3.1092 \(\int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=202 \[ \frac {x \left (c^3+3 i c^2 d+3 c d^2-3 i d^3\right )}{2 a (c-i d)^2 (c+i d)^3}+\frac {d^2 (3 c-i d) \log (c \cos (e+f x)+d \sin (e+f x))}{a f (-d+i c)^3 (c-i d)^2}+\frac {d (c-3 i d)}{2 a f (c-i d) (c+i d)^2 (c+d \tan (e+f x))}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) (c+d \tan (e+f x))} \]

[Out]

1/2*(c^3+3*I*c^2*d+3*c*d^2-3*I*d^3)*x/a/(c-I*d)^2/(c+I*d)^3+(3*c-I*d)*d^2*ln(c*cos(f*x+e)+d*sin(f*x+e))/a/(I*c
-d)^3/(c-I*d)^2/f+1/2*(c-3*I*d)*d/a/(c-I*d)/(c+I*d)^2/f/(c+d*tan(f*x+e))-1/2/(I*c-d)/f/(a+I*a*tan(f*x+e))/(c+d
*tan(f*x+e))

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Rubi [A]  time = 0.32, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3552, 3529, 3531, 3530} \[ \frac {x \left (3 i c^2 d+c^3+3 c d^2-3 i d^3\right )}{2 a (c-i d)^2 (c+i d)^3}+\frac {d^2 (3 c-i d) \log (c \cos (e+f x)+d \sin (e+f x))}{a f (-d+i c)^3 (c-i d)^2}+\frac {d (c-3 i d)}{2 a f (c-i d) (c+i d)^2 (c+d \tan (e+f x))}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) (c+d \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c^3 + (3*I)*c^2*d + 3*c*d^2 - (3*I)*d^3)*x)/(2*a*(c - I*d)^2*(c + I*d)^3) + ((3*c - I*d)*d^2*Log[c*Cos[e + f
*x] + d*Sin[e + f*x]])/(a*(I*c - d)^3*(c - I*d)^2*f) + ((c - (3*I)*d)*d)/(2*a*(c - I*d)*(c + I*d)^2*f*(c + d*T
an[e + f*x])) - 1/(2*(I*c - d)*f*(a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x]))

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 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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 3.02, size = 385, normalized size = 1.91 \[ \frac {\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (\frac {2 x \left (c^3+3 i c^2 d+3 c d^2-3 i d^3\right ) (\cos (e)+i \sin (e))}{(c-i d)^2}+\frac {4 i d^3 (c+i d) (\cos (e)+i \sin (e)) \sin (f x)}{f (c-i d) (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac {2 d^2 (d+3 i c) \left (\cos \left (\frac {e}{2}\right )+i \sin \left (\frac {e}{2}\right )\right )^2 \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )}{f (c-i d)^2}-\frac {4 d^2 (3 c-i d) \left (\cos \left (\frac {e}{2}\right )+i \sin \left (\frac {e}{2}\right )\right )^2 \tan ^{-1}\left (\frac {c \sin (f x)+d \cos (f x)}{d \sin (f x)-c \cos (f x)}\right )}{f (c-i d)^2}-\frac {4 d^2 x (3 c-i d) (\cos (e)+i \sin (e))}{(c-i d)^2}+\frac {(c+i d) (\sin (e)+i \cos (e)) \cos (2 f x)}{f}+\frac {(c+i d) (\cos (e)-i \sin (e)) \sin (2 f x)}{f}\right )}{4 (c+i d)^3 (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 332, normalized size = 1.64 \[ \frac {i \, c^{4} + 2 i \, c^{2} d^{2} + i \, d^{4} + {\left (2 \, c^{4} + 4 i \, c^{3} d + 24 \, c^{2} d^{2} - 28 i \, c d^{3} - 10 \, d^{4}\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (i \, c^{4} + 2 \, c^{3} d - 6 \, c d^{3} - 9 i \, d^{4} + {\left (2 \, c^{4} + 8 i \, c^{3} d + 12 \, c^{2} d^{2} + 8 i \, c d^{3} + 10 \, d^{4}\right )} f x\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left ({\left (12 i \, c^{2} d^{2} + 16 \, c d^{3} - 4 i \, d^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (12 i \, c^{2} d^{2} - 8 \, c d^{3} + 4 i \, d^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right )}{4 \, {\left (a c^{6} + 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} + a d^{6}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (4 \, a c^{6} + 8 i \, a c^{5} d + 4 \, a c^{4} d^{2} + 16 i \, a c^{3} d^{3} - 4 \, a c^{2} d^{4} + 8 i \, a c d^{5} - 4 \, a d^{6}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

