3.232 \(\int \cosh (x) \tanh (5 x) \, dx\)

Optimal. Leaf size=82 \[ \cosh (x)-\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tanh ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \cosh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tanh ^{-1}\left (\sqrt{\frac{2}{5} \left (5+\sqrt{5}\right )} \cosh (x)\right ) \]

[Out]

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

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Rubi [A]  time = 0.116861, antiderivative size = 82, 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} \[ \cosh (x)-\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tanh ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \cosh (x)\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tanh ^{-1}\left (\sqrt{\frac{2}{5} \left (5+\sqrt{5}\right )} \cosh (x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]*Tanh[5*x],x]

[Out]

-(Sqrt[(5 + Sqrt[5])/2]*ArcTanh[2*Sqrt[2/(5 + Sqrt[5])]*Cosh[x]])/5 - (Sqrt[(5 - Sqrt[5])/2]*ArcTanh[Sqrt[(2*(
5 + Sqrt[5]))/5]*Cosh[x]])/5 + Cosh[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 \cosh (x) \tanh (5 x) \, dx &=\operatorname{Subst}\left (\int \frac{1-12 x^2+16 x^4}{5-20 x^2+16 x^4} \, dx,x,\cosh (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,\cosh (x)\right )\\ &=\cosh (x)-4 \operatorname{Subst}\left (\int \frac{1-2 x^2}{5-20 x^2+16 x^4} \, dx,x,\cosh (x)\right )\\ &=\cosh (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,\cosh (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,\cosh (x)\right )\\ &=-\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tanh ^{-1}\left (2 \sqrt{\frac{2}{5+\sqrt{5}}} \cosh (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 )} \cosh (x)\right )+\cosh (x)\\ \end{align*}

Mathematica [C]  time = 0.0275745, size = 249, normalized size = 3.04 \[ \frac{1}{4} \text{RootSum}\left [\text{$\#$1}^8-\text{$\#$1}^6+\text{$\#$1}^4-\text{$\#$1}^2+1\& ,\frac{\text{$\#$1}^6 x-\text{$\#$1}^4 x+\text{$\#$1}^2 x+2 \text{$\#$1}^6 \log \left (-\text{$\#$1} \sinh \left (\frac{x}{2}\right )+\text{$\#$1} \cosh \left (\frac{x}{2}\right )-\sinh \left (\frac{x}{2}\right )-\cosh \left (\frac{x}{2}\right )\right )-2 \text{$\#$1}^4 \log \left (-\text{$\#$1} \sinh \left (\frac{x}{2}\right )+\text{$\#$1} \cosh \left (\frac{x}{2}\right )-\sinh \left (\frac{x}{2}\right )-\cosh \left (\frac{x}{2}\right )\right )+2 \text{$\#$1}^2 \log \left (-\text{$\#$1} \sinh \left (\frac{x}{2}\right )+\text{$\#$1} \cosh \left (\frac{x}{2}\right )-\sinh \left (\frac{x}{2}\right )-\cosh \left (\frac{x}{2}\right )\right )-2 \log \left (-\text{$\#$1} \sinh \left (\frac{x}{2}\right )+\text{$\#$1} \cosh \left (\frac{x}{2}\right )-\sinh \left (\frac{x}{2}\right )-\cosh \left (\frac{x}{2}\right )\right )-x}{4 \text{$\#$1}^7-3 \text{$\#$1}^5+2 \text{$\#$1}^3-\text{$\#$1}}\& \right ]+\cosh (x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]*Tanh[5*x],x]

[Out]

Cosh[x] + RootSum[1 - #1^2 + #1^4 - #1^6 + #1^8 & , (-x - 2*Log[-Cosh[x/2] - Sinh[x/2] + Cosh[x/2]*#1 - Sinh[x
/2]*#1] + x*#1^2 + 2*Log[-Cosh[x/2] - Sinh[x/2] + Cosh[x/2]*#1 - Sinh[x/2]*#1]*#1^2 - x*#1^4 - 2*Log[-Cosh[x/2
] - Sinh[x/2] + Cosh[x/2]*#1 - Sinh[x/2]*#1]*#1^4 + x*#1^6 + 2*Log[-Cosh[x/2] - Sinh[x/2] + Cosh[x/2]*#1 - Sin
h[x/2]*#1]*#1^6)/(-#1 + 2*#1^3 - 3*#1^5 + 4*#1^7) & ]/4

