3.119 \(\int \cos (x) \sec (5 x) \, dx\)

Optimal. Leaf size=163 \[ \frac{1}{10} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \log \left (\cos (x)-\sqrt{5-2 \sqrt{5}} \sin (x)\right )-\frac{1}{10} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \log \left (\sqrt{5-2 \sqrt{5}} \sin (x)+\cos (x)\right )-\frac{1}{10} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \log \left (\cos (x)-\sqrt{5+2 \sqrt{5}} \sin (x)\right )+\frac{1}{10} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \log \left (\sqrt{5+2 \sqrt{5}} \sin (x)+\cos (x)\right ) \]

[Out]

(Sqrt[(5 - Sqrt[5])/2]*Log[Cos[x] - Sqrt[5 - 2*Sqrt[5]]*Sin[x]])/10 - (Sqrt[(5 - Sqrt[5])/2]*Log[Cos[x] + Sqrt
[5 - 2*Sqrt[5]]*Sin[x]])/10 - (Sqrt[(5 + Sqrt[5])/2]*Log[Cos[x] - Sqrt[5 + 2*Sqrt[5]]*Sin[x]])/10 + (Sqrt[(5 +
Sqrt[5])/2]*Log[Cos[x] + Sqrt[5 + 2*Sqrt[5]]*Sin[x]])/10

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Rubi [A]  time = 0.129449, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {1166, 207} \[ \frac{1}{10} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \log \left (\cos (x)-\sqrt{5-2 \sqrt{5}} \sin (x)\right )-\frac{1}{10} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \log \left (\sqrt{5-2 \sqrt{5}} \sin (x)+\cos (x)\right )-\frac{1}{10} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \log \left (\cos (x)-\sqrt{5+2 \sqrt{5}} \sin (x)\right )+\frac{1}{10} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \log \left (\sqrt{5+2 \sqrt{5}} \sin (x)+\cos (x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]*Sec[5*x],x]

[Out]

(Sqrt[(5 - Sqrt[5])/2]*Log[Cos[x] - Sqrt[5 - 2*Sqrt[5]]*Sin[x]])/10 - (Sqrt[(5 - Sqrt[5])/2]*Log[Cos[x] + Sqrt
[5 - 2*Sqrt[5]]*Sin[x]])/10 - (Sqrt[(5 + Sqrt[5])/2]*Log[Cos[x] - Sqrt[5 + 2*Sqrt[5]]*Sin[x]])/10 + (Sqrt[(5 +
Sqrt[5])/2]*Log[Cos[x] + Sqrt[5 + 2*Sqrt[5]]*Sin[x]])/10

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 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])

Rubi steps

\begin{align*} \int \cos (x) \sec (5 x) \, dx &=\operatorname{Subst}\left (\int \frac{1+x^2}{1-10 x^2+5 x^4} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \left (1-\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{-5+2 \sqrt{5}+5 x^2} \, dx,x,\tan (x)\right )+\frac{1}{2} \left (1+\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{-5-2 \sqrt{5}+5 x^2} \, dx,x,\tan (x)\right )\\ &=\frac{1}{10} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \log \left (\cos (x)-\sqrt{5-2 \sqrt{5}} \sin (x)\right )-\frac{1}{10} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \log \left (\cos (x)+\sqrt{5-2 \sqrt{5}} \sin (x)\right )-\frac{1}{10} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \log \left (\cos (x)-\sqrt{5+2 \sqrt{5}} \sin (x)\right )+\frac{1}{10} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \log \left (\cos (x)+\sqrt{5+2 \sqrt{5}} \sin (x)\right )\\ \end{align*}

Mathematica [A]  time = 0.103674, size = 84, normalized size = 0.52 \[ \frac{\sqrt{5+\sqrt{5}} \tanh ^{-1}\left (\frac{\left (5+\sqrt{5}\right ) \tan (x)}{\sqrt{10-2 \sqrt{5}}}\right )+\sqrt{5-\sqrt{5}} \tanh ^{-1}\left (\frac{\left (\sqrt{5}-5\right ) \tan (x)}{\sqrt{2 \left (5+\sqrt{5}\right )}}\right )}{5 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*Sec[5*x],x]

[Out]

(Sqrt[5 + Sqrt[5]]*ArcTanh[((5 + Sqrt[5])*Tan[x])/Sqrt[10 - 2*Sqrt[5]]] + Sqrt[5 - Sqrt[5]]*ArcTanh[((-5 + Sqr
t[5])*Tan[x])/Sqrt[2*(5 + Sqrt[5])]])/(5*Sqrt[2])

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Maple [A]  time = 0.1, size = 68, normalized size = 0.4 \begin{align*} -{\frac{ \left ( 5+\sqrt{5} \right ) \sqrt{5}}{10\,\sqrt{25+10\,\sqrt{5}}}{\it Artanh} \left ( 5\,{\frac{\tan \left ( x \right ) }{\sqrt{25+10\,\sqrt{5}}}} \right ) }-{\frac{ \left ( \sqrt{5}-5 \right ) \sqrt{5}}{10\,\sqrt{25-10\,\sqrt{5}}}{\it Artanh} \left ( 5\,{\frac{\tan \left ( x \right ) }{\sqrt{25-10\,\sqrt{5}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*sec(5*x),x)

[Out]

