3.729 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=164 \[ -\frac {6 c^4 (A+3 i B)}{a^3 f (-\tan (e+f x)+i)}+\frac {2 c^4 (-5 B+3 i A)}{a^3 f (-\tan (e+f x)+i)^2}+\frac {8 c^4 (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac {c^4 (-7 B+i A) \log (\cos (e+f x))}{a^3 f}-\frac {c^4 x (A+7 i B)}{a^3}+\frac {i B c^4 \tan (e+f x)}{a^3 f} \]

[Out]

-(A+7*I*B)*c^4*x/a^3-(I*A-7*B)*c^4*ln(cos(f*x+e))/a^3/f+8/3*(A+I*B)*c^4/a^3/f/(-tan(f*x+e)+I)^3+2*(3*I*A-5*B)*
c^4/a^3/f/(-tan(f*x+e)+I)^2-6*(A+3*I*B)*c^4/a^3/f/(-tan(f*x+e)+I)+I*B*c^4*tan(f*x+e)/a^3/f

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Rubi [A]  time = 0.21, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3588, 77} \[ -\frac {6 c^4 (A+3 i B)}{a^3 f (-\tan (e+f x)+i)}+\frac {2 c^4 (-5 B+3 i A)}{a^3 f (-\tan (e+f x)+i)^2}+\frac {8 c^4 (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac {c^4 (-7 B+i A) \log (\cos (e+f x))}{a^3 f}-\frac {c^4 x (A+7 i B)}{a^3}+\frac {i B c^4 \tan (e+f x)}{a^3 f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

-(((A + (7*I)*B)*c^4*x)/a^3) - ((I*A - 7*B)*c^4*Log[Cos[e + f*x]])/(a^3*f) + (8*(A + I*B)*c^4)/(3*a^3*f*(I - T
an[e + f*x])^3) + (2*((3*I)*A - 5*B)*c^4)/(a^3*f*(I - Tan[e + f*x])^2) - (6*(A + (3*I)*B)*c^4)/(a^3*f*(I - Tan
[e + f*x])) + (I*B*c^4*Tan[e + f*x])/(a^3*f)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^3} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) (c-i c x)^3}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left (\frac {i B c^3}{a^4}+\frac {8 (A+i B) c^3}{a^4 (-i+x)^4}+\frac {4 (-3 i A+5 B) c^3}{a^4 (-i+x)^3}-\frac {6 (A+3 i B) c^3}{a^4 (-i+x)^2}+\frac {i (A+7 i B) c^3}{a^4 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(A+7 i B) c^4 x}{a^3}-\frac {(i A-7 B) c^4 \log (\cos (e+f x))}{a^3 f}+\frac {8 (A+i B) c^4}{3 a^3 f (i-\tan (e+f x))^3}+\frac {2 (3 i A-5 B) c^4}{a^3 f (i-\tan (e+f x))^2}-\frac {6 (A+3 i B) c^4}{a^3 f (i-\tan (e+f x))}+\frac {i B c^4 \tan (e+f x)}{a^3 f}\\ \end {align*}

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Mathematica [B]  time = 9.44, size = 1239, normalized size = 7.55 \[ c^4 \left (\frac {\sec ^3(e+f x) \left (-\frac {1}{2} B \cos (3 e-f x)+\frac {1}{2} B \cos (3 e+f x)-\frac {1}{2} i B \sin (3 e-f x)+\frac {1}{2} i B \sin (3 e+f x)\right ) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^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 ) (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}+\frac {x \sec ^2(e+f x) \left (-\frac {1}{2} A \cos ^3(e)-\frac {7}{2} i B \cos ^3(e)-2 i A \sin (e) \cos ^2(e)+14 B \sin (e) \cos ^2(e)+3 A \sin ^2(e) \cos (e)+21 i B \sin ^2(e) \cos (e)+\frac {1}{2} A \cos (e)+\frac {7}{2} i B \cos (e)+2 i A \sin ^3(e)-14 B \sin ^3(e)+i A \sin (e)-7 B \sin (e)-\frac {1}{2} A \sin ^3(e) \tan (e)-\frac {7}{2} i B \sin ^3(e) \tan (e)-\frac {1}{2} A \sin (e) \tan (e)-\frac {7}{2} i B \sin (e) \tan (e)+i (A+7 i B) (\cos (3 e)+i \sin (3 e)) \tan (e)\right ) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^3}{(A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}+\frac {(A+5 i B) \cos (2 f x) \sec ^2(e+f x) (i \cos (e)-\sin (e)) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^3}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}+\frac {(3 B-i A) \cos (4 f x) \sec ^2(e+f x) \left (\frac {\cos (e)}{2}-\frac {1}{2} i \sin (e)\right ) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^3}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}+\frac {\sec ^2(e+f x) \left (-i A \cos \left (\frac {3 e}{2}\right )+7 B \cos \left (\frac {3 e}{2}\right )+A \sin \left (\frac {3 e}{2}\right )+7 i B \sin \left (\frac {3 e}{2}\right )\right ) \left (\cos \left (\frac {3 e}{2}\right ) \log (\cos (e+f x))+i \sin \left (\frac {3 e}{2}\right ) \log (\cos (e+f x))\right ) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^3}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}+\frac {(A+i B) \cos (6 f x) \sec ^2(e+f x) \left (\frac {1}{3} i \cos (3 e)+\frac {1}{3} \sin (3 e)\right ) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^3}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}+\frac {(A+7 i B) \sec ^2(e+f x) (-f x \cos (3 e)-i f x \sin (3 e)) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^3}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}+\frac {(A+5 i B) \sec ^2(e+f x) (\cos (e)+i \sin (e)) \sin (2 f x) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^3}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}+\frac {(A+3 i B) \sec ^2(e+f x) \left (\frac {1}{2} i \sin (e)-\frac {\cos (e)}{2}\right ) \sin (4 f x) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^3}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}+\frac {(A+i B) \sec ^2(e+f x) \left (\frac {1}{3} \cos (3 e)-\frac {1}{3} i \sin (3 e)\right ) \sin (6 f x) (A+B \tan (e+f x)) (\cos (f x)+i \sin (f x))^3}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

