\(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^7} \, dx\) [703]

Optimal result

Integrand size = 41, antiderivative size = 125 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^7} \, dx=-\frac {4 a^3 (A-i B)}{7 c^7 f (i+\tan (e+f x))^7}-\frac {2 a^3 (i A+2 B)}{3 c^7 f (i+\tan (e+f x))^6}+\frac {a^3 (A-5 i B)}{5 c^7 f (i+\tan (e+f x))^5}+\frac {a^3 B}{4 c^7 f (i+\tan (e+f x))^4} \]

[Out]

-4/7*a^3*(A-I*B)/c^7/f/(I+tan(f*x+e))^7-2/3*a^3*(I*A+2*B)/c^7/f/(I+tan(f*x+e))^6+1/5*a^3*(A-5*I*B)/c^7/f/(I+ta
n(f*x+e))^5+1/4*a^3*B/c^7/f/(I+tan(f*x+e))^4

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 125, 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} \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^7} \, dx=\frac {a^3 (A-5 i B)}{5 c^7 f (\tan (e+f x)+i)^5}-\frac {2 a^3 (2 B+i A)}{3 c^7 f (\tan (e+f x)+i)^6}-\frac {4 a^3 (A-i B)}{7 c^7 f (\tan (e+f x)+i)^7}+\frac {a^3 B}{4 c^7 f (\tan (e+f x)+i)^4} \]

[In]

Int[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^7,x]

[Out]

(-4*a^3*(A - I*B))/(7*c^7*f*(I + Tan[e + f*x])^7) - (2*a^3*(I*A + 2*B))/(3*c^7*f*(I + Tan[e + f*x])^6) + (a^3*
(A - (5*I)*B))/(5*c^7*f*(I + Tan[e + f*x])^5) + (a^3*B)/(4*c^7*f*(I + Tan[e + f*x])^4)

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)^8} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (\frac {4 a^2 (A-i B)}{c^8 (i+x)^8}+\frac {4 a^2 (i A+2 B)}{c^8 (i+x)^7}-\frac {a^2 (A-5 i B)}{c^8 (i+x)^6}-\frac {a^2 B}{c^8 (i+x)^5}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {4 a^3 (A-i B)}{7 c^7 f (i+\tan (e+f x))^7}-\frac {2 a^3 (i A+2 B)}{3 c^7 f (i+\tan (e+f x))^6}+\frac {a^3 (A-5 i B)}{5 c^7 f (i+\tan (e+f x))^5}+\frac {a^3 B}{4 c^7 f (i+\tan (e+f x))^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.53 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.64 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^7} \, dx=\frac {a^3 \left (-44 A-5 i B+(-112 i A-35 B) \tan (e+f x)+21 (4 A-5 i B) \tan ^2(e+f x)+105 B \tan ^3(e+f x)\right )}{420 c^7 f (i+\tan (e+f x))^7} \]

[In]

Integrate[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^7,x]

[Out]

(a^3*(-44*A - (5*I)*B + ((-112*I)*A - 35*B)*Tan[e + f*x] + 21*(4*A - (5*I)*B)*Tan[e + f*x]^2 + 105*B*Tan[e + f
*x]^3))/(420*c^7*f*(I + Tan[e + f*x])^7)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {B}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {5 i B -A}{5 \left (i+\tan \left (f x +e \right )\right )^{5}}-\frac {4 i A +8 B}{6 \left (i+\tan \left (f x +e \right )\right )^{6}}-\frac {-4 i B +4 A}{7 \left (i+\tan \left (f x +e \right )\right )^{7}}\right )}{f \,c^{7}}\)	\(89\)
default	\(\frac {a^{3} \left (\frac {B}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {5 i B -A}{5 \left (i+\tan \left (f x +e \right )\right )^{5}}-\frac {4 i A +8 B}{6 \left (i+\tan \left (f x +e \right )\right )^{6}}-\frac {-4 i B +4 A}{7 \left (i+\tan \left (f x +e \right )\right )^{7}}\right )}{f \,c^{7}}\)	\(89\)
risch	\(-\frac {a^{3} {\mathrm e}^{14 i \left (f x +e \right )} B}{224 c^{7} f}-\frac {i a^{3} {\mathrm e}^{14 i \left (f x +e \right )} A}{224 c^{7} f}-\frac {{\mathrm e}^{12 i \left (f x +e \right )} B \,a^{3}}{96 c^{7} f}-\frac {i {\mathrm e}^{12 i \left (f x +e \right )} a^{3} A}{48 c^{7} f}-\frac {3 i a^{3} A \,{\mathrm e}^{10 i \left (f x +e \right )}}{80 c^{7} f}+\frac {{\mathrm e}^{8 i \left (f x +e \right )} B \,a^{3}}{64 c^{7} f}-\frac {i {\mathrm e}^{8 i \left (f x +e \right )} a^{3} A}{32 c^{7} f}+\frac {a^{3} {\mathrm e}^{6 i \left (f x +e \right )} B}{96 c^{7} f}-\frac {i a^{3} {\mathrm e}^{6 i \left (f x +e \right )} A}{96 c^{7} f}\)	\(196\)
norman	\(\frac {\frac {a^{3} A \tan \left (f x +e \right )}{f c}+\frac {-44 i A \,a^{3}+5 B \,a^{3}}{420 c f}+\frac {B \,a^{3} \tan \left (f x +e \right )^{10}}{4 c f}+\frac {2 \left (-125 i B \,a^{3}+139 a^{3} A \right ) \tan \left (f x +e \right )^{5}}{15 c f}-\frac {6 \left (-85 i B \,a^{3}+36 a^{3} A \right ) \tan \left (f x +e \right )^{7}}{35 c f}+\frac {\left (-10 i B \,a^{3}+a^{3} A \right ) \tan \left (f x +e \right )^{9}}{5 c f}-\frac {2 \left (-5 i B \,a^{3}+16 a^{3} A \right ) \tan \left (f x +e \right )^{3}}{3 c f}-\frac {5 \left (4 i A \,a^{3}+17 B \,a^{3}\right ) \tan \left (f x +e \right )^{8}}{12 c f}+\frac {5 \left (16 i A \,a^{3}+23 B \,a^{3}\right ) \tan \left (f x +e \right )^{6}}{6 c f}-\frac {\left (172 i A \,a^{3}+95 B \,a^{3}\right ) \tan \left (f x +e \right )^{4}}{10 c f}+\frac {\left (256 i A \,a^{3}+35 B \,a^{3}\right ) \tan \left (f x +e \right )^{2}}{60 c f}}{c^{6} \left (1+\tan \left (f x +e \right )^{2}\right )^{7}}\)	\(316\)

