3.172 \(\int \frac{\tan ^3(a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=43 \[ \frac{\tan ^2\left (a+b \log \left (c x^n\right )\right )}{2 b n}+\frac{\log \left (\cos \left (a+b \log \left (c x^n\right )\right )\right )}{b n} \]

[Out]

Log[Cos[a + b*Log[c*x^n]]]/(b*n) + Tan[a + b*Log[c*x^n]]^2/(2*b*n)

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Rubi [A]  time = 0.0342154, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3473, 3475} \[ \frac{\tan ^2\left (a+b \log \left (c x^n\right )\right )}{2 b n}+\frac{\log \left (\cos \left (a+b \log \left (c x^n\right )\right )\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Int[Tan[a + b*Log[c*x^n]]^3/x,x]

[Out]

Log[Cos[a + b*Log[c*x^n]]]/(b*n) + Tan[a + b*Log[c*x^n]]^2/(2*b*n)

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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 ^3\left (a+b \log \left (c x^n\right )\right )}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \tan ^3(a+b x) \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=\frac{\tan ^2\left (a+b \log \left (c x^n\right )\right )}{2 b n}-\frac{\operatorname{Subst}\left (\int \tan (a+b x) \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=\frac{\log \left (\cos \left (a+b \log \left (c x^n\right )\right )\right )}{b n}+\frac{\tan ^2\left (a+b \log \left (c x^n\right )\right )}{2 b n}\\ \end{align*}

Mathematica [A]  time = 0.156367, size = 38, normalized size = 0.88 \[ \frac{\tan ^2\left (a+b \log \left (c x^n\right )\right )+2 \log \left (\cos \left (a+b \log \left (c x^n\right )\right )\right )}{2 b n} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + b*Log[c*x^n]]^3/x,x]

[Out]

(2*Log[Cos[a + b*Log[c*x^n]]] + Tan[a + b*Log[c*x^n]]^2)/(2*b*n)

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Maple [A]  time = 0.018, size = 47, normalized size = 1.1 \begin{align*}{\frac{ \left ( \tan \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}}{2\,bn}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2} \right ) }{2\,bn}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a+b*ln(c*x^n))^3/x,x)

[Out]

1/2*tan(a+b*ln(c*x^n))^2/b/n-1/2/n/b*ln(1+tan(a+b*ln(c*x^n))^2)

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Maxima [B]  time = 1.14161, size = 1677, normalized size = 39. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+b*log(c*x^n))^3/x,x, algorithm="maxima")

[Out]

1/2*(8*(cos(2*b*log(c))^2 + sin(2*b*log(c))^2)*cos(2*b*log(x^n) + 2*a)^2 + 8*(cos(2*b*log(c))^2 + sin(2*b*log(
c))^2)*sin(2*b*log(x^n) + 2*a)^2 + 4*((cos(4*b*log(c))*cos(2*b*log(c)) + sin(4*b*log(c))*sin(2*b*log(c)))*cos(
2*b*log(x^n) + 2*a) + (cos(2*b*log(c))*sin(4*b*log(c)) - cos(4*b*log(c))*sin(2*b*log(c)))*sin(2*b*log(x^n) + 2
*a))*cos(4*b*log(x^n) + 4*a) + 4*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) + ((cos(4*b*log(c))^2 + sin(4*b*log(c
))^2)*cos(4*b*log(x^n) + 4*a)^2 + 4*(cos(2*b*log(c))^2 + sin(2*b*log(c))^2)*cos(2*b*log(x^n) + 2*a)^2 + (cos(4
*b*log(c))^2 + sin(4*b*log(c))^2)*sin(4*b*log(x^n) + 4*a)^2 + 4*(cos(2*b*log(c))^2 + sin(2*b*log(c))^2)*sin(2*
b*log(x^n) + 2*a)^2 + 2*(2*(cos(4*b*log(c))*cos(2*b*log(c)) + sin(4*b*log(c))*sin(2*b*log(c)))*cos(2*b*log(x^n
) + 2*a) + 2*(cos(2*b*log(c))*sin(4*b*log(c)) - cos(4*b*log(c))*sin(2*b*log(c)))*sin(2*b*log(x^n) + 2*a) + cos
(4*b*log(c)))*cos(4*b*log(x^n) + 4*a) + 4*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) - 2*(2*(cos(2*b*log(c))*sin(
4*b*log(c)) - cos(4*b*log(c))*sin(2*b*log(c)))*cos(2*b*log(x^n) + 2*a) - 2*(cos(4*b*log(c))*cos(2*b*log(c)) +
sin(4*b*log(c))*sin(2*b*log(c)))*sin(2*b*log(x^n) + 2*a) + sin(4*b*log(c)))*sin(4*b*log(x^n) + 4*a) - 4*sin(2*
b*log(c))*sin(2*b*log(x^n) + 2*a) + 1)*log((cos(2*a)^2 + sin(2*a)^2)*cos(2*b*log(c))^2 + (cos(2*a)^2 + sin(2*a
)^2)*sin(2*b*log(c))^2 + 2*(cos(2*b*log(c))*cos(2*a) - sin(2*b*log(c))*sin(2*a))*cos(2*b*log(x^n)) + cos(2*b*l
og(x^n))^2 - 2*(cos(2*a)*sin(2*b*log(c)) + cos(2*b*log(c))*sin(2*a))*sin(2*b*log(x^n)) + sin(2*b*log(x^n))^2)
- 4*((cos(2*b*log(c))*sin(4*b*log(c)) - cos(4*b*log(c))*sin(2*b*log(c)))*cos(2*b*log(x^n) + 2*a) - (cos(4*b*lo
g(c))*cos(2*b*log(c)) + sin(4*b*log(c))*sin(2*b*log(c)))*sin(2*b*log(x^n) + 2*a))*sin(4*b*log(x^n) + 4*a) - 4*
sin(2*b*log(c))*sin(2*b*log(x^n) + 2*a))/((b*cos(4*b*log(c))^2 + b*sin(4*b*log(c))^2)*n*cos(4*b*log(x^n) + 4*a
)^2 + 4*b*n*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) + 4*(b*cos(2*b*log(c))^2 + b*sin(2*b*log(c))^2)*n*cos(2*b*
log(x^n) + 2*a)^2 + (b*cos(4*b*log(c))^2 + b*sin(4*b*log(c))^2)*n*sin(4*b*log(x^n) + 4*a)^2 - 4*b*n*sin(2*b*lo
g(c))*sin(2*b*log(x^n) + 2*a) + 4*(b*cos(2*b*log(c))^2 + b*sin(2*b*log(c))^2)*n*sin(2*b*log(x^n) + 2*a)^2 + b*
n + 2*(b*n*cos(4*b*log(c)) + 2*(b*cos(4*b*log(c))*cos(2*b*log(c)) + b*sin(4*b*log(c))*sin(2*b*log(c)))*n*cos(2
*b*log(x^n) + 2*a) + 2*(b*cos(2*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(2*b*log(c)))*n*sin(2*b*log(x
^n) + 2*a))*cos(4*b*log(x^n) + 4*a) - 2*(2*(b*cos(2*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(2*b*log(
c)))*n*cos(2*b*log(x^n) + 2*a) + b*n*sin(4*b*log(c)) - 2*(b*cos(4*b*log(c))*cos(2*b*log(c)) + b*sin(4*b*log(c)
)*sin(2*b*log(c)))*n*sin(2*b*log(x^n) + 2*a))*sin(4*b*log(x^n) + 4*a))

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Fricas [A]  time = 0.50836, size = 209, normalized size = 4.86 \begin{align*} \frac{{\left (\cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + 1\right )} \log \left (\frac{1}{2} \, \cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + \frac{1}{2}\right ) + 2}{2 \,{\left (b n \cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + b n\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+b*log(c*x^n))^3/x,x, algorithm="fricas")

[Out]

1/2*((cos(2*b*n*log(x) + 2*b*log(c) + 2*a) + 1)*log(1/2*cos(2*b*n*log(x) + 2*b*log(c) + 2*a) + 1/2) + 2)/(b*n*
cos(2*b*n*log(x) + 2*b*log(c) + 2*a) + b*n)

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Sympy [A]  time = 18.4742, size = 70, normalized size = 1.63 \begin{align*} \begin{cases} \log{\left (x \right )} \tan ^{3}{\left (a \right )} & \text{for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\log{\left (x \right )} \tan ^{3}{\left (a + b \log{\left (c \right )} \right )} & \text{for}\: n = 0 \\- \frac{\log{\left (\tan ^{2}{\left (a + b n \log{\left (x \right )} + b \log{\left (c \right )} \right )} + 1 \right )}}{2 b n} + \frac{\tan ^{2}{\left (a + b n \log{\left (x \right )} + b \log{\left (c \right )} \right )}}{2 b n} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+b*ln(c*x**n))**3/x,x)

[Out]

Piecewise((log(x)*tan(a)**3, Eq(b, 0) & (Eq(b, 0) | Eq(n, 0))), (log(x)*tan(a + b*log(c))**3, Eq(n, 0)), (-log
(tan(a + b*n*log(x) + b*log(c))**2 + 1)/(2*b*n) + tan(a + b*n*log(x) + b*log(c))**2/(2*b*n), True))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+b*log(c*x^n))^3/x,x, algorithm="giac")

[Out]

Timed out