3.108 \(\int \cos (x) \tan (5 x) \, dx\)

Optimal. Leaf size=84 \[ -\cos (x)+\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tanh ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \cos (x)\right )+\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tanh ^{-1}\left (\sqrt{\frac{2}{5} \left (5+\sqrt{5}\right )} \cos (x)\right ) \]

[Out]

(Sqrt[(5 + Sqrt[5])/2]*ArcTanh[2*Sqrt[2/(5 + Sqrt[5])]*Cos[x]])/5 + (Sqrt[(5 - Sqrt[5])/2]*ArcTanh[Sqrt[(2*(5
+ Sqrt[5]))/5]*Cos[x]])/5 - Cos[x]

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Rubi [A]  time = 0.0979321, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {1676, 1166, 207} \[ -\cos (x)+\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tanh ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \cos (x)\right )+\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tanh ^{-1}\left (\sqrt{\frac{2}{5} \left (5+\sqrt{5}\right )} \cos (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(5 + Sqrt[5])/2]*ArcTanh[2*Sqrt[2/(5 + Sqrt[5])]*Cos[x]])/5 + (Sqrt[(5 - Sqrt[5])/2]*ArcTanh[Sqrt[(2*(5
+ Sqrt[5]))/5]*Cos[x]])/5 - Cos[x]

Rule 1676

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

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) \tan (5 x) \, dx &=-\operatorname{Subst}\left (\int \frac{1-12 x^2+16 x^4}{5-20 x^2+16 x^4} \, dx,x,\cos (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (1-\frac{4 \left (1-2 x^2\right )}{5-20 x^2+16 x^4}\right ) \, dx,x,\cos (x)\right )\\ &=-\cos (x)+4 \operatorname{Subst}\left (\int \frac{1-2 x^2}{5-20 x^2+16 x^4} \, dx,x,\cos (x)\right )\\ &=-\cos (x)-\frac{1}{5} \left (4 \left (5-\sqrt{5}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-10+2 \sqrt{5}+16 x^2} \, dx,x,\cos (x)\right )-\frac{1}{5} \left (4 \left (5+\sqrt{5}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-10-2 \sqrt{5}+16 x^2} \, dx,x,\cos (x)\right )\\ &=\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tanh ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \cos (x)\right )+\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tanh ^{-1}\left (\sqrt{\frac{2}{5} \left (5+\sqrt{5}\right )} \cos (x)\right )-\cos (x)\\ \end{align*}

Mathematica [B]  time = 0.597829, size = 215, normalized size = 2.56 \[ -\cos (x)+\frac{\left (1+\sqrt{5}\right ) \tanh ^{-1}\left (\frac{4-\left (\sqrt{5}-1\right ) \tan \left (\frac{x}{2}\right )}{\sqrt{2 \left (5+\sqrt{5}\right )}}\right )}{\sqrt{10 \left (5+\sqrt{5}\right )}}+\frac{\left (1+\sqrt{5}\right ) \tanh ^{-1}\left (\frac{\left (\sqrt{5}-1\right ) \tan \left (\frac{x}{2}\right )+4}{\sqrt{2 \left (5+\sqrt{5}\right )}}\right )}{\sqrt{10 \left (5+\sqrt{5}\right )}}+\frac{\left (\sqrt{5}-1\right ) \tanh ^{-1}\left (\frac{4-\left (1+\sqrt{5}\right ) \tan \left (\frac{x}{2}\right )}{\sqrt{10-2 \sqrt{5}}}\right )}{\sqrt{50-10 \sqrt{5}}}+\frac{\left (\sqrt{5}-1\right ) \tanh ^{-1}\left (\frac{\left (1+\sqrt{5}\right ) \tan \left (\frac{x}{2}\right )+4}{\sqrt{10-2 \sqrt{5}}}\right )}{\sqrt{50-10 \sqrt{5}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 + Sqrt[5])*ArcTanh[(4 - (-1 + Sqrt[5])*Tan[x/2])/Sqrt[2*(5 + Sqrt[5])]])/Sqrt[10*(5 + Sqrt[5])] + ((1 + Sq
rt[5])*ArcTanh[(4 + (-1 + Sqrt[5])*Tan[x/2])/Sqrt[2*(5 + Sqrt[5])]])/Sqrt[10*(5 + Sqrt[5])] + ((-1 + Sqrt[5])*
ArcTanh[(4 - (1 + Sqrt[5])*Tan[x/2])/Sqrt[10 - 2*Sqrt[5]]])/Sqrt[50 - 10*Sqrt[5]] + ((-1 + Sqrt[5])*ArcTanh[(4
+ (1 + Sqrt[5])*Tan[x/2])/Sqrt[10 - 2*Sqrt[5]]])/Sqrt[50 - 10*Sqrt[5]] - Cos[x]

