3.78 \(\int \sin (x) \tan (6 x) \, dx\)

Optimal. Leaf size=89 \[ -\sin (x)+\frac{\tanh ^{-1}\left (\sqrt{2} \sin (x)\right )}{3 \sqrt{2}}+\frac{1}{6} \sqrt{2-\sqrt{3}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2-\sqrt{3}}}\right )+\frac{1}{6} \sqrt{2+\sqrt{3}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2+\sqrt{3}}}\right ) \]

[Out]

ArcTanh[Sqrt[2]*Sin[x]]/(3*Sqrt[2]) + (Sqrt[2 - Sqrt[3]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[3]]])/6 + (Sqrt[2 +
Sqrt[3]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[3]]])/6 - Sin[x]

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Rubi [A]  time = 0.270688, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {12, 6742, 2073, 207, 1166} \[ -\sin (x)+\frac{\tanh ^{-1}\left (\sqrt{2} \sin (x)\right )}{3 \sqrt{2}}+\frac{1}{6} \sqrt{2-\sqrt{3}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2-\sqrt{3}}}\right )+\frac{1}{6} \sqrt{2+\sqrt{3}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2+\sqrt{3}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]*Tan[6*x],x]

[Out]

ArcTanh[Sqrt[2]*Sin[x]]/(3*Sqrt[2]) + (Sqrt[2 - Sqrt[3]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[3]]])/6 + (Sqrt[2 +
Sqrt[3]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[3]]])/6 - Sin[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 2073

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(PP /. x ->
x^2)^p*Q^q, x], x] /;  !SumQ[NonfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x^2] && PolyQ[Q, x] && ILtQ[p,
0]

Rule 207

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

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps

\begin{align*} \int \sin (x) \tan (6 x) \, dx &=\operatorname{Subst}\left (\int \frac{2 x^2 \left (3-16 x^2+16 x^4\right )}{1-18 x^2+48 x^4-32 x^6} \, dx,x,\sin (x)\right )\\ &=2 \operatorname{Subst}\left (\int \frac{x^2 \left (3-16 x^2+16 x^4\right )}{1-18 x^2+48 x^4-32 x^6} \, dx,x,\sin (x)\right )\\ &=2 \operatorname{Subst}\left (\int \left (-\frac{1}{2}+\frac{1-12 x^2+16 x^4}{2 \left (1-18 x^2+48 x^4-32 x^6\right )}\right ) \, dx,x,\sin (x)\right )\\ &=-\sin (x)+\operatorname{Subst}\left (\int \frac{1-12 x^2+16 x^4}{1-18 x^2+48 x^4-32 x^6} \, dx,x,\sin (x)\right )\\ &=-\sin (x)+\operatorname{Subst}\left (\int \left (-\frac{1}{3 \left (-1+2 x^2\right )}-\frac{2 \left (-1+8 x^2\right )}{3 \left (1-16 x^2+16 x^4\right )}\right ) \, dx,x,\sin (x)\right )\\ &=-\sin (x)-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-1+2 x^2} \, dx,x,\sin (x)\right )-\frac{2}{3} \operatorname{Subst}\left (\int \frac{-1+8 x^2}{1-16 x^2+16 x^4} \, dx,x,\sin (x)\right )\\ &=\frac{\tanh ^{-1}\left (\sqrt{2} \sin (x)\right )}{3 \sqrt{2}}-\sin (x)-\frac{1}{3} \left (4 \left (2-\sqrt{3}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-8+4 \sqrt{3}+16 x^2} \, dx,x,\sin (x)\right )-\frac{1}{3} \left (4 \left (2+\sqrt{3}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-8-4 \sqrt{3}+16 x^2} \, dx,x,\sin (x)\right )\\ &=\frac{\tanh ^{-1}\left (\sqrt{2} \sin (x)\right )}{3 \sqrt{2}}+\frac{1}{6} \sqrt{2-\sqrt{3}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2-\sqrt{3}}}\right )+\frac{1}{6} \sqrt{2+\sqrt{3}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2+\sqrt{3}}}\right )-\sin (x)\\ \end{align*}

Mathematica [A]  time = 0.133556, size = 84, normalized size = 0.94 \[ \frac{1}{6} \left (-6 \sin (x)+\sqrt{2} \tanh ^{-1}\left (\sqrt{2} \sin (x)\right )+\sqrt{2-\sqrt{3}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2-\sqrt{3}}}\right )+\sqrt{2+\sqrt{3}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2+\sqrt{3}}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]*Tan[6*x],x]

[Out]

(Sqrt[2]*ArcTanh[Sqrt[2]*Sin[x]] + Sqrt[2 - Sqrt[3]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[3]]] + Sqrt[2 + Sqrt[3]]
*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[3]]] - 6*Sin[x])/6

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Maple [B]  time = 0.281, size = 256, normalized size = 2.9 \begin{align*}{\frac{ \left ( -3+2\,\sqrt{3} \right ) \sqrt{3}}{6\,\sqrt{6}-6\,\sqrt{2}}{\it Artanh} \left ( 8\,{\frac{\sin \left ( x \right ) }{2\,\sqrt{6}-2\,\sqrt{2}}} \right ) }+{\frac{ \left ( 3+2\,\sqrt{3} \right ) \sqrt{3}}{6\,\sqrt{6}+6\,\sqrt{2}}{\it Artanh} \left ( 8\,{\frac{\sin \left ( x \right ) }{2\,\sqrt{6}+2\,\sqrt{2}}} \right ) }+{\frac{{\it Artanh} \left ( \sin \left ( x \right ) \sqrt{2} \right ) \sqrt{2}}{6}}-{\frac{4}{6\,\sqrt{6}-6\,\sqrt{2}}{\it Artanh} \left ( 8\,{\frac{\sin \left ( x \right ) }{2\,\sqrt{6}-2\,\sqrt{2}}} \right ) }-{\frac{4}{6\,\sqrt{6}+6\,\sqrt{2}}{\it Artanh} \left ( 8\,{\frac{\sin \left ( x \right ) }{2\,\sqrt{6}+2\,\sqrt{2}}} \right ) }-\sin \left ( x \right ) +{\frac{ \left ( 3+2\,\sqrt{3} \right ) \sqrt{3}}{18\,\sqrt{6}-18\,\sqrt{2}}{\it Artanh} \left ( 8\,{\frac{\sin \left ( x \right ) }{2\,\sqrt{6}-2\,\sqrt{2}}} \right ) }+{\frac{ \left ( -3+2\,\sqrt{3} \right ) \sqrt{3}}{18\,\sqrt{6}+18\,\sqrt{2}}{\it Artanh} \left ( 8\,{\frac{\sin \left ( x \right ) }{2\,\sqrt{6}+2\,\sqrt{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)*tan(6*x),x)

[Out]

1/3*(-3+2*3^(1/2))*3^(1/2)/(2*6^(1/2)-2*2^(1/2))*arctanh(8*sin(x)/(2*6^(1/2)-2*2^(1/2)))+1/3*(3+2*3^(1/2))*3^(
1/2)/(2*6^(1/2)+2*2^(1/2))*arctanh(8*sin(x)/(2*6^(1/2)+2*2^(1/2)))+1/6*arctanh(sin(x)*2^(1/2))*2^(1/2)-4/3/(2*
6^(1/2)-2*2^(1/2))*arctanh(8*sin(x)/(2*6^(1/2)-2*2^(1/2)))-4/3/(2*6^(1/2)+2*2^(1/2))*arctanh(8*sin(x)/(2*6^(1/
2)+2*2^(1/2)))-sin(x)+1/9*(3+2*3^(1/2))*3^(1/2)/(2*6^(1/2)-2*2^(1/2))*arctanh(8*sin(x)/(2*6^(1/2)-2*2^(1/2)))+
1/9*(-3+2*3^(1/2))*3^(1/2)/(2*6^(1/2)+2*2^(1/2))*arctanh(8*sin(x)/(2*6^(1/2)+2*2^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{24} \, \sqrt{2} \log \left (2 \, \cos \left (x\right )^{2} + 2 \, \sin \left (x\right )^{2} + 2 \, \sqrt{2} \cos \left (x\right ) + 2 \, \sqrt{2} \sin \left (x\right ) + 2\right ) - \frac{1}{24} \, \sqrt{2} \log \left (2 \, \cos \left (x\right )^{2} + 2 \, \sin \left (x\right )^{2} + 2 \, \sqrt{2} \cos \left (x\right ) - 2 \, \sqrt{2} \sin \left (x\right ) + 2\right ) + \frac{1}{24} \, \sqrt{2} \log \left (2 \, \cos \left (x\right )^{2} + 2 \, \sin \left (x\right )^{2} - 2 \, \sqrt{2} \cos \left (x\right ) + 2 \, \sqrt{2} \sin \left (x\right ) + 2\right ) - \frac{1}{24} \, \sqrt{2} \log \left (2 \, \cos \left (x\right )^{2} + 2 \, \sin \left (x\right )^{2} - 2 \, \sqrt{2} \cos \left (x\right ) - 2 \, \sqrt{2} \sin \left (x\right ) + 2\right ) + \int -\frac{{\left (2 \, \cos \left (7 \, x\right ) - \cos \left (5 \, x\right ) - \cos \left (3 \, x\right ) + 2 \, \cos \left (x\right )\right )} \cos \left (8 \, x\right ) - 2 \,{\left (\cos \left (4 \, x\right ) - 1\right )} \cos \left (7 \, x\right ) +{\left (\cos \left (4 \, x\right ) - 1\right )} \cos \left (5 \, x\right ) +{\left (\cos \left (3 \, x\right ) - 2 \, \cos \left (x\right )\right )} \cos \left (4 \, x\right ) +{\left (2 \, \sin \left (7 \, x\right ) - \sin \left (5 \, x\right ) - \sin \left (3 \, x\right ) + 2 \, \sin \left (x\right )\right )} \sin \left (8 \, x\right ) +{\left (\sin \left (3 \, x\right ) - 2 \, \sin \left (x\right )\right )} \sin \left (4 \, x\right ) - 2 \, \sin \left (7 \, x\right ) \sin \left (4 \, x\right ) + \sin \left (5 \, x\right ) \sin \left (4 \, x\right ) - \cos \left (3 \, x\right ) + 2 \, \cos \left (x\right )}{3 \,{\left (2 \,{\left (\cos \left (4 \, x\right ) - 1\right )} \cos \left (8 \, x\right ) - \cos \left (8 \, x\right )^{2} - \cos \left (4 \, x\right )^{2} - \sin \left (8 \, x\right )^{2} + 2 \, \sin \left (8 \, x\right ) \sin \left (4 \, x\right ) - \sin \left (4 \, x\right )^{2} + 2 \, \cos \left (4 \, x\right ) - 1\right )}}\,{d x} - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(6*x),x, algorithm="maxima")

