Integrand size = 13, antiderivative size = 44 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {1}{2} \text {arctanh}(\sin (5 x))+\frac {1}{2} \sin (5 x)+\frac {1}{6} \sin ^3(5 x)+\frac {1}{10} \sin ^3(5 x) \tan ^2(5 x) \] Output:
-1/2*arctanh(sin(5*x))+1/2*sin(5*x)+1/6*sin(5*x)^3+1/10*sin(5*x)^3*tan(5*x )^2
Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.18 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {1}{2} \text {arctanh}(\sin (5 x))+\frac {1}{2} \sec (5 x) \tan (5 x)-\frac {1}{3} \sin (5 x) \tan ^2(5 x)-\frac {1}{15} \sin ^3(5 x) \tan ^2(5 x) \] Input:
Integrate[Sin[5*x]^3*Tan[5*x]^3,x]
Output:
-1/2*ArcTanh[Sin[5*x]] + (Sec[5*x]*Tan[5*x])/2 - (Sin[5*x]*Tan[5*x]^2)/3 - (Sin[5*x]^3*Tan[5*x]^2)/15
Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20, 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.
\(\displaystyle \int \sin ^3(5 x) \tan ^3(5 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (5 x)^3 \tan (5 x)^3dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle \frac {1}{5} \int \frac {\sin ^6(5 x)}{\left (1-\sin ^2(5 x)\right )^2}d\sin (5 x)\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{5} \left (\frac {\sin ^5(5 x)}{2 \left (1-\sin ^2(5 x)\right )}-\frac {5}{2} \int \frac {\sin ^4(5 x)}{1-\sin ^2(5 x)}d\sin (5 x)\right )\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {1}{5} \left (\frac {\sin ^5(5 x)}{2 \left (1-\sin ^2(5 x)\right )}-\frac {5}{2} \int \left (-\sin ^2(5 x)+\frac {1}{1-\sin ^2(5 x)}-1\right )d\sin (5 x)\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\frac {\sin ^5(5 x)}{2 \left (1-\sin ^2(5 x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sin (5 x))-\frac {1}{3} \sin ^3(5 x)-\sin (5 x)\right )\right )\) |
Input:
Int[Sin[5*x]^3*Tan[5*x]^3,x]
Output:
(Sin[5*x]^5/(2*(1 - Sin[5*x]^2)) - (5*(ArcTanh[Sin[5*x]] - Sin[5*x] - Sin[ 5*x]^3/3))/2)/5
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 = 0.67 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {\sin \left (5 x \right )^{7}}{10 \cos \left (5 x \right )^{2}}+\frac {\sin \left (5 x \right )^{5}}{10}+\frac {\sin \left (5 x \right )^{3}}{6}+\frac {\sin \left (5 x \right )}{2}-\frac {\ln \left (\sec \left (5 x \right )+\tan \left (5 x \right )\right )}{2}\) | \(50\) |
default | \(\frac {\sin \left (5 x \right )^{7}}{10 \cos \left (5 x \right )^{2}}+\frac {\sin \left (5 x \right )^{5}}{10}+\frac {\sin \left (5 x \right )^{3}}{6}+\frac {\sin \left (5 x \right )}{2}-\frac {\ln \left (\sec \left (5 x \right )+\tan \left (5 x \right )\right )}{2}\) | \(50\) |
risch | \(\frac {i {\mathrm e}^{15 i x}}{120}-\frac {9 i {\mathrm e}^{5 i x}}{40}+\frac {9 i {\mathrm e}^{-5 i x}}{40}-\frac {i {\mathrm e}^{-15 i x}}{120}-\frac {i \left ({\mathrm e}^{15 i x}-{\mathrm e}^{5 i x}\right )}{5 \left ({\mathrm e}^{10 i x}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{5 i x}+i\right )}{2}+\frac {\ln \left ({\mathrm e}^{5 i x}-i\right )}{2}\) | \(81\) |
Input:
int(sin(5*x)^3*tan(5*x)^3,x,method=_RETURNVERBOSE)
Output:
1/10*sin(5*x)^7/cos(5*x)^2+1/10*sin(5*x)^5+1/6*sin(5*x)^3+1/2*sin(5*x)-1/2 *ln(sec(5*x)+tan(5*x))
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.48 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {15 \, \cos \left (5 \, x\right )^{2} \log \left (\sin \left (5 \, x\right ) + 1\right ) - 15 \, \cos \left (5 \, x\right )^{2} \log \left (-\sin \left (5 \, x\right ) + 1\right ) + 2 \, {\left (2 \, \cos \left (5 \, x\right )^{4} - 14 \, \cos \left (5 \, x\right )^{2} - 3\right )} \sin \left (5 \, x\right )}{60 \, \cos \left (5 \, x\right )^{2}} \] Input:
