3.1097 \(\int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=104 \[ \frac {i a}{f (c-i d)^2 (c+d \tan (e+f x))}-\frac {a}{2 f (d+i c) (c+d \tan (e+f x))^2}-\frac {a \log (c \cos (e+f x)+d \sin (e+f x))}{f (d+i c)^3}+\frac {a x}{(c-i d)^3} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3529, 3531, 3530} \[ \frac {i a}{f (c-i d)^2 (c+d \tan (e+f x))}-\frac {a}{2 f (d+i c) (c+d \tan (e+f x))^2}-\frac {a \log (c \cos (e+f x)+d \sin (e+f x))}{f (d+i c)^3}+\frac {a x}{(c-i d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [B]  time = 4.22, size = 315, normalized size = 3.03 \[ \frac {\cos (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x)) \left (-\frac {(\cos (e)-i \sin (e)) \tan ^{-1}\left (\frac {\left (d^3-3 c^2 d\right ) \cos (2 e+f x)+c \left (c^2-3 d^2\right ) \sin (2 e+f x)}{c \left (c^2-3 d^2\right ) \cos (2 e+f x)-d \left (d^2-3 c^2\right ) \sin (2 e+f x)}\right )}{f}+\frac {d^2 (c-i d) (\sin (e)+i \cos (e))}{2 f (c+i d) (c \cos (e+f x)+d \sin (e+f x))^2}+\frac {d (c-i d) (d-2 i c) (\cos (e)-i \sin (e)) \sin (f x)}{f (c+i d) (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}-\frac {i (\cos (e)-i \sin (e)) \log \left ((c \cos (e+f x)+d \sin (e+f x))^2\right )}{2 f}+2 x (\cos (e)-i \sin (e))\right )}{(c-i d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.49, size = 273, normalized size = 2.62 \[ \frac {4 i \, a c d - 2 \, a d^{2} - 4 \, {\left (-i \, a c d - a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a c^{2} + 2 i \, a c d - a d^{2} + {\left (a c^{2} - 2 i \, a c d - a d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (a c^{2} + a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right )}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (2 i \, c^{5} + 6 \, c^{4} d - 4 i \, c^{3} d^{2} + 4 \, c^{2} d^{3} - 6 i \, c d^{4} - 2 \, d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c^{5} + c^{4} d + 2 i \, c^{3} d^{2} + 2 \, c^{2} d^{3} + i \, c d^{4} + d^{5}\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*I*a*c*d - 2*a*d^2 - 4*(-I*a*c*d - a*d^2)*e^(2*I*f*x + 2*I*e) + (a*c^2 + 2*I*a*c*d - a*d^2 + (a*c^2 - 2*I*a*
c*d - a*d^2)*e^(4*I*f*x + 4*I*e) + 2*(a*c^2 + a*d^2)*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)))/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f*e^(4*I*f*x + 4*I*e) +
(2*I*c^5 + 6*c^4*d - 4*I*c^3*d^2 + 4*c^2*d^3 - 6*I*c*d^4 - 2*d^5)*f*e^(2*I*f*x + 2*I*e) + (I*c^5 + c^4*d + 2*
I*c^3*d^2 + 2*c^2*d^3 + I*c*d^4 + d^5)*f)

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giac [B]  time = 0.83, size = 364, normalized size = 3.50 \[ \frac {2 \, {\left (\frac {a \log \left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}{2 i \, c^{3} + 6 \, c^{2} d - 6 i \, c d^{2} - 2 \, d^{3}} - \frac {a \log \left (-i \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}} + \frac {3 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 4 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 12 i \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 16 i \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 12 i \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a c^{4}}{{\left (-4 i \, c^{5} - 12 \, c^{4} d + 12 i \, c^{3} d^{2} + 4 \, c^{2} d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}^{2}}\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a*log(c*tan(1/2*f*x + 1/2*e)^2 - 2*d*tan(1/2*f*x + 1/2*e) - c)/(2*I*c^3 + 6*c^2*d - 6*I*c*d^2 - 2*d^3) - a*
log(-I*tan(1/2*f*x + 1/2*e) + 1)/(I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3) + (3*a*c^4*tan(1/2*f*x + 1/2*e)^4 - 4*a*c
^3*d*tan(1/2*f*x + 1/2*e)^3 - 12*I*a*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 - 4*a*c*d^3*tan(1/2*f*x + 1/2*e)^3 - 6*a*c
^4*tan(1/2*f*x + 1/2*e)^2 + 16*I*a*c*d^3*tan(1/2*f*x + 1/2*e)^2 + 4*a*d^4*tan(1/2*f*x + 1/2*e)^2 + 4*a*c^3*d*t
an(1/2*f*x + 1/2*e) + 12*I*a*c^2*d^2*tan(1/2*f*x + 1/2*e) + 4*a*c*d^3*tan(1/2*f*x + 1/2*e) + 3*a*c^4)/((-4*I*c
^5 - 12*c^4*d + 12*I*c^3*d^2 + 4*c^2*d^3)*(c*tan(1/2*f*x + 1/2*e)^2 - 2*d*tan(1/2*f*x + 1/2*e) - c)^2))/f

