\(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx\) [698]

Optimal result

Integrand size = 41, antiderivative size = 123 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx=\frac {a^3 (A-5 i B) x}{c^2}-\frac {a^3 (i A+5 B) \log (\cos (e+f x))}{c^2 f}+\frac {i a^3 B \tan (e+f x)}{c^2 f}+\frac {2 a^3 (i A+B)}{c^2 f (i+\tan (e+f x))^2}-\frac {4 a^3 (A-2 i B)}{c^2 f (i+\tan (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 123, 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} \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx=-\frac {4 a^3 (A-2 i B)}{c^2 f (\tan (e+f x)+i)}+\frac {2 a^3 (B+i A)}{c^2 f (\tan (e+f x)+i)^2}-\frac {a^3 (5 B+i A) \log (\cos (e+f x))}{c^2 f}+\frac {a^3 x (A-5 i B)}{c^2}+\frac {i a^3 B \tan (e+f x)}{c^2 f} \]

[In]

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

[Out]

(a^3*(A - (5*I)*B)*x)/c^2 - (a^3*(I*A + 5*B)*Log[Cos[e + f*x]])/(c^2*f) + (I*a^3*B*Tan[e + f*x])/(c^2*f) + (2*
a^3*(I*A + B))/(c^2*f*(I + Tan[e + f*x])^2) - (4*a^3*(A - (2*I)*B))/(c^2*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)^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (\frac {i a^2 B}{c^3}-\frac {4 i a^2 (A-i B)}{c^3 (i+x)^3}+\frac {4 a^2 (A-2 i B)}{c^3 (i+x)^2}+\frac {a^2 (i A+5 B)}{c^3 (i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {a^3 (A-5 i B) x}{c^2}-\frac {a^3 (i A+5 B) \log (\cos (e+f x))}{c^2 f}+\frac {i a^3 B \tan (e+f x)}{c^2 f}+\frac {2 a^3 (i A+B)}{c^2 f (i+\tan (e+f x))^2}-\frac {4 a^3 (A-2 i B)}{c^2 f (i+\tan (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.60 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx=\frac {\frac {B (a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2}+\frac {a^3 (i A+5 B) \left (\log (i+\tan (e+f x))+\frac {-2+4 i \tan (e+f x)}{(i+\tan (e+f x))^2}\right )}{c^2}}{f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.54

method	result	size
risch	\(-\frac {{\mathrm e}^{4 i \left (f x +e \right )} a^{3} B}{2 c^{2} f}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )} a^{3} A}{2 c^{2} f}+\frac {3 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{3} B}{c^{2} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} a^{3} A}{c^{2} f}+\frac {10 i a^{3} B e}{c^{2} f}-\frac {2 a^{3} A e}{c^{2} f}-\frac {2 B \,a^{3}}{f \,c^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {5 a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) B}{c^{2} f}-\frac {i a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) A}{c^{2} f}\)	\(189\)
derivativedivides	\(\frac {i a^{3} B \tan \left (f x +e \right )}{c^{2} f}+\frac {8 i a^{3} B}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )}-\frac {4 a^{3} A}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )}+\frac {i a^{3} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{2}}+\frac {5 a^{3} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{2}}+\frac {a^{3} A \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{2}}-\frac {5 i a^{3} B \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{2}}+\frac {2 i a^{3} A}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {2 a^{3} B}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}\)	\(200\)
default	\(\frac {i a^{3} B \tan \left (f x +e \right )}{c^{2} f}+\frac {8 i a^{3} B}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )}-\frac {4 a^{3} A}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )}+\frac {i a^{3} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{2}}+\frac {5 a^{3} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{2}}+\frac {a^{3} A \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{2}}-\frac {5 i a^{3} B \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{2}}+\frac {2 i a^{3} A}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {2 a^{3} B}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}\)	\(200\)
norman	\(\frac {\frac {\left (-5 i B \,a^{3}+a^{3} A \right ) x}{c}+\frac {2 i A \,a^{3}+6 B \,a^{3}}{c f}+\frac {\left (-5 i B \,a^{3}+a^{3} A \right ) x \tan \left (f x +e \right )^{4}}{c}+\frac {i B \,a^{3} \tan \left (f x +e \right )^{5}}{c f}+\frac {2 \left (-5 i B \,a^{3}+a^{3} A \right ) x \tan \left (f x +e \right )^{2}}{c}-\frac {2 \left (-5 i B \,a^{3}+2 a^{3} A \right ) \tan \left (f x +e \right )^{3}}{c f}+\frac {2 \left (3 i A \,a^{3}+5 B \,a^{3}\right ) \tan \left (f x +e \right )^{2}}{c f}+\frac {5 i a^{3} \tan \left (f x +e \right ) B}{f c}}{c \left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (i A \,a^{3}+5 B \,a^{3}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 c^{2} f}\)	\(244\)

