\(\int \cos (x) \sec (6 x) \, dx\) [120]

Optimal result

Integrand size = 7, antiderivative size = 85 \[ \int \cos (x) \sec (6 x) \, dx=-\frac {\text {arctanh}\left (\sqrt {2} \sin (x)\right )}{3 \sqrt {2}}+\frac {\text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {3}}}\right )}{6 \sqrt {2-\sqrt {3}}}+\frac {\text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {3}}}\right )}{6 \sqrt {2+\sqrt {3}}} \]

[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))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4441, 2082, 213, 1180} \[ \int \cos (x) \sec (6 x) \, dx=-\frac {\text {arctanh}\left (\sqrt {2} \sin (x)\right )}{3 \sqrt {2}}+\frac {\text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {3}}}\right )}{6 \sqrt {2-\sqrt {3}}}+\frac {\text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {3}}}\right )}{6 \sqrt {2+\sqrt {3}}} \]

[In]

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

[Out]

-1/3*ArcTanh[Sqrt[2]*Sin[x]]/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 213

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

Rule 1180

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 2082

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 4441

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95 \[ \int \cos (x) \sec (6 x) \, dx=\frac {1}{6} \left (-\sqrt {2} \text {arctanh}\left (\sqrt {2} \sin (x)\right )+\sqrt {2+\sqrt {3}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {3}}}\right )+\sqrt {2-\sqrt {3}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {3}}}\right )\right ) \]

[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

Maple [A] (verified)

Time = 1.81 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {\operatorname {arctanh}\left (\sin \left (x \right ) \sqrt {2}\right ) \sqrt {2}}{6}+\frac {2 \,\operatorname {arctanh}\left (\frac {8 \sin \left (x \right )}{2 \sqrt {6}+2 \sqrt {2}}\right )}{3 \left (2 \sqrt {6}+2 \sqrt {2}\right )}+\frac {2 \,\operatorname {arctanh}\left (\frac {8 \sin \left (x \right )}{2 \sqrt {6}-2 \sqrt {2}}\right )}{3 \left (2 \sqrt {6}-2 \sqrt {2}\right )}\)	\(80\)
risch	\(\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {2}\, {\mathrm e}^{i x}-1\right )}{12}-\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {2}\, {\mathrm e}^{i x}-1\right )}{12}+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (331776 \textit {\_Z}^{4}-2304 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}+\left (-13824 i \textit {\_R}^{3}+96 i \textit {\_R} \right ) {\mathrm e}^{i x}-1\right )\right )\)	\(95\)

[In]

int(cos(x)*sec(6*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (67) = 134\).

Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.81 \[ \int \cos (x) \sec (6 x) \, dx=-\frac {1}{12} \, \sqrt {\sqrt {3} + 2} \log \left (\sqrt {\sqrt {3} + 2} {\left (\sqrt {3} - 2\right )} + 2 \, \sin \left (x\right )\right ) + \frac {1}{12} \, \sqrt {\sqrt {3} + 2} \log \left (\sqrt {\sqrt {3} + 2} {\left (\sqrt {3} - 2\right )} - 2 \, \sin \left (x\right )\right ) + \frac {1}{12} \, \sqrt {-\sqrt {3} + 2} \log \left ({\left (\sqrt {3} + 2\right )} \sqrt {-\sqrt {3} + 2} + 2 \, \sin \left (x\right )\right ) - \frac {1}{12} \, \sqrt {-\sqrt {3} + 2} \log \left ({\left (\sqrt {3} + 2\right )} \sqrt {-\sqrt {3} + 2} - 2 \, \sin \left (x\right )\right ) + \frac {1}{12} \, \sqrt {2} \log \left (-\frac {2 \, \cos \left (x\right )^{2} + 2 \, \sqrt {2} \sin \left (x\right ) - 3}{2 \, \cos \left (x\right )^{2} - 1}\right ) \]

[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))

Sympy [F]

\[ \int \cos (x) \sec (6 x) \, dx=\int \cos {\left (x \right )} \sec {\left (6 x \right )}\, dx \]

[In]

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

[Out]

Integral(cos(x)*sec(6*x), x)

Maxima [F]

\[ \int \cos (x) \sec (6 x) \, dx=\int { \cos \left (x\right ) \sec \left (6 \, x\right ) \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.55 \[ \int \cos (x) \sec (6 x) \, dx=\frac {1}{24} \, {\left (\sqrt {6} - \sqrt {2}\right )} \log \left ({\left | \frac {1}{4} \, \sqrt {6} + \frac {1}{4} \, \sqrt {2} + \sin \left (x\right ) \right |}\right ) + \frac {1}{24} \, {\left (\sqrt {6} + \sqrt {2}\right )} \log \left ({\left | \frac {1}{4} \, \sqrt {6} - \frac {1}{4} \, \sqrt {2} + \sin \left (x\right ) \right |}\right ) - \frac {1}{24} \, {\left (\sqrt {6} + \sqrt {2}\right )} \log \left ({\left | -\frac {1}{4} \, \sqrt {6} + \frac {1}{4} \, \sqrt {2} + \sin \left (x\right ) \right |}\right ) - \frac {1}{24} \, {\left (\sqrt {6} - \sqrt {2}\right )} \log \left ({\left | -\frac {1}{4} \, \sqrt {6} - \frac {1}{4} \, \sqrt {2} + \sin \left (x\right ) \right |}\right ) + \frac {1}{12} \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (x\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (x\right ) \right |}}\right ) \]

[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)))

Mupad [B] (verification not implemented)

Time = 26.14 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.39 \[ \int \cos (x) \sec (6 x) \, dx=\mathrm {atanh}\left (\frac {5\,\sqrt {2}\,\sin \left (x\right )}{2097152\,\left (\frac {\sqrt {2}\,\sqrt {6}}{4194304}+\frac {1}{1048576}\right )}+\frac {3\,\sqrt {6}\,\sin \left (x\right )}{2097152\,\left (\frac {\sqrt {2}\,\sqrt {6}}{4194304}+\frac {1}{1048576}\right )}\right )\,\left (\frac {\sqrt {2}}{12}+\frac {\sqrt {6}}{12}\right )-\mathrm {atanh}\left (\frac {5\,\sqrt {2}\,\sin \left (x\right )}{2097152\,\left (\frac {\sqrt {2}\,\sqrt {6}}{4194304}-\frac {1}{1048576}\right )}-\frac {3\,\sqrt {6}\,\sin \left (x\right )}{2097152\,\left (\frac {\sqrt {2}\,\sqrt {6}}{4194304}-\frac {1}{1048576}\right )}\right )\,\left (\frac {\sqrt {2}}{12}-\frac {\sqrt {6}}{12}\right )-\frac {\sqrt {2}\,\mathrm {atanh}\left (\sqrt {2}\,\sin \left (x\right )\right )}{6} \]

[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