3.120 \(\int \cos (x) \sec (6 x) \, dx\)

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

[Out]

-1/6*arctanh(sin(x)*2^(1/2))*2^(1/2)+1/6*arctanh(2*sin(x)/(1/2*6^(1/2)-1/2*2^(1/2)))/(1/2*6^(1/2)-1/2*2^(1/2))
+1/6*arctanh(2*sin(x)/(1/2*6^(1/2)+1/2*2^(1/2)))/(1/2*6^(1/2)+1/2*2^(1/2))

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Rubi [A]  time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00, 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 = {4356, 2057, 207, 1166} \[ -\frac {\tanh ^{-1}\left (\sqrt {2} \sin (x)\right )}{3 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {3}}}\right )}{6 \sqrt {2-\sqrt {3}}}+\frac {\tanh ^{-1}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {3}}}\right )}{6 \sqrt {2+\sqrt {3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 4356

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 81, normalized size = 0.95 \[ \frac {1}{6} \left (-\sqrt {2} \tanh ^{-1}\left (\sqrt {2} \sin (x)\right )+\sqrt {2+\sqrt {3}} \tanh ^{-1}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {3}}}\right )+\sqrt {2-\sqrt {3}} \tanh ^{-1}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {3}}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*Sec[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

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fricas [B]  time = 4.13, size = 154, normalized size = 1.81 \[ -\frac {1}{12} \, \sqrt {\sqrt {3} + 2} \log \left (\sqrt {\sqrt {3} + 2} {\left (\sqrt {3} - 2\right )} + 2 \, \sin \relax (x)\right ) + \frac {1}{12} \, \sqrt {\sqrt {3} + 2} \log \left (\sqrt {\sqrt {3} + 2} {\left (\sqrt {3} - 2\right )} - 2 \, \sin \relax (x)\right ) + \frac {1}{12} \, \sqrt {-\sqrt {3} + 2} \log \left ({\left (\sqrt {3} + 2\right )} \sqrt {-\sqrt {3} + 2} + 2 \, \sin \relax (x)\right ) - \frac {1}{12} \, \sqrt {-\sqrt {3} + 2} \log \left ({\left (\sqrt {3} + 2\right )} \sqrt {-\sqrt {3} + 2} - 2 \, \sin \relax (x)\right ) + \frac {1}{12} \, \sqrt {2} \log \left (-\frac {2 \, \cos \relax (x)^{2} + 2 \, \sqrt {2} \sin \relax (x) - 3}{2 \, \cos \relax (x)^{2} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 132, normalized size = 1.55 \[ \frac {1}{24} \, {\left (\sqrt {6} - \sqrt {2}\right )} \log \left ({\left | \frac {1}{4} \, \sqrt {6} + \frac {1}{4} \, \sqrt {2} + \sin \relax (x) \right |}\right ) + \frac {1}{24} \, {\left (\sqrt {6} + \sqrt {2}\right )} \log \left ({\left | \frac {1}{4} \, \sqrt {6} - \frac {1}{4} \, \sqrt {2} + \sin \relax (x) \right |}\right ) - \frac {1}{24} \, {\left (\sqrt {6} + \sqrt {2}\right )} \log \left ({\left | -\frac {1}{4} \, \sqrt {6} + \frac {1}{4} \, \sqrt {2} + \sin \relax (x) \right |}\right ) - \frac {1}{24} \, {\left (\sqrt {6} - \sqrt {2}\right )} \log \left ({\left | -\frac {1}{4} \, \sqrt {6} - \frac {1}{4} \, \sqrt {2} + \sin \relax (x) \right |}\right ) + \frac {1}{12} \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \relax (x) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \relax (x) \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(sqrt(6) - sqrt(2))*log(abs(1/4*sqrt(6) + 1/4*sqrt(2) + sin(x))) + 1/24*(sqrt(6) + sqrt(2))*log(abs(1/4*s
qrt(6) - 1/4*sqrt(2) + sin(x))) - 1/24*(sqrt(6) + sqrt(2))*log(abs(-1/4*sqrt(6) + 1/4*sqrt(2) + sin(x))) - 1/2
4*(sqrt(6) - sqrt(2))*log(abs(-1/4*sqrt(6) - 1/4*sqrt(2) + sin(x))) + 1/12*sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(
x))/abs(2*sqrt(2) + 4*sin(x)))