-1/10*(5+5^(1/2))*5^(1/2)/(25+10*5^(1/2))^(1/2)*arctanh(5*tan(x)/(25+10*5^(1/2))^(1/2))-1/10*(5^(1/2)-5)*5^(1/
2)/(25-10*5^(1/2))^(1/2)*arctanh(5*tan(x)/(25-10*5^(1/2))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos \left (x\right ) \sec \left (5 \, x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*sec(5*x),x, algorithm="maxima")

[Out]

integrate(cos(x)*sec(5*x), x)

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Fricas [B]  time = 2.80953, size = 759, normalized size = 4.66 \begin{align*} -\frac{1}{40} \, \sqrt{2} \sqrt{\sqrt{5} + 5} \log \left ({\left (\sqrt{5} \sqrt{2} - \sqrt{2}\right )} \sqrt{\sqrt{5} + 5} \cos \left (x\right ) \sin \left (x\right ) + 2 \,{\left (\sqrt{5} + 1\right )} \cos \left (x\right )^{2} - \sqrt{5} - 5\right ) + \frac{1}{40} \, \sqrt{2} \sqrt{\sqrt{5} + 5} \log \left (-{\left (\sqrt{5} \sqrt{2} - \sqrt{2}\right )} \sqrt{\sqrt{5} + 5} \cos \left (x\right ) \sin \left (x\right ) + 2 \,{\left (\sqrt{5} + 1\right )} \cos \left (x\right )^{2} - \sqrt{5} - 5\right ) - \frac{1}{40} \, \sqrt{2} \sqrt{-\sqrt{5} + 5} \log \left ({\left (\sqrt{5} \sqrt{2} + \sqrt{2}\right )} \sqrt{-\sqrt{5} + 5} \cos \left (x\right ) \sin \left (x\right ) + 2 \,{\left (\sqrt{5} - 1\right )} \cos \left (x\right )^{2} - \sqrt{5} + 5\right ) + \frac{1}{40} \, \sqrt{2} \sqrt{-\sqrt{5} + 5} \log \left (-{\left (\sqrt{5} \sqrt{2} + \sqrt{2}\right )} \sqrt{-\sqrt{5} + 5} \cos \left (x\right ) \sin \left (x\right ) + 2 \,{\left (\sqrt{5} - 1\right )} \cos \left (x\right )^{2} - \sqrt{5} + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*sec(5*x),x, algorithm="fricas")

[Out]

-1/40*sqrt(2)*sqrt(sqrt(5) + 5)*log((sqrt(5)*sqrt(2) - sqrt(2))*sqrt(sqrt(5) + 5)*cos(x)*sin(x) + 2*(sqrt(5) +
1)*cos(x)^2 - sqrt(5) - 5) + 1/40*sqrt(2)*sqrt(sqrt(5) + 5)*log(-(sqrt(5)*sqrt(2) - sqrt(2))*sqrt(sqrt(5) + 5
)*cos(x)*sin(x) + 2*(sqrt(5) + 1)*cos(x)^2 - sqrt(5) - 5) - 1/40*sqrt(2)*sqrt(-sqrt(5) + 5)*log((sqrt(5)*sqrt(
2) + sqrt(2))*sqrt(-sqrt(5) + 5)*cos(x)*sin(x) + 2*(sqrt(5) - 1)*cos(x)^2 - sqrt(5) + 5) + 1/40*sqrt(2)*sqrt(-
sqrt(5) + 5)*log(-(sqrt(5)*sqrt(2) + sqrt(2))*sqrt(-sqrt(5) + 5)*cos(x)*sin(x) + 2*(sqrt(5) - 1)*cos(x)^2 - sq
rt(5) + 5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos{\left (x \right )} \sec{\left (5 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*sec(5*x),x)

[Out]

Integral(cos(x)*sec(5*x), x)

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Giac [A]  time = 1.35687, size = 142, normalized size = 0.87 \begin{align*} -\frac{1}{20} \, \sqrt{-2 \, \sqrt{5} + 10} \log \left ({\left | \sqrt{\frac{2}{5} \, \sqrt{5} + 1} + \tan \left (x\right ) \right |}\right ) + \frac{1}{20} \, \sqrt{-2 \, \sqrt{5} + 10} \log \left ({\left | -\sqrt{\frac{2}{5} \, \sqrt{5} + 1} + \tan \left (x\right ) \right |}\right ) + \frac{1}{20} \, \sqrt{2 \, \sqrt{5} + 10} \log \left ({\left | \sqrt{-\frac{2}{5} \, \sqrt{5} + 1} + \tan \left (x\right ) \right |}\right ) - \frac{1}{20} \, \sqrt{2 \, \sqrt{5} + 10} \log \left ({\left | -\sqrt{-\frac{2}{5} \, \sqrt{5} + 1} + \tan \left (x\right ) \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*sec(5*x),x, algorithm="giac")

[Out]

-1/20*sqrt(-2*sqrt(5) + 10)*log(abs(sqrt(2/5*sqrt(5) + 1) + tan(x))) + 1/20*sqrt(-2*sqrt(5) + 10)*log(abs(-sqr
t(2/5*sqrt(5) + 1) + tan(x))) + 1/20*sqrt(2*sqrt(5) + 10)*log(abs(sqrt(-2/5*sqrt(5) + 1) + tan(x))) - 1/20*sqr
t(2*sqrt(5) + 10)*log(abs(-sqrt(-2/5*sqrt(5) + 1) + tan(x)))