\(\int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx\) [683]

Optimal result

Integrand size = 41, antiderivative size = 93 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=-\frac {a^2 (A-3 i B) x}{c}+\frac {a^2 (i A+3 B) \log (\cos (e+f x))}{c f}-\frac {i a^2 B \tan (e+f x)}{c f}+\frac {2 a^2 (A-i B)}{c f (i+\tan (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 93, 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} \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=\frac {2 a^2 (A-i B)}{c f (\tan (e+f x)+i)}+\frac {a^2 (3 B+i A) \log (\cos (e+f x))}{c f}-\frac {a^2 x (A-3 i B)}{c}-\frac {i a^2 B \tan (e+f x)}{c f} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 5.53 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.90 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=\frac {B (a+i a \tan (e+f x))^2}{f (c-i c \tan (e+f x))}-\frac {a^2 (i A+3 B) \left (\log (i+\tan (e+f x))+\frac {2 i}{i+\tan (e+f x)}\right )}{c f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.57

method	result	size
risch	\(-\frac {{\mathrm e}^{2 i \left (f x +e \right )} a^{2} B}{c f}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )} A \,a^{2}}{c f}-\frac {6 i a^{2} B e}{c f}+\frac {2 a^{2} A e}{c f}+\frac {2 a^{2} B}{f c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) B}{c f}+\frac {i a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) A}{c f}\)	\(146\)
derivativedivides	\(-\frac {i a^{2} B \tan \left (f x +e \right )}{c f}-\frac {2 i a^{2} B}{f c \left (i+\tan \left (f x +e \right )\right )}+\frac {2 a^{2} A}{f c \left (i+\tan \left (f x +e \right )\right )}-\frac {i a^{2} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f c}-\frac {3 a^{2} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f c}-\frac {a^{2} A \arctan \left (\tan \left (f x +e \right )\right )}{f c}+\frac {3 i a^{2} B \arctan \left (\tan \left (f x +e \right )\right )}{f c}\)	\(154\)
default	\(-\frac {i a^{2} B \tan \left (f x +e \right )}{c f}-\frac {2 i a^{2} B}{f c \left (i+\tan \left (f x +e \right )\right )}+\frac {2 a^{2} A}{f c \left (i+\tan \left (f x +e \right )\right )}-\frac {i a^{2} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f c}-\frac {3 a^{2} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f c}-\frac {a^{2} A \arctan \left (\tan \left (f x +e \right )\right )}{f c}+\frac {3 i a^{2} B \arctan \left (\tan \left (f x +e \right )\right )}{f c}\)	\(154\)
norman	\(\frac {\frac {\left (-3 i B \,a^{2}+2 A \,a^{2}\right ) \tan \left (f x +e \right )}{c f}-\frac {\left (-3 i B \,a^{2}+A \,a^{2}\right ) x}{c}-\frac {2 i A \,a^{2}+2 B \,a^{2}}{c f}-\frac {\left (-3 i B \,a^{2}+A \,a^{2}\right ) x \tan \left (f x +e \right )^{2}}{c}-\frac {i B \,a^{2} \tan \left (f x +e \right )^{3}}{c f}}{1+\tan \left (f x +e \right )^{2}}-\frac {\left (i A \,a^{2}+3 B \,a^{2}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 c f}\)	\(165\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.19 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=\frac {{\left (-i \, A - B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-i \, A - B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, B a^{2} + {\left ({\left (i \, A + 3 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A + 3 \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.44 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=\frac {2 B a^{2}}{c f e^{2 i e} e^{2 i f x} + c f} + \frac {i a^{2} \left (A - 3 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \begin {cases} \frac {\left (- i A a^{2} e^{2 i e} - B a^{2} e^{2 i e}\right ) e^{2 i f x}}{c f} & \text {for}\: c f \neq 0 \\\frac {x \left (2 A a^{2} e^{2 i e} - 2 i B a^{2} e^{2 i e}\right )}{c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

2*B*a**2/(c*f*exp(2*I*e)*exp(2*I*f*x) + c*f) + I*a**2*(A - 3*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(c*f) + Piec
ewise(((-I*A*a**2*exp(2*I*e) - B*a**2*exp(2*I*e))*exp(2*I*f*x)/(c*f), Ne(c*f, 0)), (x*(2*A*a**2*exp(2*I*e) - 2
*I*B*a**2*exp(2*I*e))/c, True))

Maxima [F(-2)]

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

[In]

integrate((a+I*a*tan(f*x+e))^2*(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. 271 vs. \(2 (83) = 166\).

Time = 0.47 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.91 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=\frac {\frac {{\left (i \, A a^{2} + 3 \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c} + \frac {2 \, {\left (-i \, A a^{2} - 3 \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c} - \frac {{\left (-i \, A a^{2} - 3 \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c} - \frac {i \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 i \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i \, A a^{2} - 3 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c} - \frac {-3 i \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 10 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 22 i \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 i \, A a^{2} + 9 \, B a^{2}}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{2}}}{f} \]

[In]

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

[Out]

((I*A*a^2 + 3*B*a^2)*log(tan(1/2*f*x + 1/2*e) + 1)/c + 2*(-I*A*a^2 - 3*B*a^2)*log(tan(1/2*f*x + 1/2*e) + I)/c
- (-I*A*a^2 - 3*B*a^2)*log(tan(1/2*f*x + 1/2*e) - 1)/c - (I*A*a^2*tan(1/2*f*x + 1/2*e)^2 + 3*B*a^2*tan(1/2*f*x
+ 1/2*e)^2 - 2*I*B*a^2*tan(1/2*f*x + 1/2*e) - I*A*a^2 - 3*B*a^2)/((tan(1/2*f*x + 1/2*e)^2 - 1)*c) - (-3*I*A*a
^2*tan(1/2*f*x + 1/2*e)^2 - 9*B*a^2*tan(1/2*f*x + 1/2*e)^2 + 10*A*a^2*tan(1/2*f*x + 1/2*e) - 22*I*B*a^2*tan(1/
2*f*x + 1/2*e) + 3*I*A*a^2 + 9*B*a^2)/(c*(tan(1/2*f*x + 1/2*e) + I)^2))/f

Mupad [B] (verification not implemented)

Time = 8.40 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.13 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{c-i c \tan (e+f x)} \, dx=-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (\frac {3\,B\,a^2}{c}+\frac {A\,a^2\,1{}\mathrm {i}}{c}\right )}{f}+\frac {\frac {A\,a^2+B\,a^2\,1{}\mathrm {i}}{c}+\frac {A\,a^2-B\,a^2\,3{}\mathrm {i}}{c}}{f\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {B\,a^2\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{c\,f} \]

[In]

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

[Out]

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