3.1071 \(\int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=153 \[ -\frac {2 a^3 (c-i d)^2 \tan (e+f x)}{f}-\frac {4 i a^3 (c-i d)^2 \log (\cos (e+f x))}{f}+4 a^3 x (c-i d)^2+\frac {2 c d (a+i a \tan (e+f x))^3}{3 f}+\frac {i a (c-i d)^2 (a+i a \tan (e+f x))^2}{2 f}-\frac {i d^2 (a+i a \tan (e+f x))^4}{4 a f} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 153, 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 = {3543, 3527, 3478, 3477, 3475} \[ -\frac {2 a^3 (c-i d)^2 \tan (e+f x)}{f}-\frac {4 i a^3 (c-i d)^2 \log (\cos (e+f x))}{f}+4 a^3 x (c-i d)^2+\frac {2 c d (a+i a \tan (e+f x))^3}{3 f}+\frac {i a (c-i d)^2 (a+i a \tan (e+f x))^2}{2 f}-\frac {i d^2 (a+i a \tan (e+f x))^4}{4 a f} \]

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

Rule 3475

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

Rule 3477

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

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3527

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

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

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

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Mathematica [B]  time = 9.23, size = 948, normalized size = 6.20 \[ \frac {\left (\frac {1}{3} \cos (3 e)-\frac {1}{3} i \sin (3 e)\right ) \left (-3 \sin (f x) d^2-2 i c \sin (f x) d\right ) (i \tan (e+f x) a+a)^3}{f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) (\cos (f x)+i \sin (f x))^3}+\frac {\cos ^2(e+f x) \left (\frac {1}{3} \cos (3 e)-\frac {1}{3} i \sin (3 e)\right ) \left (-9 \sin (f x) c^2+26 i d \sin (f x) c+15 d^2 \sin (f x)\right ) (i \tan (e+f x) a+a)^3}{f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) (\cos (f x)+i \sin (f x))^3}+\frac {x \cos ^3(e+f x) \left (2 c^2 \cos ^3(e)-2 d^2 \cos ^3(e)-4 i c d \cos ^3(e)-8 i c^2 \sin (e) \cos ^2(e)+8 i d^2 \sin (e) \cos ^2(e)-16 c d \sin (e) \cos ^2(e)-2 c^2 \cos (e)+2 d^2 \cos (e)-12 c^2 \sin ^2(e) \cos (e)+12 d^2 \sin ^2(e) \cos (e)+24 i c d \sin ^2(e) \cos (e)+4 i c d \cos (e)+8 i c^2 \sin ^3(e)-8 i d^2 \sin ^3(e)+16 c d \sin ^3(e)+4 i c^2 \sin (e)-4 i d^2 \sin (e)+8 c d \sin (e)+2 c^2 \sin ^3(e) \tan (e)-2 d^2 \sin ^3(e) \tan (e)-4 i c d \sin ^3(e) \tan (e)+2 c^2 \sin (e) \tan (e)-2 d^2 \sin (e) \tan (e)-4 i c d \sin (e) \tan (e)+i (c-i d)^2 (4 \cos (3 e)-4 i \sin (3 e)) \tan (e)\right ) (i \tan (e+f x) a+a)^3}{(\cos (f x)+i \sin (f x))^3}+\frac {\cos ^3(e+f x) \left (\cos \left (\frac {3 e}{2}\right ) c^2-i \sin \left (\frac {3 e}{2}\right ) c^2-2 i d \cos \left (\frac {3 e}{2}\right ) c-2 d \sin \left (\frac {3 e}{2}\right ) c-d^2 \cos \left (\frac {3 e}{2}\right )+i d^2 \sin \left (\frac {3 e}{2}\right )\right ) \left (-2 i \cos \left (\frac {3 e}{2}\right ) \log \left (\cos ^2(e+f x)\right )-2 \sin \left (\frac {3 e}{2}\right ) \log \left (\cos ^2(e+f x)\right )\right ) (i \tan (e+f x) a+a)^3}{f (\cos (f x)+i \sin (f x))^3}+\frac {\cos (e+f x) \left (3 \cos (e) c^2-18 i d \cos (e) c+4 d \sin (e) c-15 d^2 \cos (e)-6 i d^2 \sin (e)\right ) \left (-\frac {1}{6} i \cos (3 e)-\frac {1}{6} \sin (3 e)\right ) (i \tan (e+f x) a+a)^3}{f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) (\cos (f x)+i \sin (f x))^3}+\frac {\sec (e+f x) \left (-\frac {1}{4} i \cos (3 e) d^2-\frac {1}{4} \sin (3 e) d^2\right ) (i \tan (e+f x) a+a)^3}{f (\cos (f x)+i \sin (f x))^3}+\frac {(c-i d)^2 \cos ^3(e+f x) (4 f x \cos (3 e)-4 i f x \sin (3 e)) (i \tan (e+f x) a+a)^3}{f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]^3*(c^2*Cos[(3*e)/2] - (2*I)*c*d*Cos[(3*e)/2] - d^2*Cos[(3*e)/2] - I*c^2*Sin[(3*e)/2] - 2*c*d*Sin
[(3*e)/2] + I*d^2*Sin[(3*e)/2])*((-2*I)*Cos[(3*e)/2]*Log[Cos[e + f*x]^2] - 2*Log[Cos[e + f*x]^2]*Sin[(3*e)/2])
*(a + I*a*Tan[e + f*x])^3)/(f*(Cos[f*x] + I*Sin[f*x])^3) + (Cos[e + f*x]*(3*c^2*Cos[e] - (18*I)*c*d*Cos[e] - 1
5*d^2*Cos[e] + 4*c*d*Sin[e] - (6*I)*d^2*Sin[e])*((-1/6*I)*Cos[3*e] - Sin[3*e]/6)*(a + I*a*Tan[e + f*x])^3)/(f*
(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(Cos[f*x] + I*Sin[f*x])^3) + (Sec[e + f*x]*((-1/4*I)*d^2*Cos[3*e]
- (d^2*Sin[3*e])/4)*(a + I*a*Tan[e + f*x])^3)/(f*(Cos[f*x] + I*Sin[f*x])^3) + ((c - I*d)^2*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)/(f*(Cos[f*x] + I*Sin[f*x])^3) + ((Cos[3*e]/3 - (I/
3)*Sin[3*e])*((-2*I)*c*d*Sin[f*x] - 3*d^2*Sin[f*x])*(a + I*a*Tan[e + f*x])^3)/(f*(Cos[e/2] - Sin[e/2])*(Cos[e/
2] + Sin[e/2])*(Cos[f*x] + I*Sin[f*x])^3) + (Cos[e + f*x]^2*(Cos[3*e]/3 - (I/3)*Sin[3*e])*(-9*c^2*Sin[f*x] + (
26*I)*c*d*Sin[f*x] + 15*d^2*Sin[f*x])*(a + I*a*Tan[e + f*x])^3)/(f*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])
*(Cos[f*x] + I*Sin[f*x])^3) + (x*Cos[e + f*x]^3*(-2*c^2*Cos[e] + (4*I)*c*d*Cos[e] + 2*d^2*Cos[e] + 2*c^2*Cos[e
]^3 - (4*I)*c*d*Cos[e]^3 - 2*d^2*Cos[e]^3 + (4*I)*c^2*Sin[e] + 8*c*d*Sin[e] - (4*I)*d^2*Sin[e] - (8*I)*c^2*Cos
[e]^2*Sin[e] - 16*c*d*Cos[e]^2*Sin[e] + (8*I)*d^2*Cos[e]^2*Sin[e] - 12*c^2*Cos[e]*Sin[e]^2 + (24*I)*c*d*Cos[e]
*Sin[e]^2 + 12*d^2*Cos[e]*Sin[e]^2 + (8*I)*c^2*Sin[e]^3 + 16*c*d*Sin[e]^3 - (8*I)*d^2*Sin[e]^3 + 2*c^2*Sin[e]*
Tan[e] - (4*I)*c*d*Sin[e]*Tan[e] - 2*d^2*Sin[e]*Tan[e] + 2*c^2*Sin[e]^3*Tan[e] - (4*I)*c*d*Sin[e]^3*Tan[e] - 2
*d^2*Sin[e]^3*Tan[e] + I*(c - I*d)^2*(4*Cos[3*e] - (4*I)*Sin[3*e])*Tan[e])*(a + I*a*Tan[e + f*x])^3)/(Cos[f*x]
+ I*Sin[f*x])^3

