3.718 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=128 \[ \frac{4 c^3 (A+2 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac{2 c^3 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}+\frac{c^3 (-5 B+i A) \log (\cos (e+f x))}{a^2 f}+\frac{c^3 x (A+5 i B)}{a^2}-\frac{i B c^3 \tan (e+f x)}{a^2 f} \]

[Out]

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

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Rubi [A]  time = 0.180117, antiderivative size = 128, normalized size of antiderivative = 1., 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} \[ \frac{4 c^3 (A+2 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac{2 c^3 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}+\frac{c^3 (-5 B+i A) \log (\cos (e+f x))}{a^2 f}+\frac{c^3 x (A+5 i B)}{a^2}-\frac{i B c^3 \tan (e+f x)}{a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))))

Rubi steps

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

Mathematica [B]  time = 6.79035, size = 413, normalized size = 3.23 \[ -\frac{c^3 \sec (e) \sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (i (A+5 i B) \cos ^3(e) \log \left (\cos ^2(e+f x)\right )+\cos (e) \left (\cos (2 e) (2 f x (A+5 i B)+(A+i B) \sin (4 f x)+i (A+i B) \cos (4 f x))+2 i A f x \sin (2 e)-i A \sin (2 e) \sin (4 f x)+A \sin (2 e) \cos (4 f x)-i A \sin ^2(e) \log \left (\cos ^2(e+f x)\right )-2 A f x-2 A \sin (2 f x)-2 i A \cos (2 f x)-10 B f x \sin (2 e)+B \sin (2 e) \sin (4 f x)+i B \sin (2 e) \cos (4 f x)+5 B \sin ^2(e) \log \left (\cos ^2(e+f x)\right )-10 i B f x-6 i B \sin (2 f x)+6 B \cos (2 f x)\right )-2 (A+5 i B) \sin (e) \cos ^2(e) \log \left (\cos ^2(e+f x)\right )+2 (A+5 i B) \cos (e) (\cos (2 e)+i \sin (2 e)) \tan ^{-1}(\tan (f x))+(\cos (e)+i \sin (e)) \sec (e+f x) (2 \cos (e) (f x (5 B-i A) \sin (2 e+f x)+i \sin (f x) (A f x+B (-1+5 i f x)))+B \cos (e-f x)-B \cos (e+f x))\right )}{2 a^2 f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^3*Sec[e]*Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*(I*(A + (5*I)*B)*Cos[e]^3*Log[Cos[e + f*x]^2] - 2*(A + (
5*I)*B)*Cos[e]^2*Log[Cos[e + f*x]^2]*Sin[e] + 2*(A + (5*I)*B)*ArcTan[Tan[f*x]]*Cos[e]*(Cos[2*e] + I*Sin[2*e])
+ Cos[e]*(-2*A*f*x - (10*I)*B*f*x - (2*I)*A*Cos[2*f*x] + 6*B*Cos[2*f*x] - I*A*Log[Cos[e + f*x]^2]*Sin[e]^2 + 5
*B*Log[Cos[e + f*x]^2]*Sin[e]^2 + (2*I)*A*f*x*Sin[2*e] - 10*B*f*x*Sin[2*e] + A*Cos[4*f*x]*Sin[2*e] + I*B*Cos[4
*f*x]*Sin[2*e] - 2*A*Sin[2*f*x] - (6*I)*B*Sin[2*f*x] - I*A*Sin[2*e]*Sin[4*f*x] + B*Sin[2*e]*Sin[4*f*x] + Cos[2
*e]*(2*(A + (5*I)*B)*f*x + I*(A + I*B)*Cos[4*f*x] + (A + I*B)*Sin[4*f*x])) + Sec[e + f*x]*(Cos[e] + I*Sin[e])*
(B*Cos[e - f*x] - B*Cos[e + f*x] + 2*Cos[e]*(I*(A*f*x + B*(-1 + (5*I)*f*x))*Sin[f*x] + ((-I)*A + 5*B)*f*x*Sin[
2*e + f*x]))))/(2*a^2*f*(-I + Tan[e + f*x])^2)

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Maple [A]  time = 0.044, size = 160, normalized size = 1.3 \begin{align*}{\frac{-iB{c}^{3}\tan \left ( fx+e \right ) }{{a}^{2}f}}-{\frac{8\,i{c}^{3}B}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) }}-4\,{\frac{A{c}^{3}}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{iA{c}^{3}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{{a}^{2}f}}+5\,{\frac{B{c}^{3}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{{a}^{2}f}}-{\frac{2\,i{c}^{3}A}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+2\,{\frac{B{c}^{3}}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}} \end{align*}

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.1135, size = 460, normalized size = 3.59 \begin{align*} \frac{4 \,{\left (A + 5 i \, B\right )} c^{3} f x e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-i \, A + 5 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A - B\right )} c^{3} +{\left (4 \,{\left (A + 5 i \, B\right )} c^{3} f x +{\left (-2 i \, A + 10 \, B\right )} c^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left ({\left (2 i \, A - 10 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (2 i \, A - 10 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{2 \,{\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \end{align*}

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

[Out]

1/2*(4*(A + 5*I*B)*c^3*f*x*e^(6*I*f*x + 6*I*e) + (-I*A + 5*B)*c^3*e^(2*I*f*x + 2*I*e) + (I*A - B)*c^3 + (4*(A
+ 5*I*B)*c^3*f*x + (-2*I*A + 10*B)*c^3)*e^(4*I*f*x + 4*I*e) + ((2*I*A - 10*B)*c^3*e^(6*I*f*x + 6*I*e) + (2*I*A
- 10*B)*c^3*e^(4*I*f*x + 4*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x
+ 4*I*e))

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Sympy [A]  time = 8.06495, size = 269, normalized size = 2.1 \begin{align*} \frac{2 B c^{3} e^{- 2 i e}}{a^{2} f \left (e^{2 i f x} + e^{- 2 i e}\right )} + \frac{c^{3} \left (i A - 5 B\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} + \frac{\left (\begin{cases} 2 A c^{3} x e^{4 i e} - \frac{i A c^{3} e^{2 i e} e^{- 2 i f x}}{f} + \frac{i A c^{3} e^{- 4 i f x}}{2 f} + 10 i B c^{3} x e^{4 i e} + \frac{3 B c^{3} e^{2 i e} e^{- 2 i f x}}{f} - \frac{B c^{3} e^{- 4 i f x}}{2 f} & \text{for}\: f \neq 0 \\x \left (2 A c^{3} e^{4 i e} - 2 A c^{3} e^{2 i e} + 2 A c^{3} + 10 i B c^{3} e^{4 i e} - 6 i B c^{3} e^{2 i e} + 2 i B c^{3}\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i e}}{a^{2}} \end{align*}

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

[Out]

2*B*c**3*exp(-2*I*e)/(a**2*f*(exp(2*I*f*x) + exp(-2*I*e))) + c**3*(I*A - 5*B)*log(exp(2*I*f*x) + exp(-2*I*e))/
(a**2*f) + Piecewise((2*A*c**3*x*exp(4*I*e) - I*A*c**3*exp(2*I*e)*exp(-2*I*f*x)/f + I*A*c**3*exp(-4*I*f*x)/(2*
f) + 10*I*B*c**3*x*exp(4*I*e) + 3*B*c**3*exp(2*I*e)*exp(-2*I*f*x)/f - B*c**3*exp(-4*I*f*x)/(2*f), Ne(f, 0)), (
x*(2*A*c**3*exp(4*I*e) - 2*A*c**3*exp(2*I*e) + 2*A*c**3 + 10*I*B*c**3*exp(4*I*e) - 6*I*B*c**3*exp(2*I*e) + 2*I
*B*c**3), True))*exp(-4*I*e)/a**2

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Giac [B]  time = 1.94927, size = 485, normalized size = 3.79 \begin{align*} \frac{\frac{12 \,{\left (-i \, A c^{3} + 5 \, B c^{3}\right )} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}{a^{2}} + \frac{6 \,{\left (i \, A c^{3} - 5 \, B c^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \,{\left (-i \, A c^{3} + 5 \, B c^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac{6 \,{\left (i \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 5 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 i \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i \, A c^{3} + 5 \, B c^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} a^{2}} - \frac{-25 i \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 125 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 100 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 548 i \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 198 i \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 894 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 100 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 548 i \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 25 i \, A c^{3} + 125 \, B c^{3}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}^{4}}}{6 \, f} \end{align*}

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

[Out]

1/6*(12*(-I*A*c^3 + 5*B*c^3)*log(tan(1/2*f*x + 1/2*e) - I)/a^2 + 6*(I*A*c^3 - 5*B*c^3)*log(abs(tan(1/2*f*x + 1
/2*e) + 1))/a^2 - 6*(-I*A*c^3 + 5*B*c^3)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 - 6*(I*A*c^3*tan(1/2*f*x + 1/2
*e)^2 - 5*B*c^3*tan(1/2*f*x + 1/2*e)^2 - 2*I*B*c^3*tan(1/2*f*x + 1/2*e) - I*A*c^3 + 5*B*c^3)/((tan(1/2*f*x + 1
/2*e)^2 - 1)*a^2) - (-25*I*A*c^3*tan(1/2*f*x + 1/2*e)^4 + 125*B*c^3*tan(1/2*f*x + 1/2*e)^4 - 100*A*c^3*tan(1/2
*f*x + 1/2*e)^3 - 548*I*B*c^3*tan(1/2*f*x + 1/2*e)^3 + 198*I*A*c^3*tan(1/2*f*x + 1/2*e)^2 - 894*B*c^3*tan(1/2*
f*x + 1/2*e)^2 + 100*A*c^3*tan(1/2*f*x + 1/2*e) + 548*I*B*c^3*tan(1/2*f*x + 1/2*e) - 25*I*A*c^3 + 125*B*c^3)/(
a^2*(tan(1/2*f*x + 1/2*e) - I)^4))/f