(I*c^4 + 2*I*c^2*d^2 + I*d^4 + (2*c^4 + 4*I*c^3*d + 24*c^2*d^2 - 28*I*c*d^3 - 10*d^4)*f*x*e^(4*I*f*x + 4*I*e)
+ (I*c^4 + 2*c^3*d - 6*c*d^3 - 9*I*d^4 + (2*c^4 + 8*I*c^3*d + 12*c^2*d^2 + 8*I*c*d^3 + 10*d^4)*f*x)*e^(2*I*f*x
+ 2*I*e) + ((12*I*c^2*d^2 + 16*c*d^3 - 4*I*d^4)*e^(4*I*f*x + 4*I*e) + (12*I*c^2*d^2 - 8*c*d^3 + 4*I*d^4)*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*(a*c^6 + 3*a*c^4*d^2 + 3*a*c^2*d
^4 + a*d^6)*f*e^(4*I*f*x + 4*I*e) + (4*a*c^6 + 8*I*a*c^5*d + 4*a*c^4*d^2 + 16*I*a*c^3*d^3 - 4*a*c^2*d^4 + 8*I*
a*c*d^5 - 4*a*d^6)*f*e^(2*I*f*x + 2*I*e))

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giac [A]  time = 0.59, size = 343, normalized size = 1.70 \[ \frac {\frac {16 \, {\left (3 \, c d^{3} - i \, d^{4}\right )} \log \left (i \, d \tan \left (f x + e\right ) + i \, c\right )}{-2 i \, a c^{5} d + 2 \, a c^{4} d^{2} - 4 i \, a c^{3} d^{3} + 4 \, a c^{2} d^{4} - 2 i \, a c d^{5} + 2 \, a d^{6}} - \frac {16 \, {\left (i \, c - 5 \, d\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{8 \, a c^{3} + 24 i \, a c^{2} d - 24 \, a c d^{2} - 8 i \, a d^{3}} - \frac {16 \, \log \left (-i \, \tan \left (f x + e\right ) + 1\right )}{8 i \, a c^{2} + 16 \, a c d - 8 i \, a d^{2}} + \frac {i \, c^{2} d \tan \left (f x + e\right )^{2} - 2 \, c d^{2} \tan \left (f x + e\right )^{2} - i \, d^{3} \tan \left (f x + e\right )^{2} + i \, c^{3} \tan \left (f x + e\right ) + 3 \, c^{2} d \tan \left (f x + e\right ) - 15 i \, c d^{2} \tan \left (f x + e\right ) - 13 \, d^{3} \tan \left (f x + e\right ) + 5 \, c^{3} - 6 i \, c^{2} d - 13 \, c d^{2} + 8 i \, d^{3}}{{\left (a c^{4} + 2 \, a c^{2} d^{2} + a d^{4}\right )} {\left (d \tan \left (f x + e\right )^{2} + c \tan \left (f x + e\right ) - i \, d \tan \left (f x + e\right ) - i \, c\right )}}}{8 \, f} \]

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))^2,x, algorithm="giac")

[Out]

1/8*(16*(3*c*d^3 - I*d^4)*log(I*d*tan(f*x + e) + I*c)/(-2*I*a*c^5*d + 2*a*c^4*d^2 - 4*I*a*c^3*d^3 + 4*a*c^2*d^
4 - 2*I*a*c*d^5 + 2*a*d^6) - 16*(I*c - 5*d)*log(tan(f*x + e) - I)/(8*a*c^3 + 24*I*a*c^2*d - 24*a*c*d^2 - 8*I*a
*d^3) - 16*log(-I*tan(f*x + e) + 1)/(8*I*a*c^2 + 16*a*c*d - 8*I*a*d^2) + (I*c^2*d*tan(f*x + e)^2 - 2*c*d^2*tan
(f*x + e)^2 - I*d^3*tan(f*x + e)^2 + I*c^3*tan(f*x + e) + 3*c^2*d*tan(f*x + e) - 15*I*c*d^2*tan(f*x + e) - 13*
d^3*tan(f*x + e) + 5*c^3 - 6*I*c^2*d - 13*c*d^2 + 8*I*d^3)/((a*c^4 + 2*a*c^2*d^2 + a*d^4)*(d*tan(f*x + e)^2 +
c*tan(f*x + e) - I*d*tan(f*x + e) - I*c)))/f

