\(\int \sin ^3(5 x) \tan ^3(5 x) \, dx\) [890]

Optimal result

Integrand size = 13, antiderivative size = 44 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {1}{2} \text {arctanh}(\sin (5 x))+\frac {1}{2} \sin (5 x)+\frac {1}{6} \sin ^3(5 x)+\frac {1}{10} \sin ^3(5 x) \tan ^2(5 x) \]

[Out]

-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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 44, 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} \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {1}{2} \text {arctanh}(\sin (5 x))+\frac {1}{6} \sin ^3(5 x)+\frac {1}{2} \sin (5 x)+\frac {1}{10} \sin ^3(5 x) \tan ^2(5 x) \]

[In]

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

[Out]

-1/2*ArcTanh[Sin[5*x]] + Sin[5*x]/2 + Sin[5*x]^3/6 + (Sin[5*x]^3*Tan[5*x]^2)/10

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}{5} \text {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^2} \, dx,x,\sin (5 x)\right ) \\ & = \frac {1}{10} \sin ^3(5 x) \tan ^2(5 x)-\frac {1}{2} \text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (5 x)\right ) \\ & = \frac {1}{10} \sin ^3(5 x) \tan ^2(5 x)-\frac {1}{2} \text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (5 x)\right ) \\ & = \frac {1}{2} \sin (5 x)+\frac {1}{6} \sin ^3(5 x)+\frac {1}{10} \sin ^3(5 x) \tan ^2(5 x)-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (5 x)\right ) \\ & = -\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) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.18 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {1}{2} \text {arctanh}(\sin (5 x))+\frac {1}{2} \sec (5 x) \tan (5 x)-\frac {1}{3} \sin (5 x) \tan ^2(5 x)-\frac {1}{15} \sin ^3(5 x) \tan ^2(5 x) \]

[In]

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

[Out]

-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

Maple [A] (verified)

Time = 3.52 (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	\(\frac {i {\mathrm e}^{15 i x}}{120}-\frac {9 i {\mathrm e}^{5 i x}}{40}+\frac {9 i {\mathrm e}^{-5 i x}}{40}-\frac {i {\mathrm e}^{-15 i x}}{120}-\frac {i \left ({\mathrm e}^{15 i x}-{\mathrm e}^{5 i x}\right )}{5 \left ({\mathrm e}^{10 i x}+1\right )^{2}}-\frac {\ln \left (i+{\mathrm e}^{5 i x}\right )}{2}+\frac {\ln \left ({\mathrm e}^{5 i x}-i\right )}{2}\)	\(81\)

[In]

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

[Out]

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))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.48 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {15 \, \cos \left (5 \, x\right )^{2} \log \left (\sin \left (5 \, x\right ) + 1\right ) - 15 \, \cos \left (5 \, x\right )^{2} \log \left (-\sin \left (5 \, x\right ) + 1\right ) + 2 \, {\left (2 \, \cos \left (5 \, x\right )^{4} - 14 \, \cos \left (5 \, x\right )^{2} - 3\right )} \sin \left (5 \, x\right )}{60 \, \cos \left (5 \, x\right )^{2}} \]

[In]

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

[Out]

-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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {\log {\left (\sin {\left (5 x \right )} - 1 \right )}}{4} - \frac {\log {\left (\sin {\left (5 x \right )} + 1 \right )}}{4} + \frac {\sin ^{3}{\left (5 x \right )}}{15} + \frac {2 \sin {\left (5 x \right )}}{5} - \frac {\sin {\left (5 x \right )}}{5 \cdot \left (2 \sin ^{2}{\left (5 x \right )} - 2\right )} \]

[In]

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

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {1}{15} \, \sin \left (5 \, x\right )^{3} - \frac {\sin \left (5 \, x\right )}{10 \, {\left (\sin \left (5 \, x\right )^{2} - 1\right )}} - \frac {1}{4} \, \log \left (\sin \left (5 \, x\right ) + 1\right ) + \frac {1}{4} \, \log \left (\sin \left (5 \, x\right ) - 1\right ) + \frac {2}{5} \, \sin \left (5 \, x\right ) \]

[In]

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

[Out]

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
)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {1}{15} \, \sin \left (5 \, x\right )^{3} - \frac {\sin \left (5 \, x\right )}{10 \, {\left (\sin \left (5 \, x\right )^{2} - 1\right )}} - \frac {1}{4} \, \log \left (\sin \left (5 \, x\right ) + 1\right ) + \frac {1}{4} \, \log \left (-\sin \left (5 \, x\right ) + 1\right ) + \frac {2}{5} \, \sin \left (5 \, x\right ) \]

[In]

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

[Out]

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)

Mupad [B] (verification not implemented)

Time = 26.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.57 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {5\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^9+\frac {20\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^7}{3}-\frac {22\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^5}{3}+\frac {20\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^3}{3}+5\,\mathrm {tan}\left (\frac {5\,x}{2}\right )}{5\,{\left ({\mathrm {tan}\left (\frac {5\,x}{2}\right )}^2-1\right )}^2\,{\left ({\mathrm {tan}\left (\frac {5\,x}{2}\right )}^2+1\right )}^3}-\mathrm {atanh}\left (\mathrm {tan}\left (\frac {5\,x}{2}\right )\right ) \]

[In]

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

[Out]

(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))