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

Optimal. Leaf size=135 \[ \frac{c^3 (-B+i A) (1-i \tan (e+f x))^3}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{4 i B c^3}{a^3 f (-\tan (e+f x)+i)}-\frac{2 B c^3}{a^3 f (-\tan (e+f x)+i)^2}+\frac{B c^3 \log (\cos (e+f x))}{a^3 f}-\frac{i B c^3 x}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.156834, antiderivative size = 135, 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} \[ \frac{c^3 (-B+i A) (1-i \tan (e+f x))^3}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{4 i B c^3}{a^3 f (-\tan (e+f x)+i)}-\frac{2 B c^3}{a^3 f (-\tan (e+f x)+i)^2}+\frac{B c^3 \log (\cos (e+f x))}{a^3 f}-\frac{i B c^3 x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 3.60465, size = 145, normalized size = 1.07 \[ \frac{c^3 \sec ^3(e+f x) (-\cos (3 (e+f x)) (A-6 i B \log (\cos (e+f x))-6 B f x+i B)+i A \sin (3 (e+f x))+9 B \sin (e+f x)-B \sin (3 (e+f x))+6 i B f x \sin (3 (e+f x))-3 i B \cos (e+f x)-6 B \sin (3 (e+f x)) \log (\cos (e+f x)))}{6 a^3 f (\tan (e+f x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 164, normalized size = 1.2 \begin{align*}{\frac{5\,i{c}^{3}B}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{A{c}^{3}}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{B{c}^{3}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{f{a}^{3}}}+{\frac{2\,i{c}^{3}A}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-4\,{\frac{B{c}^{3}}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{4\,i}{3}}{c}^{3}B}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}-{\frac{4\,A{c}^{3}}{3\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.10689, size = 279, normalized size = 2.07 \begin{align*} \frac{{\left (-12 i \, B c^{3} f x e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, B c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 6 \, B c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, B c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, A - B\right )} c^{3}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{6 \, a^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(-12*I*B*c^3*f*x*e^(6*I*f*x + 6*I*e) + 6*B*c^3*e^(6*I*f*x + 6*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 6*B*c^3*
e^(4*I*f*x + 4*I*e) + 3*B*c^3*e^(2*I*f*x + 2*I*e) + (I*A - B)*c^3)*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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Sympy [A]  time = 3.60864, size = 260, normalized size = 1.93 \begin{align*} - \frac{2 i B c^{3} x}{a^{3}} + \frac{B c^{3} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{3} f} + \begin{cases} \frac{\left (- 12 B a^{6} c^{3} f^{2} e^{10 i e} e^{- 2 i f x} + 6 B a^{6} c^{3} f^{2} e^{8 i e} e^{- 4 i f x} + \left (2 i A a^{6} c^{3} f^{2} e^{6 i e} - 2 B a^{6} c^{3} f^{2} e^{6 i e}\right ) e^{- 6 i f x}\right ) e^{- 12 i e}}{12 a^{9} f^{3}} & \text{for}\: 12 a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (\frac{2 i B c^{3}}{a^{3}} + \frac{\left (A c^{3} - 2 i B c^{3} e^{6 i e} + 2 i B c^{3} e^{4 i e} - 2 i B c^{3} e^{2 i e} + i B c^{3}\right ) e^{- 6 i e}}{a^{3}}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*B*c**3*x/a**3 + B*c**3*log(exp(2*I*f*x) + exp(-2*I*e))/(a**3*f) + Piecewise(((-12*B*a**6*c**3*f**2*exp(10
*I*e)*exp(-2*I*f*x) + 6*B*a**6*c**3*f**2*exp(8*I*e)*exp(-4*I*f*x) + (2*I*A*a**6*c**3*f**2*exp(6*I*e) - 2*B*a**
6*c**3*f**2*exp(6*I*e))*exp(-6*I*f*x))*exp(-12*I*e)/(12*a**9*f**3), Ne(12*a**9*f**3*exp(12*I*e), 0)), (x*(2*I*
B*c**3/a**3 + (A*c**3 - 2*I*B*c**3*exp(6*I*e) + 2*I*B*c**3*exp(4*I*e) - 2*I*B*c**3*exp(2*I*e) + I*B*c**3)*exp(
-6*I*e)/a**3), True))

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Giac [B]  time = 1.54282, size = 348, normalized size = 2.58 \begin{align*} -\frac{\frac{60 \, B c^{3} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - i\right )}{a^{3}} - \frac{30 \, B c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac{30 \, B c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} - \frac{147 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 60 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 942 i \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 2445 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 200 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 3620 i \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2445 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 60 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 942 i \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 147 \, B c^{3}}{a^{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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

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