c^4*(((A + (5*I)*B)*Cos[2*f*x]*Sec[e + f*x]^2*(I*Cos[e] - Sin[e])*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x
]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + (((-I)*A + 3*B)*Cos[4*f*x]*Sec[e + f*x]^2
*(Cos[e]/2 - (I/2)*Sin[e])*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x]
)*(a + I*a*Tan[e + f*x])^3) + (Sec[e + f*x]^2*((-I)*A*Cos[(3*e)/2] + 7*B*Cos[(3*e)/2] + A*Sin[(3*e)/2] + (7*I)
*B*Sin[(3*e)/2])*(Cos[(3*e)/2]*Log[Cos[e + f*x]] + I*Log[Cos[e + f*x]]*Sin[(3*e)/2])*(Cos[f*x] + I*Sin[f*x])^3
*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + ((A + I*B)*Cos[6*f*x]*
Sec[e + f*x]^2*((I/3)*Cos[3*e] + Sin[3*e]/3)*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x
] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + ((A + (7*I)*B)*Sec[e + f*x]^2*(-(f*x*Cos[3*e]) - I*f*x*Sin[3*e
])*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])
^3) + ((A + (5*I)*B)*Sec[e + f*x]^2*(Cos[e] + I*Sin[e])*(Cos[f*x] + I*Sin[f*x])^3*Sin[2*f*x]*(A + B*Tan[e + f*
x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + ((A + (3*I)*B)*Sec[e + f*x]^2*(-1/2*Cos[
e] + (I/2)*Sin[e])*(Cos[f*x] + I*Sin[f*x])^3*Sin[4*f*x]*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f
*x])*(a + I*a*Tan[e + f*x])^3) + ((A + I*B)*Sec[e + f*x]^2*(Cos[3*e]/3 - (I/3)*Sin[3*e])*(Cos[f*x] + I*Sin[f*x
])^3*Sin[6*f*x]*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + (Sec[e
+ f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*(-1/2*(B*Cos[3*e - f*x]) + (B*Cos[3*e + f*x])/2 - (I/2)*B*Sin[3*e - f*x] +
(I/2)*B*Sin[3*e + f*x])*(A + B*Tan[e + f*x]))/(f*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(A*Cos[e + f*x] +
B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3) + (x*Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^3*((A*Cos[e])/2 + ((7*I
)/2)*B*Cos[e] - (A*Cos[e]^3)/2 - ((7*I)/2)*B*Cos[e]^3 + I*A*Sin[e] - 7*B*Sin[e] - (2*I)*A*Cos[e]^2*Sin[e] + 14
*B*Cos[e]^2*Sin[e] + 3*A*Cos[e]*Sin[e]^2 + (21*I)*B*Cos[e]*Sin[e]^2 + (2*I)*A*Sin[e]^3 - 14*B*Sin[e]^3 - (A*Si
n[e]*Tan[e])/2 - ((7*I)/2)*B*Sin[e]*Tan[e] - (A*Sin[e]^3*Tan[e])/2 - ((7*I)/2)*B*Sin[e]^3*Tan[e] + I*(A + (7*I
)*B)*(Cos[3*e] + I*Sin[3*e])*Tan[e])*(A + B*Tan[e + f*x]))/((A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e +
f*x])^3))

