Integrand size = 13, antiderivative size = 54 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=-\frac {7}{12} \text {arctanh}(\sin (6 x))+\frac {7}{12} \sin (6 x)+\frac {7}{36} \sin ^3(6 x)+\frac {7}{60} \sin ^5(6 x)+\frac {1}{12} \sin ^5(6 x) \tan ^2(6 x) \] Output:
-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
Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.26 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=-\frac {7}{12} \text {arctanh}(\sin (6 x))+\frac {7}{12} \sec (6 x) \tan (6 x)-\frac {7}{18} \sin (6 x) \tan ^2(6 x)-\frac {7}{90} \sin ^3(6 x) \tan ^2(6 x)-\frac {1}{30} \sin ^5(6 x) \tan ^2(6 x) \] Input:
Integrate[Sin[6*x]^5*Tan[6*x]^3,x]
Output:
(-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
Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3072, 252, 254, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^5(6 x) \tan ^3(6 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (6 x)^5 \tan (6 x)^3dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle \frac {1}{6} \int \frac {\sin ^8(6 x)}{\left (1-\sin ^2(6 x)\right )^2}d\sin (6 x)\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{6} \left (\frac {\sin ^7(6 x)}{2 \left (1-\sin ^2(6 x)\right )}-\frac {7}{2} \int \frac {\sin ^6(6 x)}{1-\sin ^2(6 x)}d\sin (6 x)\right )\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {1}{6} \left (\frac {\sin ^7(6 x)}{2 \left (1-\sin ^2(6 x)\right )}-\frac {7}{2} \int \left (-\sin ^4(6 x)-\sin ^2(6 x)+\frac {1}{1-\sin ^2(6 x)}-1\right )d\sin (6 x)\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} \left (\frac {\sin ^7(6 x)}{2 \left (1-\sin ^2(6 x)\right )}-\frac {7}{2} \left (\text {arctanh}(\sin (6 x))-\frac {1}{5} \sin ^5(6 x)-\frac {1}{3} \sin ^3(6 x)-\sin (6 x)\right )\right )\) |
Input:
Int[Sin[6*x]^5*Tan[6*x]^3,x]
Output:
(Sin[6*x]^7/(2*(1 - Sin[6*x]^2)) - (7*(ArcTanh[Sin[6*x]] - Sin[6*x] - Sin[ 6*x]^3/3 - Sin[6*x]^5/5))/2)/6
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[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]
Time = 9.58 (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 | \(\frac {11 i {\mathrm e}^{18 i x}}{576}-\frac {29 i {\mathrm e}^{6 i x}}{96}+\frac {29 i {\mathrm e}^{-6 i x}}{96}-\frac {11 i {\mathrm e}^{-18 i x}}{576}-\frac {i \left ({\mathrm e}^{18 i x}-{\mathrm e}^{6 i x}\right )}{6 \left ({\mathrm e}^{12 i x}+1\right )^{2}}-\frac {7 \ln \left ({\mathrm e}^{6 i x}+i\right )}{12}+\frac {7 \ln \left ({\mathrm e}^{6 i x}-i\right )}{12}+\frac {\sin \left (30 x \right )}{480}\) | \(87\) |
Input:
int(sin(6*x)^5*tan(6*x)^3,x,method=_RETURNVERBOSE)
Output:
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))
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.35 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=-\frac {105 \, \cos \left (6 \, x\right )^{2} \log \left (\sin \left (6 \, x\right ) + 1\right ) - 105 \, \cos \left (6 \, x\right )^{2} \log \left (-\sin \left (6 \, x\right ) + 1\right ) - 2 \, {\left (6 \, \cos \left (6 \, x\right )^{6} - 32 \, \cos \left (6 \, x\right )^{4} + 116 \, \cos \left (6 \, x\right )^{2} + 15\right )} \sin \left (6 \, x\right )}{360 \, \cos \left (6 \, x\right )^{2}} \] Input:
integrate(sin(6*x)^5*tan(6*x)^3,x, algorithm="fricas")
Output:
-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
Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.13 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=\frac {7 \log {\left (\sin {\left (6 x \right )} - 1 \right )}}{24} - \frac {7 \log {\left (\sin {\left (6 x \right )} + 1 \right )}}{24} + \frac {\sin ^{5}{\left (6 x \right )}}{30} + \frac {\sin ^{3}{\left (6 x \right )}}{9} + \frac {\sin {\left (6 x \right )}}{2} - \frac {\sin {\left (6 x \right )}}{6 \cdot \left (2 \sin ^{2}{\left (6 x \right )} - 2\right )} \] Input:
integrate(sin(6*x)**5*tan(6*x)**3,x)
Output:
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))
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.06 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=\frac {1}{30} \, \sin \left (6 \, x\right )^{5} + \frac {1}{9} \, \sin \left (6 \, x\right )^{3} - \frac {\sin \left (6 \, x\right )}{12 \, {\left (\sin \left (6 \, x\right )^{2} - 1\right )}} - \frac {7}{24} \, \log \left (\sin \left (6 \, x\right ) + 1\right ) + \frac {7}{24} \, \log \left (\sin \left (6 \, x\right ) - 1\right ) + \frac {1}{2} \, \sin \left (6 \, x\right ) \] Input:
integrate(sin(6*x)^5*tan(6*x)^3,x, algorithm="maxima")
Output:
1/30*sin(6*x)^5 + 1/9*sin(6*x)^3 - 1/12*sin(6*x)/(sin(6*x)^2 - 1) - 7/24*l og(sin(6*x) + 1) + 7/24*log(sin(6*x) - 1) + 1/2*sin(6*x)
Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=\frac {1}{30} \, \sin \left (6 \, x\right )^{5} + \frac {1}{9} \, \sin \left (6 \, x\right )^{3} - \frac {\sin \left (6 \, x\right )}{12 \, {\left (\sin \left (6 \, x\right )^{2} - 1\right )}} - \frac {7}{24} \, \log \left (\sin \left (6 \, x\right ) + 1\right ) + \frac {7}{24} \, \log \left (-\sin \left (6 \, x\right ) + 1\right ) + \frac {1}{2} \, \sin \left (6 \, x\right ) \] Input:
integrate(sin(6*x)^5*tan(6*x)^3,x, algorithm="giac")
Output:
1/30*sin(6*x)^5 + 1/9*sin(6*x)^3 - 1/12*sin(6*x)/(sin(6*x)^2 - 1) - 7/24*l og(sin(6*x) + 1) + 7/24*log(-sin(6*x) + 1) + 1/2*sin(6*x)
Time = 21.98 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.57 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=\frac {7\,{\mathrm {tan}\left (3\,x\right )}^{13}+\frac {70\,{\mathrm {tan}\left (3\,x\right )}^{11}}{3}+\frac {77\,{\mathrm {tan}\left (3\,x\right )}^9}{5}-\frac {412\,{\mathrm {tan}\left (3\,x\right )}^7}{15}+\frac {77\,{\mathrm {tan}\left (3\,x\right )}^5}{5}+\frac {70\,{\mathrm {tan}\left (3\,x\right )}^3}{3}+7\,\mathrm {tan}\left (3\,x\right )}{6\,{\left ({\mathrm {tan}\left (3\,x\right )}^2-1\right )}^2\,{\left ({\mathrm {tan}\left (3\,x\right )}^2+1\right )}^5}-\frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (3\,x\right )\right )}{6} \] Input:
int(sin(6*x)^5*tan(6*x)^3,x)
Output:
(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)^11)/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
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.70 \[ \int \sin ^5(6 x) \tan ^3(6 x) \, dx=\frac {105 \,\mathrm {log}\left (\tan \left (3 x \right )-1\right ) \sin \left (6 x \right )^{2}-105 \,\mathrm {log}\left (\tan \left (3 x \right )-1\right )-105 \,\mathrm {log}\left (\tan \left (3 x \right )+1\right ) \sin \left (6 x \right )^{2}+105 \,\mathrm {log}\left (\tan \left (3 x \right )+1\right )+6 \sin \left (6 x \right )^{7}+14 \sin \left (6 x \right )^{5}+70 \sin \left (6 x \right )^{3}-105 \sin \left (6 x \right )}{180 \sin \left (6 x \right )^{2}-180} \] Input:
int(sin(6*x)^5*tan(6*x)^3,x)
Output:
(105*log(tan(3*x) - 1)*sin(6*x)**2 - 105*log(tan(3*x) - 1) - 105*log(tan(3 *x) + 1)*sin(6*x)**2 + 105*log(tan(3*x) + 1) + 6*sin(6*x)**7 + 14*sin(6*x) **5 + 70*sin(6*x)**3 - 105*sin(6*x))/(180*(sin(6*x)**2 - 1))