[In]

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

-1/2/c^2/f*exp(4*I*(f*x+e))*a^3*B-1/2*I/c^2/f*exp(4*I*(f*x+e))*a^3*A+3/c^2/f*exp(2*I*(f*x+e))*a^3*B+I/c^2/f*ex
p(2*I*(f*x+e))*a^3*A+10*I*a^3/c^2/f*B*e-2*a^3/c^2/f*A*e-2/f/c^2*B*a^3/(exp(2*I*(f*x+e))+1)-5*a^3/c^2/f*ln(exp(
2*I*(f*x+e))+1)*B-I*a^3/c^2/f*ln(exp(2*I*(f*x+e))+1)*A

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.12 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx=\frac {{\left (-i \, A - B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (i \, A + 5 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, {\left (-i \, A - 3 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, B a^{3} - 2 \, {\left ({\left (i \, A + 5 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A + 5 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{2 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.92 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx=- \frac {2 B a^{3}}{c^{2} f e^{2 i e} e^{2 i f x} + c^{2} f} - \frac {i a^{3} \left (A - 5 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{2} f} + \begin {cases} \frac {\left (2 i A a^{3} c^{2} f e^{2 i e} + 6 B a^{3} c^{2} f e^{2 i e}\right ) e^{2 i f x} + \left (- i A a^{3} c^{2} f e^{4 i e} - B a^{3} c^{2} f e^{4 i e}\right ) e^{4 i f x}}{2 c^{4} f^{2}} & \text {for}\: c^{4} f^{2} \neq 0 \\\frac {x \left (2 A a^{3} e^{4 i e} - 2 A a^{3} e^{2 i e} - 2 i B a^{3} e^{4 i e} + 6 i B a^{3} e^{2 i e}\right )}{c^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**2,x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^2,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. 341 vs. \(2 (109) = 218\).

Time = 0.64 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.77 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx=\frac {\frac {6 \, {\left (-i \, A a^{3} - 5 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{2}} + \frac {12 \, {\left (i \, A a^{3} + 5 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{2}} - \frac {6 \, {\left (i \, A a^{3} + 5 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{2}} - \frac {6 \, {\left (-i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i \, A a^{3} + 5 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2}} - \frac {25 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 125 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 100 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 548 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 198 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 894 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 100 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 548 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 25 i \, A a^{3} + 125 \, B a^{3}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{4}}}{6 \, f} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.10 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.48 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^2} \, dx=-\frac {a^3\,\left (7\,B\,\mathrm {tan}\left (e+f\,x\right )+B\,6{}\mathrm {i}+A\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}-2\,A+B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,2{}\mathrm {i}+B\,{\mathrm {tan}\left (e+f\,x\right )}^3-A\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )+B\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,5{}\mathrm {i}+A\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+10\,B\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )+A\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2-B\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2\,5{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^2\,f\,{\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \]

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3)/(c - c*tan(e + f*x)*1i)^2,x)

[Out]

-(a^3*(B*6i - 2*A + A*tan(e + f*x)*4i + 7*B*tan(e + f*x) + B*tan(e + f*x)^2*2i + B*tan(e + f*x)^3 - A*log(tan(
e + f*x) + 1i) + B*log(tan(e + f*x) + 1i)*5i + A*log(tan(e + f*x) + 1i)*tan(e + f*x)*2i + 10*B*log(tan(e + f*x
) + 1i)*tan(e + f*x) + A*log(tan(e + f*x) + 1i)*tan(e + f*x)^2 - B*log(tan(e + f*x) + 1i)*tan(e + f*x)^2*5i)*1
i)/(c^2*f*(tan(e + f*x)*1i - 1)^2)