3.931 \(\int \frac{(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=101 \[ -\frac{a^4 \tan (e+f x)}{c^2 f}+\frac{12 i a^4}{f \left (c^2-i c^2 \tan (e+f x)\right )}-\frac{6 i a^4 \log (\cos (e+f x))}{c^2 f}+\frac{6 a^4 x}{c^2}-\frac{4 i a^4}{f (c-i c \tan (e+f x))^2} \]

[Out]

(6*a^4*x)/c^2 - ((6*I)*a^4*Log[Cos[e + f*x]])/(c^2*f) - (a^4*Tan[e + f*x])/(c^2*f) - ((4*I)*a^4)/(f*(c - I*c*T
an[e + f*x])^2) + ((12*I)*a^4)/(f*(c^2 - I*c^2*Tan[e + f*x]))

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Rubi [A]  time = 0.136019, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3522, 3487, 43} \[ -\frac{a^4 \tan (e+f x)}{c^2 f}+\frac{12 i a^4}{f \left (c^2-i c^2 \tan (e+f x)\right )}-\frac{6 i a^4 \log (\cos (e+f x))}{c^2 f}+\frac{6 a^4 x}{c^2}-\frac{4 i a^4}{f (c-i c \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^4/(c - I*c*Tan[e + f*x])^2,x]

[Out]

(6*a^4*x)/c^2 - ((6*I)*a^4*Log[Cos[e + f*x]])/(c^2*f) - (a^4*Tan[e + f*x])/(c^2*f) - ((4*I)*a^4)/(f*(c - I*c*T
an[e + f*x])^2) + ((12*I)*a^4)/(f*(c^2 - I*c^2*Tan[e + f*x]))

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

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))^4}{(c-i c \tan (e+f x))^2} \, dx &=\left (a^4 c^4\right ) \int \frac{\sec ^8(e+f x)}{(c-i c \tan (e+f x))^6} \, dx\\ &=\frac{\left (i a^4\right ) \operatorname{Subst}\left (\int \frac{(c-x)^3}{(c+x)^3} \, dx,x,-i c \tan (e+f x)\right )}{c^3 f}\\ &=\frac{\left (i a^4\right ) \operatorname{Subst}\left (\int \left (-1+\frac{8 c^3}{(c+x)^3}-\frac{12 c^2}{(c+x)^2}+\frac{6 c}{c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^3 f}\\ &=\frac{6 a^4 x}{c^2}-\frac{6 i a^4 \log (\cos (e+f x))}{c^2 f}-\frac{a^4 \tan (e+f x)}{c^2 f}-\frac{4 i a^4}{f (c-i c \tan (e+f x))^2}+\frac{12 i a^4}{f \left (c^2-i c^2 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [B]  time = 2.68934, size = 374, normalized size = 3.7 \[ \frac{a^4 \sec (e) \sec (e+f x) (\cos (2 (e+3 f x))+i \sin (2 (e+3 f x))) \left (-6 i f x \sin (2 e+f x)+3 \sin (2 e+f x)-6 i f x \sin (2 e+3 f x)-\sin (2 e+3 f x)-6 i f x \sin (4 e+3 f x)+\sin (4 e+3 f x)+6 f x \cos (2 e+3 f x)-3 i \cos (2 e+3 f x)+6 f x \cos (4 e+3 f x)-i \cos (4 e+3 f x)-3 i \cos (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )+\cos (f x) \left (-3 i \log \left (\cos ^2(e+f x)\right )+6 f x+7 i\right )+\cos (2 e+f x) \left (-3 i \log \left (\cos ^2(e+f x)\right )+6 f x+9 i\right )-3 i \cos (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )-3 \sin (f x) \log \left (\cos ^2(e+f x)\right )-3 \sin (2 e+f x) \log \left (\cos ^2(e+f x)\right )-3 \sin (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )-3 \sin (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )-6 i f x \sin (f x)+\sin (f x)\right )}{4 c^2 f (\cos (f x)+i \sin (f x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^4/(c - I*c*Tan[e + f*x])^2,x]

[Out]