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fricas [B]  time = 0.43, size = 352, normalized size = 2.30 \[ \frac {-18 i \, a^{3} c^{2} - 52 \, a^{3} c d + 30 i \, a^{3} d^{2} + {\left (-24 i \, a^{3} c^{2} - 96 \, a^{3} c d + 72 i \, a^{3} d^{2}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-66 i \, a^{3} c^{2} - 228 \, a^{3} c d + 138 i \, a^{3} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-60 i \, a^{3} c^{2} - 184 \, a^{3} c d + 108 i \, a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-12 i \, a^{3} c^{2} - 24 \, a^{3} c d + 12 i \, a^{3} d^{2} + {\left (-12 i \, a^{3} c^{2} - 24 \, a^{3} c d + 12 i \, a^{3} d^{2}\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-48 i \, a^{3} c^{2} - 96 \, a^{3} c d + 48 i \, a^{3} d^{2}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-72 i \, a^{3} c^{2} - 144 \, a^{3} c d + 72 i \, a^{3} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-48 i \, a^{3} c^{2} - 96 \, a^{3} c d + 48 i \, a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{3 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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]

1/3*(-18*I*a^3*c^2 - 52*a^3*c*d + 30*I*a^3*d^2 + (-24*I*a^3*c^2 - 96*a^3*c*d + 72*I*a^3*d^2)*e^(6*I*f*x + 6*I*
e) + (-66*I*a^3*c^2 - 228*a^3*c*d + 138*I*a^3*d^2)*e^(4*I*f*x + 4*I*e) + (-60*I*a^3*c^2 - 184*a^3*c*d + 108*I*
a^3*d^2)*e^(2*I*f*x + 2*I*e) + (-12*I*a^3*c^2 - 24*a^3*c*d + 12*I*a^3*d^2 + (-12*I*a^3*c^2 - 24*a^3*c*d + 12*I
*a^3*d^2)*e^(8*I*f*x + 8*I*e) + (-48*I*a^3*c^2 - 96*a^3*c*d + 48*I*a^3*d^2)*e^(6*I*f*x + 6*I*e) + (-72*I*a^3*c
^2 - 144*a^3*c*d + 72*I*a^3*d^2)*e^(4*I*f*x + 4*I*e) + (-48*I*a^3*c^2 - 96*a^3*c*d + 48*I*a^3*d^2)*e^(2*I*f*x
+ 2*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*
I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)

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giac [B]  time = 1.55, size = 670, normalized size = 4.38 \[ \frac {-12 i \, a^{3} c^{2} e^{\left (8 i \, f x + 8 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 24 \, a^{3} c d e^{\left (8 i \, f x + 8 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + 12 i \, a^{3} d^{2} e^{\left (8 i \, f x + 8 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 48 i \, a^{3} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 96 \, a^{3} c d e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + 48 i \, a^{3} d^{2} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 72 i \, a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 144 \, a^{3} c d e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + 72 i \, a^{3} d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 48 i \, a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 96 \, a^{3} c d e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + 48 i \, a^{3} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 24 i \, a^{3} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 96 \, a^{3} c d e^{\left (6 i \, f x + 6 i \, e\right )} + 72 i \, a^{3} d^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - 66 i \, a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 228 \, a^{3} c d e^{\left (4 i \, f x + 4 i \, e\right )} + 138 i \, a^{3} d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 60 i \, a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 184 \, a^{3} c d e^{\left (2 i \, f x + 2 i \, e\right )} + 108 i \, a^{3} d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 12 i \, a^{3} c^{2} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 24 \, a^{3} c d \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) + 12 i \, a^{3} d^{2} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 18 i \, a^{3} c^{2} - 52 \, a^{3} c d + 30 i \, a^{3} d^{2}}{3 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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]

