\(\int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx\) [687]

Optimal result

Integrand size = 41, antiderivative size = 95 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\frac {2 a^2 (A-i B)}{5 c^5 f (i+\tan (e+f x))^5}+\frac {a^2 (i A+3 B)}{4 c^5 f (i+\tan (e+f x))^4}+\frac {i a^2 B}{3 c^5 f (i+\tan (e+f x))^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 95, 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} \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\frac {a^2 (3 B+i A)}{4 c^5 f (\tan (e+f x)+i)^4}+\frac {2 a^2 (A-i B)}{5 c^5 f (\tan (e+f x)+i)^5}+\frac {i a^2 B}{3 c^5 f (\tan (e+f x)+i)^3} \]

[In]

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

[Out]

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

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)^6} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (-\frac {2 a (A-i B)}{c^6 (i+x)^6}-\frac {i a (A-3 i B)}{c^6 (i+x)^5}-\frac {i a B}{c^6 (i+x)^4}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {2 a^2 (A-i B)}{5 c^5 f (i+\tan (e+f x))^5}+\frac {a^2 (i A+3 B)}{4 c^5 f (i+\tan (e+f x))^4}+\frac {i a^2 B}{3 c^5 f (i+\tan (e+f x))^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.36 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\frac {a^2 \left (9 A+i B+5 (3 i A+B) \tan (e+f x)+20 i B \tan ^2(e+f x)\right )}{60 c^5 f (i+\tan (e+f x))^5} \]

[In]

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

[Out]

(a^2*(9*A + I*B + 5*((3*I)*A + B)*Tan[e + f*x] + (20*I)*B*Tan[e + f*x]^2))/(60*c^5*f*(I + Tan[e + f*x])^5)

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {2 i B -2 A}{5 \left (i+\tan \left (f x +e \right )\right )^{5}}+\frac {i B}{3 \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {-i A -3 B}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}\right )}{f \,c^{5}}\)	\(69\)
default	\(\frac {a^{2} \left (-\frac {2 i B -2 A}{5 \left (i+\tan \left (f x +e \right )\right )^{5}}+\frac {i B}{3 \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {-i A -3 B}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}\right )}{f \,c^{5}}\)	\(69\)
risch	\(-\frac {a^{2} {\mathrm e}^{10 i \left (f x +e \right )} B}{80 c^{5} f}-\frac {i a^{2} {\mathrm e}^{10 i \left (f x +e \right )} A}{80 c^{5} f}-\frac {{\mathrm e}^{8 i \left (f x +e \right )} B \,a^{2}}{64 c^{5} f}-\frac {3 i {\mathrm e}^{8 i \left (f x +e \right )} A \,a^{2}}{64 c^{5} f}+\frac {{\mathrm e}^{6 i \left (f x +e \right )} B \,a^{2}}{48 c^{5} f}-\frac {i {\mathrm e}^{6 i \left (f x +e \right )} A \,a^{2}}{16 c^{5} f}+\frac {a^{2} {\mathrm e}^{4 i \left (f x +e \right )} B}{32 c^{5} f}-\frac {i a^{2} {\mathrm e}^{4 i \left (f x +e \right )} A}{32 c^{5} f}\)	\(174\)
norman	\(\frac {\frac {A \,a^{2} \tan \left (f x +e \right )}{c f}+\frac {-9 i A \,a^{2}+B \,a^{2}}{60 c f}-\frac {\left (-7 i B \,a^{2}+12 A \,a^{2}\right ) \tan \left (f x +e \right )^{3}}{3 c f}+\frac {\left (i A \,a^{2}+7 B \,a^{2}\right ) \tan \left (f x +e \right )^{6}}{4 c f}+\frac {7 \left (-8 i B \,a^{2}+3 A \,a^{2}\right ) \tan \left (f x +e \right )^{5}}{15 c f}+\frac {\left (33 i A \,a^{2}+7 B \,a^{2}\right ) \tan \left (f x +e \right )^{2}}{12 c f}-\frac {\left (39 i A \,a^{2}+49 B \,a^{2}\right ) \tan \left (f x +e \right )^{4}}{12 c f}+\frac {i B \,a^{2} \tan \left (f x +e \right )^{7}}{3 c f}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{5} c^{4}}\)	\(227\)

[In]

int((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=-\frac {12 \, {\left (i \, A + B\right )} a^{2} e^{\left (10 i \, f x + 10 i \, e\right )} + 15 \, {\left (3 i \, A + B\right )} a^{2} e^{\left (8 i \, f x + 8 i \, e\right )} + 20 \, {\left (3 i \, A - B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 30 \, {\left (i \, A - B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )}}{960 \, c^{5} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^5,x, algorithm="fricas")

[Out]

-1/960*(12*(I*A + B)*a^2*e^(10*I*f*x + 10*I*e) + 15*(3*I*A + B)*a^2*e^(8*I*f*x + 8*I*e) + 20*(3*I*A - B)*a^2*e
^(6*I*f*x + 6*I*e) + 30*(I*A - B)*a^2*e^(4*I*f*x + 4*I*e))/(c^5*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. 332 vs. \(2 (76) = 152\).

