3.7 \(\int \frac{\sinh ^5(x)}{a+b \cosh ^2(x)} \, dx\)

Optimal. Leaf size=54 \[ -\frac{(a+2 b) \cosh (x)}{b^2}+\frac{(a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \cosh (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{5/2}}+\frac{\cosh ^3(x)}{3 b} \]

[Out]

((a + b)^2*ArcTan[(Sqrt[b]*Cosh[x])/Sqrt[a]])/(Sqrt[a]*b^(5/2)) - ((a + 2*b)*Cosh[x])/b^2 + Cosh[x]^3/(3*b)

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Rubi [A]  time = 0.0868472, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3190, 390, 205} \[ -\frac{(a+2 b) \cosh (x)}{b^2}+\frac{(a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \cosh (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{5/2}}+\frac{\cosh ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^5/(a + b*Cosh[x]^2),x]

[Out]

((a + b)^2*ArcTan[(Sqrt[b]*Cosh[x])/Sqrt[a]])/(Sqrt[a]*b^(5/2)) - ((a + 2*b)*Cosh[x])/b^2 + Cosh[x]^3/(3*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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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 ^5(x)}{a+b \cosh ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{a+b x^2} \, dx,x,\cosh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{a+2 b}{b^2}+\frac{x^2}{b}+\frac{a^2+2 a b+b^2}{b^2 \left (a+b x^2\right )}\right ) \, dx,x,\cosh (x)\right )\\ &=-\frac{(a+2 b) \cosh (x)}{b^2}+\frac{\cosh ^3(x)}{3 b}+\frac{(a+b)^2 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\cosh (x)\right )}{b^2}\\ &=\frac{(a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \cosh (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{5/2}}-\frac{(a+2 b) \cosh (x)}{b^2}+\frac{\cosh ^3(x)}{3 b}\\ \end{align*}

Mathematica [C]  time = 0.174367, size = 120, normalized size = 2.22 \[ \frac{-3 \sqrt{b} (4 a+7 b) \cosh (x)+\frac{12 (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a+b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a}}\right )}{\sqrt{a}}+\frac{12 (a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a+b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a}}\right )}{\sqrt{a}}+b^{3/2} \cosh (3 x)}{12 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^5/(a + b*Cosh[x]^2),x]

[Out]

((12*(a + b)^2*ArcTan[(Sqrt[b] - I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a]])/Sqrt[a] + (12*(a + b)^2*ArcTan[(Sqrt[b] +
I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a]])/Sqrt[a] - 3*Sqrt[b]*(4*a + 7*b)*Cosh[x] + b^(3/2)*Cosh[3*x])/(12*b^(5/2))

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Maple [B]  time = 0.025, size = 214, normalized size = 4. \begin{align*}{\frac{1}{3\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}-{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{a}{{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{3}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{{a}^{2}}{{b}^{2}}\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}}}}+2\,{\frac{a}{\sqrt{ab}b}\arctan \left ( 1/4\,{\frac{2\, \left ( a+b \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,a+2\,b}{\sqrt{ab}}} \right ) }+{\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}{3\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{a}{{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3}{2\,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)^5/(a+b*cosh(x)^2),x)

[Out]

1/3/b/(tanh(1/2*x)+1)^3-1/2/b/(tanh(1/2*x)+1)^2-1/b^2/(tanh(1/2*x)+1)*a-3/2/b/(tanh(1/2*x)+1)+1/b^2/(a*b)^(1/2
)*arctan(1/4*(2*(a+b)*tanh(1/2*x)^2-2*a+2*b)/(a*b)^(1/2))*a^2+2/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)*tanh(1/2*x)^2-2*a+2*b)/(a*b)^(1/2))-1/3/b/(tanh(1/
2*x)-1)^3-1/2/b/(tanh(1/2*x)-1)^2+1/b^2/(tanh(1/2*x)-1)*a+3/2/b/(tanh(1/2*x)-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (b e^{\left (6 \, x\right )} - 3 \,{\left (4 \, a + 7 \, b\right )} e^{\left (4 \, x\right )} - 3 \,{\left (4 \, a + 7 \, b\right )} e^{\left (2 \, x\right )} + b\right )} e^{\left (-3 \, x\right )}}{24 \, b^{2}} + \frac{1}{32} \, \int \frac{64 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (3 \, x\right )} -{\left (a^{2} + 2 \, a b + b^{2}\right )} e^{x}\right )}}{b^{3} e^{\left (4 \, x\right )} + b^{3} + 2 \,{\left (2 \, a b^{2} + b^{3}\right )} e^{\left (2 \, x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^5/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