integrate(sin(5*x)^3*tan(5*x)^3,x, algorithm="fricas")
Output:
-1/60*(15*cos(5*x)^2*log(sin(5*x) + 1) - 15*cos(5*x)^2*log(-sin(5*x) + 1) + 2*(2*cos(5*x)^4 - 14*cos(5*x)^2 - 3)*sin(5*x))/cos(5*x)^2
Time = 0.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {\log {\left (\sin {\left (5 x \right )} - 1 \right )}}{4} - \frac {\log {\left (\sin {\left (5 x \right )} + 1 \right )}}{4} + \frac {\sin ^{3}{\left (5 x \right )}}{15} + \frac {2 \sin {\left (5 x \right )}}{5} - \frac {\sin {\left (5 x \right )}}{5 \cdot \left (2 \sin ^{2}{\left (5 x \right )} - 2\right )} \] Input:
integrate(sin(5*x)**3*tan(5*x)**3,x)
Output:
log(sin(5*x) - 1)/4 - log(sin(5*x) + 1)/4 + sin(5*x)**3/15 + 2*sin(5*x)/5 - sin(5*x)/(5*(2*sin(5*x)**2 - 2))
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {1}{15} \, \sin \left (5 \, x\right )^{3} - \frac {\sin \left (5 \, x\right )}{10 \, {\left (\sin \left (5 \, x\right )^{2} - 1\right )}} - \frac {1}{4} \, \log \left (\sin \left (5 \, x\right ) + 1\right ) + \frac {1}{4} \, \log \left (\sin \left (5 \, x\right ) - 1\right ) + \frac {2}{5} \, \sin \left (5 \, x\right ) \] Input:
integrate(sin(5*x)^3*tan(5*x)^3,x, algorithm="maxima")
Output:
1/15*sin(5*x)^3 - 1/10*sin(5*x)/(sin(5*x)^2 - 1) - 1/4*log(sin(5*x) + 1) + 1/4*log(sin(5*x) - 1) + 2/5*sin(5*x)
Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {1}{15} \, \sin \left (5 \, x\right )^{3} - \frac {\sin \left (5 \, x\right )}{10 \, {\left (\sin \left (5 \, x\right )^{2} - 1\right )}} - \frac {1}{4} \, \log \left (\sin \left (5 \, x\right ) + 1\right ) + \frac {1}{4} \, \log \left (-\sin \left (5 \, x\right ) + 1\right ) + \frac {2}{5} \, \sin \left (5 \, x\right ) \] Input:
integrate(sin(5*x)^3*tan(5*x)^3,x, algorithm="giac")
Output:
1/15*sin(5*x)^3 - 1/10*sin(5*x)/(sin(5*x)^2 - 1) - 1/4*log(sin(5*x) + 1) + 1/4*log(-sin(5*x) + 1) + 2/5*sin(5*x)
Time = 17.45 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.57 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {5\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^9+\frac {20\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^7}{3}-\frac {22\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^5}{3}+\frac {20\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^3}{3}+5\,\mathrm {tan}\left (\frac {5\,x}{2}\right )}{5\,{\left ({\mathrm {tan}\left (\frac {5\,x}{2}\right )}^2-1\right )}^2\,{\left ({\mathrm {tan}\left (\frac {5\,x}{2}\right )}^2+1\right )}^3}-\mathrm {atanh}\left (\mathrm {tan}\left (\frac {5\,x}{2}\right )\right ) \] Input:
int(sin(5*x)^3*tan(5*x)^3,x)
Output:
(5*tan((5*x)/2) + (20*tan((5*x)/2)^3)/3 - (22*tan((5*x)/2)^5)/3 + (20*tan( (5*x)/2)^7)/3 + 5*tan((5*x)/2)^9)/(5*(tan((5*x)/2)^2 - 1)^2*(tan((5*x)/2)^ 2 + 1)^3) - atanh(tan((5*x)/2))
Time = 0.17 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.91 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {15 \,\mathrm {log}\left (\tan \left (\frac {5 x}{2}\right )-1\right ) \sin \left (5 x \right )^{2}-15 \,\mathrm {log}\left (\tan \left (\frac {5 x}{2}\right )-1\right )-15 \,\mathrm {log}\left (\tan \left (\frac {5 x}{2}\right )+1\right ) \sin \left (5 x \right )^{2}+15 \,\mathrm {log}\left (\tan \left (\frac {5 x}{2}\right )+1\right )+2 \sin \left (5 x \right )^{5}+10 \sin \left (5 x \right )^{3}-15 \sin \left (5 x \right )}{30 \sin \left (5 x \right )^{2}-30} \] Input:
int(sin(5*x)^3*tan(5*x)^3,x)
Output:
(15*log(tan((5*x)/2) - 1)*sin(5*x)**2 - 15*log(tan((5*x)/2) - 1) - 15*log( tan((5*x)/2) + 1)*sin(5*x)**2 + 15*log(tan((5*x)/2) + 1) + 2*sin(5*x)**5 + 10*sin(5*x)**3 - 15*sin(5*x))/(30*(sin(5*x)**2 - 1))