3.1089 \(\int \frac{(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=142 \[ -\frac{a^3 (-d+i c) (c-3 i d) \log (c \cos (e+f x)+d \sin (e+f x))}{d^2 f (c-i d)^2}+\frac{(c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) (c+d \tan (e+f x))}+\frac{4 a^3 x}{(c-i d)^2}+\frac{i a^3 \log (\cos (e+f x))}{d^2 f} \]

[Out]

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

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Rubi [A]  time = 0.369006, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3553, 3589, 3475, 3531, 3530} \[ -\frac{a^3 (-d+i c) (c-3 i d) \log (c \cos (e+f x)+d \sin (e+f x))}{d^2 f (c-i d)^2}+\frac{(c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) (c+d \tan (e+f x))}+\frac{4 a^3 x}{(c-i d)^2}+\frac{i a^3 \log (\cos (e+f x))}{d^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3589

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [B]  time = 8.64622, size = 1936, normalized size = 13.63 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*Cos[3*e]*Cos[e + f*x]^3*Log[Cos[e + f*x]^2]*(a + I*a*Tan[e + f*x])^3)/(d^2*f*(Cos[f*x] + I*Sin[f*x])^3)
+ (Cos[e + f*x]^3*(c^2*Cos[(3*e)/2] - (2*I)*c*d*Cos[(3*e)/2] + 3*d^2*Cos[(3*e)/2] - I*c^2*Sin[(3*e)/2] - 2*c*
d*Sin[(3*e)/2] - (3*I)*d^2*Sin[(3*e)/2])*((ArcTan[(2*c*d*Cos[4*e + f*x] - c^2*Sin[4*e + f*x] + d^2*Sin[4*e + f
*x])/(c^2*Cos[4*e + f*x] - d^2*Cos[4*e + f*x] + 2*c*d*Sin[4*e + f*x])]*Cos[(3*e)/2])/d^2 - (I*ArcTan[(2*c*d*Co
s[4*e + f*x] - c^2*Sin[4*e + f*x] + d^2*Sin[4*e + f*x])/(c^2*Cos[4*e + f*x] - d^2*Cos[4*e + f*x] + 2*c*d*Sin[4
*e + f*x])]*Sin[(3*e)/2])/d^2)*(a + I*a*Tan[e + f*x])^3)/((c - I*d)^2*f*(Cos[f*x] + I*Sin[f*x])^3) + (Cos[e +
f*x]^3*(c^2*Cos[(3*e)/2] - (2*I)*c*d*Cos[(3*e)/2] + 3*d^2*Cos[(3*e)/2] - I*c^2*Sin[(3*e)/2] - 2*c*d*Sin[(3*e)/
2] - (3*I)*d^2*Sin[(3*e)/2])*(((-I/2)*Cos[(3*e)/2]*Log[(c*Cos[e + f*x] + d*Sin[e + f*x])^2])/d^2 - (Log[(c*Cos
[e + f*x] + d*Sin[e + f*x])^2]*Sin[(3*e)/2])/(2*d^2))*(a + I*a*Tan[e + f*x])^3)/((c - I*d)^2*f*(Cos[f*x] + I*S
in[f*x])^3) + (Cos[e + f*x]^3*Log[Cos[e + f*x]^2]*Sin[3*e]*(a + I*a*Tan[e + f*x])^3)/(2*d^2*f*(Cos[f*x] + I*Si
n[f*x])^3) + (Cos[e + f*x]^3*(4*f*x*Cos[3*e] - (4*I)*f*x*Sin[3*e])*(a + I*a*Tan[e + f*x])^3)/((c - I*d)^2*f*(C
os[f*x] + I*Sin[f*x])^3) + (Cos[e + f*x]^3*(Cos[3*e]/d - (I*Sin[3*e])/d)*(I*c^2*Sin[f*x] - 2*c*d*Sin[f*x] - I*
d^2*Sin[f*x])*(a + I*a*Tan[e + f*x])^3)/((c - I*d)*f*(c*Cos[e] + d*Sin[e])*(Cos[f*x] + I*Sin[f*x])^3*(c*Cos[e
+ f*x] + d*Sin[e + f*x])) + (x*Cos[e + f*x]^3*(Cos[e]/(2*d^2) - Cos[e]^3/(2*d^2) - (I*Sin[e])/d^2 + ((2*I)*Cos
[e]^2*Sin[e])/d^2 + (3*Cos[e]*Sin[e]^2)/d^2 - ((2*I)*Sin[e]^3)/d^2 + (5*c*Cos[e]^4)/((c - I*d)^2*(c*Cos[e] + d
*Sin[e])) + (c^3*Cos[e]^4)/((c - I*d)^2*d^2*(c*Cos[e] + d*Sin[e])) - (I*c^2*Cos[e]^4)/((c - I*d)^2*d*(c*Cos[e]
+ d*Sin[e])) + ((3*I)*d*Cos[e]^4)/((c - I*d)^2*(c*Cos[e] + d*Sin[e])) - ((20*I)*c*Cos[e]^3*Sin[e])/((c - I*d)
^2*(c*Cos[e] + d*Sin[e])) - ((4*I)*c^3*Cos[e]^3*Sin[e])/((c - I*d)^2*d^2*(c*Cos[e] + d*Sin[e])) - (4*c^2*Cos[e
]^3*Sin[e])/((c - I*d)^2*d*(c*Cos[e] + d*Sin[e])) + (12*d*Cos[e]^3*Sin[e])/((c - I*d)^2*(c*Cos[e] + d*Sin[e]))
- (30*c*Cos[e]^2*Sin[e]^2)/((c - I*d)^2*(c*Cos[e] + d*Sin[e])) - (6*c^3*Cos[e]^2*Sin[e]^2)/((c - I*d)^2*d^2*(
c*Cos[e] + d*Sin[e])) + ((6*I)*c^2*Cos[e]^2*Sin[e]^2)/((c - I*d)^2*d*(c*Cos[e] + d*Sin[e])) - ((18*I)*d*Cos[e]
^2*Sin[e]^2)/((c - I*d)^2*(c*Cos[e] + d*Sin[e])) + ((20*I)*c*Cos[e]*Sin[e]^3)/((c - I*d)^2*(c*Cos[e] + d*Sin[e
])) + ((4*I)*c^3*Cos[e]*Sin[e]^3)/((c - I*d)^2*d^2*(c*Cos[e] + d*Sin[e])) + (4*c^2*Cos[e]*Sin[e]^3)/((c - I*d)
^2*d*(c*Cos[e] + d*Sin[e])) - (12*d*Cos[e]*Sin[e]^3)/((c - I*d)^2*(c*Cos[e] + d*Sin[e])) + (5*c*Sin[e]^4)/((c
- I*d)^2*(c*Cos[e] + d*Sin[e])) + (c^3*Sin[e]^4)/((c - I*d)^2*d^2*(c*Cos[e] + d*Sin[e])) - (I*c^2*Sin[e]^4)/((
c - I*d)^2*d*(c*Cos[e] + d*Sin[e])) + ((3*I)*d*Sin[e]^4)/((c - I*d)^2*(c*Cos[e] + d*Sin[e])) + ((-5*c - (3*I)*
d + c*Cos[2*e] - (3*I)*d*Cos[2*e] + I*c*Sin[2*e] + 3*d*Sin[2*e])*(Cos[3*e] - I*Sin[3*e]))/((c - I*d)^2*(c + I*
d + c*Cos[2*e] - I*d*Cos[2*e] + I*c*Sin[2*e] + d*Sin[2*e])) + ((1 - Cos[2*e] - I*Sin[2*e])*(Cos[3*e]/d^2 - (I*
Sin[3*e])/d^2))/(1 + Cos[2*e] + I*Sin[2*e]) + ((-c^3 + c^3*Cos[2*e] + I*c^3*Sin[2*e])*(Cos[3*e]/d^2 - (I*Sin[3
*e])/d^2))/((c - I*d)^2*(c + I*d + c*Cos[2*e] - I*d*Cos[2*e] + I*c*Sin[2*e] + d*Sin[2*e])) + ((-c^2 + 3*c^2*Co
s[2*e] + (3*I)*c^2*Sin[2*e])*(((-I)*Cos[3*e])/d - Sin[3*e]/d))/((c - I*d)^2*(c + I*d + c*Cos[2*e] - I*d*Cos[2*
e] + I*c*Sin[2*e] + d*Sin[2*e])) - (Sin[e]*Tan[e])/(2*d^2) - (Sin[e]^3*Tan[e])/(2*d^2))*(a + I*a*Tan[e + f*x])
^3)/(Cos[f*x] + I*Sin[f*x])^3

