3.351 \(\int \frac{\cosh ^4(x)}{1-\sinh ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ -\frac{5 x}{2}+2 \sqrt{2} \tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )-\frac{1}{2} \sinh (x) \cosh (x) \]

[Out]

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

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Rubi [A]  time = 0.0589306, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {3191, 414, 522, 206} \[ -\frac{5 x}{2}+2 \sqrt{2} \tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )-\frac{1}{2} \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(1 - Sinh[x]^2),x]

[Out]

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

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\cosh ^4(x)}{1-\sinh ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1-2 x^2\right ) \left (1-x^2\right )^2} \, dx,x,\tanh (x)\right )\\ &=-\frac{1}{2} \cosh (x) \sinh (x)-\frac{1}{2} \operatorname{Subst}\left (\int \frac{-3-2 x^2}{\left (1-2 x^2\right ) \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=-\frac{1}{2} \cosh (x) \sinh (x)-\frac{5}{2} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (x)\right )+4 \operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\tanh (x)\right )\\ &=-\frac{5 x}{2}+2 \sqrt{2} \tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )-\frac{1}{2} \cosh (x) \sinh (x)\\ \end{align*}

Mathematica [A]  time = 0.0686895, size = 32, normalized size = 1.07 \[ -2 \left (\frac{5 x}{4}+\frac{1}{8} \sinh (2 x)-\sqrt{2} \tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^4/(1 - Sinh[x]^2),x]

[Out]

-2*((5*x)/4 - Sqrt[2]*ArcTanh[Sqrt[2]*Tanh[x]] + Sinh[2*x]/8)

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Maple [B]  time = 0.032, size = 98, normalized size = 3.3 \begin{align*} 2\,\sqrt{2}{\it Artanh} \left ( 1/4\, \left ( 2\,\tanh \left ( x/2 \right ) +2 \right ) \sqrt{2} \right ) +{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{5}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+2\,\sqrt{2}{\it Artanh} \left ( 1/4\, \left ( 2\,\tanh \left ( x/2 \right ) -2 \right ) \sqrt{2} \right ) -{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{5}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(1-sinh(x)^2),x)

[Out]

2*2^(1/2)*arctanh(1/4*(2*tanh(1/2*x)+2)*2^(1/2))+1/2/(tanh(1/2*x)+1)^2-1/2/(tanh(1/2*x)+1)-5/2*ln(tanh(1/2*x)+
1)+2*2^(1/2)*arctanh(1/4*(2*tanh(1/2*x)-2)*2^(1/2))-1/2/(tanh(1/2*x)-1)^2-1/2/(tanh(1/2*x)-1)+5/2*ln(tanh(1/2*
x)-1)

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Maxima [B]  time = 1.53231, size = 101, normalized size = 3.37 \begin{align*} \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{\left (-x\right )} + 1}{\sqrt{2} + e^{\left (-x\right )} - 1}\right ) - \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{\left (-x\right )} - 1}{\sqrt{2} + e^{\left (-x\right )} + 1}\right ) - \frac{5}{2} \, x - \frac{1}{8} \, e^{\left (2 \, x\right )} + \frac{1}{8} \, e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(1-sinh(x)^2),x, algorithm="maxima")

[Out]

sqrt(2)*log(-(sqrt(2) - e^(-x) + 1)/(sqrt(2) + e^(-x) - 1)) - sqrt(2)*log(-(sqrt(2) - e^(-x) - 1)/(sqrt(2) + e
^(-x) + 1)) - 5/2*x - 1/8*e^(2*x) + 1/8*e^(-2*x)

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Fricas [B]  time = 1.51314, size = 552, normalized size = 18.4 \begin{align*} -\frac{\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 20 \, x \cosh \left (x\right )^{2} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 10 \, x\right )} \sinh \left (x\right )^{2} - 8 \,{\left (\sqrt{2} \cosh \left (x\right )^{2} + 2 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt{2} \sinh \left (x\right )^{2}\right )} \log \left (-\frac{3 \,{\left (2 \, \sqrt{2} - 3\right )} \cosh \left (x\right )^{2} - 4 \,{\left (3 \, \sqrt{2} - 4\right )} \cosh \left (x\right ) \sinh \left (x\right ) + 3 \,{\left (2 \, \sqrt{2} - 3\right )} \sinh \left (x\right )^{2} - 2 \, \sqrt{2} + 3}{\cosh \left (x\right )^{2} + \sinh \left (x\right )^{2} - 3}\right ) + 4 \,{\left (\cosh \left (x\right )^{3} + 10 \, x \cosh \left (x\right )\right )} \sinh \left (x\right ) - 1}{8 \,{\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(1-sinh(x)^2),x, algorithm="fricas")

[Out]

-1/8*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 20*x*cosh(x)^2 + 2*(3*cosh(x)^2 + 10*x)*sinh(x)^2 - 8*(sqr
t(2)*cosh(x)^2 + 2*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2)*log(-(3*(2*sqrt(2) - 3)*cosh(x)^2 - 4*(3*sqrt(
2) - 4)*cosh(x)*sinh(x) + 3*(2*sqrt(2) - 3)*sinh(x)^2 - 2*sqrt(2) + 3)/(cosh(x)^2 + sinh(x)^2 - 3)) + 4*(cosh(
x)^3 + 10*x*cosh(x))*sinh(x) - 1)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)

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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(cosh(x)**4/(1-sinh(x)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.15401, size = 82, normalized size = 2.73 \begin{align*} \frac{1}{8} \,{\left (10 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )} - \sqrt{2} \log \left (\frac{{\left | -4 \, \sqrt{2} + 2 \, e^{\left (2 \, x\right )} - 6 \right |}}{{\left | 4 \, \sqrt{2} + 2 \, e^{\left (2 \, x\right )} - 6 \right |}}\right ) - \frac{5}{2} \, x - \frac{1}{8} \, e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(1-sinh(x)^2),x, algorithm="giac")

[Out]

1/8*(10*e^(2*x) + 1)*e^(-2*x) - sqrt(2)*log(abs(-4*sqrt(2) + 2*e^(2*x) - 6)/abs(4*sqrt(2) + 2*e^(2*x) - 6)) -
5/2*x - 1/8*e^(2*x)