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Maple [A]  time = 0.036, size = 72, normalized size = 0.9 \begin{align*} -\cos \left ( x \right ) +{\frac{ \left ( \sqrt{5}-1 \right ) \sqrt{5}}{5\,\sqrt{10-2\,\sqrt{5}}}{\it Artanh} \left ( 4\,{\frac{\cos \left ( x \right ) }{\sqrt{10-2\,\sqrt{5}}}} \right ) }+{\frac{ \left ( \sqrt{5}+1 \right ) \sqrt{5}}{5\,\sqrt{10+2\,\sqrt{5}}}{\it Artanh} \left ( 4\,{\frac{\cos \left ( x \right ) }{\sqrt{10+2\,\sqrt{5}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(x)+1/5*(5^(1/2)-1)*5^(1/2)/(10-2*5^(1/2))^(1/2)*arctanh(4*cos(x)/(10-2*5^(1/2))^(1/2))+1/5*(5^(1/2)+1)*5^
(1/2)/(10+2*5^(1/2))^(1/2)*arctanh(4*cos(x)/(10+2*5^(1/2))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(x) - integrate(((sin(7*x) - sin(5*x) + sin(3*x) - sin(x))*cos(8*x) + (sin(6*x) - sin(4*x) + sin(2*x))*cos
(7*x) + (sin(5*x) - sin(3*x) + sin(x))*cos(6*x) + (sin(4*x) - sin(2*x))*cos(5*x) + (sin(3*x) - sin(x))*cos(4*x
) - (cos(7*x) - cos(5*x) + cos(3*x) - cos(x))*sin(8*x) - (cos(6*x) - cos(4*x) + cos(2*x) - 1)*sin(7*x) - (cos(
5*x) - cos(3*x) + cos(x))*sin(6*x) - (cos(4*x) - cos(2*x) + 1)*sin(5*x) - (cos(3*x) - cos(x))*sin(4*x) - (cos(
2*x) - 1)*sin(3*x) + cos(3*x)*sin(2*x) - cos(x)*sin(2*x) + cos(2*x)*sin(x) - sin(x))/(2*(cos(6*x) - cos(4*x) +
cos(2*x) - 1)*cos(8*x) - cos(8*x)^2 + 2*(cos(4*x) - cos(2*x) + 1)*cos(6*x) - cos(6*x)^2 + 2*(cos(2*x) - 1)*co
s(4*x) - cos(4*x)^2 - cos(2*x)^2 + 2*(sin(6*x) - sin(4*x) + sin(2*x))*sin(8*x) - sin(8*x)^2 + 2*(sin(4*x) - si
n(2*x))*sin(6*x) - sin(6*x)^2 - sin(4*x)^2 + 2*sin(4*x)*sin(2*x) - sin(2*x)^2 + 2*cos(2*x) - 1), x)

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Fricas [B]  time = 2.63353, size = 421, normalized size = 5.01 \begin{align*} \frac{1}{20} \, \sqrt{2} \sqrt{\sqrt{5} + 5} \log \left (\sqrt{2} \sqrt{\sqrt{5} + 5} + 4 \, \cos \left (x\right )\right ) - \frac{1}{20} \, \sqrt{2} \sqrt{\sqrt{5} + 5} \log \left (\sqrt{2} \sqrt{\sqrt{5} + 5} - 4 \, \cos \left (x\right )\right ) + \frac{1}{20} \, \sqrt{2} \sqrt{-\sqrt{5} + 5} \log \left (\sqrt{2} \sqrt{-\sqrt{5} + 5} + 4 \, \cos \left (x\right )\right ) - \frac{1}{20} \, \sqrt{2} \sqrt{-\sqrt{5} + 5} \log \left (\sqrt{2} \sqrt{-\sqrt{5} + 5} - 4 \, \cos \left (x\right )\right ) - \cos \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*sqrt(2)*sqrt(sqrt(5) + 5)*log(sqrt(2)*sqrt(sqrt(5) + 5) + 4*cos(x)) - 1/20*sqrt(2)*sqrt(sqrt(5) + 5)*log(
sqrt(2)*sqrt(sqrt(5) + 5) - 4*cos(x)) + 1/20*sqrt(2)*sqrt(-sqrt(5) + 5)*log(sqrt(2)*sqrt(-sqrt(5) + 5) + 4*cos
(x)) - 1/20*sqrt(2)*sqrt(-sqrt(5) + 5)*log(sqrt(2)*sqrt(-sqrt(5) + 5) - 4*cos(x)) - cos(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(cos(x)*tan(5*x),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(x)*tan(5*x), x)