[In]

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^7,x,method=_RETURNVERBOSE)

[Out]

1/f*a^3/c^7*(1/4*B/(I+tan(f*x+e))^4-1/5*(-A+5*I*B)/(I+tan(f*x+e))^5-1/6*(4*I*A+8*B)/(I+tan(f*x+e))^6-1/7*(-4*I
*B+4*A)/(I+tan(f*x+e))^7)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.83 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^7} \, dx=-\frac {30 \, {\left (i \, A + B\right )} a^{3} e^{\left (14 i \, f x + 14 i \, e\right )} + 70 \, {\left (2 i \, A + B\right )} a^{3} e^{\left (12 i \, f x + 12 i \, e\right )} + 252 i \, A a^{3} e^{\left (10 i \, f x + 10 i \, e\right )} + 105 \, {\left (2 i \, A - B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} + 70 \, {\left (i \, A - B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )}}{6720 \, c^{7} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^7,x, algorithm="fricas")

[Out]

-1/6720*(30*(I*A + B)*a^3*e^(14*I*f*x + 14*I*e) + 70*(2*I*A + B)*a^3*e^(12*I*f*x + 12*I*e) + 252*I*A*a^3*e^(10
*I*f*x + 10*I*e) + 105*(2*I*A - B)*a^3*e^(8*I*f*x + 8*I*e) + 70*(I*A - B)*a^3*e^(6*I*f*x + 6*I*e))/(c^7*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. 379 vs. \(2 (104) = 208\).

Time = 0.66 (sec) , antiderivative size = 379, normalized size of antiderivative = 3.03 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^7} \, dx=\begin {cases} \frac {- 396361728 i A a^{3} c^{28} f^{4} e^{10 i e} e^{10 i f x} + \left (- 110100480 i A a^{3} c^{28} f^{4} e^{6 i e} + 110100480 B a^{3} c^{28} f^{4} e^{6 i e}\right ) e^{6 i f x} + \left (- 330301440 i A a^{3} c^{28} f^{4} e^{8 i e} + 165150720 B a^{3} c^{28} f^{4} e^{8 i e}\right ) e^{8 i f x} + \left (- 220200960 i A a^{3} c^{28} f^{4} e^{12 i e} - 110100480 B a^{3} c^{28} f^{4} e^{12 i e}\right ) e^{12 i f x} + \left (- 47185920 i A a^{3} c^{28} f^{4} e^{14 i e} - 47185920 B a^{3} c^{28} f^{4} e^{14 i e}\right ) e^{14 i f x}}{10569646080 c^{35} f^{5}} & \text {for}\: c^{35} f^{5} \neq 0 \\\frac {x \left (A a^{3} e^{14 i e} + 4 A a^{3} e^{12 i e} + 6 A a^{3} e^{10 i e} + 4 A a^{3} e^{8 i e} + A a^{3} e^{6 i e} - i B a^{3} e^{14 i e} - 2 i B a^{3} e^{12 i e} + 2 i B a^{3} e^{8 i e} + i B a^{3} e^{6 i e}\right )}{16 c^{7}} & \text {otherwise} \end {cases} \]

[In]

integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**7,x)

[Out]