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Maple [B]  time = 0.035, size = 438, normalized size = 3.1 \begin{align*}{\frac{2\,i{a}^{3}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{2\,i{a}^{3}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{8\,i{a}^{3}\arctan \left ( \tan \left ( fx+e \right ) \right ) cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-4\,{\frac{{a}^{3}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) cd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+4\,{\frac{{a}^{3}\arctan \left ( \tan \left ( fx+e \right ) \right ){c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-4\,{\frac{{a}^{3}\arctan \left ( \tan \left ( fx+e \right ) \right ){d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}-{\frac{i{a}^{3}{c}^{3}}{f{d}^{2} \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}+{\frac{3\,i{a}^{3}c}{f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}+3\,{\frac{{a}^{3}{c}^{2}}{fd \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}-{\frac{{a}^{3}d}{f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) }}-{\frac{i{a}^{3}\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{4}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}{d}^{2}}}-{\frac{6\,i{a}^{3}\ln \left ( c+d\tan \left ( fx+e \right ) \right ){c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+{\frac{3\,i{a}^{3}{d}^{2}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) }{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}}+8\,{\frac{{a}^{3}d\ln \left ( c+d\tan \left ( fx+e \right ) \right ) c}{f \left ({c}^{2}+{d}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f*a^3/(c^2+d^2)^2*ln(1+tan(f*x+e)^2)*c^2-2*I/f*a^3/(c^2+d^2)^2*ln(1+tan(f*x+e)^2)*d^2+8*I/f*a^3/(c^2+d^2)^
2*arctan(tan(f*x+e))*c*d-4/f*a^3/(c^2+d^2)^2*ln(1+tan(f*x+e)^2)*c*d+4/f*a^3/(c^2+d^2)^2*arctan(tan(f*x+e))*c^2
-4/f*a^3/(c^2+d^2)^2*arctan(tan(f*x+e))*d^2-I/f*a^3/d^2/(c^2+d^2)/(c+d*tan(f*x+e))*c^3+3*I/f*a^3/(c^2+d^2)/(c+
d*tan(f*x+e))*c+3/f*a^3/d/(c^2+d^2)/(c+d*tan(f*x+e))*c^2-1/f*a^3*d/(c^2+d^2)/(c+d*tan(f*x+e))-I/f*a^3/(c^2+d^2
)^2/d^2*ln(c+d*tan(f*x+e))*c^4-6*I/f*a^3/(c^2+d^2)^2*ln(c+d*tan(f*x+e))*c^2+3*I/f*a^3/(c^2+d^2)^2*d^2*ln(c+d*t
an(f*x+e))+8/f*a^3/(c^2+d^2)^2*d*ln(c+d*tan(f*x+e))*c

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Maxima [A]  time = 1.53669, size = 333, normalized size = 2.35 \begin{align*} \frac{\frac{8 \,{\left (a^{3} c^{2} + 2 i \, a^{3} c d - a^{3} d^{2}\right )}{\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac{2 \,{\left (-i \, a^{3} c^{4} - 6 i \, a^{3} c^{2} d^{2} + 8 \, a^{3} c d^{3} + 3 i \, a^{3} d^{4}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{4} d^{2} + 2 \, c^{2} d^{4} + d^{6}} + \frac{{\left (4 i \, a^{3} c^{2} - 8 \, a^{3} c d - 4 i \, a^{3} d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac{2 \,{\left (-i \, a^{3} c^{3} + 3 \, a^{3} c^{2} d + 3 i \, a^{3} c d^{2} - a^{3} d^{3}\right )}}{c^{3} d^{2} + c d^{4} +{\left (c^{2} d^{3} + d^{5}\right )} \tan \left (f x + e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.0973, size = 666, normalized size = 4.69 \begin{align*} \frac{2 i \, a^{3} c^{2} d - 4 \, a^{3} c d^{2} - 2 i \, a^{3} d^{3} -{\left (a^{3} c^{3} - i \, a^{3} c^{2} d + 5 \, a^{3} c d^{2} + 3 i \, a^{3} d^{3} +{\left (a^{3} c^{3} - 3 i \, a^{3} c^{2} d + a^{3} c d^{2} - 3 i \, a^{3} d^{3}\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 (a^{3} c^{3} - i \, a^{3} c^{2} d + a^{3} c d^{2} - i \, a^{3} d^{3} +{\left (a^{3} c^{3} - 3 i \, a^{3} c^{2} d - 3 \, a^{3} c d^{2} + i \, a^{3} d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{{\left (-i \, c^{3} d^{2} - 3 \, c^{2} d^{3} + 3 i \, c d^{4} + d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, c^{3} d^{2} - c^{2} d^{3} - i \, c d^{4} - d^{5}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.61024, size = 533, normalized size = 3.75 \begin{align*} \frac{\frac{8 \, a^{3} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{-i \, c^{2} - 2 \, c d + i \, d^{2}} + \frac{i \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{d^{2}} + \frac{i \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{d^{2}} + \frac{2 \,{\left (-i \, a^{3} c^{2} - 2 \, a^{3} c d - 3 i \, a^{3} d^{2}\right )} \log \left ({\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 |}\right )}{2 \, c^{2} d^{2} - 4 i \, c d^{3} - 2 \, d^{4}} - \frac{2 \,{\left (-i \, a^{3} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, a^{3} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 i \, a^{3} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 4 i \, a^{3} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, a^{3} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 8 i \, a^{3} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, a^{3} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i \, a^{3} c^{4} + 2 \, a^{3} c^{3} d + 3 i \, a^{3} c^{2} d^{2}\right )}}{{\left (2 \, c^{3} d^{2} - 4 i \, c^{2} d^{3} - 2 \, c d^{4}\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 )}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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