3.892 \(\int \sin ^5(6 x) \tan ^3(6 x) \, dx\)

Optimal. Leaf size=54 \[ \frac{7}{60} \sin ^5(6 x)+\frac{7}{36} \sin ^3(6 x)+\frac{7}{12} \sin (6 x)+\frac{1}{12} \sin ^5(6 x) \tan ^2(6 x)-\frac{7}{12} \tanh ^{-1}(\sin (6 x)) \]

[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

________________________________________________________________________________________

Rubi [A]  time = 0.0411704, antiderivative size = 54, 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} \[ \frac{7}{60} \sin ^5(6 x)+\frac{7}{36} \sin ^3(6 x)+\frac{7}{12} \sin (6 x)+\frac{1}{12} \sin ^5(6 x) \tan ^2(6 x)-\frac{7}{12} \tanh ^{-1}(\sin (6 x)) \]

Antiderivative was successfully verified.

[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 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 ^5(6 x) \tan ^3(6 x) \, dx &=\frac{1}{6} \operatorname{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} \operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (6 x)\right )\\ &=-\frac{7}{12} \tanh ^{-1}(\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]  time = 0.0917743, size = 68, normalized size = 1.26 \[ -\frac{1}{30} \sin ^5(6 x) \tan ^2(6 x)-\frac{7}{90} \sin ^3(6 x) \tan ^2(6 x)-\frac{7}{18} \sin (6 x) \tan ^2(6 x)-\frac{7}{12} \tanh ^{-1}(\sin (6 x))+\frac{7}{12} \tan (6 x) \sec (6 x) \]

Antiderivative was successfully verified.

[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]  time = 0.038, size = 58, normalized size = 1.1 \begin{align*}{\frac{ \left ( \sin \left ( 6\,x \right ) \right ) ^{9}}{12\, \left ( \cos \left ( 6\,x \right ) \right ) ^{2}}}+{\frac{ \left ( \sin \left ( 6\,x \right ) \right ) ^{7}}{12}}+{\frac{7\, \left ( \sin \left ( 6\,x \right ) \right ) ^{5}}{60}}+{\frac{7\, \left ( \sin \left ( 6\,x \right ) \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

________________________________________________________________________________________

Maxima [A]  time = 0.941967, size = 77, normalized size = 1.43 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [A]  time = 2.55227, size = 211, normalized size = 3.91 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0.1208, size = 61, normalized size = 1.13 \begin{align*} \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 \left (2 \sin ^{2}{\left (6 x \right )} - 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

Giac [B]  time = 2.93253, size = 890, normalized size = 16.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/360*(105*log(2*(tan(3*x)^2 + 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^14 - 105*log(2*(tan(3*x)^2 - 2*tan(
3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^14 + 315*log(2*(tan(3*x)^2 + 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^1
2 - 315*log(2*(tan(3*x)^2 - 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^12 - 420*tan(3*x)^13 + 105*log(2*(tan(3
*x)^2 + 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^10 - 105*log(2*(tan(3*x)^2 - 2*tan(3*x) + 1)/(tan(3*x)^2 +
1))*tan(3*x)^10 - 1400*tan(3*x)^11 - 525*log(2*(tan(3*x)^2 + 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^8 + 52
5*log(2*(tan(3*x)^2 - 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^8 - 924*tan(3*x)^9 - 525*log(2*(tan(3*x)^2 +
2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^6 + 525*log(2*(tan(3*x)^2 - 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3
*x)^6 + 1648*tan(3*x)^7 + 105*log(2*(tan(3*x)^2 + 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^4 - 105*log(2*(ta
n(3*x)^2 - 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^4 - 924*tan(3*x)^5 + 315*log(2*(tan(3*x)^2 + 2*tan(3*x)
+ 1)/(tan(3*x)^2 + 1))*tan(3*x)^2 - 315*log(2*(tan(3*x)^2 - 2*tan(3*x) + 1)/(tan(3*x)^2 + 1))*tan(3*x)^2 - 140
0*tan(3*x)^3 + 105*log(2*(tan(3*x)^2 + 2*tan(3*x) + 1)/(tan(3*x)^2 + 1)) - 105*log(2*(tan(3*x)^2 - 2*tan(3*x)
+ 1)/(tan(3*x)^2 + 1)) - 420*tan(3*x))/(tan(3*x)^14 + 3*tan(3*x)^12 + tan(3*x)^10 - 5*tan(3*x)^8 - 5*tan(3*x)^
6 + tan(3*x)^4 + 3*tan(3*x)^2 + 1)