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maple [A]  time = 0.26, size = 80, normalized size = 0.94 \[ \frac {2 \arctanh \left (\frac {8 \sin \relax (x )}{2 \sqrt {6}-2 \sqrt {2}}\right )}{3 \left (2 \sqrt {6}-2 \sqrt {2}\right )}+\frac {2 \arctanh \left (\frac {8 \sin \relax (x )}{2 \sqrt {6}+2 \sqrt {2}}\right )}{3 \left (2 \sqrt {6}+2 \sqrt {2}\right )}-\frac {\arctanh \left (\sin \relax (x ) \sqrt {2}\right ) \sqrt {2}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{24} \, \sqrt {2} \log \left (2 \, \cos \relax (x)^{2} + 2 \, \sin \relax (x)^{2} + 2 \, \sqrt {2} \cos \relax (x) + 2 \, \sqrt {2} \sin \relax (x) + 2\right ) + \frac {1}{24} \, \sqrt {2} \log \left (2 \, \cos \relax (x)^{2} + 2 \, \sin \relax (x)^{2} + 2 \, \sqrt {2} \cos \relax (x) - 2 \, \sqrt {2} \sin \relax (x) + 2\right ) - \frac {1}{24} \, \sqrt {2} \log \left (2 \, \cos \relax (x)^{2} + 2 \, \sin \relax (x)^{2} - 2 \, \sqrt {2} \cos \relax (x) + 2 \, \sqrt {2} \sin \relax (x) + 2\right ) + \frac {1}{24} \, \sqrt {2} \log \left (2 \, \cos \relax (x)^{2} + 2 \, \sin \relax (x)^{2} - 2 \, \sqrt {2} \cos \relax (x) - 2 \, \sqrt {2} \sin \relax (x) + 2\right ) + \int -\frac {{\left (\cos \left (7 \, x\right ) + \cos \left (5 \, x\right ) + \cos \left (3 \, x\right ) + \cos \relax (x)\right )} \cos \left (8 \, x\right ) - {\left (\cos \left (4 \, x\right ) - 1\right )} \cos \left (7 \, x\right ) - {\left (\cos \left (4 \, x\right ) - 1\right )} \cos \left (5 \, x\right ) - {\left (\cos \left (3 \, x\right ) + \cos \relax (x)\right )} \cos \left (4 \, x\right ) + {\left (\sin \left (7 \, x\right ) + \sin \left (5 \, x\right ) + \sin \left (3 \, x\right ) + \sin \relax (x)\right )} \sin \left (8 \, x\right ) - {\left (\sin \left (3 \, x\right ) + \sin \relax (x)\right )} \sin \left (4 \, x\right ) - \sin \left (7 \, x\right ) \sin \left (4 \, x\right ) - \sin \left (5 \, x\right ) \sin \left (4 \, x\right ) + \cos \left (3 \, x\right ) + \cos \relax (x)}{3 \, {\left (2 \, {\left (\cos \left (4 \, x\right ) - 1\right )} \cos \left (8 \, x\right ) - \cos \left (8 \, x\right )^{2} - \cos \left (4 \, x\right )^{2} - \sin \left (8 \, x\right )^{2} + 2 \, \sin \left (8 \, x\right ) \sin \left (4 \, x\right ) - \sin \left (4 \, x\right )^{2} + 2 \, \cos \left (4 \, x\right ) - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*sec(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*sq
rt(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*((cos(7*x) + cos(5*x) + cos(3*x) + cos(x))*cos(8*x) - (cos(4*x) - 1)*cos(7*x) -
(cos(4*x) - 1)*cos(5*x) - (cos(3*x) + cos(x))*cos(4*x) + (sin(7*x) + sin(5*x) + sin(3*x) + sin(x))*sin(8*x) -
(sin(3*x) + sin(x))*sin(4*x) - sin(7*x)*sin(4*x) - sin(5*x)*sin(4*x) + cos(3*x) + cos(x))/(2*(cos(4*x) - 1)*co
s(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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mupad [B]  time = 2.29, size = 118, normalized size = 1.39 \[ \mathrm {atanh}\left (\frac {5\,\sqrt {2}\,\sin \relax (x)}{2097152\,\left (\frac {\sqrt {2}\,\sqrt {6}}{4194304}+\frac {1}{1048576}\right )}+\frac {3\,\sqrt {6}\,\sin \relax (x)}{2097152\,\left (\frac {\sqrt {2}\,\sqrt {6}}{4194304}+\frac {1}{1048576}\right )}\right )\,\left (\frac {\sqrt {2}}{12}+\frac {\sqrt {6}}{12}\right )-\mathrm {atanh}\left (\frac {5\,\sqrt {2}\,\sin \relax (x)}{2097152\,\left (\frac {\sqrt {2}\,\sqrt {6}}{4194304}-\frac {1}{1048576}\right )}-\frac {3\,\sqrt {6}\,\sin \relax (x)}{2097152\,\left (\frac {\sqrt {2}\,\sqrt {6}}{4194304}-\frac {1}{1048576}\right )}\right )\,\left (\frac {\sqrt {2}}{12}-\frac {\sqrt {6}}{12}\right )-\frac {\sqrt {2}\,\mathrm {atanh}\left (\sqrt {2}\,\sin \relax (x)\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/cos(6*x),x)

[Out]

atanh((5*2^(1/2)*sin(x))/(2097152*((2^(1/2)*6^(1/2))/4194304 + 1/1048576)) + (3*6^(1/2)*sin(x))/(2097152*((2^(
1/2)*6^(1/2))/4194304 + 1/1048576)))*(2^(1/2)/12 + 6^(1/2)/12) - atanh((5*2^(1/2)*sin(x))/(2097152*((2^(1/2)*6
^(1/2))/4194304 - 1/1048576)) - (3*6^(1/2)*sin(x))/(2097152*((2^(1/2)*6^(1/2))/4194304 - 1/1048576)))*(2^(1/2)
/12 - 6^(1/2)/12) - (2^(1/2)*atanh(2^(1/2)*sin(x)))/6

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos {\relax (x )} \sec {\left (6 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(x)*sec(6*x), x)

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