\(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx\) [731]

Optimal result

Integrand size = 41, antiderivative size = 99 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {2 (A+i B) c^2}{3 a^3 f (i-\tan (e+f x))^3}+\frac {(i A-3 B) c^2}{2 a^3 f (i-\tan (e+f x))^2}-\frac {i B c^2}{a^3 f (i-\tan (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 99, 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} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {c^2 (-3 B+i A)}{2 a^3 f (-\tan (e+f x)+i)^2}+\frac {2 c^2 (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac {i B c^2}{a^3 f (-\tan (e+f x)+i)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 5.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {c^2 \left (-A-i B+3 (i A+B) \tan (e+f x)+6 i B \tan ^2(e+f x)\right )}{6 a^3 f (-i+\tan (e+f x))^3} \]

[In]

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

[Out]

(c^2*(-A - I*B + 3*(I*A + B)*Tan[e + f*x] + (6*I)*B*Tan[e + f*x]^2))/(6*a^3*f*(-I + Tan[e + f*x])^3)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {c^{2} \left (\frac {i B}{-i+\tan \left (f x +e \right )}-\frac {-i A +3 B}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {2 i B +2 A}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}\right )}{f \,a^{3}}\)	\(69\)
default	\(\frac {c^{2} \left (\frac {i B}{-i+\tan \left (f x +e \right )}-\frac {-i A +3 B}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {2 i B +2 A}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}\right )}{f \,a^{3}}\)	\(69\)
risch	\(\frac {c^{2} {\mathrm e}^{-4 i \left (f x +e \right )} B}{8 a^{3} f}+\frac {i c^{2} {\mathrm e}^{-4 i \left (f x +e \right )} A}{8 a^{3} f}-\frac {c^{2} {\mathrm e}^{-6 i \left (f x +e \right )} B}{12 a^{3} f}+\frac {i c^{2} {\mathrm e}^{-6 i \left (f x +e \right )} A}{12 a^{3} f}\)	\(88\)
norman	\(\frac {-\frac {2 i A \,c^{2} \tan \left (f x +e \right )^{2}}{a f}+\frac {c^{2} A \tan \left (f x +e \right )}{a f}+\frac {i c^{2} B \tan \left (f x +e \right )^{5}}{a f}-\frac {-i A \,c^{2}+c^{2} B}{6 a f}-\frac {5 \left (i c^{2} B +c^{2} A \right ) \tan \left (f x +e \right )^{3}}{3 a f}-\frac {\left (-i A \,c^{2}+5 c^{2} B \right ) \tan \left (f x +e \right )^{4}}{2 a f}}{a^{2} \left (1+\tan \left (f x +e \right )^{2}\right )^{3}}\)	\(157\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.49 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=-\frac {{\left (3 \, {\left (-i \, A - B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (-i \, A + B\right )} c^{2}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{24 \, a^{3} f} \]

[In]

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

[Out]

-1/24*(3*(-I*A - B)*c^2*e^(2*I*f*x + 2*I*e) + 2*(-I*A + B)*c^2)*e^(-6*I*f*x - 6*I*e)/(a^3*f)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (78) = 156\).

Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (\left (8 i A a^{3} c^{2} f e^{4 i e} - 8 B a^{3} c^{2} f e^{4 i e}\right ) e^{- 6 i f x} + \left (12 i A a^{3} c^{2} f e^{6 i e} + 12 B a^{3} c^{2} f e^{6 i e}\right ) e^{- 4 i f x}\right ) e^{- 10 i e}}{96 a^{6} f^{2}} & \text {for}\: a^{6} f^{2} e^{10 i e} \neq 0 \\\frac {x \left (A c^{2} e^{2 i e} + A c^{2} - i B c^{2} e^{2 i e} + i B c^{2}\right ) e^{- 6 i e}}{2 a^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((((8*I*A*a**3*c**2*f*exp(4*I*e) - 8*B*a**3*c**2*f*exp(4*I*e))*exp(-6*I*f*x) + (12*I*A*a**3*c**2*f*ex
p(6*I*e) + 12*B*a**3*c**2*f*exp(6*I*e))*exp(-4*I*f*x))*exp(-10*I*e)/(96*a**6*f**2), Ne(a**6*f**2*exp(10*I*e),
0)), (x*(A*c**2*exp(2*I*e) + A*c**2 - I*B*c**2*exp(2*I*e) + I*B*c**2)*exp(-6*I*e)/(2*a**3), True))

Maxima [F(-2)]

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

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,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. 156 vs. \(2 (77) = 154\).

Time = 0.73 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.58 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=-\frac {2 \, {\left (3 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 3 i \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 8 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 i \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 i \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}} \]

[In]

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

[Out]

-2/3*(3*A*c^2*tan(1/2*f*x + 1/2*e)^5 - 3*I*A*c^2*tan(1/2*f*x + 1/2*e)^4 - 3*B*c^2*tan(1/2*f*x + 1/2*e)^4 - 8*A
*c^2*tan(1/2*f*x + 1/2*e)^3 - 2*I*B*c^2*tan(1/2*f*x + 1/2*e)^3 + 3*I*A*c^2*tan(1/2*f*x + 1/2*e)^2 + 3*B*c^2*ta
n(1/2*f*x + 1/2*e)^2 + 3*A*c^2*tan(1/2*f*x + 1/2*e))/(a^3*f*(tan(1/2*f*x + 1/2*e) - I)^6)

Mupad [B] (verification not implemented)

Time = 8.52 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {c^2\,\left (-B+A\,1{}\mathrm {i}\right )}{6}+\frac {c^2\,\mathrm {tan}\left (e+f\,x\right )\,\left (3\,A-B\,3{}\mathrm {i}\right )}{6}+B\,c^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{a^3\,f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )} \]

[In]

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

[Out]

((c^2*(A*1i - B))/6 + (c^2*tan(e + f*x)*(3*A - B*3i))/6 + B*c^2*tan(e + f*x)^2)/(a^3*f*(tan(e + f*x)*3i - 3*ta
n(e + f*x)^2 - tan(e + f*x)^3*1i + 1))