3.939 \(\int \frac{(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=134 \[ \frac{a^5 \tan (e+f x)}{c^3 f}-\frac{24 i a^5}{f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac{16 i a^5 c^5}{f \left (c^4-i c^4 \tan (e+f x)\right )^2}+\frac{8 i a^5 \log (\cos (e+f x))}{c^3 f}-\frac{8 a^5 x}{c^3}-\frac{16 i a^5}{3 f (c-i c \tan (e+f x))^3} \]

[Out]

(-8*a^5*x)/c^3 + ((8*I)*a^5*Log[Cos[e + f*x]])/(c^3*f) + (a^5*Tan[e + f*x])/(c^3*f) - (((16*I)/3)*a^5)/(f*(c -
I*c*Tan[e + f*x])^3) - ((24*I)*a^5)/(f*(c^3 - I*c^3*Tan[e + f*x])) + ((16*I)*a^5*c^5)/(f*(c^4 - I*c^4*Tan[e +
f*x])^2)

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Rubi [A]  time = 0.145938, antiderivative size = 134, 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} \[ \frac{a^5 \tan (e+f x)}{c^3 f}-\frac{24 i a^5}{f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac{16 i a^5 c^5}{f \left (c^4-i c^4 \tan (e+f x)\right )^2}+\frac{8 i a^5 \log (\cos (e+f x))}{c^3 f}-\frac{8 a^5 x}{c^3}-\frac{16 i a^5}{3 f (c-i c \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^5/(c - I*c*Tan[e + f*x])^3,x]

[Out]

(-8*a^5*x)/c^3 + ((8*I)*a^5*Log[Cos[e + f*x]])/(c^3*f) + (a^5*Tan[e + f*x])/(c^3*f) - (((16*I)/3)*a^5)/(f*(c -
I*c*Tan[e + f*x])^3) - ((24*I)*a^5)/(f*(c^3 - I*c^3*Tan[e + f*x])) + ((16*I)*a^5*c^5)/(f*(c^4 - I*c^4*Tan[e +
f*x])^2)

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

Mathematica [B]  time = 8.55067, size = 923, normalized size = 6.89 \[ \frac{\cos ^4(e+f x) \left (\frac{\cos (5 e)}{c^3}-\frac{i \sin (5 e)}{c^3}\right ) \sin (f x) (i \tan (e+f x) a+a)^5}{f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{e}{2}\right )+\sin \left (\frac{e}{2}\right )\right ) (\cos (f x)+i \sin (f x))^5}+\frac{\cos ^5(e+f x) \left (\frac{6 \cos (3 e)}{c^3}-\frac{6 i \sin (3 e)}{c^3}\right ) \sin (2 f x) (i \tan (e+f x) a+a)^5}{f (\cos (f x)+i \sin (f x))^5}+\frac{\cos ^5(e+f x) \left (\frac{2 i \sin (e)}{c^3}-\frac{2 \cos (e)}{c^3}\right ) \sin (4 f x) (i \tan (e+f x) a+a)^5}{f (\cos (f x)+i \sin (f x))^5}+\frac{\cos ^5(e+f x) \left (\frac{2 \cos (e)}{3 c^3}+\frac{2 i \sin (e)}{3 c^3}\right ) \sin (6 f x) (i \tan (e+f x) a+a)^5}{f (\cos (f x)+i \sin (f x))^5}+\frac{x \cos ^5(e+f x) \left (-\frac{4 \cos ^5(e)}{c^3}+\frac{24 i \sin (e) \cos ^4(e)}{c^3}+\frac{60 \sin ^2(e) \cos ^3(e)}{c^3}+\frac{4 \cos ^3(e)}{c^3}-\frac{80 i \sin ^3(e) \cos ^2(e)}{c^3}-\frac{16 i \sin (e) \cos ^2(e)}{c^3}-\frac{60 \sin ^4(e) \cos (e)}{c^3}-\frac{24 \sin ^2(e) \cos (e)}{c^3}+\frac{24 i \sin ^5(e)}{c^3}+\frac{16 i \sin ^3(e)}{c^3}+\frac{4 \sin ^5(e) \tan (e)}{c^3}+\frac{4 \sin ^3(e) \tan (e)}{c^3}-i \left (\frac{8 \cos (5 e)}{c^3}-\frac{8 i \sin (5 e)}{c^3}\right ) \tan (e)\right ) (i \tan (e+f x) a+a)^5}{(\cos (f x)+i \sin (f x))^5}-\frac{8 x \cos (5 e) \cos ^5(e+f x) (i \tan (e+f x) a+a)^5}{c^3 (\cos (f x)+i \sin (f x))^5}+\frac{4 i \cos (5 e) \cos ^5(e+f x) \log \left (\cos ^2(e+f x)\right ) (i \tan (e+f x) a+a)^5}{c^3 f (\cos (f x)+i \sin (f x))^5}+\frac{\cos (6 f x) \cos ^5(e+f x) \left (\frac{2 \sin (e)}{3 c^3}-\frac{2 i \cos (e)}{3 c^3}\right ) (i \tan (e+f x) a+a)^5}{f (\cos (f x)+i \sin (f x))^5}+\frac{\cos (4 f x) \cos ^5(e+f x) \left (\frac{2 i \cos (e)}{c^3}+\frac{2 \sin (e)}{c^3}\right ) (i \tan (e+f x) a+a)^5}{f (\cos (f x)+i \sin (f x))^5}+\frac{\cos (2 f x) \cos ^5(e+f x) \left (-\frac{6 i \cos (3 e)}{c^3}-\frac{6 \sin (3 e)}{c^3}\right ) (i \tan (e+f x) a+a)^5}{f (\cos (f x)+i \sin (f x))^5}+\frac{8 i x \cos ^5(e+f x) \sin (5 e) (i \tan (e+f x) a+a)^5}{c^3 (\cos (f x)+i \sin (f x))^5}+\frac{4 \cos ^5(e+f x) \log \left (\cos ^2(e+f x)\right ) \sin (5 e) (i \tan (e+f x) a+a)^5}{c^3 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])^3,x]

