3.46 \(\int \frac{\tan ^5(c+d x)}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=109 \[ -\frac{\tan ^4(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{5 i \tan ^3(c+d x)}{6 a d}+\frac{\tan ^2(c+d x)}{a d}+\frac{5 i \tan (c+d x)}{2 a d}+\frac{2 \log (\cos (c+d x))}{a d}-\frac{5 i x}{2 a} \]

[Out]

(((-5*I)/2)*x)/a + (2*Log[Cos[c + d*x]])/(a*d) + (((5*I)/2)*Tan[c + d*x])/(a*d) + Tan[c + d*x]^2/(a*d) - (((5*
I)/6)*Tan[c + d*x]^3)/(a*d) - Tan[c + d*x]^4/(2*d*(a + I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.123876, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3550, 3528, 3525, 3475} \[ -\frac{\tan ^4(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{5 i \tan ^3(c+d x)}{6 a d}+\frac{\tan ^2(c+d x)}{a d}+\frac{5 i \tan (c+d x)}{2 a d}+\frac{2 \log (\cos (c+d x))}{a d}-\frac{5 i x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x]),x]

[Out]

(((-5*I)/2)*x)/a + (2*Log[Cos[c + d*x]])/(a*d) + (((5*I)/2)*Tan[c + d*x])/(a*d) + Tan[c + d*x]^2/(a*d) - (((5*
I)/6)*Tan[c + d*x]^3)/(a*d) - Tan[c + d*x]^4/(2*d*(a + I*a*Tan[c + d*x]))

Rule 3550

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\tan ^5(c+d x)}{a+i a \tan (c+d x)} \, dx &=-\frac{\tan ^4(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \tan ^3(c+d x) (4 a-5 i a \tan (c+d x)) \, dx}{2 a^2}\\ &=-\frac{5 i \tan ^3(c+d x)}{6 a d}-\frac{\tan ^4(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \tan ^2(c+d x) (5 i a+4 a \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{\tan ^2(c+d x)}{a d}-\frac{5 i \tan ^3(c+d x)}{6 a d}-\frac{\tan ^4(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \tan (c+d x) (-4 a+5 i a \tan (c+d x)) \, dx}{2 a^2}\\ &=-\frac{5 i x}{2 a}+\frac{5 i \tan (c+d x)}{2 a d}+\frac{\tan ^2(c+d x)}{a d}-\frac{5 i \tan ^3(c+d x)}{6 a d}-\frac{\tan ^4(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{2 \int \tan (c+d x) \, dx}{a}\\ &=-\frac{5 i x}{2 a}+\frac{2 \log (\cos (c+d x))}{a d}+\frac{5 i \tan (c+d x)}{2 a d}+\frac{\tan ^2(c+d x)}{a d}-\frac{5 i \tan ^3(c+d x)}{6 a d}-\frac{\tan ^4(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end{align*}

Mathematica [B]  time = 3.43854, size = 235, normalized size = 2.16 \[ \frac{\sec (c+d x) (\cos (d x)+i \sin (d x)) \left (6 d x \sin (c)+3 \sin (c) \sin (2 d x)+3 i \sin (c) \cos (2 d x)-4 i \sin (d x) \sec ^3(c+d x)+2 i \sin (c) \sec ^2(c+d x)+28 i \sin (d x) \sec (c+d x)+12 i \sin (c) \log \left (\cos ^2(c+d x)\right )+24 (\sin (c)-i \cos (c)) \tan ^{-1}(\tan (d x))+4 \tan (c) \sin (d x) \sec ^3(c+d x)+4 \sin (c) \tan (c) \sec ^2(c+d x)-28 \tan (c) \sin (d x) \sec (c+d x)+3 \cos (c) \left (2 \sec ^2(c+d x)+4 \log \left (\cos ^2(c+d x)\right )-2 i d x+i \sin (2 d x)-\cos (2 d x)\right )\right )}{12 d (a+i a \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x]),x]

[Out]

(Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])*(6*d*x*Sin[c] + (3*I)*Cos[2*d*x]*Sin[c] + (12*I)*Log[Cos[c + d*x]^2]*Sin
[c] + (2*I)*Sec[c + d*x]^2*Sin[c] + 24*ArcTan[Tan[d*x]]*((-I)*Cos[c] + Sin[c]) + (28*I)*Sec[c + d*x]*Sin[d*x]
- (4*I)*Sec[c + d*x]^3*Sin[d*x] + 3*Cos[c]*((-2*I)*d*x - Cos[2*d*x] + 4*Log[Cos[c + d*x]^2] + 2*Sec[c + d*x]^2
+ I*Sin[2*d*x]) + 3*Sin[c]*Sin[2*d*x] + 4*Sec[c + d*x]^2*Sin[c]*Tan[c] - 28*Sec[c + d*x]*Sin[d*x]*Tan[c] + 4*
Sec[c + d*x]^3*Sin[d*x]*Tan[c]))/(12*d*(a + I*a*Tan[c + d*x]))

