\(\int \sin ^5(6 x) \tan ^3(6 x) \, dx\) [892]

Optimal result

Integrand size = 13, antiderivative size = 54 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=-\frac {7}{12} \text {arctanh}(\sin (6 x))+\frac {7}{12} \sin (6 x)+\frac {7}{36} \sin ^3(6 x)+\frac {7}{60} \sin ^5(6 x)+\frac {1}{12} \sin ^5(6 x) \tan ^2(6 x) \]

[Out]

-7/12*arctanh(sin(6*x))+7/12*sin(6*x)+7/36*sin(6*x)^3+7/60*sin(6*x)^5+1/12*sin(6*x)^5*tan(6*x)^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2672, 294, 308, 212} \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=-\frac {7}{12} \text {arctanh}(\sin (6 x))+\frac {7}{60} \sin ^5(6 x)+\frac {7}{36} \sin ^3(6 x)+\frac {7}{12} \sin (6 x)+\frac {1}{12} \sin ^5(6 x) \tan ^2(6 x) \]

[In]

Int[Sin[6*x]^5*Tan[6*x]^3,x]

[Out]

(-7*ArcTanh[Sin[6*x]])/12 + (7*Sin[6*x])/12 + (7*Sin[6*x]^3)/36 + (7*Sin[6*x]^5)/60 + (Sin[6*x]^5*Tan[6*x]^2)/
12

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} \text {Subst}\left (\int \frac {x^8}{\left (1-x^2\right )^2} \, dx,x,\sin (6 x)\right ) \\ & = \frac {1}{12} \sin ^5(6 x) \tan ^2(6 x)-\frac {7}{12} \text {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\sin (6 x)\right ) \\ & = \frac {1}{12} \sin ^5(6 x) \tan ^2(6 x)-\frac {7}{12} \text {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\sin (6 x)\right ) \\ & = \frac {7}{12} \sin (6 x)+\frac {7}{36} \sin ^3(6 x)+\frac {7}{60} \sin ^5(6 x)+\frac {1}{12} \sin ^5(6 x) \tan ^2(6 x)-\frac {7}{12} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (6 x)\right ) \\ & = -\frac {7}{12} \text {arctanh}(\sin (6 x))+\frac {7}{12} \sin (6 x)+\frac {7}{36} \sin ^3(6 x)+\frac {7}{60} \sin ^5(6 x)+\frac {1}{12} \sin ^5(6 x) \tan ^2(6 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.26 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=-\frac {7}{12} \text {arctanh}(\sin (6 x))+\frac {7}{12} \sec (6 x) \tan (6 x)-\frac {7}{18} \sin (6 x) \tan ^2(6 x)-\frac {7}{90} \sin ^3(6 x) \tan ^2(6 x)-\frac {1}{30} \sin ^5(6 x) \tan ^2(6 x) \]

[In]

Integrate[Sin[6*x]^5*Tan[6*x]^3,x]

[Out]

(-7*ArcTanh[Sin[6*x]])/12 + (7*Sec[6*x]*Tan[6*x])/12 - (7*Sin[6*x]*Tan[6*x]^2)/18 - (7*Sin[6*x]^3*Tan[6*x]^2)/
90 - (Sin[6*x]^5*Tan[6*x]^2)/30

Maple [A] (verified)

Time = 35.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(\frac {\sin \left (6 x \right )^{9}}{12 \cos \left (6 x \right )^{2}}+\frac {\sin \left (6 x \right )^{7}}{12}+\frac {7 \sin \left (6 x \right )^{5}}{60}+\frac {7 \sin \left (6 x \right )^{3}}{36}+\frac {7 \sin \left (6 x \right )}{12}-\frac {7 \ln \left (\sec \left (6 x \right )+\tan \left (6 x \right )\right )}{12}\)	\(58\)
default	\(\frac {\sin \left (6 x \right )^{9}}{12 \cos \left (6 x \right )^{2}}+\frac {\sin \left (6 x \right )^{7}}{12}+\frac {7 \sin \left (6 x \right )^{5}}{60}+\frac {7 \sin \left (6 x \right )^{3}}{36}+\frac {7 \sin \left (6 x \right )}{12}-\frac {7 \ln \left (\sec \left (6 x \right )+\tan \left (6 x \right )\right )}{12}\)	\(58\)
risch	\(\frac {11 i {\mathrm e}^{18 i x}}{576}-\frac {29 i {\mathrm e}^{6 i x}}{96}+\frac {29 i {\mathrm e}^{-6 i x}}{96}-\frac {11 i {\mathrm e}^{-18 i x}}{576}-\frac {i \left ({\mathrm e}^{18 i x}-{\mathrm e}^{6 i x}\right )}{6 \left ({\mathrm e}^{12 i x}+1\right )^{2}}+\frac {7 \ln \left ({\mathrm e}^{6 i x}-i\right )}{12}-\frac {7 \ln \left (i+{\mathrm e}^{6 i x}\right )}{12}+\frac {\sin \left (30 x \right )}{480}\)	\(87\)