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maple [B]  time = 0.27, size = 493, normalized size = 4.74 \[ \frac {3 i a \arctan \left (\tan \left (f x +e \right )\right ) c^{2} d}{f \left (c^{2}+d^{2}\right )^{3}}-\frac {i a \arctan \left (\tan \left (f x +e \right )\right ) d^{3}}{f \left (c^{2}+d^{2}\right )^{3}}-\frac {2 a c d}{f \left (c^{2}+d^{2}\right )^{2} \left (c +d \tan \left (f x +e \right )\right )}-\frac {i a \ln \left (c +d \tan \left (f x +e \right )\right ) c^{3}}{f \left (c^{2}+d^{2}\right )^{3}}-\frac {a d}{2 f \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{2}}+\frac {i a \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c^{3}}{2 f \left (c^{2}+d^{2}\right )^{3}}-\frac {3 i a \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c \,d^{2}}{2 f \left (c^{2}+d^{2}\right )^{3}}+\frac {3 a \ln \left (c +d \tan \left (f x +e \right )\right ) c^{2} d}{f \left (c^{2}+d^{2}\right )^{3}}-\frac {a \ln \left (c +d \tan \left (f x +e \right )\right ) d^{3}}{f \left (c^{2}+d^{2}\right )^{3}}-\frac {3 a \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c^{2} d}{2 f \left (c^{2}+d^{2}\right )^{3}}+\frac {a \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) d^{3}}{2 f \left (c^{2}+d^{2}\right )^{3}}-\frac {i a \,d^{2}}{f \left (c^{2}+d^{2}\right )^{2} \left (c +d \tan \left (f x +e \right )\right )}+\frac {i a \,c^{2}}{f \left (c^{2}+d^{2}\right )^{2} \left (c +d \tan \left (f x +e \right )\right )}+\frac {3 i a \ln \left (c +d \tan \left (f x +e \right )\right ) c \,d^{2}}{f \left (c^{2}+d^{2}\right )^{3}}+\frac {i a c}{2 f \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )^{2}}+\frac {a \arctan \left (\tan \left (f x +e \right )\right ) c^{3}}{f \left (c^{2}+d^{2}\right )^{3}}-\frac {3 a \arctan \left (\tan \left (f x +e \right )\right ) c \,d^{2}}{f \left (c^{2}+d^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*I/f*a/(c^2+d^2)^3*arctan(tan(f*x+e))*c^2*d-I/f*a/(c^2+d^2)^3*arctan(tan(f*x+e))*d^3-2/f*a/(c^2+d^2)^2/(c+d*t
an(f*x+e))*c*d-I/f*a/(c^2+d^2)^3*ln(c+d*tan(f*x+e))*c^3-1/2/f*a/(c^2+d^2)/(c+d*tan(f*x+e))^2*d+1/2*I/f*a/(c^2+
d^2)^3*ln(1+tan(f*x+e)^2)*c^3-3/2*I/f*a/(c^2+d^2)^3*ln(1+tan(f*x+e)^2)*c*d^2+3/f*a/(c^2+d^2)^3*ln(c+d*tan(f*x+
e))*c^2*d-1/f*a/(c^2+d^2)^3*ln(c+d*tan(f*x+e))*d^3-3/2/f*a/(c^2+d^2)^3*ln(1+tan(f*x+e)^2)*c^2*d+1/2/f*a/(c^2+d
^2)^3*ln(1+tan(f*x+e)^2)*d^3-I/f*a/(c^2+d^2)^2/(c+d*tan(f*x+e))*d^2+I/f*a/(c^2+d^2)^2/(c+d*tan(f*x+e))*c^2+3*I
/f*a/(c^2+d^2)^3*ln(c+d*tan(f*x+e))*c*d^2+1/2*I/f*a/(c^2+d^2)/(c+d*tan(f*x+e))^2*c+1/f*a/(c^2+d^2)^3*arctan(ta
n(f*x+e))*c^3-3/f*a/(c^2+d^2)^3*arctan(tan(f*x+e))*c*d^2

