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

Optimal. Leaf size=129 \[ -\frac{a^3 (B+i A) (1+i \tan (e+f x))^3}{6 c^3 f (1-i \tan (e+f x))^3}-\frac{4 i a^3 B}{c^3 f (\tan (e+f x)+i)}-\frac{2 a^3 B}{c^3 f (\tan (e+f x)+i)^2}+\frac{a^3 B \log (\cos (e+f x))}{c^3 f}+\frac{i a^3 B x}{c^3} \]

[Out]

(I*a^3*B*x)/c^3 + (a^3*B*Log[Cos[e + f*x]])/(c^3*f) - (a^3*(I*A + B)*(1 + I*Tan[e + f*x])^3)/(6*c^3*f*(1 - I*T
an[e + f*x])^3) - (2*a^3*B)/(c^3*f*(I + Tan[e + f*x])^2) - ((4*I)*a^3*B)/(c^3*f*(I + Tan[e + f*x]))

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Rubi [A]  time = 0.153866, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.073, Rules used = {3588, 78, 43} \[ -\frac{a^3 (B+i A) (1+i \tan (e+f x))^3}{6 c^3 f (1-i \tan (e+f x))^3}-\frac{4 i a^3 B}{c^3 f (\tan (e+f x)+i)}-\frac{2 a^3 B}{c^3 f (\tan (e+f x)+i)^2}+\frac{a^3 B \log (\cos (e+f x))}{c^3 f}+\frac{i a^3 B x}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*a^3*B*x)/c^3 + (a^3*B*Log[Cos[e + f*x]])/(c^3*f) - (a^3*(I*A + B)*(1 + I*Tan[e + f*x])^3)/(6*c^3*f*(1 - I*T
an[e + f*x])^3) - (2*a^3*B)/(c^3*f*(I + Tan[e + f*x])^2) - ((4*I)*a^3*B)/(c^3*f*(I + Tan[e + f*x]))

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^3} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^2 (A+B x)}{(c-i c x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a^3 (i A+B) (1+i \tan (e+f x))^3}{6 c^3 f (1-i \tan (e+f x))^3}+\frac{(i a B) \operatorname{Subst}\left (\int \frac{(a+i a x)^2}{(c-i c x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a^3 (i A+B) (1+i \tan (e+f x))^3}{6 c^3 f (1-i \tan (e+f x))^3}+\frac{(i a B) \operatorname{Subst}\left (\int \left (-\frac{4 i a^2}{c^3 (i+x)^3}+\frac{4 a^2}{c^3 (i+x)^2}+\frac{i a^2}{c^3 (i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i a^3 B x}{c^3}+\frac{a^3 B \log (\cos (e+f x))}{c^3 f}-\frac{a^3 (i A+B) (1+i \tan (e+f x))^3}{6 c^3 f (1-i \tan (e+f x))^3}-\frac{2 a^3 B}{c^3 f (i+\tan (e+f x))^2}-\frac{4 i a^3 B}{c^3 f (i+\tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 4.28417, size = 167, normalized size = 1.29 \[ \frac{a^3 (\cos (3 (e+2 f x))+i \sin (3 (e+2 f x))) \left (\cos (3 (e+f x)) \left (-i A+3 B \log \left (\cos ^2(e+f x)\right )+6 i B f x-B\right )+A \sin (3 (e+f x))+9 i B \sin (e+f x)-i B \sin (3 (e+f x))+6 B f x \sin (3 (e+f x))-3 B \cos (e+f x)-3 i B \sin (3 (e+f x)) \log \left (\cos ^2(e+f x)\right )\right )}{6 c^3 f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(-3*B*Cos[e + f*x] + Cos[3*(e + f*x)]*((-I)*A - B + (6*I)*B*f*x + 3*B*Log[Cos[e + f*x]^2]) + (9*I)*B*Sin[
e + f*x] + A*Sin[3*(e + f*x)] - I*B*Sin[3*(e + f*x)] + 6*B*f*x*Sin[3*(e + f*x)] - (3*I)*B*Log[Cos[e + f*x]^2]*
Sin[3*(e + f*x)])*(Cos[3*(e + 2*f*x)] + I*Sin[3*(e + 2*f*x)]))/(6*c^3*f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]  time = 0.052, size = 164, normalized size = 1.3 \begin{align*}{\frac{{\frac{4\,i}{3}}{a}^{3}B}{f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}-{\frac{4\,A{a}^{3}}{3\,f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}-{\frac{5\,iB{a}^{3}}{f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{A{a}^{3}}{f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{B{a}^{3}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{c}^{3}}}-{\frac{2\,i{a}^{3}A}{f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-4\,{\frac{B{a}^{3}}{f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.454, size = 201, normalized size = 1.56 \begin{align*} \frac{{\left (-i \, A - B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, B a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 6 \, B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 \, B a^{3} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{6 \, c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.66081, size = 214, normalized size = 1.66 \begin{align*} \frac{B a^{3} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{3} f} + \begin{cases} \frac{6 B a^{3} c^{6} f^{2} e^{4 i e} e^{4 i f x} - 12 B a^{3} c^{6} f^{2} e^{2 i e} e^{2 i f x} + \left (- 2 i A a^{3} c^{6} f^{2} e^{6 i e} - 2 B a^{3} c^{6} f^{2} e^{6 i e}\right ) e^{6 i f x}}{12 c^{9} f^{3}} & \text{for}\: 12 c^{9} f^{3} \neq 0 \\\frac{x \left (A a^{3} e^{6 i e} - i B a^{3} e^{6 i e} + 2 i B a^{3} e^{4 i e} - 2 i B a^{3} e^{2 i e}\right )}{c^{3}} & \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))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**3,x)