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maple [A]  time = 0.32, size = 281, normalized size = 1.39 \[ \frac {i \ln \left (\tan \left (f x +e \right )+i\right )}{4 f a \left (i d -c \right )^{2}}+\frac {3 i d^{2} \ln \left (c +d \tan \left (f x +e \right )\right ) c}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3}}+\frac {d^{3} \ln \left (c +d \tan \left (f x +e \right )\right )}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3}}-\frac {i d^{2} c^{2}}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3} \left (c +d \tan \left (f x +e \right )\right )}-\frac {i d^{4}}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3} \left (c +d \tan \left (f x +e \right )\right )}+\frac {1}{2 f a \left (i d +c \right )^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {i \ln \left (\tan \left (f x +e \right )-i\right ) c}{4 f a \left (i d +c \right )^{3}}+\frac {5 \ln \left (\tan \left (f x +e \right )-i\right ) d}{4 f a \left (i d +c \right )^{3}} \]

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))^2,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 8.77, size = 1334, normalized size = 6.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log(tan(e + f*x)*(a*d^4 - a*c^2*d^2 + a*c*d^3*2i)*(c*d^3*6i + 9*d^4 - c^2*d^2) - (a*d^4 - a*c^2*d^2 + a
*c*d^3*2i)*(11*c*d^3 + c^3*d - d^4*6i) - root(a^3*c^5*d^5*e^3*192i + a^3*c^7*d^3*e^3*128i + a^3*c^3*d^7*e^3*12
8i + 48*a^3*c^8*d^2*e^3 - 48*a^3*c^2*d^8*e^3 + 32*a^3*c^6*d^4*e^3 - 32*a^3*c^4*d^6*e^3 + a^3*c^9*d*e^3*32i + a
^3*c*d^9*e^3*32i - 16*a^3*d^10*e^3 + 16*a^3*c^10*e^3 + 135*a*c^2*d^4*e + a*c^3*d^3*e*12i - 3*a*c^4*d^2*e - a*c
*d^5*e*90i + a*c^5*d*e*6i - 21*a*d^6*e + a*c^6*e - c^2*d^2*3i + 14*c*d^3 - d^4*5i, e, k)*((a*d^4 - a*c^2*d^2 +
a*c*d^3*2i)*(2*a*c^6 - 6*a*d^6 - 10*a*c^2*d^4 + a*c^3*d^3*16i - 2*a*c^4*d^2 + a*c*d^5*8i + a*c^5*d*8i) + root
(a^3*c^5*d^5*e^3*192i + a^3*c^7*d^3*e^3*128i + a^3*c^3*d^7*e^3*128i + 48*a^3*c^8*d^2*e^3 - 48*a^3*c^2*d^8*e^3
+ 32*a^3*c^6*d^4*e^3 - 32*a^3*c^4*d^6*e^3 + a^3*c^9*d*e^3*32i + a^3*c*d^9*e^3*32i - 16*a^3*d^10*e^3 + 16*a^3*c
^10*e^3 + 135*a*c^2*d^4*e + a*c^3*d^3*e*12i - 3*a*c^4*d^2*e - a*c*d^5*e*90i + a*c^5*d*e*6i - 21*a*d^6*e + a*c^
6*e - c^2*d^2*3i + 14*c*d^3 - d^4*5i, e, k)*((a*d^4 - a*c^2*d^2 + a*c*d^3*2i)*(32*a^2*c^7*d - 32*a^2*c*d^7 + a
^2*c^2*d^6*64i - 32*a^2*c^3*d^5 + a^2*c^4*d^4*128i + 32*a^2*c^5*d^3 + a^2*c^6*d^2*64i) + tan(e + f*x)*(a*d^4 -
a*c^2*d^2 + a*c*d^3*2i)*(a^2*c*d^7*48i - 24*a^2*d^8 - 8*a^2*c^8 - a^2*c^7*d*16i - 16*a^2*c^2*d^6 + a^2*c^3*d^
5*80i + 32*a^2*c^4*d^4 + a^2*c^5*d^3*16i + 16*a^2*c^6*d^2)) + tan(e + f*x)*(a*d^4 - a*c^2*d^2 + a*c*d^3*2i)*(a
*d^6*12i + a*c^2*d^4*40i - 8*a*c^3*d^3 + a*c^4*d^2*28i - 12*a*c*d^5 + 4*a*c^5*d)))*root(a^3*c^5*d^5*e^3*192i +
a^3*c^7*d^3*e^3*128i + a^3*c^3*d^7*e^3*128i + 48*a^3*c^8*d^2*e^3 - 48*a^3*c^2*d^8*e^3 + 32*a^3*c^6*d^4*e^3 -
32*a^3*c^4*d^6*e^3 + a^3*c^9*d*e^3*32i + a^3*c*d^9*e^3*32i - 16*a^3*d^10*e^3 + 16*a^3*c^10*e^3 + 135*a*c^2*d^4
*e + a*c^3*d^3*e*12i - 3*a*c^4*d^2*e - a*c*d^5*e*90i + a*c^5*d*e*6i - 21*a*d^6*e + a*c^6*e - c^2*d^2*3i + 14*c
*d^3 - d^4*5i, e, k), k, 1, 3)/f - ((c*d*2i - 2*c^2 + 4*d^2)/(4*a*d*(c*d^2 + c^2*d*1i + c^3 + d^3*1i)) - (tan(
e + f*x)*(c - d*3i))/(2*a*(c*d^2 + c^2*d*1i + c^3 + d^3*1i)))/(f*(tan(e + f*x)*(c/d - 1i) - (c*1i)/d + tan(e +
f*x)^2))