1/3*(-12*I*a^3*c^2*e^(8*I*f*x + 8*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 24*a^3*c*d*e^(8*I*f*x + 8*I*e)*log(e^(2*
I*f*x + 2*I*e) + 1) + 12*I*a^3*d^2*e^(8*I*f*x + 8*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 48*I*a^3*c^2*e^(6*I*f*x
+ 6*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 96*a^3*c*d*e^(6*I*f*x + 6*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) + 48*I*a^3
*d^2*e^(6*I*f*x + 6*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 72*I*a^3*c^2*e^(4*I*f*x + 4*I*e)*log(e^(2*I*f*x + 2*I*
e) + 1) - 144*a^3*c*d*e^(4*I*f*x + 4*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) + 72*I*a^3*d^2*e^(4*I*f*x + 4*I*e)*log(
e^(2*I*f*x + 2*I*e) + 1) - 48*I*a^3*c^2*e^(2*I*f*x + 2*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 96*a^3*c*d*e^(2*I*f
*x + 2*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) + 48*I*a^3*d^2*e^(2*I*f*x + 2*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 24*
I*a^3*c^2*e^(6*I*f*x + 6*I*e) - 96*a^3*c*d*e^(6*I*f*x + 6*I*e) + 72*I*a^3*d^2*e^(6*I*f*x + 6*I*e) - 66*I*a^3*c
^2*e^(4*I*f*x + 4*I*e) - 228*a^3*c*d*e^(4*I*f*x + 4*I*e) + 138*I*a^3*d^2*e^(4*I*f*x + 4*I*e) - 60*I*a^3*c^2*e^
(2*I*f*x + 2*I*e) - 184*a^3*c*d*e^(2*I*f*x + 2*I*e) + 108*I*a^3*d^2*e^(2*I*f*x + 2*I*e) - 12*I*a^3*c^2*log(e^(
2*I*f*x + 2*I*e) + 1) - 24*a^3*c*d*log(e^(2*I*f*x + 2*I*e) + 1) + 12*I*a^3*d^2*log(e^(2*I*f*x + 2*I*e) + 1) -
18*I*a^3*c^2 - 52*a^3*c*d + 30*I*a^3*d^2)/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x +
4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)

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maple [B]  time = 0.02, size = 290, normalized size = 1.90 \[ -\frac {i a^{3} d^{2} \left (\tan ^{4}\left (f x +e \right )\right )}{4 f}-\frac {2 i a^{3} \left (\tan ^{3}\left (f x +e \right )\right ) c d}{3 f}-\frac {i a^{3} \left (\tan ^{2}\left (f x +e \right )\right ) c^{2}}{2 f}+\frac {2 i a^{3} \left (\tan ^{2}\left (f x +e \right )\right ) d^{2}}{f}-\frac {a^{3} d^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{f}+\frac {8 i a^{3} c d \tan \left (f x +e \right )}{f}-\frac {3 a^{3} \left (\tan ^{2}\left (f x +e \right )\right ) c d}{f}-\frac {3 a^{3} c^{2} \tan \left (f x +e \right )}{f}+\frac {4 a^{3} \tan \left (f x +e \right ) d^{2}}{f}+\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c^{2}}{f}-\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) d^{2}}{f}+\frac {4 a^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c d}{f}-\frac {8 i a^{3} \arctan \left (\tan \left (f x +e \right )\right ) c d}{f}+\frac {4 a^{3} \arctan \left (\tan \left (f x +e \right )\right ) c^{2}}{f}-\frac {4 a^{3} \arctan \left (\tan \left (f x +e \right )\right ) d^{2}}{f} \]

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]

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

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maxima [A]  time = 0.41, size = 181, normalized size = 1.18 \[ -\frac {3 i \, a^{3} d^{2} \tan \left (f x + e\right )^{4} + 4 \, {\left (2 i \, a^{3} c d + 3 \, a^{3} d^{2}\right )} \tan \left (f x + e\right )^{3} - {\left (-6 i \, a^{3} c^{2} - 36 \, a^{3} c d + 24 i \, a^{3} d^{2}\right )} \tan \left (f x + e\right )^{2} - 12 \, {\left (4 \, a^{3} c^{2} - 8 i \, a^{3} c d - 4 \, a^{3} d^{2}\right )} {\left (f x + e\right )} - 12 \, {\left (2 i \, a^{3} c^{2} + 4 \, a^{3} c d - 2 i \, a^{3} d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + {\left (36 \, a^{3} c^{2} - 96 i \, a^{3} c d - 48 \, a^{3} d^{2}\right )} \tan \left (f x + e\right )}{12 \, f} \]

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]

-1/12*(3*I*a^3*d^2*tan(f*x + e)^4 + 4*(2*I*a^3*c*d + 3*a^3*d^2)*tan(f*x + e)^3 - (-6*I*a^3*c^2 - 36*a^3*c*d +
24*I*a^3*d^2)*tan(f*x + e)^2 - 12*(4*a^3*c^2 - 8*I*a^3*c*d - 4*a^3*d^2)*(f*x + e) - 12*(2*I*a^3*c^2 + 4*a^3*c*
d - 2*I*a^3*d^2)*log(tan(f*x + e)^2 + 1) + (36*a^3*c^2 - 96*I*a^3*c*d - 48*a^3*d^2)*tan(f*x + e))/f

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mupad [B]  time = 4.92, size = 217, normalized size = 1.42 \[ \frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {a^3\,d^2\,1{}\mathrm {i}}{2}-\frac {a^3\,\left (c^2\,1{}\mathrm {i}+4\,c\,d-d^2\,1{}\mathrm {i}\right )}{2}+a^3\,d\,\left (d+c\,1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{f}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {a^3\,d^2}{3}+\frac {2\,a^3\,d\,\left (d+c\,1{}\mathrm {i}\right )}{3}\right )}{f}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (a^3\,c^2\,4{}\mathrm {i}+8\,a^3\,c\,d-a^3\,d^2\,4{}\mathrm {i}\right )}{f}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a^3\,d^2+a^3\,\left (c^2\,1{}\mathrm {i}+4\,c\,d-d^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}-2\,a^3\,c\,\left (c-d\,1{}\mathrm {i}\right )+2\,a^3\,d\,\left (d+c\,1{}\mathrm {i}\right )\right )}{f}-\frac {a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}}{4\,f} \]

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]

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

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sympy [B]  time = 1.13, size = 314, normalized size = 2.05 \[ - \frac {4 i a^{3} \left (c - i d\right )^{2} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{f} + \frac {18 i a^{3} c^{2} + 52 a^{3} c d - 30 i a^{3} d^{2} + \left (60 i a^{3} c^{2} e^{2 i e} + 184 a^{3} c d e^{2 i e} - 108 i a^{3} d^{2} e^{2 i e}\right ) e^{2 i f x} + \left (66 i a^{3} c^{2} e^{4 i e} + 228 a^{3} c d e^{4 i e} - 138 i a^{3} d^{2} e^{4 i e}\right ) e^{4 i f x} + \left (24 i a^{3} c^{2} e^{6 i e} + 96 a^{3} c d e^{6 i e} - 72 i a^{3} d^{2} e^{6 i e}\right ) e^{6 i f x}}{- 3 f e^{8 i e} e^{8 i f x} - 12 f e^{6 i e} e^{6 i f x} - 18 f e^{4 i e} e^{4 i f x} - 12 f e^{2 i e} e^{2 i f x} - 3 f} \]

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)

[Out]

-4*I*a**3*(c - I*d)**2*log(exp(2*I*f*x) + exp(-2*I*e))/f + (18*I*a**3*c**2 + 52*a**3*c*d - 30*I*a**3*d**2 + (6
0*I*a**3*c**2*exp(2*I*e) + 184*a**3*c*d*exp(2*I*e) - 108*I*a**3*d**2*exp(2*I*e))*exp(2*I*f*x) + (66*I*a**3*c**
2*exp(4*I*e) + 228*a**3*c*d*exp(4*I*e) - 138*I*a**3*d**2*exp(4*I*e))*exp(4*I*f*x) + (24*I*a**3*c**2*exp(6*I*e)
+ 96*a**3*c*d*exp(6*I*e) - 72*I*a**3*d**2*exp(6*I*e))*exp(6*I*f*x))/(-3*f*exp(8*I*e)*exp(8*I*f*x) - 12*f*exp(
6*I*e)*exp(6*I*f*x) - 18*f*exp(4*I*e)*exp(4*I*f*x) - 12*f*exp(2*I*e)*exp(2*I*f*x) - 3*f)

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