3.89 \(\int \sec (6 x) \sin (x) \, dx\)

Optimal. Leaf size=85 \[ -\frac{\tanh ^{-1}\left (\sqrt{2} \cos (x)\right )}{3 \sqrt{2}}+\frac{\tanh ^{-1}\left (\frac{2 \cos (x)}{\sqrt{2-\sqrt{3}}}\right )}{6 \sqrt{2-\sqrt{3}}}+\frac{\tanh ^{-1}\left (\frac{2 \cos (x)}{\sqrt{2+\sqrt{3}}}\right )}{6 \sqrt{2+\sqrt{3}}} \]

[Out]

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

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Rubi [A]  time = 0.0631543, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {4357, 2057, 207, 1166} \[ -\frac{\tanh ^{-1}\left (\sqrt{2} \cos (x)\right )}{3 \sqrt{2}}+\frac{\tanh ^{-1}\left (\frac{2 \cos (x)}{\sqrt{2-\sqrt{3}}}\right )}{6 \sqrt{2-\sqrt{3}}}+\frac{\tanh ^{-1}\left (\frac{2 \cos (x)}{\sqrt{2+\sqrt{3}}}\right )}{6 \sqrt{2+\sqrt{3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4357

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[d/(
b*c), Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a
+ b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])

Rule 2057

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(u /. x -> x^2)^p, x], x
] /;  !SumQ[NonfreeFactors[u, x]]] /; PolyQ[P, x^2] && 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 \sec (6 x) \sin (x) \, dx &=-\operatorname{Subst}\left (\int \frac{1}{-1+18 x^2-48 x^4+32 x^6} \, dx,x,\cos (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-\frac{1}{3 \left (-1+2 x^2\right )}+\frac{4 \left (-1+2 x^2\right )}{3 \left (1-16 x^2+16 x^4\right )}\right ) \, dx,x,\cos (x)\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-1+2 x^2} \, dx,x,\cos (x)\right )-\frac{4}{3} \operatorname{Subst}\left (\int \frac{-1+2 x^2}{1-16 x^2+16 x^4} \, dx,x,\cos (x)\right )\\ &=-\frac{\tanh ^{-1}\left (\sqrt{2} \cos (x)\right )}{3 \sqrt{2}}-\frac{4}{3} \operatorname{Subst}\left (\int \frac{1}{-8-4 \sqrt{3}+16 x^2} \, dx,x,\cos (x)\right )-\frac{4}{3} \operatorname{Subst}\left (\int \frac{1}{-8+4 \sqrt{3}+16 x^2} \, dx,x,\cos (x)\right )\\ &=-\frac{\tanh ^{-1}\left (\sqrt{2} \cos (x)\right )}{3 \sqrt{2}}+\frac{\tanh ^{-1}\left (\frac{2 \cos (x)}{\sqrt{2-\sqrt{3}}}\right )}{6 \sqrt{2-\sqrt{3}}}+\frac{\tanh ^{-1}\left (\frac{2 \cos (x)}{\sqrt{2+\sqrt{3}}}\right )}{6 \sqrt{2+\sqrt{3}}}\\ \end{align*}

