3.24 \(\int \frac{\cosh ^3(x)}{a+b \cosh ^2(x)} \, dx\)

Optimal. Leaf size=38 \[ \frac{\sinh (x)}{b}-\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{b^{3/2} \sqrt{a+b}} \]

[Out]

-((a*ArcTan[(Sqrt[b]*Sinh[x])/Sqrt[a + b]])/(b^(3/2)*Sqrt[a + b])) + Sinh[x]/b

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Rubi [A]  time = 0.0592605, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3186, 388, 205} \[ \frac{\sinh (x)}{b}-\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{b^{3/2} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(a + b*Cosh[x]^2),x]

[Out]

-((a*ArcTan[(Sqrt[b]*Sinh[x])/Sqrt[a + b]])/(b^(3/2)*Sqrt[a + b])) + Sinh[x]/b

Rule 3186

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

Rule 388

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.028578, size = 38, normalized size = 1. \[ \frac{\sinh (x)}{b}-\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{b^{3/2} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^3/(a + b*Cosh[x]^2),x]

[Out]

-((a*ArcTan[(Sqrt[b]*Sinh[x])/Sqrt[a + b]])/(b^(3/2)*Sqrt[a + b])) + Sinh[x]/b

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Maple [B]  time = 0.033, size = 96, normalized size = 2.5 \begin{align*} -{a\arctan \left ({\frac{1}{2} \left ( 2\,\tanh \left ( x/2 \right ) \sqrt{a+b}+2\,\sqrt{a} \right ){\frac{1}{\sqrt{b}}}} \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a+b}}}}-{a\arctan \left ({\frac{1}{2} \left ( 2\,\tanh \left ( x/2 \right ) \sqrt{a+b}-2\,\sqrt{a} \right ){\frac{1}{\sqrt{b}}}} \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a+b}}}}-{\frac{1}{b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(a+b*cosh(x)^2),x)

[Out]

-a/b^(3/2)/(a+b)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*(a+b)^(1/2)+2*a^(1/2))/b^(1/2))-a/b^(3/2)/(a+b)^(1/2)*arctan(
1/2*(2*tanh(1/2*x)*(a+b)^(1/2)-2*a^(1/2))/b^(1/2))-1/b/(tanh(1/2*x)-1)-1/b/(tanh(1/2*x)+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}}{2 \, b} - \frac{1}{8} \, \int \frac{16 \,{\left (a e^{\left (3 \, x\right )} + a e^{x}\right )}}{b^{2} e^{\left (4 \, x\right )} + b^{2} + 2 \,{\left (2 \, a b + b^{2}\right )} e^{\left (2 \, x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

1/2*(e^(2*x) - 1)*e^(-x)/b - 1/8*integrate(16*(a*e^(3*x) + a*e^x)/(b^2*e^(4*x) + b^2 + 2*(2*a*b + b^2)*e^(2*x)
), x)

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Fricas [B]  time = 1.98394, size = 1443, normalized size = 37.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(a+b*cosh(x)^2),x, algorithm="fricas")

[Out]

[1/2*((a*b + b^2)*cosh(x)^2 + 2*(a*b + b^2)*cosh(x)*sinh(x) + (a*b + b^2)*sinh(x)^2 - sqrt(-a*b - b^2)*(a*cosh
(x) + a*sinh(x))*log((b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 - 2*(2*a + 3*b)*cosh(x)^2 + 2*(3*b*cos
h(x)^2 - 2*a - 3*b)*sinh(x)^2 + 4*(b*cosh(x)^3 - (2*a + 3*b)*cosh(x))*sinh(x) + 4*(cosh(x)^3 + 3*cosh(x)*sinh(
x)^2 + sinh(x)^3 + (3*cosh(x)^2 - 1)*sinh(x) - cosh(x))*sqrt(-a*b - b^2) + b)/(b*cosh(x)^4 + 4*b*cosh(x)*sinh(
x)^3 + b*sinh(x)^4 + 2*(2*a + b)*cosh(x)^2 + 2*(3*b*cosh(x)^2 + 2*a + b)*sinh(x)^2 + 4*(b*cosh(x)^3 + (2*a + b
)*cosh(x))*sinh(x) + b)) - a*b - b^2)/((a*b^2 + b^3)*cosh(x) + (a*b^2 + b^3)*sinh(x)), 1/2*((a*b + b^2)*cosh(x
)^2 + 2*(a*b + b^2)*cosh(x)*sinh(x) + (a*b + b^2)*sinh(x)^2 - 2*sqrt(a*b + b^2)*(a*cosh(x) + a*sinh(x))*arctan
(1/2*(b*cosh(x)^3 + 3*b*cosh(x)*sinh(x)^2 + b*sinh(x)^3 + (4*a + 3*b)*cosh(x) + (3*b*cosh(x)^2 + 4*a + 3*b)*si
nh(x))/sqrt(a*b + b^2)) - 2*sqrt(a*b + b^2)*(a*cosh(x) + a*sinh(x))*arctan(1/2*sqrt(a*b + b^2)*(cosh(x) + sinh
(x))/(a + b)) - a*b - b^2)/((a*b^2 + b^3)*cosh(x) + (a*b^2 + b^3)*sinh(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(cosh(x)**3/(a+b*cosh(x)**2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(a+b*cosh(x)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError