\(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx\) [707]

Optimal result

Integrand size = 41, antiderivative size = 121 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {4 (A+2 i B) c^3 x}{a}-\frac {4 (i A-2 B) c^3 \log (\cos (e+f x))}{a f}-\frac {4 (A+i B) c^3}{a f (i-\tan (e+f x))}+\frac {(A+4 i B) c^3 \tan (e+f x)}{a f}+\frac {B c^3 \tan ^2(e+f x)}{2 a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3669, 78} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {c^3 (A+4 i B) \tan (e+f x)}{a f}-\frac {4 c^3 (A+i B)}{a f (-\tan (e+f x)+i)}-\frac {4 c^3 (-2 B+i A) \log (\cos (e+f x))}{a f}-\frac {4 c^3 x (A+2 i B)}{a}+\frac {B c^3 \tan ^2(e+f x)}{2 a f} \]

[In]

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

[Out]

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

Rule 78

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 3669

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

Mathematica [A] (verified)

Time = 5.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {c^3 \left (14 A+27 i B+8 (A+2 i B) \log (i-\tan (e+f x))+(4 i A-11 B+8 i (A+2 i B) \log (i-\tan (e+f x))) \tan (e+f x)+(2 A+7 i B) \tan ^2(e+f x)+B \tan ^3(e+f x)\right )}{2 a f (-i+\tan (e+f x))} \]

[In]

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

[Out]

(c^3*(14*A + (27*I)*B + 8*(A + (2*I)*B)*Log[I - Tan[e + f*x]] + ((4*I)*A - 11*B + (8*I)*(A + (2*I)*B)*Log[I -
Tan[e + f*x]])*Tan[e + f*x] + (2*A + (7*I)*B)*Tan[e + f*x]^2 + B*Tan[e + f*x]^3))/(2*a*f*(-I + Tan[e + f*x]))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.58

method	result	size
derivativedivides	\(\frac {c^{3} A \tan \left (f x +e \right )}{f a}+\frac {4 i c^{3} \tan \left (f x +e \right ) B}{f a}+\frac {B \,c^{3} \tan \left (f x +e \right )^{2}}{2 a f}+\frac {4 i c^{3} B}{f a \left (-i+\tan \left (f x +e \right )\right )}+\frac {4 c^{3} A}{f a \left (-i+\tan \left (f x +e \right )\right )}-\frac {4 c^{3} A \arctan \left (\tan \left (f x +e \right )\right )}{f a}+\frac {2 i c^{3} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f a}-\frac {8 i c^{3} B \arctan \left (\tan \left (f x +e \right )\right )}{f a}-\frac {4 c^{3} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f a}\)	\(191\)
default	\(\frac {c^{3} A \tan \left (f x +e \right )}{f a}+\frac {4 i c^{3} \tan \left (f x +e \right ) B}{f a}+\frac {B \,c^{3} \tan \left (f x +e \right )^{2}}{2 a f}+\frac {4 i c^{3} B}{f a \left (-i+\tan \left (f x +e \right )\right )}+\frac {4 c^{3} A}{f a \left (-i+\tan \left (f x +e \right )\right )}-\frac {4 c^{3} A \arctan \left (\tan \left (f x +e \right )\right )}{f a}+\frac {2 i c^{3} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f a}-\frac {8 i c^{3} B \arctan \left (\tan \left (f x +e \right )\right )}{f a}-\frac {4 c^{3} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f a}\)	\(191\)
norman	\(\frac {\frac {\left (4 i c^{3} B +c^{3} A \right ) \tan \left (f x +e \right )^{3}}{a f}+\frac {\left (8 i c^{3} B +5 c^{3} A \right ) \tan \left (f x +e \right )}{a f}-\frac {4 \left (2 i c^{3} B +c^{3} A \right ) x}{a}-\frac {-8 i c^{3} A +9 c^{3} B}{2 a f}-\frac {4 \left (2 i c^{3} B +c^{3} A \right ) x \tan \left (f x +e \right )^{2}}{a}+\frac {c^{3} B \tan \left (f x +e \right )^{4}}{2 a f}}{1+\tan \left (f x +e \right )^{2}}-\frac {2 \left (-i c^{3} A +2 c^{3} B \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{a f}\)	\(192\)
risch	\(-\frac {2 c^{3} {\mathrm e}^{-2 i \left (f x +e \right )} B}{a f}+\frac {2 i c^{3} {\mathrm e}^{-2 i \left (f x +e \right )} A}{a f}-\frac {16 i c^{3} B x}{a}-\frac {8 c^{3} A x}{a}-\frac {16 i c^{3} B e}{f a}-\frac {8 c^{3} A e}{f a}-\frac {2 c^{3} \left (-i A \,{\mathrm e}^{2 i \left (f x +e \right )}+3 B \,{\mathrm e}^{2 i \left (f x +e \right )}-i A +4 B \right )}{a f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {8 c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) B}{f a}-\frac {4 i c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) A}{f a}\)	\(199\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (107) = 214\).