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fricas [A]  time = 1.20, size = 199, normalized size = 1.21 \[ -\frac {12 \, {\left (A + 7 i \, B\right )} c^{4} f x e^{\left (8 i \, f x + 8 i \, e\right )} - {\left (3 i \, A - 21 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-i \, A + 7 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (2 i \, A - 2 \, B\right )} c^{4} + {\left (12 \, {\left (A + 7 i \, B\right )} c^{4} f x - {\left (6 i \, A - 42 \, B\right )} c^{4}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left ({\left (-6 i \, A + 42 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-6 i \, A + 42 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{6 \, {\left (a^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} + a^{3} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/6*(12*(A + 7*I*B)*c^4*f*x*e^(8*I*f*x + 8*I*e) - (3*I*A - 21*B)*c^4*e^(4*I*f*x + 4*I*e) - (-I*A + 7*B)*c^4*e
^(2*I*f*x + 2*I*e) - (2*I*A - 2*B)*c^4 + (12*(A + 7*I*B)*c^4*f*x - (6*I*A - 42*B)*c^4)*e^(6*I*f*x + 6*I*e) - (
(-6*I*A + 42*B)*c^4*e^(8*I*f*x + 8*I*e) + (-6*I*A + 42*B)*c^4*e^(6*I*f*x + 6*I*e))*log(e^(2*I*f*x + 2*I*e) + 1
))/(a^3*f*e^(8*I*f*x + 8*I*e) + a^3*f*e^(6*I*f*x + 6*I*e))

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giac [B]  time = 4.93, size = 429, normalized size = 2.62 \[ \frac {\frac {30 \, {\left (-i \, A c^{4} + 7 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3}} + \frac {60 \, {\left (i \, A c^{4} - 7 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{3}} - \frac {30 \, {\left (i \, A c^{4} - 7 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{3}} - \frac {30 \, {\left (-i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i \, A c^{4} - 7 \, B c^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{3}} - \frac {147 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 1029 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 1002 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 6534 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 2445 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 17115 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3820 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 23860 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2445 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 17115 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1002 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6534 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 147 i \, A c^{4} + 1029 \, B c^{4}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}}}{30 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

1/30*(30*(-I*A*c^4 + 7*B*c^4)*log(tan(1/2*f*x + 1/2*e) + 1)/a^3 + 60*(I*A*c^4 - 7*B*c^4)*log(tan(1/2*f*x + 1/2
*e) - I)/a^3 - 30*(I*A*c^4 - 7*B*c^4)*log(tan(1/2*f*x + 1/2*e) - 1)/a^3 - 30*(-I*A*c^4*tan(1/2*f*x + 1/2*e)^2
+ 7*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 2*I*B*c^4*tan(1/2*f*x + 1/2*e) + I*A*c^4 - 7*B*c^4)/((tan(1/2*f*x + 1/2*e)^
2 - 1)*a^3) - (147*I*A*c^4*tan(1/2*f*x + 1/2*e)^6 - 1029*B*c^4*tan(1/2*f*x + 1/2*e)^6 + 1002*A*c^4*tan(1/2*f*x
+ 1/2*e)^5 + 6534*I*B*c^4*tan(1/2*f*x + 1/2*e)^5 - 2445*I*A*c^4*tan(1/2*f*x + 1/2*e)^4 + 17115*B*c^4*tan(1/2*
f*x + 1/2*e)^4 - 3820*A*c^4*tan(1/2*f*x + 1/2*e)^3 - 23860*I*B*c^4*tan(1/2*f*x + 1/2*e)^3 + 2445*I*A*c^4*tan(1
/2*f*x + 1/2*e)^2 - 17115*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 1002*A*c^4*tan(1/2*f*x + 1/2*e) + 6534*I*B*c^4*tan(1/
2*f*x + 1/2*e) - 147*I*A*c^4 + 1029*B*c^4)/(a^3*(tan(1/2*f*x + 1/2*e) - I)^6))/f

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maple [A]  time = 0.23, size = 207, normalized size = 1.26 \[ \frac {i B \,c^{4} \tan \left (f x +e \right )}{a^{3} f}-\frac {8 c^{4} A}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {8 i c^{4} B}{3 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {18 i c^{4} B}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}+\frac {6 c^{4} A}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}+\frac {i c^{4} A \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{3}}-\frac {7 c^{4} B \ln \left (\tan \left (f x +e \right )-i\right )}{f \,a^{3}}+\frac {6 i c^{4} A}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {10 c^{4} B}{f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^3,x)

[Out]

I*B*c^4*tan(f*x+e)/a^3/f-8/3/f*c^4/a^3/(tan(f*x+e)-I)^3*A-8/3*I/f*c^4/a^3/(tan(f*x+e)-I)^3*B+18*I/f*c^4/a^3/(t
an(f*x+e)-I)*B+6/f*c^4/a^3/(tan(f*x+e)-I)*A+I/f*c^4/a^3*A*ln(tan(f*x+e)-I)-7/f*c^4/a^3*B*ln(tan(f*x+e)-I)+6*I/
f*c^4/a^3/(tan(f*x+e)-I)^2*A-10/f*c^4/a^3/(tan(f*x+e)-I)^2*B