Time = 0.41 (sec) , antiderivative size = 332, normalized size of antiderivative = 3.49 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\begin {cases} \frac {\left (- 245760 i A a^{2} c^{15} f^{3} e^{4 i e} + 245760 B a^{2} c^{15} f^{3} e^{4 i e}\right ) e^{4 i f x} + \left (- 491520 i A a^{2} c^{15} f^{3} e^{6 i e} + 163840 B a^{2} c^{15} f^{3} e^{6 i e}\right ) e^{6 i f x} + \left (- 368640 i A a^{2} c^{15} f^{3} e^{8 i e} - 122880 B a^{2} c^{15} f^{3} e^{8 i e}\right ) e^{8 i f x} + \left (- 98304 i A a^{2} c^{15} f^{3} e^{10 i e} - 98304 B a^{2} c^{15} f^{3} e^{10 i e}\right ) e^{10 i f x}}{7864320 c^{20} f^{4}} & \text {for}\: c^{20} f^{4} \neq 0 \\\frac {x \left (A a^{2} e^{10 i e} + 3 A a^{2} e^{8 i e} + 3 A a^{2} e^{6 i e} + A a^{2} e^{4 i e} - i B a^{2} e^{10 i e} - i B a^{2} e^{8 i e} + i B a^{2} e^{6 i e} + i B a^{2} e^{4 i e}\right )}{8 c^{5}} & \text {otherwise} \end {cases} \]

[In]

integrate((a+I*a*tan(f*x+e))**2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**5,x)

[Out]

Piecewise((((-245760*I*A*a**2*c**15*f**3*exp(4*I*e) + 245760*B*a**2*c**15*f**3*exp(4*I*e))*exp(4*I*f*x) + (-49
1520*I*A*a**2*c**15*f**3*exp(6*I*e) + 163840*B*a**2*c**15*f**3*exp(6*I*e))*exp(6*I*f*x) + (-368640*I*A*a**2*c*
*15*f**3*exp(8*I*e) - 122880*B*a**2*c**15*f**3*exp(8*I*e))*exp(8*I*f*x) + (-98304*I*A*a**2*c**15*f**3*exp(10*I
*e) - 98304*B*a**2*c**15*f**3*exp(10*I*e))*exp(10*I*f*x))/(7864320*c**20*f**4), Ne(c**20*f**4, 0)), (x*(A*a**2
*exp(10*I*e) + 3*A*a**2*exp(8*I*e) + 3*A*a**2*exp(6*I*e) + A*a**2*exp(4*I*e) - I*B*a**2*exp(10*I*e) - I*B*a**2
*exp(8*I*e) + I*B*a**2*exp(6*I*e) + I*B*a**2*exp(4*I*e))/(8*c**5), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^5,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. 292 vs. \(2 (77) = 154\).

Time = 1.13 (sec) , antiderivative size = 292, normalized size of antiderivative = 3.07 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=-\frac {2 \, {\left (15 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 45 i \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 15 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 150 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 10 i \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 225 i \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 55 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 306 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 24 i \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 225 i \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 55 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 150 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 10 i \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 45 i \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{15 \, c^{5} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{10}} \]

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^5,x, algorithm="giac")

[Out]

-2/15*(15*A*a^2*tan(1/2*f*x + 1/2*e)^9 + 45*I*A*a^2*tan(1/2*f*x + 1/2*e)^8 - 15*B*a^2*tan(1/2*f*x + 1/2*e)^8 -
150*A*a^2*tan(1/2*f*x + 1/2*e)^7 - 10*I*B*a^2*tan(1/2*f*x + 1/2*e)^7 - 225*I*A*a^2*tan(1/2*f*x + 1/2*e)^6 + 5
5*B*a^2*tan(1/2*f*x + 1/2*e)^6 + 306*A*a^2*tan(1/2*f*x + 1/2*e)^5 + 24*I*B*a^2*tan(1/2*f*x + 1/2*e)^5 + 225*I*
A*a^2*tan(1/2*f*x + 1/2*e)^4 - 55*B*a^2*tan(1/2*f*x + 1/2*e)^4 - 150*A*a^2*tan(1/2*f*x + 1/2*e)^3 - 10*I*B*a^2
*tan(1/2*f*x + 1/2*e)^3 - 45*I*A*a^2*tan(1/2*f*x + 1/2*e)^2 + 15*B*a^2*tan(1/2*f*x + 1/2*e)^2 + 15*A*a^2*tan(1
/2*f*x + 1/2*e))/(c^5*f*(tan(1/2*f*x + 1/2*e) + I)^10)

Mupad [B] (verification not implemented)

Time = 8.65 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.14 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\frac {\frac {a^2\,\left (9\,A+B\,1{}\mathrm {i}\right )}{60}+\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )\,\left (5\,B+A\,15{}\mathrm {i}\right )}{60}+\frac {B\,a^2\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}}{3}}{c^5\,f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5+{\mathrm {tan}\left (e+f\,x\right )}^4\,5{}\mathrm {i}-10\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,10{}\mathrm {i}+5\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )} \]

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^2)/(c - c*tan(e + f*x)*1i)^5,x)

[Out]

((a^2*(9*A + B*1i))/60 + (a^2*tan(e + f*x)*(A*15i + 5*B))/60 + (B*a^2*tan(e + f*x)^2*1i)/3)/(c^5*f*(5*tan(e +
f*x) - tan(e + f*x)^2*10i - 10*tan(e + f*x)^3 + tan(e + f*x)^4*5i + tan(e + f*x)^5 + 1i))