Mathematica [C]  time = 9.28038, size = 678, normalized size = 7.98 \[ \left (\frac{1}{6}+\frac{i}{6}\right ) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \sec \left (\frac{x}{2}\right ) \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )-\left (\frac{1}{6}+\frac{i}{6}\right ) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \sec \left (\frac{x}{2}\right ) \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )\right )+\frac{\left (1+\sqrt{2}\right ) \left (x-\log \left (\sec ^2\left (\frac{x}{2}\right )\right )+2 \sqrt{3} \tanh ^{-1}\left (\frac{\left (2+\sqrt{2}\right ) \tan \left (\frac{x}{2}\right )+2}{\sqrt{6}}\right )+\log \left (-\sec ^2\left (\frac{x}{2}\right ) \left (2 \sin (x)-2 \cos (x)+\sqrt{2}\right )\right )\right )}{12 \left (2+\sqrt{2}\right )}-\frac{x-\log \left (\sec ^2\left (\frac{x}{2}\right )\right )-2 \sqrt{3} \tanh ^{-1}\left (\frac{\left (\sqrt{2}-1\right ) \tan \left (\frac{x}{2}\right )+\sqrt{2}}{\sqrt{3}}\right )+\log \left (\sec ^2\left (\frac{x}{2}\right ) \left (-\sqrt{2} \sin (x)+\sqrt{2} \cos (x)+1\right )\right )}{12 \sqrt{2}}+\frac{\left (\sqrt{6} \sin (x)+1\right ) \left (\left (2+\sqrt{6}\right ) \sin (x)-\left (2+\sqrt{6}\right ) \cos (x)+\sqrt{6}+3\right ) \left (2 \left (\sqrt{2}+\sqrt{3}\right ) \tanh ^{-1}\left (\frac{\left (2+\sqrt{6}\right ) \tan \left (\frac{x}{2}\right )+2}{\sqrt{2}}\right )+\left (3+\sqrt{6}\right ) \left (x-\log \left (\sec ^2\left (\frac{x}{2}\right )\right )+\log \left (-\sec ^2\left (\frac{x}{2}\right ) \left (2 \sin (x)-2 \cos (x)+\sqrt{6}\right )\right )\right )\right )}{12 \left (-2 \left (4 \left (5+2 \sqrt{6}\right ) \sin (x)-6 \sin (2 x)+5 \sqrt{6}+12\right )+\left (12+5 \sqrt{6}\right ) \cos (2 x)+2 \left (5 \sqrt{6} \sin (x)+2 \sqrt{6}+5\right ) \cos (x)\right )}+\frac{\left (\sqrt{2}-2 \sqrt{3} \sin (x)\right ) \left (\left (\sqrt{6}-2\right ) \sin (x)-\left (\sqrt{6}-2\right ) \cos (x)+\sqrt{6}-3\right ) \left (\left (3 \sqrt{2}-2 \sqrt{3}\right ) \left (x-\log \left (\sec ^2\left (\frac{x}{2}\right )\right )+\log \left (-\sec ^2\left (\frac{x}{2}\right ) \left (-\sqrt{2} \sin (x)+\sqrt{2} \cos (x)+\sqrt{3}\right )\right )\right )-2 \left (\sqrt{6}-2\right ) \tanh ^{-1}\left (\left (\sqrt{2}-\sqrt{3}\right ) \tan \left (\frac{x}{2}\right )+\sqrt{2}\right )\right )}{24 \left (-2 \left (4 \left (2 \sqrt{6}-5\right ) \sin (x)+6 \sin (2 x)+5 \sqrt{6}-12\right )+\left (5 \sqrt{6}-12\right ) \cos (2 x)+2 \left (5 \sqrt{6} \sin (x)+2 \sqrt{6}-5\right ) \cos (x)\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(1/6 + I/6)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*Sec[x/2]*(Cos[x/2] + Sin[x/2])] - (1/6 + I/6)*(-1)^(3/4)*
ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[x/2]*(Cos[x/2] - Sin[x/2])] + ((1 + Sqrt[2])*(x + 2*Sqrt[3]*ArcTanh[(2 + (2
+ Sqrt[2])*Tan[x/2])/Sqrt[6]] - Log[Sec[x/2]^2] + Log[-(Sec[x/2]^2*(Sqrt[2] - 2*Cos[x] + 2*Sin[x]))]))/(12*(2
+ Sqrt[2])) - (x - 2*Sqrt[3]*ArcTanh[(Sqrt[2] + (-1 + Sqrt[2])*Tan[x/2])/Sqrt[3]] - Log[Sec[x/2]^2] + Log[Sec
[x/2]^2*(1 + Sqrt[2]*Cos[x] - Sqrt[2]*Sin[x])])/(12*Sqrt[2]) + ((2*(Sqrt[2] + Sqrt[3])*ArcTanh[(2 + (2 + Sqrt[
6])*Tan[x/2])/Sqrt[2]] + (3 + Sqrt[6])*(x - Log[Sec[x/2]^2] + Log[-(Sec[x/2]^2*(Sqrt[6] - 2*Cos[x] + 2*Sin[x])
)]))*(1 + Sqrt[6]*Sin[x])*(3 + Sqrt[6] - (2 + Sqrt[6])*Cos[x] + (2 + Sqrt[6])*Sin[x]))/(12*((12 + 5*Sqrt[6])*C
os[2*x] + 2*Cos[x]*(5 + 2*Sqrt[6] + 5*Sqrt[6]*Sin[x]) - 2*(12 + 5*Sqrt[6] + 4*(5 + 2*Sqrt[6])*Sin[x] - 6*Sin[2
*x]))) + ((-2*(-2 + Sqrt[6])*ArcTanh[Sqrt[2] + (Sqrt[2] - Sqrt[3])*Tan[x/2]] + (3*Sqrt[2] - 2*Sqrt[3])*(x - Lo
g[Sec[x/2]^2] + Log[-(Sec[x/2]^2*(Sqrt[3] + Sqrt[2]*Cos[x] - Sqrt[2]*Sin[x]))]))*(Sqrt[2] - 2*Sqrt[3]*Sin[x])*
(-3 + Sqrt[6] - (-2 + Sqrt[6])*Cos[x] + (-2 + Sqrt[6])*Sin[x]))/(24*((-12 + 5*Sqrt[6])*Cos[2*x] + 2*Cos[x]*(-5
+ 2*Sqrt[6] + 5*Sqrt[6]*Sin[x]) - 2*(-12 + 5*Sqrt[6] + 4*(-5 + 2*Sqrt[6])*Sin[x] + 6*Sin[2*x])))