[In]

int(sin(6*x)^5*tan(6*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/12*sin(6*x)^9/cos(6*x)^2+1/12*sin(6*x)^7+7/60*sin(6*x)^5+7/36*sin(6*x)^3+7/12*sin(6*x)-7/12*ln(sec(6*x)+tan(
6*x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.35 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=-\frac {105 \, \cos \left (6 \, x\right )^{2} \log \left (\sin \left (6 \, x\right ) + 1\right ) - 105 \, \cos \left (6 \, x\right )^{2} \log \left (-\sin \left (6 \, x\right ) + 1\right ) - 2 \, {\left (6 \, \cos \left (6 \, x\right )^{6} - 32 \, \cos \left (6 \, x\right )^{4} + 116 \, \cos \left (6 \, x\right )^{2} + 15\right )} \sin \left (6 \, x\right )}{360 \, \cos \left (6 \, x\right )^{2}} \]

[In]

integrate(sin(6*x)^5*tan(6*x)^3,x, algorithm="fricas")

[Out]

-1/360*(105*cos(6*x)^2*log(sin(6*x) + 1) - 105*cos(6*x)^2*log(-sin(6*x) + 1) - 2*(6*cos(6*x)^6 - 32*cos(6*x)^4
+ 116*cos(6*x)^2 + 15)*sin(6*x))/cos(6*x)^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.13 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=\frac {7 \log {\left (\sin {\left (6 x \right )} - 1 \right )}}{24} - \frac {7 \log {\left (\sin {\left (6 x \right )} + 1 \right )}}{24} + \frac {\sin ^{5}{\left (6 x \right )}}{30} + \frac {\sin ^{3}{\left (6 x \right )}}{9} + \frac {\sin {\left (6 x \right )}}{2} - \frac {\sin {\left (6 x \right )}}{6 \cdot \left (2 \sin ^{2}{\left (6 x \right )} - 2\right )} \]

[In]

integrate(sin(6*x)**5*tan(6*x)**3,x)

[Out]

7*log(sin(6*x) - 1)/24 - 7*log(sin(6*x) + 1)/24 + sin(6*x)**5/30 + sin(6*x)**3/9 + sin(6*x)/2 - sin(6*x)/(6*(2
*sin(6*x)**2 - 2))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.06 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=\frac {1}{30} \, \sin \left (6 \, x\right )^{5} + \frac {1}{9} \, \sin \left (6 \, x\right )^{3} - \frac {\sin \left (6 \, x\right )}{12 \, {\left (\sin \left (6 \, x\right )^{2} - 1\right )}} - \frac {7}{24} \, \log \left (\sin \left (6 \, x\right ) + 1\right ) + \frac {7}{24} \, \log \left (\sin \left (6 \, x\right ) - 1\right ) + \frac {1}{2} \, \sin \left (6 \, x\right ) \]

[In]

integrate(sin(6*x)^5*tan(6*x)^3,x, algorithm="maxima")

[Out]

1/30*sin(6*x)^5 + 1/9*sin(6*x)^3 - 1/12*sin(6*x)/(sin(6*x)^2 - 1) - 7/24*log(sin(6*x) + 1) + 7/24*log(sin(6*x)
- 1) + 1/2*sin(6*x)

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=\frac {1}{30} \, \sin \left (6 \, x\right )^{5} + \frac {1}{9} \, \sin \left (6 \, x\right )^{3} - \frac {\sin \left (6 \, x\right )}{12 \, {\left (\sin \left (6 \, x\right )^{2} - 1\right )}} - \frac {7}{24} \, \log \left (\sin \left (6 \, x\right ) + 1\right ) + \frac {7}{24} \, \log \left (-\sin \left (6 \, x\right ) + 1\right ) + \frac {1}{2} \, \sin \left (6 \, x\right ) \]

[In]

integrate(sin(6*x)^5*tan(6*x)^3,x, algorithm="giac")

[Out]

1/30*sin(6*x)^5 + 1/9*sin(6*x)^3 - 1/12*sin(6*x)/(sin(6*x)^2 - 1) - 7/24*log(sin(6*x) + 1) + 7/24*log(-sin(6*x
) + 1) + 1/2*sin(6*x)

Mupad [B] (verification not implemented)

Time = 31.82 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.57 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=\frac {7\,{\mathrm {tan}\left (3\,x\right )}^{13}+\frac {70\,{\mathrm {tan}\left (3\,x\right )}^{11}}{3}+\frac {77\,{\mathrm {tan}\left (3\,x\right )}^9}{5}-\frac {412\,{\mathrm {tan}\left (3\,x\right )}^7}{15}+\frac {77\,{\mathrm {tan}\left (3\,x\right )}^5}{5}+\frac {70\,{\mathrm {tan}\left (3\,x\right )}^3}{3}+7\,\mathrm {tan}\left (3\,x\right )}{6\,{\left ({\mathrm {tan}\left (3\,x\right )}^2-1\right )}^2\,{\left ({\mathrm {tan}\left (3\,x\right )}^2+1\right )}^5}-\frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (3\,x\right )\right )}{6} \]

[In]

int(sin(6*x)^5*tan(6*x)^3,x)

[Out]

(7*tan(3*x) + (70*tan(3*x)^3)/3 + (77*tan(3*x)^5)/5 - (412*tan(3*x)^7)/15 + (77*tan(3*x)^9)/5 + (70*tan(3*x)^1
1)/3 + 7*tan(3*x)^13)/(6*(tan(3*x)^2 - 1)^2*(tan(3*x)^2 + 1)^5) - (7*atanh(tan(3*x)))/6