Piecewise(((-396361728*I*A*a**3*c**28*f**4*exp(10*I*e)*exp(10*I*f*x) + (-110100480*I*A*a**3*c**28*f**4*exp(6*I
*e) + 110100480*B*a**3*c**28*f**4*exp(6*I*e))*exp(6*I*f*x) + (-330301440*I*A*a**3*c**28*f**4*exp(8*I*e) + 1651
50720*B*a**3*c**28*f**4*exp(8*I*e))*exp(8*I*f*x) + (-220200960*I*A*a**3*c**28*f**4*exp(12*I*e) - 110100480*B*a
**3*c**28*f**4*exp(12*I*e))*exp(12*I*f*x) + (-47185920*I*A*a**3*c**28*f**4*exp(14*I*e) - 47185920*B*a**3*c**28
*f**4*exp(14*I*e))*exp(14*I*f*x))/(10569646080*c**35*f**5), Ne(c**35*f**5, 0)), (x*(A*a**3*exp(14*I*e) + 4*A*a
**3*exp(12*I*e) + 6*A*a**3*exp(10*I*e) + 4*A*a**3*exp(8*I*e) + A*a**3*exp(6*I*e) - I*B*a**3*exp(14*I*e) - 2*I*
B*a**3*exp(12*I*e) + 2*I*B*a**3*exp(8*I*e) + I*B*a**3*exp(6*I*e))/(16*c**7), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^7} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^7,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. 428 vs. \(2 (103) = 206\).

Time = 1.07 (sec) , antiderivative size = 428, normalized size of antiderivative = 3.42 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^7} \, dx=-\frac {2 \, {\left (105 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{13} + 420 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 105 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 2170 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{11} - 70 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{11} - 5180 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 875 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 11431 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 700 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 15904 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 2380 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 19436 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 1340 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 15904 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 2380 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 11431 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 700 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 5180 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 875 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2170 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 70 i \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 420 i \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 105 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 105 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{105 \, c^{7} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{14}} \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^7,x, algorithm="giac")

[Out]

-2/105*(105*A*a^3*tan(1/2*f*x + 1/2*e)^13 + 420*I*A*a^3*tan(1/2*f*x + 1/2*e)^12 - 105*B*a^3*tan(1/2*f*x + 1/2*
e)^12 - 2170*A*a^3*tan(1/2*f*x + 1/2*e)^11 - 70*I*B*a^3*tan(1/2*f*x + 1/2*e)^11 - 5180*I*A*a^3*tan(1/2*f*x + 1
/2*e)^10 + 875*B*a^3*tan(1/2*f*x + 1/2*e)^10 + 11431*A*a^3*tan(1/2*f*x + 1/2*e)^9 + 700*I*B*a^3*tan(1/2*f*x +
1/2*e)^9 + 15904*I*A*a^3*tan(1/2*f*x + 1/2*e)^8 - 2380*B*a^3*tan(1/2*f*x + 1/2*e)^8 - 19436*A*a^3*tan(1/2*f*x
+ 1/2*e)^7 - 1340*I*B*a^3*tan(1/2*f*x + 1/2*e)^7 - 15904*I*A*a^3*tan(1/2*f*x + 1/2*e)^6 + 2380*B*a^3*tan(1/2*f
*x + 1/2*e)^6 + 11431*A*a^3*tan(1/2*f*x + 1/2*e)^5 + 700*I*B*a^3*tan(1/2*f*x + 1/2*e)^5 + 5180*I*A*a^3*tan(1/2
*f*x + 1/2*e)^4 - 875*B*a^3*tan(1/2*f*x + 1/2*e)^4 - 2170*A*a^3*tan(1/2*f*x + 1/2*e)^3 - 70*I*B*a^3*tan(1/2*f*
x + 1/2*e)^3 - 420*I*A*a^3*tan(1/2*f*x + 1/2*e)^2 + 105*B*a^3*tan(1/2*f*x + 1/2*e)^2 + 105*A*a^3*tan(1/2*f*x +
1/2*e))/(c^7*f*(tan(1/2*f*x + 1/2*e) + I)^14)

Mupad [B] (verification not implemented)

Time = 8.94 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.21 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^7} \, dx=\frac {\frac {a^3\,\left (44\,A+B\,5{}\mathrm {i}\right )}{420}+\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )\,\left (35\,B+A\,112{}\mathrm {i}\right )}{420}-\frac {B\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^3}{4}-\frac {a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (84\,A-B\,105{}\mathrm {i}\right )}{420}}{c^7\,f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^7-{\mathrm {tan}\left (e+f\,x\right )}^6\,7{}\mathrm {i}+21\,{\mathrm {tan}\left (e+f\,x\right )}^5+{\mathrm {tan}\left (e+f\,x\right )}^4\,35{}\mathrm {i}-35\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,21{}\mathrm {i}+7\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )} \]

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3)/(c - c*tan(e + f*x)*1i)^7,x)

[Out]

((a^3*(44*A + B*5i))/420 + (a^3*tan(e + f*x)*(A*112i + 35*B))/420 - (B*a^3*tan(e + f*x)^3)/4 - (a^3*tan(e + f*
x)^2*(84*A - B*105i))/420)/(c^7*f*(7*tan(e + f*x) - tan(e + f*x)^2*21i - 35*tan(e + f*x)^3 + tan(e + f*x)^4*35
i + 21*tan(e + f*x)^5 - tan(e + f*x)^6*7i - tan(e + f*x)^7 + 1i))