[Out]

B*a**3*log(exp(2*I*f*x) + exp(-2*I*e))/(c**3*f) + Piecewise(((6*B*a**3*c**6*f**2*exp(4*I*e)*exp(4*I*f*x) - 12*
B*a**3*c**6*f**2*exp(2*I*e)*exp(2*I*f*x) + (-2*I*A*a**3*c**6*f**2*exp(6*I*e) - 2*B*a**3*c**6*f**2*exp(6*I*e))*
exp(6*I*f*x))/(12*c**9*f**3), Ne(12*c**9*f**3, 0)), (x*(A*a**3*exp(6*I*e) - I*B*a**3*exp(6*I*e) + 2*I*B*a**3*e
xp(4*I*e) - 2*I*B*a**3*exp(2*I*e))/c**3, True))

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Giac [B]  time = 1.63729, size = 348, normalized size = 2.7 \begin{align*} -\frac{\frac{60 \, B a^{3} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c^{3}} - \frac{30 \, B a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{3}} - \frac{30 \, B a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{3}} - \frac{147 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 60 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 942 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 2445 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 200 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3620 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2445 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 60 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 942 i \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 147 \, B a^{3}}{c^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{6}}}{30 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(60*B*a^3*log(tan(1/2*f*x + 1/2*e) + I)/c^3 - 30*B*a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^3 - 30*B*a^3
*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c^3 - (147*B*a^3*tan(1/2*f*x + 1/2*e)^6 - 60*A*a^3*tan(1/2*f*x + 1/2*e)^5
+ 942*I*B*a^3*tan(1/2*f*x + 1/2*e)^5 - 2445*B*a^3*tan(1/2*f*x + 1/2*e)^4 + 200*A*a^3*tan(1/2*f*x + 1/2*e)^3 -
3620*I*B*a^3*tan(1/2*f*x + 1/2*e)^3 + 2445*B*a^3*tan(1/2*f*x + 1/2*e)^2 - 60*A*a^3*tan(1/2*f*x + 1/2*e) + 942*
I*B*a^3*tan(1/2*f*x + 1/2*e) - 147*B*a^3)/(c^3*(tan(1/2*f*x + 1/2*e) + I)^6))/f