Time = 0.27 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.81 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=-\frac {2 \, {\left (4 \, {\left (A + 2 i \, B\right )} c^{3} f x e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-i \, A + B\right )} c^{3} + 2 \, {\left (4 \, {\left (A + 2 i \, B\right )} c^{3} f x + {\left (-i \, A + 2 \, B\right )} c^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (4 \, {\left (A + 2 i \, B\right )} c^{3} f x + 3 \, {\left (-i \, A + 2 \, B\right )} c^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left ({\left (i \, A - 2 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, {\left (i \, A - 2 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (i \, A - 2 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{a f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.19 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {2 i A c^{3} - 8 B c^{3} + \left (2 i A c^{3} e^{2 i e} - 6 B c^{3} e^{2 i e}\right ) e^{2 i f x}}{a f e^{4 i e} e^{4 i f x} + 2 a f e^{2 i e} e^{2 i f x} + a f} + \begin {cases} \frac {\left (2 i A c^{3} - 2 B c^{3}\right ) e^{- 2 i e} e^{- 2 i f x}}{a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (- \frac {- 8 A c^{3} - 16 i B c^{3}}{a} + \frac {\left (- 8 A c^{3} e^{2 i e} + 4 A c^{3} - 16 i B c^{3} e^{2 i e} + 4 i B c^{3}\right ) e^{- 2 i e}}{a}\right ) & \text {otherwise} \end {cases} - \frac {4 i c^{3} \left (A + 2 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} + \frac {x \left (- 8 A c^{3} - 16 i B c^{3}\right )}{a} \]

[In]

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

[Out]

(2*I*A*c**3 - 8*B*c**3 + (2*I*A*c**3*exp(2*I*e) - 6*B*c**3*exp(2*I*e))*exp(2*I*f*x))/(a*f*exp(4*I*e)*exp(4*I*f
*x) + 2*a*f*exp(2*I*e)*exp(2*I*f*x) + a*f) + Piecewise(((2*I*A*c**3 - 2*B*c**3)*exp(-2*I*e)*exp(-2*I*f*x)/(a*f
), Ne(a*f*exp(2*I*e), 0)), (x*(-(-8*A*c**3 - 16*I*B*c**3)/a + (-8*A*c**3*exp(2*I*e) + 4*A*c**3 - 16*I*B*c**3*e
xp(2*I*e) + 4*I*B*c**3)*exp(-2*I*e)/a), True)) - 4*I*c**3*(A + 2*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(a*f) +
x*(-8*A*c**3 - 16*I*B*c**3)/a

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 307 vs. \(2 (107) = 214\).

Time = 0.60 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.54 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {2 \, {\left (\frac {2 \, {\left (-i \, A c^{3} + 2 \, B c^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a} - \frac {4 \, {\left (-i \, A c^{3} + 2 \, B c^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a} - \frac {2 \, {\left (i \, A c^{3} - 2 \, B c^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a} - \frac {5 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 8 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 2 i \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 7 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 10 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 14 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 i \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 7 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - i \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{2} a}\right )}}{f} \]

[In]

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

[Out]

2*(2*(-I*A*c^3 + 2*B*c^3)*log(tan(1/2*f*x + 1/2*e) + 1)/a - 4*(-I*A*c^3 + 2*B*c^3)*log(tan(1/2*f*x + 1/2*e) -
I)/a - 2*(I*A*c^3 - 2*B*c^3)*log(tan(1/2*f*x + 1/2*e) - 1)/a - (5*A*c^3*tan(1/2*f*x + 1/2*e)^5 + 8*I*B*c^3*tan
(1/2*f*x + 1/2*e)^5 - 2*I*A*c^3*tan(1/2*f*x + 1/2*e)^4 + 7*B*c^3*tan(1/2*f*x + 1/2*e)^4 - 10*A*c^3*tan(1/2*f*x
+ 1/2*e)^3 - 14*I*B*c^3*tan(1/2*f*x + 1/2*e)^3 + 2*I*A*c^3*tan(1/2*f*x + 1/2*e)^2 - 7*B*c^3*tan(1/2*f*x + 1/2
*e)^2 + 5*A*c^3*tan(1/2*f*x + 1/2*e) + 8*I*B*c^3*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^3 - I*tan(1/2*f*
x + 1/2*e)^2 - tan(1/2*f*x + 1/2*e) + I)^2*a))/f

Mupad [B] (verification not implemented)

Time = 8.48 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-\frac {8\,B\,c^3}{a}+\frac {A\,c^3\,4{}\mathrm {i}}{a}\right )}{f}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {c^3\,\left (A+B\,2{}\mathrm {i}\right )}{a}+\frac {B\,c^3\,2{}\mathrm {i}}{a}\right )}{f}+\frac {\frac {4\,B\,c^3}{a}+\frac {\left (4\,A\,c^3+B\,c^3\,8{}\mathrm {i}\right )\,1{}\mathrm {i}}{a}}{f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {B\,c^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,a\,f} \]

[In]

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

[Out]

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