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sympy [A]  time = 27.61, size = 515, normalized size = 2.55 \[ - \frac {2 i d^{3}}{i a c^{5} f - a c^{4} d f + 2 i a c^{3} d^{2} f - 2 a c^{2} d^{3} f + i a c d^{4} f - a d^{5} f + \left (i a c^{5} f e^{2 i e} + a c^{4} d f e^{2 i e} + 2 i a c^{3} d^{2} f e^{2 i e} + 2 a c^{2} d^{3} f e^{2 i e} + i a c d^{4} f e^{2 i e} + a d^{5} f e^{2 i e}\right ) e^{2 i f x}} + \frac {x \left (- c - 5 i d\right )}{- 2 a c^{3} - 6 i a c^{2} d + 6 a c d^{2} + 2 i a d^{3}} + \begin {cases} \frac {i e^{- 2 i f x}}{4 a c^{2} f e^{2 i e} + 8 i a c d f e^{2 i e} - 4 a d^{2} f e^{2 i e}} & \text {for}\: 4 a c^{2} f e^{2 i e} + 8 i a c d f e^{2 i e} - 4 a d^{2} f e^{2 i e} \neq 0 \\x \left (- \frac {i c - 5 d}{2 i a c^{3} - 6 a c^{2} d - 6 i a c d^{2} + 2 a d^{3}} + \frac {- c e^{2 i e} - c - 5 i d e^{2 i e} - i d}{- 2 a c^{3} e^{2 i e} - 6 i a c^{2} d e^{2 i e} + 6 a c d^{2} e^{2 i e} + 2 i a d^{3} e^{2 i e}}\right ) & \text {otherwise} \end {cases} + \frac {i d^{2} \left (3 c - i d\right ) \log {\left (\frac {i c - d}{i c e^{2 i e} + d e^{2 i e}} + e^{2 i f x} \right )}}{a f \left (c - i d\right )^{2} \left (c + i d\right )^{3}} \]

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))**2,x)

[Out]

-2*I*d**3/(I*a*c**5*f - a*c**4*d*f + 2*I*a*c**3*d**2*f - 2*a*c**2*d**3*f + I*a*c*d**4*f - a*d**5*f + (I*a*c**5
*f*exp(2*I*e) + a*c**4*d*f*exp(2*I*e) + 2*I*a*c**3*d**2*f*exp(2*I*e) + 2*a*c**2*d**3*f*exp(2*I*e) + I*a*c*d**4
*f*exp(2*I*e) + a*d**5*f*exp(2*I*e))*exp(2*I*f*x)) + x*(-c - 5*I*d)/(-2*a*c**3 - 6*I*a*c**2*d + 6*a*c*d**2 + 2
*I*a*d**3) + Piecewise((I*exp(-2*I*f*x)/(4*a*c**2*f*exp(2*I*e) + 8*I*a*c*d*f*exp(2*I*e) - 4*a*d**2*f*exp(2*I*e
)), Ne(4*a*c**2*f*exp(2*I*e) + 8*I*a*c*d*f*exp(2*I*e) - 4*a*d**2*f*exp(2*I*e), 0)), (x*(-(I*c - 5*d)/(2*I*a*c*
*3 - 6*a*c**2*d - 6*I*a*c*d**2 + 2*a*d**3) + (-c*exp(2*I*e) - c - 5*I*d*exp(2*I*e) - I*d)/(-2*a*c**3*exp(2*I*e
) - 6*I*a*c**2*d*exp(2*I*e) + 6*a*c*d**2*exp(2*I*e) + 2*I*a*d**3*exp(2*I*e))), True)) + I*d**2*(3*c - I*d)*log
((I*c - d)/(I*c*exp(2*I*e) + d*exp(2*I*e)) + exp(2*I*f*x))/(a*f*(c - I*d)**2*(c + I*d)**3)

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