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Maple [A]  time = 0.061, size = 70, normalized size = 0.9 \begin{align*} \cosh \left ( x \right ) -{\frac{ \left ( \sqrt{5}-1 \right ) \sqrt{5}}{5\,\sqrt{10-2\,\sqrt{5}}}{\it Artanh} \left ( 4\,{\frac{\cosh \left ( x \right ) }{\sqrt{10-2\,\sqrt{5}}}} \right ) }-{\frac{\sqrt{5} \left ( \sqrt{5}+1 \right ) }{5\,\sqrt{10+2\,\sqrt{5}}}{\it Artanh} \left ( 4\,{\frac{\cosh \left ( x \right ) }{\sqrt{10+2\,\sqrt{5}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)*tanh(5*x),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (e^{\left (2 \, x\right )} + 1\right )} e^{\left (-x\right )} + \frac{1}{2} \, \int \frac{2 \,{\left (e^{\left (7 \, x\right )} - e^{\left (5 \, x\right )} + e^{\left (3 \, x\right )} - e^{x}\right )}}{e^{\left (8 \, x\right )} - e^{\left (6 \, x\right )} + e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*tanh(5*x),x, algorithm="maxima")

[Out]

1/2*(e^(2*x) + 1)*e^(-x) + 1/2*integrate(2*(e^(7*x) - e^(5*x) + e^(3*x) - e^x)/(e^(8*x) - e^(6*x) + e^(4*x) -
e^(2*x) + 1), x)

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Fricas [B]  time = 2.15884, size = 1019, normalized size = 12.43 \begin{align*} -\frac{{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{\sqrt{5} + 5} \log \left (2 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} +{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{\sqrt{5} + 5} + 2\right ) -{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{\sqrt{5} + 5} \log \left (2 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} -{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{\sqrt{5} + 5} + 2\right ) +{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{-\sqrt{5} + 5} \log \left (2 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} +{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{-\sqrt{5} + 5} + 2\right ) -{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{-\sqrt{5} + 5} \log \left (2 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} -{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{-\sqrt{5} + 5} + 2\right ) - 10 \, \cosh \left (x\right )^{2} - 20 \, \cosh \left (x\right ) \sinh \left (x\right ) - 10 \, \sinh \left (x\right )^{2} - 10}{20 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*tanh(5*x),x, algorithm="fricas")

[Out]

-1/20*((sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(sqrt(5) + 5)*log(2*cosh(x)^2 + 4*cosh(x)*sinh(x) + 2*sinh(x)^2
+ (sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(sqrt(5) + 5) + 2) - (sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(sqrt(
5) + 5)*log(2*cosh(x)^2 + 4*cosh(x)*sinh(x) + 2*sinh(x)^2 - (sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(sqrt(5) +
5) + 2) + (sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(-sqrt(5) + 5)*log(2*cosh(x)^2 + 4*cosh(x)*sinh(x) + 2*sinh
(x)^2 + (sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(-sqrt(5) + 5) + 2) - (sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt
(-sqrt(5) + 5)*log(2*cosh(x)^2 + 4*cosh(x)*sinh(x) + 2*sinh(x)^2 - (sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(-s
qrt(5) + 5) + 2) - 10*cosh(x)^2 - 20*cosh(x)*sinh(x) - 10*sinh(x)^2 - 10)/(cosh(x) + sinh(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*tanh(5*x),x)

[Out]

Integral(cosh(x)*tanh(5*x), x)

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Giac [B]  time = 1.23672, size = 171, normalized size = 2.09 \begin{align*} -\frac{1}{20} \, \sqrt{2 \, \sqrt{5} + 10} \log \left (\sqrt{\frac{1}{2} \, \sqrt{5} + \frac{5}{2}} + e^{\left (-x\right )} + e^{x}\right ) + \frac{1}{20} \, \sqrt{2 \, \sqrt{5} + 10} \log \left (-\sqrt{\frac{1}{2} \, \sqrt{5} + \frac{5}{2}} + e^{\left (-x\right )} + e^{x}\right ) - \frac{1}{20} \, \sqrt{-2 \, \sqrt{5} + 10} \log \left (\sqrt{-\frac{1}{2} \, \sqrt{5} + \frac{5}{2}} + e^{\left (-x\right )} + e^{x}\right ) + \frac{1}{20} \, \sqrt{-2 \, \sqrt{5} + 10} \log \left (-\sqrt{-\frac{1}{2} \, \sqrt{5} + \frac{5}{2}} + e^{\left (-x\right )} + e^{x}\right ) + \frac{1}{2} \, e^{\left (-x\right )} + \frac{1}{2} \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*tanh(5*x),x, algorithm="giac")

[Out]

-1/20*sqrt(2*sqrt(5) + 10)*log(sqrt(1/2*sqrt(5) + 5/2) + e^(-x) + e^x) + 1/20*sqrt(2*sqrt(5) + 10)*log(-sqrt(1
/2*sqrt(5) + 5/2) + e^(-x) + e^x) - 1/20*sqrt(-2*sqrt(5) + 10)*log(sqrt(-1/2*sqrt(5) + 5/2) + e^(-x) + e^x) +
1/20*sqrt(-2*sqrt(5) + 10)*log(-sqrt(-1/2*sqrt(5) + 5/2) + e^(-x) + e^x) + 1/2*e^(-x) + 1/2*e^x