3.948 \(\int \frac{(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=146 \[ \frac{8 i a^5}{f \left (c^4-i c^4 \tan (e+f x)\right )}-\frac{12 i a^5}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac{32 i a^5 c^5}{3 f \left (c^3-i c^3 \tan (e+f x)\right )^3}-\frac{i a^5 \log (\cos (e+f x))}{c^4 f}+\frac{a^5 x}{c^4}-\frac{4 i a^5}{f (c-i c \tan (e+f x))^4} \]

[Out]

(a^5*x)/c^4 - (I*a^5*Log[Cos[e + f*x]])/(c^4*f) - ((4*I)*a^5)/(f*(c - I*c*Tan[e + f*x])^4) - ((12*I)*a^5)/(f*(
c^2 - I*c^2*Tan[e + f*x])^2) + (((32*I)/3)*a^5*c^5)/(f*(c^3 - I*c^3*Tan[e + f*x])^3) + ((8*I)*a^5)/(f*(c^4 - I
*c^4*Tan[e + f*x]))

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Rubi [A]  time = 0.14569, antiderivative size = 146, 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} \[ \frac{8 i a^5}{f \left (c^4-i c^4 \tan (e+f x)\right )}-\frac{12 i a^5}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac{32 i a^5 c^5}{3 f \left (c^3-i c^3 \tan (e+f x)\right )^3}-\frac{i a^5 \log (\cos (e+f x))}{c^4 f}+\frac{a^5 x}{c^4}-\frac{4 i a^5}{f (c-i c \tan (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^5/(c - I*c*Tan[e + f*x])^4,x]

[Out]

(a^5*x)/c^4 - (I*a^5*Log[Cos[e + f*x]])/(c^4*f) - ((4*I)*a^5)/(f*(c - I*c*Tan[e + f*x])^4) - ((12*I)*a^5)/(f*(
c^2 - I*c^2*Tan[e + f*x])^2) + (((32*I)/3)*a^5*c^5)/(f*(c^3 - I*c^3*Tan[e + f*x])^3) + ((8*I)*a^5)/(f*(c^4 - I
*c^4*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))^5}{(c-i c \tan (e+f x))^4} \, dx &=\left (a^5 c^5\right ) \int \frac{\sec ^{10}(e+f x)}{(c-i c \tan (e+f x))^9} \, dx\\ &=\frac{\left (i a^5\right ) \operatorname{Subst}\left (\int \frac{(c-x)^4}{(c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{c^4 f}\\ &=\frac{\left (i a^5\right ) \operatorname{Subst}\left (\int \left (\frac{16 c^4}{(c+x)^5}-\frac{32 c^3}{(c+x)^4}+\frac{24 c^2}{(c+x)^3}-\frac{8 c}{(c+x)^2}+\frac{1}{c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^4 f}\\ &=\frac{a^5 x}{c^4}-\frac{i a^5 \log (\cos (e+f x))}{c^4 f}-\frac{4 i a^5}{f (c-i c \tan (e+f x))^4}+\frac{32 i a^5}{3 c f (c-i c \tan (e+f x))^3}-\frac{12 i a^5}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac{8 i a^5}{f \left (c^4-i c^4 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.64362, size = 151, normalized size = 1.03 \[ \frac{a^5 (\cos (4 e+9 f x)+i \sin (4 e+9 f x)) \left (8 \sin (2 (e+f x))-12 i f x \sin (4 (e+f x))+3 \sin (4 (e+f x))+16 i \cos (2 (e+f x))+3 \cos (4 (e+f x)) \left (-2 i \log \left (\cos ^2(e+f x)\right )+4 f x-i\right )-6 \sin (4 (e+f x)) \log \left (\cos ^2(e+f x)\right )-6 i\right )}{12 c^4 f (\cos (f x)+i \sin (f x))^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^5/(c - I*c*Tan[e + f*x])^4,x]

[Out]

(a^5*(-6*I + (16*I)*Cos[2*(e + f*x)] + 3*Cos[4*(e + f*x)]*(-I + 4*f*x - (2*I)*Log[Cos[e + f*x]^2]) + 8*Sin[2*(
e + f*x)] + 3*Sin[4*(e + f*x)] - (12*I)*f*x*Sin[4*(e + f*x)] - 6*Log[Cos[e + f*x]^2]*Sin[4*(e + f*x)])*(Cos[4*
e + 9*f*x] + I*Sin[4*e + 9*f*x]))/(12*c^4*f*(Cos[f*x] + I*Sin[f*x])^5)

