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4}}}+{\frac{2\,i{d}^{4}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) }{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4}}}-4\,{\frac{{d}^{3}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) c}{af \left ( id-c \right ) ^{3} \left ( c+id \right ) ^{4}}} \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,x)

[Out]

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

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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,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.26446, size = 1751, normalized size = 6.41 \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,x, algorithm="fricas")

[Out]

-(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6 - (2*I*c^6 - 4*c^5*d + 50*I*c^4*d^2 + 120*c^3*d^3 - 130*I*c^2*d^4 - 68*c*d
^5 + 14*I*d^6)*f*x*e^(6*I*f*x + 6*I*e) + (c^6 - 4*I*c^5*d - 5*c^4*d^2 + 32*I*c^3*d^3 - 5*c^2*d^4 + 36*I*c*d^5
+ d^6 - (4*I*c^6 - 16*c^5*d + 76*I*c^4*d^2 + 64*c^3*d^3 + 44*I*c^2*d^4 + 80*c*d^5 - 28*I*d^6)*f*x)*e^(4*I*f*x
+ 4*I*e) + (2*c^6 - 4*I*c^5*d + 2*c^4*d^2 + 24*I*c^3*d^3 - 58*c^2*d^4 - 20*I*c*d^5 - 10*d^6 - (2*I*c^6 - 12*c^
5*d + 18*I*c^4*d^2 - 24*c^3*d^3 + 30*I*c^2*d^4 - 12*c*d^5 + 14*I*d^6)*f*x)*e^(2*I*f*x + 2*I*e) + ((24*c^4*d^2
- 64*I*c^3*d^3 - 64*c^2*d^4 + 32*I*c*d^5 + 8*d^6)*e^(6*I*f*x + 6*I*e) + (48*c^4*d^2 - 32*I*c^3*d^3 + 32*c^2*d^
4 - 32*I*c*d^5 - 16*d^6)*e^(4*I*f*x + 4*I*e) + (24*c^4*d^2 + 32*I*c^3*d^3 + 8*d^6)*e^(2*I*f*x + 2*I*e))*log(((
I*c + d)*e^(2*I*f*x + 2*I*e) + I*c - d)/(I*c + d)))/((4*I*a*c^9 + 4*a*c^8*d + 16*I*a*c^7*d^2 + 16*a*c^6*d^3 +
24*I*a*c^5*d^4 + 24*a*c^4*d^5 + 16*I*a*c^3*d^6 + 16*a*c^2*d^7 + 4*I*a*c*d^8 + 4*a*d^9)*f*e^(6*I*f*x + 6*I*e) +
(8*I*a*c^9 - 8*a*c^8*d + 32*I*a*c^7*d^2 - 32*a*c^6*d^3 + 48*I*a*c^5*d^4 - 48*a*c^4*d^5 + 32*I*a*c^3*d^6 - 32*
a*c^2*d^7 + 8*I*a*c*d^8 - 8*a*d^9)*f*e^(4*I*f*x + 4*I*e) + (4*I*a*c^9 - 12*a*c^8*d - 32*a*c^6*d^3 - 24*I*a*c^5
*d^4 - 24*a*c^4*d^5 - 32*I*a*c^3*d^6 - 12*I*a*c*d^8 + 4*a*d^9)*f*e^(2*I*f*x + 2*I*e))

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

[Out]