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Maple [A]  time = 0.096, size = 80, normalized size = 0.9 \begin{align*}{\frac{2}{6\,\sqrt{6}-6\,\sqrt{2}}{\it Artanh} \left ( 8\,{\frac{\cos \left ( x \right ) }{2\,\sqrt{6}-2\,\sqrt{2}}} \right ) }+{\frac{2}{6\,\sqrt{6}+6\,\sqrt{2}}{\it Artanh} \left ( 8\,{\frac{\cos \left ( x \right ) }{2\,\sqrt{6}+2\,\sqrt{2}}} \right ) }-{\frac{{\it Artanh} \left ( \cos \left ( x \right ) \sqrt{2} \right ) \sqrt{2}}{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/(2*6^(1/2)-2*2^(1/2))*arctanh(8*cos(x)/(2*6^(1/2)-2*2^(1/2)))+2/3/(2*6^(1/2)+2*2^(1/2))*arctanh(8*cos(x)/(
2*6^(1/2)+2*2^(1/2)))-1/6*arctanh(cos(x)*2^(1/2))*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 \, \sqrt{2} \sin \left (2 \, x\right ) \sin \left (x\right ) + 2 \,{\left (\sqrt{2} \cos \left (x\right ) + 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (x\right )^{2} + 2 \, \sqrt{2} \cos \left (x\right ) + 1\right ) + \frac{1}{24} \, \sqrt{2} \log \left (-2 \, \sqrt{2} \sin \left (2 \, x\right ) \sin \left (x\right ) - 2 \,{\left (\sqrt{2} \cos \left (x\right ) - 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (x\right )^{2} - 2 \, \sqrt{2} \cos \left (x\right ) + 1\right ) - \int \frac{{\left (\sin \left (7 \, x\right ) - \sin \left (5 \, x\right ) + \sin \left (3 \, x\right ) - \sin \left (x\right )\right )} \cos \left (8 \, x\right ) -{\left (\sin \left (3 \, x\right ) - \sin \left (x\right )\right )} \cos \left (4 \, x\right ) -{\left (\cos \left (7 \, x\right ) - \cos \left (5 \, x\right ) + \cos \left (3 \, x\right ) - \cos \left (x\right )\right )} \sin \left (8 \, x\right ) -{\left (\cos \left (4 \, x\right ) - 1\right )} \sin \left (7 \, x\right ) +{\left (\cos \left (4 \, x\right ) - 1\right )} \sin \left (5 \, x\right ) +{\left (\cos \left (3 \, x\right ) - \cos \left (x\right )\right )} \sin \left (4 \, x\right ) + \cos \left (7 \, x\right ) \sin \left (4 \, x\right ) - \cos \left (5 \, x\right ) \sin \left (4 \, x\right ) + \sin \left (3 \, x\right ) - \sin \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.99748, size = 498, normalized size = 5.86 \begin{align*} -\frac{1}{12} \, \sqrt{\sqrt{3} + 2} \log \left (\sqrt{\sqrt{3} + 2}{\left (\sqrt{3} - 2\right )} + 2 \, \cos \left (x\right )\right ) + \frac{1}{12} \, \sqrt{\sqrt{3} + 2} \log \left (\sqrt{\sqrt{3} + 2}{\left (\sqrt{3} - 2\right )} - 2 \, \cos \left (x\right )\right ) + \frac{1}{12} \, \sqrt{-\sqrt{3} + 2} \log \left ({\left (\sqrt{3} + 2\right )} \sqrt{-\sqrt{3} + 2} + 2 \, \cos \left (x\right )\right ) - \frac{1}{12} \, \sqrt{-\sqrt{3} + 2} \log \left ({\left (\sqrt{3} + 2\right )} \sqrt{-\sqrt{3} + 2} - 2 \, \cos \left (x\right )\right ) + \frac{1}{12} \, \sqrt{2} \log \left (\frac{2 \, \cos \left (x\right )^{2} - 2 \, \sqrt{2} \cos \left (x\right ) + 1}{2 \, \cos \left (x\right )^{2} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(x)*sec(6*x), x)