[Out]

(-8*x*Cos[5*e]*Cos[e + f*x]^5*(a + I*a*Tan[e + f*x])^5)/(c^3*(Cos[f*x] + I*Sin[f*x])^5) + ((4*I)*Cos[5*e]*Cos[
e + f*x]^5*Log[Cos[e + f*x]^2]*(a + I*a*Tan[e + f*x])^5)/(c^3*f*(Cos[f*x] + I*Sin[f*x])^5) + (Cos[6*f*x]*Cos[e
+ f*x]^5*((((-2*I)/3)*Cos[e])/c^3 + (2*Sin[e])/(3*c^3))*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] + I*Sin[f*x])^
5) + (Cos[4*f*x]*Cos[e + f*x]^5*(((2*I)*Cos[e])/c^3 + (2*Sin[e])/c^3)*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] +
I*Sin[f*x])^5) + (Cos[2*f*x]*Cos[e + f*x]^5*(((-6*I)*Cos[3*e])/c^3 - (6*Sin[3*e])/c^3)*(a + I*a*Tan[e + f*x])
^5)/(f*(Cos[f*x] + I*Sin[f*x])^5) + ((8*I)*x*Cos[e + f*x]^5*Sin[5*e]*(a + I*a*Tan[e + f*x])^5)/(c^3*(Cos[f*x]
+ I*Sin[f*x])^5) + (4*Cos[e + f*x]^5*Log[Cos[e + f*x]^2]*Sin[5*e]*(a + I*a*Tan[e + f*x])^5)/(c^3*f*(Cos[f*x] +
I*Sin[f*x])^5) + (Cos[e + f*x]^4*(Cos[5*e]/c^3 - (I*Sin[5*e])/c^3)*Sin[f*x]*(a + I*a*Tan[e + f*x])^5)/(f*(Cos
[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(Cos[f*x] + I*Sin[f*x])^5) + (Cos[e + f*x]^5*((6*Cos[3*e])/c^3 - ((6*I
)*Sin[3*e])/c^3)*Sin[2*f*x]*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] + I*Sin[f*x])^5) + (Cos[e + f*x]^5*((-2*Cos
[e])/c^3 + ((2*I)*Sin[e])/c^3)*Sin[4*f*x]*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] + I*Sin[f*x])^5) + (Cos[e + f
*x]^5*((2*Cos[e])/(3*c^3) + (((2*I)/3)*Sin[e])/c^3)*Sin[6*f*x]*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] + I*Sin[
f*x])^5) + (x*Cos[e + f*x]^5*((4*Cos[e]^3)/c^3 - (4*Cos[e]^5)/c^3 - ((16*I)*Cos[e]^2*Sin[e])/c^3 + ((24*I)*Cos
[e]^4*Sin[e])/c^3 - (24*Cos[e]*Sin[e]^2)/c^3 + (60*Cos[e]^3*Sin[e]^2)/c^3 + ((16*I)*Sin[e]^3)/c^3 - ((80*I)*Co
s[e]^2*Sin[e]^3)/c^3 - (60*Cos[e]*Sin[e]^4)/c^3 + ((24*I)*Sin[e]^5)/c^3 + (4*Sin[e]^3*Tan[e])/c^3 + (4*Sin[e]^
5*Tan[e])/c^3 - I*((8*Cos[5*e])/c^3 - ((8*I)*Sin[5*e])/c^3)*Tan[e])*(a + I*a*Tan[e + f*x])^5)/(Cos[f*x] + I*Si
n[f*x])^5