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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((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^4/(a+I*a*tan(f*x+e))^3,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 = 11.01, size = 266, normalized size = 1.62 \[ -\frac {c^4\,\left (25\,B\,\mathrm {tan}\left (e+f\,x\right )-\frac {B\,32{}\mathrm {i}}{3}-A\,\mathrm {tan}\left (e+f\,x\right )\,6{}\mathrm {i}-\frac {8\,A}{3}-A\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-B\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}+6\,A\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,15{}\mathrm {i}+3\,B\,{\mathrm {tan}\left (e+f\,x\right )}^3+B\,{\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+3\,A\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+A\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}+B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,21{}\mathrm {i}-7\,B\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-A\,\mathrm {tan}\left (e+f\,x\right )\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}+21\,B\,\mathrm {tan}\left (e+f\,x\right )\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{a^3\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^4)/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

-(c^4*(25*B*tan(e + f*x) - (B*32i)/3 - A*tan(e + f*x)*6i - (8*A)/3 - A*log(- tan(e + f*x)*1i - 1) - B*log(- ta
n(e + f*x)*1i - 1)*7i + 6*A*tan(e + f*x)^2 + B*tan(e + f*x)^2*15i + 3*B*tan(e + f*x)^3 + B*tan(e + f*x)^4*1i +
3*A*tan(e + f*x)^2*log(- tan(e + f*x)*1i - 1) + A*tan(e + f*x)^3*log(- tan(e + f*x)*1i - 1)*1i + B*tan(e + f*
x)^2*log(- tan(e + f*x)*1i - 1)*21i - 7*B*tan(e + f*x)^3*log(- tan(e + f*x)*1i - 1) - A*tan(e + f*x)*log(- tan
(e + f*x)*1i - 1)*3i + 21*B*tan(e + f*x)*log(- tan(e + f*x)*1i - 1))*1i)/(a^3*f*(tan(e + f*x)*1i + 1)^3)

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sympy [A]  time = 1.40, size = 408, normalized size = 2.49 \[ \frac {2 B c^{4}}{- a^{3} f e^{2 i e} e^{2 i f x} - a^{3} f} + \begin {cases} - \frac {\left (\left (- 2 i A a^{6} c^{4} f^{2} e^{6 i e} + 2 B a^{6} c^{4} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (3 i A a^{6} c^{4} f^{2} e^{8 i e} - 9 B a^{6} c^{4} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (- 6 i A a^{6} c^{4} f^{2} e^{10 i e} + 30 B a^{6} c^{4} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{6 a^{9} f^{3}} & \text {for}\: 6 a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {- 2 A c^{4} - 14 i B c^{4}}{a^{3}} + \frac {i \left (2 i A c^{4} e^{6 i e} - 2 i A c^{4} e^{4 i e} + 2 i A c^{4} e^{2 i e} - 2 i A c^{4} - 14 B c^{4} e^{6 i e} + 10 B c^{4} e^{4 i e} - 6 B c^{4} e^{2 i e} + 2 B c^{4}\right ) e^{- 6 i e}}{a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {i c^{4} \left (A + 7 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{3} f} - \frac {x \left (2 A c^{4} + 14 i B c^{4}\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**4/(a+I*a*tan(f*x+e))**3,x)

[Out]

2*B*c**4/(-a**3*f*exp(2*I*e)*exp(2*I*f*x) - a**3*f) + Piecewise((-((-2*I*A*a**6*c**4*f**2*exp(6*I*e) + 2*B*a**
6*c**4*f**2*exp(6*I*e))*exp(-6*I*f*x) + (3*I*A*a**6*c**4*f**2*exp(8*I*e) - 9*B*a**6*c**4*f**2*exp(8*I*e))*exp(
-4*I*f*x) + (-6*I*A*a**6*c**4*f**2*exp(10*I*e) + 30*B*a**6*c**4*f**2*exp(10*I*e))*exp(-2*I*f*x))*exp(-12*I*e)/
(6*a**9*f**3), Ne(6*a**9*f**3*exp(12*I*e), 0)), (x*(-(-2*A*c**4 - 14*I*B*c**4)/a**3 + I*(2*I*A*c**4*exp(6*I*e)
- 2*I*A*c**4*exp(4*I*e) + 2*I*A*c**4*exp(2*I*e) - 2*I*A*c**4 - 14*B*c**4*exp(6*I*e) + 10*B*c**4*exp(4*I*e) -
6*B*c**4*exp(2*I*e) + 2*B*c**4)*exp(-6*I*e)/a**3), True)) - I*c**4*(A + 7*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))
/(a**3*f) - x*(2*A*c**4 + 14*I*B*c**4)/a**3

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