(a^4*Sec[e]*Sec[e + f*x]*(Cos[2*(e + 3*f*x)] + I*Sin[2*(e + 3*f*x)])*((-3*I)*Cos[2*e + 3*f*x] + 6*f*x*Cos[2*e
+ 3*f*x] - I*Cos[4*e + 3*f*x] + 6*f*x*Cos[4*e + 3*f*x] + Cos[f*x]*(7*I + 6*f*x - (3*I)*Log[Cos[e + f*x]^2]) +
Cos[2*e + f*x]*(9*I + 6*f*x - (3*I)*Log[Cos[e + f*x]^2]) - (3*I)*Cos[2*e + 3*f*x]*Log[Cos[e + f*x]^2] - (3*I)*
Cos[4*e + 3*f*x]*Log[Cos[e + f*x]^2] + Sin[f*x] - (6*I)*f*x*Sin[f*x] - 3*Log[Cos[e + f*x]^2]*Sin[f*x] + 3*Sin[
2*e + f*x] - (6*I)*f*x*Sin[2*e + f*x] - 3*Log[Cos[e + f*x]^2]*Sin[2*e + f*x] - Sin[2*e + 3*f*x] - (6*I)*f*x*Si
n[2*e + 3*f*x] - 3*Log[Cos[e + f*x]^2]*Sin[2*e + 3*f*x] + Sin[4*e + 3*f*x] - (6*I)*f*x*Sin[4*e + 3*f*x] - 3*Lo
g[Cos[e + f*x]^2]*Sin[4*e + 3*f*x]))/(4*c^2*f*(Cos[f*x] + I*Sin[f*x])^4)

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Maple [A]  time = 0.028, size = 86, normalized size = 0.9 \begin{align*} -{\frac{{a}^{4}\tan \left ( fx+e \right ) }{{c}^{2}f}}-12\,{\frac{{a}^{4}}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{4\,i{a}^{4}}{{c}^{2}f \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{6\,i{a}^{4}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{{c}^{2}f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^2,x)

[Out]

-a^4*tan(f*x+e)/c^2/f-12/f*a^4/c^2/(tan(f*x+e)+I)+4*I/f*a^4/c^2/(tan(f*x+e)+I)^2+6*I/f*a^4/c^2*ln(tan(f*x+e)+I
)

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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))^4/(c-I*c*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.3803, size = 281, normalized size = 2.78 \begin{align*} \frac{-i \, a^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 i \, a^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, a^{4} +{\left (-6 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 6 i \, a^{4}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.26356, size = 139, normalized size = 1.38 \begin{align*} - \frac{6 i a^{4} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{2} f} - \frac{2 i a^{4} e^{- 2 i e}}{c^{2} f \left (e^{2 i f x} + e^{- 2 i e}\right )} + \frac{\begin{cases} - \frac{i a^{4} e^{4 i e} e^{4 i f x}}{f} + \frac{4 i a^{4} e^{2 i e} e^{2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (4 a^{4} e^{4 i e} - 8 a^{4} e^{2 i e}\right ) & \text{otherwise} \end{cases}}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**4/(c-I*c*tan(f*x+e))**2,x)

[Out]

-6*I*a**4*log(exp(2*I*f*x) + exp(-2*I*e))/(c**2*f) - 2*I*a**4*exp(-2*I*e)/(c**2*f*(exp(2*I*f*x) + exp(-2*I*e))
) + Piecewise((-I*a**4*exp(4*I*e)*exp(4*I*f*x)/f + 4*I*a**4*exp(2*I*e)*exp(2*I*f*x)/f, Ne(f, 0)), (x*(4*a**4*e
xp(4*I*e) - 8*a**4*exp(2*I*e)), True))/c**2

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Giac [B]  time = 1.5818, size = 296, normalized size = 2.93 \begin{align*} -\frac{-\frac{12 i \, a^{4} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c^{2}} + \frac{6 i \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{2}} + \frac{6 i \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{2}} - \frac{2 \,{\left (3 i \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 3 i \, a^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} c^{2}} + \frac{25 i \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 108 \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 182 i \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 108 \, a^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 25 i \, a^{4}}{c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{4}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-(-12*I*a^4*log(tan(1/2*f*x + 1/2*e) + I)/c^2 + 6*I*a^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^2 + 6*I*a^4*log(a
bs(tan(1/2*f*x + 1/2*e) - 1))/c^2 - 2*(3*I*a^4*tan(1/2*f*x + 1/2*e)^2 + a^4*tan(1/2*f*x + 1/2*e) - 3*I*a^4)/((
tan(1/2*f*x + 1/2*e)^2 - 1)*c^2) + (25*I*a^4*tan(1/2*f*x + 1/2*e)^4 - 108*a^4*tan(1/2*f*x + 1/2*e)^3 - 182*I*a
^4*tan(1/2*f*x + 1/2*e)^2 + 108*a^4*tan(1/2*f*x + 1/2*e) + 25*I*a^4)/(c^2*(tan(1/2*f*x + 1/2*e) + I)^4))/f