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Maple [A]  time = 0.049, size = 107, normalized size = 0.8 \begin{align*}{\frac{{a}^{5}\tan \left ( fx+e \right ) }{{c}^{3}f}}+24\,{\frac{{a}^{5}}{{c}^{3}f \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{16\,i{a}^{5}}{{c}^{3}f \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}-{\frac{8\,i{a}^{5}\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{{c}^{3}f}}-{\frac{16\,{a}^{5}}{3\,{c}^{3}f \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))^3,x)

[Out]

a^5*tan(f*x+e)/c^3/f+24/f*a^5/c^3/(tan(f*x+e)+I)-16*I/f*a^5/c^3/(tan(f*x+e)+I)^2-8*I/f*a^5/c^3*ln(tan(f*x+e)+I
)-16/3/f*a^5/c^3/(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))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.46296, size = 333, normalized size = 2.49 \begin{align*} \frac{-2 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 )} - 12 i \, a^{5} e^{\left (4 i \, f x + 4 i \, e\right )} - 18 i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, a^{5} +{\left (24 i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, a^{5}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{3 \,{\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )}} \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))^3,x, algorithm="fricas")

[Out]

1/3*(-2*I*a^5*e^(8*I*f*x + 8*I*e) + 4*I*a^5*e^(6*I*f*x + 6*I*e) - 12*I*a^5*e^(4*I*f*x + 4*I*e) - 18*I*a^5*e^(2
*I*f*x + 2*I*e) + 6*I*a^5 + (24*I*a^5*e^(2*I*f*x + 2*I*e) + 24*I*a^5)*log(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f*e^(
2*I*f*x + 2*I*e) + c^3*f)

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Sympy [A]  time = 2.08988, size = 178, normalized size = 1.33 \begin{align*} \frac{8 i a^{5} \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{3} f} + \frac{2 i a^{5} e^{- 2 i e}}{c^{3} f \left (e^{2 i f x} + e^{- 2 i e}\right )} + \frac{\begin{cases} - \frac{2 i a^{5} e^{6 i e} e^{6 i f x}}{3 f} + \frac{2 i a^{5} e^{4 i e} e^{4 i f x}}{f} - \frac{6 i a^{5} e^{2 i e} e^{2 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (4 a^{5} e^{6 i e} - 8 a^{5} e^{4 i e} + 12 a^{5} e^{2 i e}\right ) & \text{otherwise} \end{cases}}{c^{3}} \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))**3,x)

[Out]

8*I*a**5*log(exp(2*I*f*x) + exp(-2*I*e))/(c**3*f) + 2*I*a**5*exp(-2*I*e)/(c**3*f*(exp(2*I*f*x) + exp(-2*I*e)))
+ Piecewise((-2*I*a**5*exp(6*I*e)*exp(6*I*f*x)/(3*f) + 2*I*a**5*exp(4*I*e)*exp(4*I*f*x)/f - 6*I*a**5*exp(2*I*
e)*exp(2*I*f*x)/f, Ne(f, 0)), (x*(4*a**5*exp(6*I*e) - 8*a**5*exp(4*I*e) + 12*a**5*exp(2*I*e)), True))/c**3

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Giac [B]  time = 1.66698, size = 343, normalized size = 2.56 \begin{align*} -\frac{2 \,{\left (\frac{120 i \, a^{5} \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}{c^{3}} - \frac{60 i \, a^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{3}} - \frac{60 i \, a^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{3}} - \frac{15 \,{\left (-4 i \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 i \, a^{5}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} c^{3}} + \frac{-294 i \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 1884 \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 4890 i \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 6920 \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 4890 i \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1884 \, a^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 294 i \, a^{5}}{c^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + i\right )}^{6}}\right )}}{15 \, 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))^3,x, algorithm="giac")

[Out]

-2/15*(120*I*a^5*log(tan(1/2*f*x + 1/2*e) + I)/c^3 - 60*I*a^5*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^3 - 60*I*a^
5*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c^3 - 15*(-4*I*a^5*tan(1/2*f*x + 1/2*e)^2 - a^5*tan(1/2*f*x + 1/2*e) + 4*
I*a^5)/((tan(1/2*f*x + 1/2*e)^2 - 1)*c^3) + (-294*I*a^5*tan(1/2*f*x + 1/2*e)^6 + 1884*a^5*tan(1/2*f*x + 1/2*e)
^5 + 4890*I*a^5*tan(1/2*f*x + 1/2*e)^4 - 6920*a^5*tan(1/2*f*x + 1/2*e)^3 - 4890*I*a^5*tan(1/2*f*x + 1/2*e)^2 +
1884*a^5*tan(1/2*f*x + 1/2*e) + 294*I*a^5)/(c^3*(tan(1/2*f*x + 1/2*e) + I)^6))/f