\(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx\) [718]

Optimal result

Integrand size = 41, antiderivative size = 128 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {(A+5 i B) c^3 x}{a^2}+\frac {(i A-5 B) c^3 \log (\cos (e+f x))}{a^2 f}-\frac {2 (i A-B) c^3}{a^2 f (i-\tan (e+f x))^2}+\frac {4 (A+2 i B) c^3}{a^2 f (i-\tan (e+f x))}-\frac {i B c^3 \tan (e+f x)}{a^2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 128, 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} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {4 c^3 (A+2 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac {2 c^3 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}+\frac {c^3 (-5 B+i A) \log (\cos (e+f x))}{a^2 f}+\frac {c^3 x (A+5 i B)}{a^2}-\frac {i B c^3 \tan (e+f x)}{a^2 f} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 5.64 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.72 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {\frac {B (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2}+\frac {(-i A+5 B) c^3 \left (\log (i-\tan (e+f x))+\frac {-2-4 i \tan (e+f x)}{(-i+\tan (e+f x))^2}\right )}{a^2}}{f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.56

method	result	size
derivativedivides	\(-\frac {i B \,c^{3} \tan \left (f x +e \right )}{a^{2} f}-\frac {8 i c^{3} B}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}-\frac {4 c^{3} A}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}-\frac {2 i c^{3} A}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {2 c^{3} B}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i c^{3} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2}}+\frac {5 c^{3} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2}}+\frac {c^{3} A \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}+\frac {5 i c^{3} B \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}\)	\(200\)
default	\(-\frac {i B \,c^{3} \tan \left (f x +e \right )}{a^{2} f}-\frac {8 i c^{3} B}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}-\frac {4 c^{3} A}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )}-\frac {2 i c^{3} A}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {2 c^{3} B}{f \,a^{2} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i c^{3} A \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2}}+\frac {5 c^{3} B \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2}}+\frac {c^{3} A \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}+\frac {5 i c^{3} B \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}\)	\(200\)
risch	\(\frac {3 c^{3} {\mathrm e}^{-2 i \left (f x +e \right )} B}{a^{2} f}-\frac {i c^{3} {\mathrm e}^{-2 i \left (f x +e \right )} A}{a^{2} f}-\frac {c^{3} {\mathrm e}^{-4 i \left (f x +e \right )} B}{2 a^{2} f}+\frac {i c^{3} {\mathrm e}^{-4 i \left (f x +e \right )} A}{2 a^{2} f}+\frac {10 i c^{3} B x}{a^{2}}+\frac {2 c^{3} A x}{a^{2}}+\frac {10 i c^{3} B e}{a^{2} f}+\frac {2 c^{3} A e}{a^{2} f}+\frac {2 c^{3} B}{f \,a^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {5 c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) B}{a^{2} f}+\frac {i c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) A}{a^{2} f}\)	\(210\)
norman	\(\frac {\frac {\left (5 i c^{3} B +c^{3} A \right ) x}{a}+\frac {-2 i c^{3} A +6 c^{3} B}{a f}+\frac {\left (5 i c^{3} B +c^{3} A \right ) x \tan \left (f x +e \right )^{4}}{a}+\frac {2 \left (5 i c^{3} B +c^{3} A \right ) x \tan \left (f x +e \right )^{2}}{a}-\frac {2 \left (5 i c^{3} B +2 c^{3} A \right ) \tan \left (f x +e \right )^{3}}{a f}+\frac {2 \left (-3 i c^{3} A +5 c^{3} B \right ) \tan \left (f x +e \right )^{2}}{a f}-\frac {5 i c^{3} \tan \left (f x +e \right ) B}{f a}-\frac {i c^{3} B \tan \left (f x +e \right )^{5}}{a f}}{a \left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (-i c^{3} A +5 c^{3} B \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 a^{2} f}\)	\(244\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {4 \, {\left (A + 5 i \, B\right )} c^{3} f x e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-i \, A + 5 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A - B\right )} c^{3} + 2 \, {\left (2 \, {\left (A + 5 i \, B\right )} c^{3} f x - {\left (i \, A - 5 \, B\right )} c^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, {\left ({\left (-i \, A + 5 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-i \, A + 5 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{2 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.41 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {2 B c^{3}}{a^{2} f e^{2 i e} e^{2 i f x} + a^{2} f} + \begin {cases} \frac {\left (\left (i A a^{2} c^{3} f e^{2 i e} - B a^{2} c^{3} f e^{2 i e}\right ) e^{- 4 i f x} + \left (- 2 i A a^{2} c^{3} f e^{4 i e} + 6 B a^{2} c^{3} f e^{4 i e}\right ) e^{- 2 i f x}\right ) e^{- 6 i e}}{2 a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {2 A c^{3} + 10 i B c^{3}}{a^{2}} + \frac {\left (2 A c^{3} e^{4 i e} - 2 A c^{3} e^{2 i e} + 2 A c^{3} + 10 i B c^{3} e^{4 i e} - 6 i B c^{3} e^{2 i e} + 2 i B c^{3}\right ) e^{- 4 i e}}{a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {i c^{3} \left (A + 5 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} + \frac {x \left (2 A c^{3} + 10 i B c^{3}\right )}{a^{2}} \]

[In]

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

[Out]

2*B*c**3/(a**2*f*exp(2*I*e)*exp(2*I*f*x) + a**2*f) + Piecewise((((I*A*a**2*c**3*f*exp(2*I*e) - B*a**2*c**3*f*e
xp(2*I*e))*exp(-4*I*f*x) + (-2*I*A*a**2*c**3*f*exp(4*I*e) + 6*B*a**2*c**3*f*exp(4*I*e))*exp(-2*I*f*x))*exp(-6*
I*e)/(2*a**4*f**2), Ne(a**4*f**2*exp(6*I*e), 0)), (x*(-(2*A*c**3 + 10*I*B*c**3)/a**2 + (2*A*c**3*exp(4*I*e) -
2*A*c**3*exp(2*I*e) + 2*A*c**3 + 10*I*B*c**3*exp(4*I*e) - 6*I*B*c**3*exp(2*I*e) + 2*I*B*c**3)*exp(-4*I*e)/a**2
), True)) + I*c**3*(A + 5*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(a**2*f) + x*(2*A*c**3 + 10*I*B*c**3)/a**2

Maxima [F(-2)]

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

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^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. \(2 (108) = 216\).

Time = 0.66 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.66 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {\frac {6 \, {\left (i \, A c^{3} - 5 \, B c^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2}} + \frac {12 \, {\left (-i \, A c^{3} + 5 \, B c^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{2}} - \frac {6 \, {\left (-i \, A c^{3} + 5 \, B c^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{2}} - \frac {6 \, {\left (i \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i \, A c^{3} + 5 \, B c^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{2}} - \frac {-25 i \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 125 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 100 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 548 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 198 i \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 894 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 100 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 548 i \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 25 i \, A c^{3} + 125 \, B c^{3}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{4}}}{6 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.52 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {c^3\,\left (6\,B-A\,2{}\mathrm {i}+4\,A\,\mathrm {tan}\left (e+f\,x\right )+B\,\mathrm {tan}\left (e+f\,x\right )\,7{}\mathrm {i}-A\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}+5\,B\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+2\,B\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}+A\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}-5\,B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+2\,A\,\mathrm {tan}\left (e+f\,x\right )\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+B\,\mathrm {tan}\left (e+f\,x\right )\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,10{}\mathrm {i}\right )}{a^2\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \]

[In]

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

[Out]

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