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maxima [B]  time = 0.43, size = 325, normalized size = 3.12 \[ \frac {\frac {2 \, {\left (a c^{3} + 3 i \, a c^{2} d - 3 \, a c d^{2} - i \, a d^{3}\right )} {\left (f x + e\right )}}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac {2 \, {\left (-i \, a c^{3} + 3 \, a c^{2} d + 3 i \, a c d^{2} - a d^{3}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac {{\left (i \, a c^{3} - 3 \, a c^{2} d - 3 i \, a c d^{2} + a d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac {3 i \, a c^{3} - 5 \, a c^{2} d - i \, a c d^{2} - a d^{3} - {\left (-2 i \, a c^{2} d + 4 \, a c d^{2} + 2 i \, a d^{3}\right )} \tan \left (f x + e\right )}{c^{6} + 2 \, c^{4} d^{2} + c^{2} d^{4} + {\left (c^{4} d^{2} + 2 \, c^{2} d^{4} + d^{6}\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left (c^{5} d + 2 \, c^{3} d^{3} + c d^{5}\right )} \tan \left (f x + e\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.37, size = 281, normalized size = 2.70 \[ -\frac {\frac {\left (3\,a\,c-a\,d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d^2\,\left (-c^2+c\,d\,2{}\mathrm {i}+d^2\right )}+\frac {a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{d\,\left (-c^2+c\,d\,2{}\mathrm {i}+d^2\right )}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2+\frac {c^2}{d^2}+\frac {2\,c\,\mathrm {tan}\left (e+f\,x\right )}{d}\right )}+\frac {a\,\mathrm {atan}\left (\frac {c^3-c^2\,d\,1{}\mathrm {i}+c\,d^2-d^3\,1{}\mathrm {i}}{{\left (c-d\,1{}\mathrm {i}\right )}^2\,\left (d+c\,1{}\mathrm {i}\right )}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (2\,c^8\,d^2+8\,c^6\,d^4+12\,c^4\,d^6+8\,c^2\,d^8+2\,d^{10}\right )\,1{}\mathrm {i}}{{\left (c-d\,1{}\mathrm {i}\right )}^2\,\left (d+c\,1{}\mathrm {i}\right )\,\left (-c^6\,d\,1{}\mathrm {i}+2\,c^5\,d^2-c^4\,d^3\,1{}\mathrm {i}+4\,c^3\,d^4+c^2\,d^5\,1{}\mathrm {i}+2\,c\,d^6+d^7\,1{}\mathrm {i}\right )}\right )\,2{}\mathrm {i}}{f\,{\left (c-d\,1{}\mathrm {i}\right )}^2\,\left (d+c\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*atan((c*d^2 - c^2*d*1i + c^3 - d^3*1i)/((c - d*1i)^2*(c*1i + d)) - (tan(e + f*x)*(2*d^10 + 8*c^2*d^8 + 12*c
^4*d^6 + 8*c^6*d^4 + 2*c^8*d^2)*1i)/((c - d*1i)^2*(c*1i + d)*(2*c*d^6 - c^6*d*1i + d^7*1i + c^2*d^5*1i + 4*c^3
*d^4 - c^4*d^3*1i + 2*c^5*d^2)))*2i)/(f*(c - d*1i)^2*(c*1i + d)) - (((3*a*c - a*d*1i)*1i)/(2*d^2*(c*d*2i - c^2
+ d^2)) + (a*tan(e + f*x)*1i)/(d*(c*d*2i - c^2 + d^2)))/(f*(tan(e + f*x)^2 + c^2/d^2 + (2*c*tan(e + f*x))/d))

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sympy [B]  time = 8.24, size = 355, normalized size = 3.41 \[ - \frac {i a \log {\left (\frac {- i c + d}{- i c e^{2 i e} - d e^{2 i e}} + e^{2 i f x} \right )}}{f \left (c - i d\right )^{3}} + \frac {4 i a c d - 2 a d^{2} + \left (4 i a c d e^{2 i e} + 4 a d^{2} e^{2 i e}\right ) e^{2 i f x}}{i c^{5} f + c^{4} d f + 2 i c^{3} d^{2} f + 2 c^{2} d^{3} f + i c d^{4} f + d^{5} f + \left (2 i c^{5} f e^{2 i e} + 6 c^{4} d f e^{2 i e} - 4 i c^{3} d^{2} f e^{2 i e} + 4 c^{2} d^{3} f e^{2 i e} - 6 i c d^{4} f e^{2 i e} - 2 d^{5} f e^{2 i e}\right ) e^{2 i f x} + \left (i c^{5} f e^{4 i e} + 5 c^{4} d f e^{4 i e} - 10 i c^{3} d^{2} f e^{4 i e} - 10 c^{2} d^{3} f e^{4 i e} + 5 i c d^{4} f e^{4 i e} + d^{5} f e^{4 i e}\right ) e^{4 i f x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a*log((-I*c + d)/(-I*c*exp(2*I*e) - d*exp(2*I*e)) + exp(2*I*f*x))/(f*(c - I*d)**3) + (4*I*a*c*d - 2*a*d**2
+ (4*I*a*c*d*exp(2*I*e) + 4*a*d**2*exp(2*I*e))*exp(2*I*f*x))/(I*c**5*f + c**4*d*f + 2*I*c**3*d**2*f + 2*c**2*d
**3*f + I*c*d**4*f + d**5*f + (2*I*c**5*f*exp(2*I*e) + 6*c**4*d*f*exp(2*I*e) - 4*I*c**3*d**2*f*exp(2*I*e) + 4*
c**2*d**3*f*exp(2*I*e) - 6*I*c*d**4*f*exp(2*I*e) - 2*d**5*f*exp(2*I*e))*exp(2*I*f*x) + (I*c**5*f*exp(4*I*e) +
5*c**4*d*f*exp(4*I*e) - 10*I*c**3*d**2*f*exp(4*I*e) - 10*c**2*d**3*f*exp(4*I*e) + 5*I*c*d**4*f*exp(4*I*e) + d*
*5*f*exp(4*I*e))*exp(4*I*f*x))

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