[Out]

1/24*sqrt(2)*log(2*cos(x)^2 + 2*sin(x)^2 + 2*sqrt(2)*cos(x) + 2*sqrt(2)*sin(x) + 2) - 1/24*sqrt(2)*log(2*cos(x
)^2 + 2*sin(x)^2 + 2*sqrt(2)*cos(x) - 2*sqrt(2)*sin(x) + 2) + 1/24*sqrt(2)*log(2*cos(x)^2 + 2*sin(x)^2 - 2*sqr
t(2)*cos(x) + 2*sqrt(2)*sin(x) + 2) - 1/24*sqrt(2)*log(2*cos(x)^2 + 2*sin(x)^2 - 2*sqrt(2)*cos(x) - 2*sqrt(2)*
sin(x) + 2) + integrate(-1/3*((2*cos(7*x) - cos(5*x) - cos(3*x) + 2*cos(x))*cos(8*x) - 2*(cos(4*x) - 1)*cos(7*
x) + (cos(4*x) - 1)*cos(5*x) + (cos(3*x) - 2*cos(x))*cos(4*x) + (2*sin(7*x) - sin(5*x) - sin(3*x) + 2*sin(x))*
sin(8*x) + (sin(3*x) - 2*sin(x))*sin(4*x) - 2*sin(7*x)*sin(4*x) + sin(5*x)*sin(4*x) - cos(3*x) + 2*cos(x))/(2*
(cos(4*x) - 1)*cos(8*x) - cos(8*x)^2 - cos(4*x)^2 - sin(8*x)^2 + 2*sin(8*x)*sin(4*x) - sin(4*x)^2 + 2*cos(4*x)
- 1), x) - sin(x)

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Fricas [B]  time = 2.78035, size = 435, normalized size = 4.89 \begin{align*} \frac{1}{12} \, \sqrt{\sqrt{3} + 2} \log \left (\sqrt{\sqrt{3} + 2} + 2 \, \sin \left (x\right )\right ) - \frac{1}{12} \, \sqrt{\sqrt{3} + 2} \log \left (\sqrt{\sqrt{3} + 2} - 2 \, \sin \left (x\right )\right ) + \frac{1}{12} \, \sqrt{-\sqrt{3} + 2} \log \left (\sqrt{-\sqrt{3} + 2} + 2 \, \sin \left (x\right )\right ) - \frac{1}{12} \, \sqrt{-\sqrt{3} + 2} \log \left (\sqrt{-\sqrt{3} + 2} - 2 \, \sin \left (x\right )\right ) + \frac{1}{12} \, \sqrt{2} \log \left (-\frac{2 \, \cos \left (x\right )^{2} - 2 \, \sqrt{2} \sin \left (x\right ) - 3}{2 \, \cos \left (x\right )^{2} - 1}\right ) - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(6*x),x, algorithm="fricas")

[Out]

1/12*sqrt(sqrt(3) + 2)*log(sqrt(sqrt(3) + 2) + 2*sin(x)) - 1/12*sqrt(sqrt(3) + 2)*log(sqrt(sqrt(3) + 2) - 2*si
n(x)) + 1/12*sqrt(-sqrt(3) + 2)*log(sqrt(-sqrt(3) + 2) + 2*sin(x)) - 1/12*sqrt(-sqrt(3) + 2)*log(sqrt(-sqrt(3)
+ 2) - 2*sin(x)) + 1/12*sqrt(2)*log(-(2*cos(x)^2 - 2*sqrt(2)*sin(x) - 3)/(2*cos(x)^2 - 1)) - sin(x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(6*x),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (x\right ) \tan \left (6 \, x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(x)*tan(6*x), x)