3.890 \(\int \sin ^3(5 x) \tan ^3(5 x) \, dx\)

Optimal. Leaf size=44 \[ \frac{1}{6} \sin ^3(5 x)+\frac{1}{2} \sin (5 x)+\frac{1}{10} \sin ^3(5 x) \tan ^2(5 x)-\frac{1}{2} \tanh ^{-1}(\sin (5 x)) \]

[Out]

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

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Rubi [A]  time = 0.0380145, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2592, 288, 302, 206} \[ \frac{1}{6} \sin ^3(5 x)+\frac{1}{2} \sin (5 x)+\frac{1}{10} \sin ^3(5 x) \tan ^2(5 x)-\frac{1}{2} \tanh ^{-1}(\sin (5 x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2592

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]

Rule 288

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 302

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 206

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

Rubi steps

\begin{align*} \int \sin ^3(5 x) \tan ^3(5 x) \, dx &=\frac{1}{5} \operatorname{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} \operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (5 x)\right )\\ &=-\frac{1}{2} \tanh ^{-1}(\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]  time = 0.0448652, size = 52, normalized size = 1.18 \[ -\frac{1}{15} \sin ^3(5 x) \tan ^2(5 x)-\frac{1}{3} \sin (5 x) \tan ^2(5 x)-\frac{1}{2} \tanh ^{-1}(\sin (5 x))+\frac{1}{2} \tan (5 x) \sec (5 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Sin[5*x]]/2 + (Sec[5*x]*Tan[5*x])/2 - (Sin[5*x]*Tan[5*x]^2)/3 - (Sin[5*x]^3*Tan[5*x]^2)/15

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Maple [A]  time = 0.025, size = 50, normalized size = 1.1 \begin{align*}{\frac{ \left ( \sin \left ( 5\,x \right ) \right ) ^{7}}{10\, \left ( \cos \left ( 5\,x \right ) \right ) ^{2}}}+{\frac{ \left ( \sin \left ( 5\,x \right ) \right ) ^{5}}{10}}+{\frac{ \left ( \sin \left ( 5\,x \right ) \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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Maxima [A]  time = 0.943373, size = 66, normalized size = 1.5 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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
)

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Fricas [A]  time = 2.51175, size = 182, normalized size = 4.14 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.115596, size = 51, normalized size = 1.16 \begin{align*} \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 \left (2 \sin ^{2}{\left (5 x \right )} - 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [B]  time = 3.06945, size = 657, normalized size = 14.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*log(2*(tan(5/2*x)^2 + 2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1))*tan(5/2*x)^10 - 15*log(2*(tan(5/2*x)^2 -
2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1))*tan(5/2*x)^10 + 15*log(2*(tan(5/2*x)^2 + 2*tan(5/2*x) + 1)/(tan(5/2*x)^
2 + 1))*tan(5/2*x)^8 - 15*log(2*(tan(5/2*x)^2 - 2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1))*tan(5/2*x)^8 - 60*tan(5/
2*x)^9 - 30*log(2*(tan(5/2*x)^2 + 2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1))*tan(5/2*x)^6 + 30*log(2*(tan(5/2*x)^2
- 2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1))*tan(5/2*x)^6 - 80*tan(5/2*x)^7 - 30*log(2*(tan(5/2*x)^2 + 2*tan(5/2*x)
+ 1)/(tan(5/2*x)^2 + 1))*tan(5/2*x)^4 + 30*log(2*(tan(5/2*x)^2 - 2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1))*tan(5/
2*x)^4 + 88*tan(5/2*x)^5 + 15*log(2*(tan(5/2*x)^2 + 2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1))*tan(5/2*x)^2 - 15*lo
g(2*(tan(5/2*x)^2 - 2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1))*tan(5/2*x)^2 - 80*tan(5/2*x)^3 + 15*log(2*(tan(5/2*x
)^2 + 2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1)) - 15*log(2*(tan(5/2*x)^2 - 2*tan(5/2*x) + 1)/(tan(5/2*x)^2 + 1)) -
60*tan(5/2*x))/(tan(5/2*x)^10 + tan(5/2*x)^8 - 2*tan(5/2*x)^6 - 2*tan(5/2*x)^4 + tan(5/2*x)^2 + 1)