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Giac [B]  time = 1.31898, size = 208, normalized size = 2.45 \begin{align*} \frac{1}{24} \,{\left (\sqrt{6} + \sqrt{2}\right )} \log \left ({\left | 8 \, \sqrt{6} - 8 \, \sqrt{2} + 32 \, \cos \left (x\right ) \right |}\right ) + \frac{1}{24} \,{\left (\sqrt{6} - \sqrt{2}\right )} \log \left ({\left | \sqrt{6} + \sqrt{2} + 4 \, \cos \left (x\right ) \right |}\right ) - \frac{1}{24} \,{\left (\sqrt{6} - \sqrt{2}\right )} \log \left ({\left | -\sqrt{6} - \sqrt{2} + 4 \, \cos \left (x\right ) \right |}\right ) - \frac{1}{24} \,{\left (\sqrt{6} + \sqrt{2}\right )} \log \left ({\left | -8 \, \sqrt{6} + 8 \, \sqrt{2} + 32 \, \cos \left (x\right ) \right |}\right ) - \frac{1}{12} \, \sqrt{2} \log \left (\frac{{\left | -4 \, \sqrt{2} - \frac{2 \,{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} - 6 \right |}}{{\left | 4 \, \sqrt{2} - \frac{2 \,{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} - 6 \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(sqrt(6) + sqrt(2))*log(abs(8*sqrt(6) - 8*sqrt(2) + 32*cos(x))) + 1/24*(sqrt(6) - sqrt(2))*log(abs(sqrt(6
) + sqrt(2) + 4*cos(x))) - 1/24*(sqrt(6) - sqrt(2))*log(abs(-sqrt(6) - sqrt(2) + 4*cos(x))) - 1/24*(sqrt(6) +
sqrt(2))*log(abs(-8*sqrt(6) + 8*sqrt(2) + 32*cos(x))) - 1/12*sqrt(2)*log(abs(-4*sqrt(2) - 2*(cos(x) - 1)/(cos(
x) + 1) - 6)/abs(4*sqrt(2) - 2*(cos(x) - 1)/(cos(x) + 1) - 6))