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Maple [A]  time = 0.046, size = 114, normalized size = 0.8 \begin{align*}{\frac{-4\,i{a}^{5}}{f{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}}-8\,{\frac{{a}^{5}}{f{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) }}+{\frac{12\,i{a}^{5}}{f{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{i{a}^{5}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{c}^{4}}}+{\frac{32\,{a}^{5}}{3\,f{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x)

[Out]

-4*I/f*a^5/c^4/(tan(f*x+e)+I)^4-8/f*a^5/c^4/(tan(f*x+e)+I)+12*I/f*a^5/c^4/(tan(f*x+e)+I)^2+I/f*a^5/c^4*ln(tan(
f*x+e)+I)+32/3/f*a^5/c^4/(tan(f*x+e)+I)^3

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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))^5/(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.406, size = 238, normalized size = 1.63 \begin{align*} \frac{-3 i \, a^{5} e^{\left (8 i \, f x + 8 i \, e\right )} + 4 i \, a^{5} e^{\left (6 i \, f x + 6 i \, e\right )} - 6 i \, a^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 12 i \, a^{5} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{12 \, c^{4} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/12*(-3*I*a^5*e^(8*I*f*x + 8*I*e) + 4*I*a^5*e^(6*I*f*x + 6*I*e) - 6*I*a^5*e^(4*I*f*x + 4*I*e) + 12*I*a^5*e^(2
*I*f*x + 2*I*e) - 12*I*a^5*log(e^(2*I*f*x + 2*I*e) + 1))/(c^4*f)

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Sympy [A]  time = 2.34123, size = 173, normalized size = 1.18 \begin{align*} - \frac{i a^{5} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{4} f} + \frac{\begin{cases} - \frac{i a^{5} e^{8 i e} e^{8 i f x}}{4 f} + \frac{i a^{5} e^{6 i e} e^{6 i f x}}{3 f} - \frac{i a^{5} e^{4 i e} e^{4 i f x}}{2 f} + \frac{i a^{5} e^{2 i e} e^{2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (2 a^{5} e^{8 i e} - 2 a^{5} e^{6 i e} + 2 a^{5} e^{4 i e} - 2 a^{5} e^{2 i e}\right ) & \text{otherwise} \end{cases}}{c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**5/(c-I*c*tan(f*x+e))**4,x)

[Out]

-I*a**5*log(exp(2*I*f*x) + exp(-2*I*e))/(c**4*f) + Piecewise((-I*a**5*exp(8*I*e)*exp(8*I*f*x)/(4*f) + I*a**5*e
xp(6*I*e)*exp(6*I*f*x)/(3*f) - I*a**5*exp(4*I*e)*exp(4*I*f*x)/(2*f) + I*a**5*exp(2*I*e)*exp(2*I*f*x)/f, Ne(f,
0)), (x*(2*a**5*exp(8*I*e) - 2*a**5*exp(6*I*e) + 2*a**5*exp(4*I*e) - 2*a**5*exp(2*I*e)), True))/c**4

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Giac [A]  time = 1.68487, size = 309, normalized size = 2.12 \begin{align*} -\frac{-\frac{840 i \, a^{5} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c^{4}} + \frac{420 i \, a^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{4}} + \frac{420 i \, a^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{4}} + \frac{2283 i \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 18264 \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 70644 i \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 136808 \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 191170 i \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 136808 \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 70644 i \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 18264 \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2283 i \, a^{5}}{c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{8}}}{420 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-1/420*(-840*I*a^5*log(tan(1/2*f*x + 1/2*e) + I)/c^4 + 420*I*a^5*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^4 + 420*
I*a^5*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c^4 + (2283*I*a^5*tan(1/2*f*x + 1/2*e)^8 - 18264*a^5*tan(1/2*f*x + 1/
2*e)^7 - 70644*I*a^5*tan(1/2*f*x + 1/2*e)^6 + 136808*a^5*tan(1/2*f*x + 1/2*e)^5 + 191170*I*a^5*tan(1/2*f*x + 1
/2*e)^4 - 136808*a^5*tan(1/2*f*x + 1/2*e)^3 - 70644*I*a^5*tan(1/2*f*x + 1/2*e)^2 + 18264*a^5*tan(1/2*f*x + 1/2
*e) + 2283*I*a^5)/(c^4*(tan(1/2*f*x + 1/2*e) + I)^8))/f