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Maple [A]  time = 0.024, size = 106, normalized size = 1. \begin{align*}{\frac{2\,i\tan \left ( dx+c \right ) }{ad}}-{\frac{{\frac{i}{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{ad}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,ad}}+{\frac{{\frac{i}{2}}}{ad \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{9\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{4\,ad}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{4\,ad}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+I*a*tan(d*x+c)),x)

[Out]

2*I/d/a*tan(d*x+c)-1/3*I/d/a*tan(d*x+c)^3+1/2*tan(d*x+c)^2/a/d+1/2*I/d/a/(tan(d*x+c)-I)-9/4/d/a*ln(tan(d*x+c)-
I)+1/4/d/a*ln(tan(d*x+c)+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(tan(d*x+c)^5/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.38229, size = 532, normalized size = 4.88 \begin{align*} \frac{-54 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \,{\left (54 i \, d x + 17\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 81 \,{\left (2 i \, d x + 1\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-54 i \, d x - 65\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 24 \,{\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 3}{12 \,{\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(-54*I*d*x*e^(8*I*d*x + 8*I*c) - 3*(54*I*d*x + 17)*e^(6*I*d*x + 6*I*c) - 81*(2*I*d*x + 1)*e^(4*I*d*x + 4*
I*c) + (-54*I*d*x - 65)*e^(2*I*d*x + 2*I*c) + 24*(e^(8*I*d*x + 8*I*c) + 3*e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x +
4*I*c) + e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 3)/(a*d*e^(8*I*d*x + 8*I*c) + 3*a*d*e^(6*I*d*x +
6*I*c) + 3*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

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Sympy [A]  time = 4.08176, size = 202, normalized size = 1.85 \begin{align*} \frac{- \frac{4 e^{- 2 i c} e^{4 i d x}}{a d} - \frac{6 e^{- 4 i c} e^{2 i d x}}{a d} - \frac{14 e^{- 6 i c}}{3 a d}}{e^{6 i d x} + 3 e^{- 2 i c} e^{4 i d x} + 3 e^{- 4 i c} e^{2 i d x} + e^{- 6 i c}} + \begin{cases} - \frac{e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text{for}\: 4 a d e^{2 i c} \neq 0 \\x \left (- \frac{\left (9 i e^{2 i c} - i\right ) e^{- 2 i c}}{2 a} + \frac{9 i}{2 a}\right ) & \text{otherwise} \end{cases} - \frac{9 i x}{2 a} + \frac{2 \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+I*a*tan(d*x+c)),x)

[Out]

(-4*exp(-2*I*c)*exp(4*I*d*x)/(a*d) - 6*exp(-4*I*c)*exp(2*I*d*x)/(a*d) - 14*exp(-6*I*c)/(3*a*d))/(exp(6*I*d*x)
+ 3*exp(-2*I*c)*exp(4*I*d*x) + 3*exp(-4*I*c)*exp(2*I*d*x) + exp(-6*I*c)) + Piecewise((-exp(-2*I*c)*exp(-2*I*d*
x)/(4*a*d), Ne(4*a*d*exp(2*I*c), 0)), (x*(-(9*I*exp(2*I*c) - I)*exp(-2*I*c)/(2*a) + 9*I/(2*a)), True)) - 9*I*x
/(2*a) + 2*log(exp(2*I*d*x) + exp(-2*I*c))/(a*d)

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Giac [A]  time = 3.60245, size = 140, normalized size = 1.28 \begin{align*} \frac{\frac{3 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a} - \frac{27 \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a} + \frac{3 \,{\left (9 \, \tan \left (d x + c\right ) - 7 i\right )}}{a{\left (\tan \left (d x + c\right ) - i\right )}} - \frac{2 \,{\left (2 i \, a^{2} \tan \left (d x + c\right )^{3} - 3 \, a^{2} \tan \left (d x + c\right )^{2} - 12 i \, a^{2} \tan \left (d x + c\right )\right )}}{a^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

1/12*(3*log(tan(d*x + c) + I)/a - 27*log(I*tan(d*x + c) + 1)/a + 3*(9*tan(d*x + c) - 7*I)/(a*(tan(d*x + c) - I
)) - 2*(2*I*a^2*tan(d*x + c)^3 - 3*a^2*tan(d*x + c)^2 - 12*I*a^2*tan(d*x + c))/a^3)/d