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

Optimal. Leaf size=95 \[ \frac{a^2 (3 B+i A)}{4 c^5 f (\tan (e+f x)+i)^4}+\frac{2 a^2 (A-i B)}{5 c^5 f (\tan (e+f x)+i)^5}+\frac{i a^2 B}{3 c^5 f (\tan (e+f x)+i)^3} \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.152285, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3588, 77} \[ \frac{a^2 (3 B+i A)}{4 c^5 f (\tan (e+f x)+i)^4}+\frac{2 a^2 (A-i B)}{5 c^5 f (\tan (e+f x)+i)^5}+\frac{i a^2 B}{3 c^5 f (\tan (e+f x)+i)^3} \]

Antiderivative was successfully verified.

[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 3588

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]

Rule 77

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]))))

Rubi steps

\begin{align*} \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 c) \operatorname{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) \operatorname{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]  time = 3.62902, size = 116, normalized size = 1.22 \[ \frac{a^2 (\cos (7 e+9 f x)+i \sin (7 e+9 f x)) (-(3 A+7 i B) (5 \sin (e+f x)+6 \sin (3 (e+f x)))+5 (B-21 i A) \cos (e+f x)+6 (3 B-7 i A) \cos (3 (e+f x)))}{960 c^5 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[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*(5*((-21*I)*A + B)*Cos[e + f*x] + 6*((-7*I)*A + 3*B)*Cos[3*(e + f*x)] - (3*A + (7*I)*B)*(5*Sin[e + f*x] +
6*Sin[3*(e + f*x)]))*(Cos[7*e + 9*f*x] + I*Sin[7*e + 9*f*x]))/(960*c^5*f*(Cos[f*x] + I*Sin[f*x])^2)

________________________________________________________________________________________

Maple [A]  time = 0.051, size = 69, normalized size = 0.7 \begin{align*}{\frac{{a}^{2}}{f{c}^{5}} \left ( -{\frac{-iA-3\,B}{4\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}}-{\frac{-2\,A+2\,iB}{5\, \left ( \tan \left ( fx+e \right ) +i \right ) ^{5}}}+{\frac{{\frac{i}{3}}B}{ \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 1.42725, size = 255, normalized size = 2.68 \begin{align*} \frac{{\left (-12 i \, A - 12 \, B\right )} a^{2} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-45 i \, A - 15 \, B\right )} a^{2} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-60 i \, A + 20 \, B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-30 i \, A + 30 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )}}{960 \, c^{5} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 - 12*B)*a^2*e^(10*I*f*x + 10*I*e) + (-45*I*A - 15*B)*a^2*e^(8*I*f*x + 8*I*e) + (-60*I*A + 20*B
)*a^2*e^(6*I*f*x + 6*I*e) + (-30*I*A + 30*B)*a^2*e^(4*I*f*x + 4*I*e))/(c^5*f)

________________________________________________________________________________________

Sympy [A]  time = 1.81099, size = 333, normalized size = 3.51 \begin{align*} \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}\: 7864320 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(7864320*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))

________________________________________________________________________________________

Giac [B]  time = 1.56321, size = 417, normalized size = 4.39 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)