1/24*(b*e^(6*x) - 3*(4*a + 7*b)*e^(4*x) - 3*(4*a + 7*b)*e^(2*x) + b)*e^(-3*x)/b^2 + 1/32*integrate(64*((a^2 +
2*a*b + b^2)*e^(3*x) - (a^2 + 2*a*b + b^2)*e^x)/(b^3*e^(4*x) + b^3 + 2*(2*a*b^2 + b^3)*e^(2*x)), x)

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Fricas [B]  time = 2.07122, size = 2957, normalized size = 54.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^5/(a+b*cosh(x)^2),x, algorithm="fricas")

[Out]

[1/24*(a*b^2*cosh(x)^6 + 6*a*b^2*cosh(x)*sinh(x)^5 + a*b^2*sinh(x)^6 - 3*(4*a^2*b + 7*a*b^2)*cosh(x)^4 + 3*(5*
a*b^2*cosh(x)^2 - 4*a^2*b - 7*a*b^2)*sinh(x)^4 + 4*(5*a*b^2*cosh(x)^3 - 3*(4*a^2*b + 7*a*b^2)*cosh(x))*sinh(x)
^3 + a*b^2 - 3*(4*a^2*b + 7*a*b^2)*cosh(x)^2 + 3*(5*a*b^2*cosh(x)^4 - 4*a^2*b - 7*a*b^2 - 6*(4*a^2*b + 7*a*b^2
)*cosh(x)^2)*sinh(x)^2 - 12*((a^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(x)^2*sinh(x) + 3*(a^2
+ 2*a*b + b^2)*cosh(x)*sinh(x)^2 + (a^2 + 2*a*b + b^2)*sinh(x)^3)*sqrt(-a*b)*log((b*cosh(x)^4 + 4*b*cosh(x)*si
nh(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) - 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)/(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)) + 6*(a*b^2*cosh(x)^5 - 2*(4*a^2*b + 7
*a*b^2)*cosh(x)^3 - (4*a^2*b + 7*a*b^2)*cosh(x))*sinh(x))/(a*b^3*cosh(x)^3 + 3*a*b^3*cosh(x)^2*sinh(x) + 3*a*b
^3*cosh(x)*sinh(x)^2 + a*b^3*sinh(x)^3), 1/24*(a*b^2*cosh(x)^6 + 6*a*b^2*cosh(x)*sinh(x)^5 + a*b^2*sinh(x)^6 -
3*(4*a^2*b + 7*a*b^2)*cosh(x)^4 + 3*(5*a*b^2*cosh(x)^2 - 4*a^2*b - 7*a*b^2)*sinh(x)^4 + 4*(5*a*b^2*cosh(x)^3
- 3*(4*a^2*b + 7*a*b^2)*cosh(x))*sinh(x)^3 + a*b^2 - 3*(4*a^2*b + 7*a*b^2)*cosh(x)^2 + 3*(5*a*b^2*cosh(x)^4 -
4*a^2*b - 7*a*b^2 - 6*(4*a^2*b + 7*a*b^2)*cosh(x)^2)*sinh(x)^2 + 24*((a^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^2 +
2*a*b + b^2)*cosh(x)^2*sinh(x) + 3*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^2 + (a^2 + 2*a*b + b^2)*sinh(x)^3)*sqrt
(a*b)*arctan(1/2*sqrt(a*b)*(cosh(x) + sinh(x))/a) - 24*((a^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^2 + 2*a*b + b^2)*
cosh(x)^2*sinh(x) + 3*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^2 + (a^2 + 2*a*b + b^2)*sinh(x)^3)*sqrt(a*b)*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*cosh(x)^2 + 4*a + b)*sinh(x)
)*sqrt(a*b)/(a*b)) + 6*(a*b^2*cosh(x)^5 - 2*(4*a^2*b + 7*a*b^2)*cosh(x)^3 - (4*a^2*b + 7*a*b^2)*cosh(x))*sinh(
x))/(a*b^3*cosh(x)^3 + 3*a*b^3*cosh(x)^2*sinh(x) + 3*a*b^3*cosh(x)*sinh(x)^2 + a*b^3*sinh(x)^3)]

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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)**5/(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)^5/(a+b*cosh(x)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError