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

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Rubi [A]  time = 0.37, antiderivative size = 142, normalized size of antiderivative = 1.00, 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 3475

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

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 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]

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*}

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Mathematica [B]  time = 9.06, 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])*(((-1/2*I)*Cos[(3*e)/2]*Log[(c*Cos[e + f*x] + d*Sin[e + f*x])^2])/d^2 - (Log[(c*C
os[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
*Sin[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*
Sin[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*
(Cos[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)*C
os[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*
Cos[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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fricas [B]  time = 0.64, size = 298, normalized size = 2.10 \[ \frac {2 i \, a^{3} c^{2} d - 4 \, a^{3} c d^{2} - 2 i \, a^{3} d^{3} - {\left (a^{3} c^{3} - i \, a^{3} c^{2} d + 5 \, a^{3} c d^{2} + 3 i \, a^{3} d^{3} + {\left (a^{3} c^{3} - 3 i \, a^{3} c^{2} d + a^{3} c d^{2} - 3 i \, a^{3} d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right ) + {\left (a^{3} c^{3} - i \, a^{3} c^{2} d + a^{3} c d^{2} - i \, a^{3} d^{3} + {\left (a^{3} c^{3} - 3 i \, a^{3} c^{2} d - 3 \, a^{3} c d^{2} + i \, a^{3} d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{{\left (-i \, c^{3} d^{2} - 3 \, c^{2} d^{3} + 3 i \, c d^{4} + d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c^{3} d^{2} - c^{2} d^{3} - i \, c d^{4} - d^{5}\right )} f} \]

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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giac [B]  time = 0.79, size = 392, normalized size = 2.76 \[ \frac {\frac {8 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{-i \, c^{2} - 2 \, c d + i \, d^{2}} + \frac {i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{d^{2}} + \frac {i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{d^{2}} + \frac {2 \, {\left (-i \, a^{3} c^{2} - 2 \, a^{3} c d - 3 i \, a^{3} d^{2}\right )} \log \left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}{2 \, c^{2} d^{2} - 4 i \, c d^{3} - 2 \, d^{4}} - \frac {2 \, {\left (-i \, a^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 i \, a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 i \, a^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 i \, a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a^{3} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i \, a^{3} c^{4} + 2 \, a^{3} c^{3} d + 3 i \, a^{3} c^{2} d^{2}\right )}}{{\left (2 \, c^{3} d^{2} - 4 i \, c^{2} d^{3} - 2 \, c d^{4}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - c\right )}}}{f} \]

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

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maple [B]  time = 0.30, size = 438, normalized size = 3.08 \[ -\frac {i a^{3} \ln \left (c +d \tan \left (f x +e \right )\right ) c^{4}}{f \left (c^{2}+d^{2}\right )^{2} d^{2}}-\frac {6 i a^{3} \ln \left (c +d \tan \left (f x +e \right )\right ) c^{2}}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {3 i a^{3} d^{2} \ln \left (c +d \tan \left (f x +e \right )\right )}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {8 a^{3} d \ln \left (c +d \tan \left (f x +e \right )\right ) c}{f \left (c^{2}+d^{2}\right )^{2}}-\frac {i a^{3} c^{3}}{f \,d^{2} \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )}+\frac {3 i a^{3} c}{f \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )}+\frac {3 a^{3} c^{2}}{f d \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )}-\frac {a^{3} d}{f \left (c^{2}+d^{2}\right ) \left (c +d \tan \left (f x +e \right )\right )}+\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c^{2}}{f \left (c^{2}+d^{2}\right )^{2}}-\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) d^{2}}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {8 i a^{3} \arctan \left (\tan \left (f x +e \right )\right ) c d}{f \left (c^{2}+d^{2}\right )^{2}}-\frac {4 a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c d}{f \left (c^{2}+d^{2}\right )^{2}}+\frac {4 a^{3} \arctan \left (\tan \left (f x +e \right )\right ) c^{2}}{f \left (c^{2}+d^{2}\right )^{2}}-\frac {4 a^{3} \arctan \left (\tan \left (f x +e \right )\right ) d^{2}}{f \left (c^{2}+d^{2}\right )^{2}} \]

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]

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

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maxima [A]  time = 0.42, size = 248, normalized size = 1.75 \[ \frac {\frac {2 \, {\left (4 \, a^{3} c^{2} + 8 i \, a^{3} c d - 4 \, a^{3} d^{2}\right )} {\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {2 \, {\left (-i \, a^{3} c^{4} - 6 i \, a^{3} c^{2} d^{2} + 8 \, a^{3} c d^{3} + 3 i \, a^{3} d^{4}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{4} d^{2} + 2 \, c^{2} d^{4} + d^{6}} + \frac {{\left (4 i \, a^{3} c^{2} - 8 \, a^{3} c d - 4 i \, a^{3} d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {2 \, {\left (-i \, a^{3} c^{3} + 3 \, a^{3} c^{2} d + 3 i \, a^{3} c d^{2} - a^{3} d^{3}\right )}}{c^{3} d^{2} + c d^{4} + {\left (c^{2} d^{3} + d^{5}\right )} \tan \left (f x + e\right )}}{2 \, f} \]

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*(2*(4*a^3*c^2 + 8*I*a^3*c*d - 4*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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mupad [B]  time = 6.82, size = 133, normalized size = 0.94 \[ -\frac {4\,a^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}{f\,\left (c^2\,1{}\mathrm {i}+2\,c\,d-d^2\,1{}\mathrm {i}\right )}+\frac {a^3\,\left (-c^2\,1{}\mathrm {i}+2\,c\,d+d^2\,1{}\mathrm {i}\right )}{d^3\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+\frac {c}{d}\right )\,\left (c-d\,1{}\mathrm {i}\right )}+\frac {a^3\,\ln \left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (c^2\,1{}\mathrm {i}+2\,c\,d+d^2\,3{}\mathrm {i}\right )}{d^2\,f\,{\left (d+c\,1{}\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 33.04, size = 382, normalized size = 2.69 \[ - \frac {i a^{3} \left (c - 3 i d\right ) \left (c + i d\right ) \log {\left (e^{2 i f x} + \frac {i a^{3} c^{2} - \frac {a^{3} c d \left (c - 3 i d\right ) \left (c + i d\right )}{\left (c - i d\right )^{2}} + a^{3} c d + \frac {i a^{3} d^{2} \left (c - 3 i d\right ) \left (c + i d\right )}{\left (c - i d\right )^{2}} + 2 i a^{3} d^{2}}{i a^{3} c^{2} e^{2 i e} + 2 a^{3} c d e^{2 i e} + i a^{3} d^{2} e^{2 i e}} \right )}}{d^{2} f \left (c - i d\right )^{2}} + \frac {i a^{3} \log {\left (\frac {i a^{3} c^{2} + 2 a^{3} c d + i a^{3} d^{2}}{i a^{3} c^{2} e^{2 i e} + 2 a^{3} c d e^{2 i e} + i a^{3} d^{2} e^{2 i e}} + e^{2 i f x} \right )}}{d^{2} f} + \frac {- 2 i a^{3} c^{2} + 4 a^{3} c d + 2 i a^{3} d^{2}}{i c^{3} d f + c^{2} d^{2} f + i c d^{3} f + d^{4} f + \left (i c^{3} d f e^{2 i e} + 3 c^{2} d^{2} f e^{2 i e} - 3 i c d^{3} f e^{2 i e} - d^{4} f e^{2 i e}\right ) e^{2 i f x}} \]

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]

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

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