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

Optimal. Leaf size=36 \[ \frac{\cosh (x)}{b}-\frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \cosh (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0646584, antiderivative size = 36, 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 = {3190, 388, 205} \[ \frac{\cosh (x)}{b}-\frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \cosh (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(a + b*Cosh[x]^2),x]

[Out]

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

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[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{\sinh ^3(x)}{a+b \cosh ^2(x)} \, dx &=-\operatorname{Subst}\left (\int \frac{1-x^2}{a+b x^2} \, dx,x,\cosh (x)\right )\\ &=\frac{\cosh (x)}{b}-\frac{(a+b) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\cosh (x)\right )}{b}\\ &=-\frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \cosh (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{3/2}}+\frac{\cosh (x)}{b}\\ \end{align*}

Mathematica [C]  time = 0.175573, size = 83, normalized size = 2.31 \[ \frac{\cosh (x)}{b}-\frac{(a+b) \left (\tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a+b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a}}\right )+\tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a+b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a}}\right )\right )}{\sqrt{a} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b)*(ArcTan[(Sqrt[b] - I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a]] + ArcTan[(Sqrt[b] + I*Sqrt[a + b]*Tanh[x/2])/S
qrt[a]]))/(Sqrt[a]*b^(3/2))) + Cosh[x]/b

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Maple [B]  time = 0.017, size = 97, normalized size = 2.7 \begin{align*} -{\frac{a}{b}\arctan \left ({\frac{1}{4} \left ( 2\, \left ( a+b \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,a+2\,b \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\arctan \left ({\frac{1}{4} \left ( 2\, \left ( a+b \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,a+2\,b \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\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(sinh(x)^3/(a+b*cosh(x)^2),x)

[Out]

-1/b/(a*b)^(1/2)*arctan(1/4*(2*(a+b)*tanh(1/2*x)^2-2*a+2*b)/(a*b)^(1/2))*a-1/(a*b)^(1/2)*arctan(1/4*(2*(a+b)*t
anh(1/2*x)^2-2*a+2*b)/(a*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 ({\left (a + b\right )} e^{\left (3 \, x\right )} -{\left (a + b\right )} 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(sinh(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 + b)*e^(3*x) - (a + b)*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 = 2.00023, size = 1305, normalized size = 36.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(a*b*cosh(x)^2 + 2*a*b*cosh(x)*sinh(x) + a*b*sinh(x)^2 - sqrt(-a*b)*((a + b)*cosh(x) + (a + b)*sinh(x))*l
og((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)*si
nh(x)^2 + 4*(b*cosh(x)^3 - (2*a - b)*cosh(x))*sinh(x) + 4*(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 + (3*co
sh(x)^2 + 1)*sinh(x) + cosh(x))*sqrt(-a*b) + 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
)/(a*b^2*cosh(x) + a*b^2*sinh(x)), 1/2*(a*b*cosh(x)^2 + 2*a*b*cosh(x)*sinh(x) + a*b*sinh(x)^2 - 2*sqrt(a*b)*((
a + b)*cosh(x) + (a + b)*sinh(x))*arctan(1/2*sqrt(a*b)*(cosh(x) + sinh(x))/a) + 2*sqrt(a*b)*((a + b)*cosh(x) +
(a + b)*sinh(x))*arctan(1/2*(b*cosh(x)^3 + 3*b*cosh(x)*sinh(x)^2 + b*sinh(x)^3 + (4*a + b)*cosh(x) + (3*b*cos
h(x)^2 + 4*a + b)*sinh(x))*sqrt(a*b)/(a*b)) + a*b)/(a*b^2*cosh(x) + a*b^2*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(sinh(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(sinh(x)^3/(a+b*cosh(x)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError