\(\int \frac {\tan ^3(a+b \log (c x^n))}{x} \, dx\) [172]

Optimal result

Integrand size = 17, antiderivative size = 43 \[ \int \frac {\tan ^3\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\log \left (\cos \left (a+b \log \left (c x^n\right )\right )\right )}{b n}+\frac {\tan ^2\left (a+b \log \left (c x^n\right )\right )}{2 b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3554, 3556} \[ \int \frac {\tan ^3\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\tan ^2\left (a+b \log \left (c x^n\right )\right )}{2 b n}+\frac {\log \left (\cos \left (a+b \log \left (c x^n\right )\right )\right )}{b n} \]

[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 3554

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 3556

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*} \text {integral}& = \frac {\text {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 {\text {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] (verified)

Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^3\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {2 \log \left (\cos \left (a+b \log \left (c x^n\right )\right )\right )+\tan ^2\left (a+b \log \left (c x^n\right )\right )}{2 b n} \]

[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)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(\frac {{\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}-\ln \left (1+{\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}\right )}{2 b n}\)	\(41\)
derivativedivides	\(\frac {\frac {{\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{2}-\frac {\ln \left (1+{\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}\right )}{2}}{n b}\)	\(42\)
default	\(\frac {\frac {{\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{2}-\frac {\ln \left (1+{\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}\right )}{2}}{n b}\)	\(42\)
risch	\(i \ln \left (x \right )-\frac {2 i a}{n b}-\frac {2 i \ln \left (c \right )}{n}-\frac {2 i \ln \left (x^{n}\right )}{n}+\frac {\pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{n}-\frac {\pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{n}-\frac {\pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{n}+\frac {\pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{n}+\frac {2 \left (x^{n}\right )^{2 i b} c^{2 i b} {\mathrm e}^{-b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}} {\mathrm e}^{b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )} {\mathrm e}^{b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}} {\mathrm e}^{-b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )} {\mathrm e}^{2 i a}}{b n {\left (\left (x^{n}\right )^{2 i b} c^{2 i b} {\mathrm e}^{-b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}} {\mathrm e}^{b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )} {\mathrm e}^{b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}} {\mathrm e}^{-b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )} {\mathrm e}^{2 i a}+1\right )}^{2}}+\frac {\ln \left (\left (x^{n}\right )^{2 i b} c^{2 i b} {\mathrm e}^{-b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}} {\mathrm e}^{b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )} {\mathrm e}^{b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}} {\mathrm e}^{-b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )} {\mathrm e}^{2 i a}+1\right )}{b n}\)	\(452\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.60 \[ \int \frac {\tan ^3\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {{\left (\cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + \frac {1}{2}\right ) + 2}{2 \, {\left (b n \cos \left (2 \, b n \log \left (x\right ) + 2 \, b \log \left (c\right ) + 2 \, a\right ) + b n\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.47 \[ \int \frac {\tan ^3\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\begin {cases} \log {\left (x \right )} \tan ^{3}{\left (a \right )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\log {\left (x \right )} \tan ^{3}{\left (a + b \log {\left (c \right )} \right )} & \text {for}\: n = 0 \\- \frac {\log {\left (\tan ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} + 1 \right )}}{2 b n} + \frac {\tan ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{2 b n} & \text {otherwise} \end {cases} \]

[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*log(c*x**n))**2 + 1)/(2*b*n) + tan(a + b*log(c*x**n))**2/(2*b*n), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1242 vs. \(2 (41) = 82\).

Time = 0.24 (sec) , antiderivative size = 1242, normalized size of antiderivative = 28.88 \[ \int \frac {\tan ^3\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Too large to display} \]

[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))

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^3\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 32.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.44 \[ \int \frac {\tan ^3\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\ln \left (x\right )\,1{}\mathrm {i}-\frac {2}{b\,n\,\left (2\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}+{\mathrm {e}}^{a\,4{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,4{}\mathrm {i}}+1\right )}+\frac {2}{b\,n\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}+1\right )}+\frac {\ln \left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}+1\right )}{b\,n} \]

[In]

int(tan(a + b*log(c*x^n))^3/x,x)

[Out]

2/(b*n*(exp(a*2i)*(c*x^n)^(b*2i) + 1)) - 2/(b*n*(2*exp(a*2i)*(c*x^n)^(b*2i) + exp(a*4i)*(c*x^n)^(b*4i) + 1)) -
log(x)*1i + log(exp(a*2i)*(c*x^n)^(b*2i) + 1)/(b*n)