Timed out

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Giac [B]  time = 1.48764, size = 676, normalized size = 2.48 \begin{align*} -\frac{2 \,{\left (\frac{{\left (i \, c - 7 \, d\right )} \log \left (i \, \tan \left (f x + e\right ) + 1\right )}{8 \, a c^{4} + 32 i \, a c^{3} d - 48 \, a c^{2} d^{2} - 32 i \, a c d^{3} + 8 \, a d^{4}} + \frac{{\left (3 \, c^{2} d^{3} - 2 i \, c d^{4} - d^{5}\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{i \, a c^{7} d - a c^{6} d^{2} + 3 i \, a c^{5} d^{3} - 3 \, a c^{4} d^{4} + 3 i \, a c^{3} d^{5} - 3 \, a c^{2} d^{6} + i \, a c d^{7} - a d^{8}} - \frac{i \, \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{8 \, a c^{3} - 24 i \, a c^{2} d - 24 \, a c d^{2} + 8 i \, a d^{3}} + \frac{-i \, c \tan \left (f x + e\right ) + 7 \, d \tan \left (f x + e\right ) - 3 \, c - 9 i \, d}{{\left (8 \, a c^{4} + 32 i \, a c^{3} d - 48 \, a c^{2} d^{2} - 32 i \, a c d^{3} + 8 \, a d^{4}\right )}{\left (\tan \left (f x + e\right ) - i\right )}} - \frac{18 \, c^{2} d^{4} \tan \left (f x + e\right )^{2} - 12 i \, c d^{5} \tan \left (f x + e\right )^{2} - 6 \, d^{6} \tan \left (f x + e\right )^{2} + 42 \, c^{3} d^{3} \tan \left (f x + e\right ) - 26 i \, c^{2} d^{4} \tan \left (f x + e\right ) - 6 \, c d^{5} \tan \left (f x + e\right ) - 2 i \, d^{6} \tan \left (f x + e\right ) + 25 \, c^{4} d^{2} - 14 i \, c^{3} d^{3} + 2 \, c^{2} d^{4} - 2 i \, c d^{5} + d^{6}}{{\left (4 i \, a c^{7} - 4 \, a c^{6} d + 12 i \, a c^{5} d^{2} - 12 \, a c^{4} d^{3} + 12 i \, a c^{3} d^{4} - 12 \, a c^{2} d^{5} + 4 i \, a c d^{6} - 4 \, a d^{7}\right )}{\left (d \tan \left (f x + e\right ) + c\right )}^{2}}\right )}}{f} \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,x, algorithm="giac")

[Out]

-2*((I*c - 7*d)*log(I*tan(f*x + e) + 1)/(8*a*c^4 + 32*I*a*c^3*d - 48*a*c^2*d^2 - 32*I*a*c*d^3 + 8*a*d^4) + (3*
c^2*d^3 - 2*I*c*d^4 - d^5)*log(abs(d*tan(f*x + e) + c))/(I*a*c^7*d - a*c^6*d^2 + 3*I*a*c^5*d^3 - 3*a*c^4*d^4 +
3*I*a*c^3*d^5 - 3*a*c^2*d^6 + I*a*c*d^7 - a*d^8) - I*log(-I*tan(f*x + e) + 1)/(8*a*c^3 - 24*I*a*c^2*d - 24*a*
c*d^2 + 8*I*a*d^3) + (-I*c*tan(f*x + e) + 7*d*tan(f*x + e) - 3*c - 9*I*d)/((8*a*c^4 + 32*I*a*c^3*d - 48*a*c^2*
d^2 - 32*I*a*c*d^3 + 8*a*d^4)*(tan(f*x + e) - I)) - (18*c^2*d^4*tan(f*x + e)^2 - 12*I*c*d^5*tan(f*x + e)^2 - 6
*d^6*tan(f*x + e)^2 + 42*c^3*d^3*tan(f*x + e) - 26*I*c^2*d^4*tan(f*x + e) - 6*c*d^5*tan(f*x + e) - 2*I*d^6*tan
(f*x + e) + 25*c^4*d^2 - 14*I*c^3*d^3 + 2*c^2*d^4 - 2*I*c*d^5 + d^6)/((4*I*a*c^7 - 4*a*c^6*d + 12*I*a*c^5*d^2
- 12*a*c^4*d^3 + 12*I*a*c^3*d^4 - 12*a*c^2*d^5 + 4*I*a*c*d^6 - 4*a*d^7)*(d*tan(f*x + e) + c)^2))/f