\(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx\) [697]

Optimal result

Integrand size = 41, antiderivative size = 119 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=-\frac {4 a^3 (A-2 i B) x}{c}+\frac {4 a^3 (i A+2 B) \log (\cos (e+f x))}{c f}+\frac {a^3 (A-4 i B) \tan (e+f x)}{c f}+\frac {a^3 B \tan ^2(e+f x)}{2 c f}+\frac {4 a^3 (A-i B)}{c f (i+\tan (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 119, 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} \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=\frac {a^3 (A-4 i B) \tan (e+f x)}{c f}+\frac {4 a^3 (A-i B)}{c f (\tan (e+f x)+i)}+\frac {4 a^3 (2 B+i A) \log (\cos (e+f x))}{c f}-\frac {4 a^3 x (A-2 i B)}{c}+\frac {a^3 B \tan ^2(e+f x)}{2 c f} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 5.68 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=\frac {a^3 \left (14 A-27 i B+8 (A-2 i B) \log (i+\tan (e+f x))+(-4 i A-11 B-8 i (A-2 i B) \log (i+\tan (e+f x))) \tan (e+f x)+(2 A-7 i B) \tan ^2(e+f x)+B \tan ^3(e+f x)\right )}{2 c f (i+\tan (e+f x))} \]

[In]

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

[Out]

(a^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*c*f*(I + Tan[e + f*x]))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.50

method	result	size
risch	\(-\frac {2 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{3} B}{c f}-\frac {2 i {\mathrm e}^{2 i \left (f x +e \right )} a^{3} A}{c f}-\frac {16 i a^{3} B e}{f c}+\frac {8 a^{3} A e}{f c}+\frac {2 a^{3} \left (i A \,{\mathrm e}^{2 i \left (f x +e \right )}+5 B \,{\mathrm e}^{2 i \left (f x +e \right )}+i A +4 B \right )}{c f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {8 a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) B}{f c}+\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) A}{f c}\)	\(178\)
derivativedivides	\(\frac {a^{3} A \tan \left (f x +e \right )}{f c}-\frac {4 i a^{3} \tan \left (f x +e \right ) B}{f c}+\frac {a^{3} B \tan \left (f x +e \right )^{2}}{2 c f}-\frac {4 a^{3} A \arctan \left (\tan \left (f x +e \right )\right )}{f c}-\frac {2 i a^{3} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f c}+\frac {8 i a^{3} B \arctan \left (\tan \left (f x +e \right )\right )}{f c}-\frac {4 a^{3} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f c}-\frac {4 i a^{3} B}{f c \left (i+\tan \left (f x +e \right )\right )}+\frac {4 a^{3} A}{f c \left (i+\tan \left (f x +e \right )\right )}\)	\(191\)
default	\(\frac {a^{3} A \tan \left (f x +e \right )}{f c}-\frac {4 i a^{3} \tan \left (f x +e \right ) B}{f c}+\frac {a^{3} B \tan \left (f x +e \right )^{2}}{2 c f}-\frac {4 a^{3} A \arctan \left (\tan \left (f x +e \right )\right )}{f c}-\frac {2 i a^{3} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f c}+\frac {8 i a^{3} B \arctan \left (\tan \left (f x +e \right )\right )}{f c}-\frac {4 a^{3} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f c}-\frac {4 i a^{3} B}{f c \left (i+\tan \left (f x +e \right )\right )}+\frac {4 a^{3} A}{f c \left (i+\tan \left (f x +e \right )\right )}\)	\(191\)
norman	\(\frac {\frac {\left (-4 i B \,a^{3}+a^{3} A \right ) \tan \left (f x +e \right )^{3}}{c f}+\frac {\left (-8 i B \,a^{3}+5 a^{3} A \right ) \tan \left (f x +e \right )}{c f}-\frac {4 \left (-2 i B \,a^{3}+a^{3} A \right ) x}{c}-\frac {8 i A \,a^{3}+9 B \,a^{3}}{2 c f}-\frac {4 \left (-2 i B \,a^{3}+a^{3} A \right ) x \tan \left (f x +e \right )^{2}}{c}+\frac {B \,a^{3} \tan \left (f x +e \right )^{4}}{2 c f}}{1+\tan \left (f x +e \right )^{2}}-\frac {2 \left (i A \,a^{3}+2 B \,a^{3}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{c f}\)	\(192\)

[In]

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

[Out]

-2/c/f*exp(2*I*(f*x+e))*a^3*B-2*I/c/f*exp(2*I*(f*x+e))*a^3*A-16*I*a^3/f/c*B*e+8*a^3/f/c*A*e+2*a^3*(I*A*exp(2*I
*(f*x+e))+5*B*exp(2*I*(f*x+e))+I*A+4*B)/c/f/(exp(2*I*(f*x+e))+1)^2+8*a^3/f/c*ln(exp(2*I*(f*x+e))+1)*B+4*I*a^3/
f/c*ln(exp(2*I*(f*x+e))+1)*A

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.38 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=-\frac {2 \, {\left ({\left (i \, A + B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, {\left (i \, A + B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 4 \, B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A - 4 \, B\right )} a^{3} + 2 \, {\left ({\left (-i \, A - 2 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-i \, A - 2 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A - 2 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{c f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.73 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=\frac {4 i a^{3} \left (A - 2 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \frac {2 i A a^{3} + 8 B a^{3} + \left (2 i A a^{3} e^{2 i e} + 10 B a^{3} e^{2 i e}\right ) e^{2 i f x}}{c f e^{4 i e} e^{4 i f x} + 2 c f e^{2 i e} e^{2 i f x} + c f} + \begin {cases} \frac {\left (- 2 i A a^{3} e^{2 i e} - 2 B a^{3} e^{2 i e}\right ) e^{2 i f x}}{c f} & \text {for}\: c f \neq 0 \\\frac {x \left (4 A a^{3} e^{2 i e} - 4 i B a^{3} e^{2 i e}\right )}{c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

4*I*a**3*(A - 2*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(c*f) + (2*I*A*a**3 + 8*B*a**3 + (2*I*A*a**3*exp(2*I*e) +
10*B*a**3*exp(2*I*e))*exp(2*I*f*x))/(c*f*exp(4*I*e)*exp(4*I*f*x) + 2*c*f*exp(2*I*e)*exp(2*I*f*x) + c*f) + Pie
cewise(((-2*I*A*a**3*exp(2*I*e) - 2*B*a**3*exp(2*I*e))*exp(2*I*f*x)/(c*f), Ne(c*f, 0)), (x*(4*A*a**3*exp(2*I*e
) - 4*I*B*a**3*exp(2*I*e))/c, True))

Maxima [F(-2)]

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

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*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. 306 vs. \(2 (107) = 214\).

Time = 0.58 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.57 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=-\frac {2 \, {\left (\frac {2 \, {\left (-i \, A a^{3} - 2 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c} + \frac {4 \, {\left (i \, A a^{3} + 2 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c} + \frac {2 \, {\left (-i \, A a^{3} - 2 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c} + \frac {5 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 8 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 2 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 7 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 10 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 14 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 7 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 8 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + i \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{2} c}\right )}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.63 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.17 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=\frac {B\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,c\,f}+\frac {\frac {4\,A\,a^3-B\,a^3\,8{}\mathrm {i}}{c}+\frac {B\,a^3\,4{}\mathrm {i}}{c}}{f\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {B\,a^3\,2{}\mathrm {i}}{c}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c}\right )}{f}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (\frac {8\,B\,a^3}{c}+\frac {A\,a^3\,4{}\mathrm {i}}{c}\right )}{f} \]

[In]

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

[Out]

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