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

Optimal. Leaf size=56 \[ \frac{a^2 \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{b^{5/2} \sqrt{a+b}}-\frac{(a-b) \sinh (x)}{b^2}+\frac{\sinh ^3(x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0729572, antiderivative size = 56, 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 = {3186, 390, 205} \[ \frac{a^2 \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{b^{5/2} \sqrt{a+b}}-\frac{(a-b) \sinh (x)}{b^2}+\frac{\sinh ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^5/(a + b*Cosh[x]^2),x]

[Out]

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

Mathematica [A]  time = 0.161387, size = 61, normalized size = 1.09 \[ -\frac{a^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \text{csch}(x)}{\sqrt{b}}\right )}{b^{5/2} \sqrt{a+b}}-\frac{(4 a-3 b) \sinh (x)}{4 b^2}+\frac{\sinh (3 x)}{12 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*ArcTan[(Sqrt[a + b]*Csch[x])/Sqrt[b]])/(b^(5/2)*Sqrt[a + b])) - ((4*a - 3*b)*Sinh[x])/(4*b^2) + Sinh[3*
x]/(12*b)

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Maple [B]  time = 0.036, size = 177, normalized size = 3.2 \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{1}{b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{{a}^{2}\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{5}{2}}}{\frac{1}{\sqrt{a+b}}}}-{{a}^{2}\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{5}{2}}}{\frac{1}{\sqrt{a+b}}}}-{\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{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)^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-1/b/(tanh(1/2*x)+1)+a^2/b^(5/2)/(a+b)
^(1/2)*arctan(1/2*(2*tanh(1/2*x)*(a+b)^(1/2)+2*a^(1/2))/b^(1/2))-a^2/b^(5/2)/(a+b)^(1/2)*arctan(1/2*(-2*tanh(1
/2*x)*(a+b)^(1/2)+2*a^(1/2))/b^(1/2))-1/3/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-
1/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 - 3 \, b\right )} e^{\left (4 \, x\right )} + 3 \,{\left (4 \, a - 3 \, b\right )} e^{\left (2 \, x\right )} - b\right )} e^{\left (-3 \, x\right )}}{24 \, b^{2}} + \frac{1}{32} \, \int \frac{64 \,{\left (a^{2} e^{\left (3 \, x\right )} + a^{2} 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(cosh(x)^5/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

1/24*(b*e^(6*x) - 3*(4*a - 3*b)*e^(4*x) + 3*(4*a - 3*b)*e^(2*x) - b)*e^(-3*x)/b^2 + 1/32*integrate(64*(a^2*e^(
3*x) + a^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 = 2.03053, size = 3143, normalized size = 56.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*((a*b^2 + b^3)*cosh(x)^6 + 6*(a*b^2 + b^3)*cosh(x)*sinh(x)^5 + (a*b^2 + b^3)*sinh(x)^6 - 3*(4*a^2*b + a*
b^2 - 3*b^3)*cosh(x)^4 - 3*(4*a^2*b + a*b^2 - 3*b^3 - 5*(a*b^2 + b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a*b^2 + b^3
)*cosh(x)^3 - 3*(4*a^2*b + a*b^2 - 3*b^3)*cosh(x))*sinh(x)^3 - a*b^2 - b^3 + 3*(4*a^2*b + a*b^2 - 3*b^3)*cosh(
x)^2 + 3*(5*(a*b^2 + b^3)*cosh(x)^4 + 4*a^2*b + a*b^2 - 3*b^3 - 6*(4*a^2*b + a*b^2 - 3*b^3)*cosh(x)^2)*sinh(x)
^2 - 12*(a^2*cosh(x)^3 + 3*a^2*cosh(x)^2*sinh(x) + 3*a^2*cosh(x)*sinh(x)^2 + a^2*sinh(x)^3)*sqrt(-a*b - b^2)*l
og((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*cosh(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)) + 6*((a*b^2 + b^3)*cosh(x)^5 - 2*(4*a^2*b + a*b^2 - 3*b^3)*cosh(x)^3 + (4*a^2*b + a*b^2 - 3*b^3)*cosh(x
))*sinh(x))/((a*b^3 + b^4)*cosh(x)^3 + 3*(a*b^3 + b^4)*cosh(x)^2*sinh(x) + 3*(a*b^3 + b^4)*cosh(x)*sinh(x)^2 +
(a*b^3 + b^4)*sinh(x)^3), 1/24*((a*b^2 + b^3)*cosh(x)^6 + 6*(a*b^2 + b^3)*cosh(x)*sinh(x)^5 + (a*b^2 + b^3)*s
inh(x)^6 - 3*(4*a^2*b + a*b^2 - 3*b^3)*cosh(x)^4 - 3*(4*a^2*b + a*b^2 - 3*b^3 - 5*(a*b^2 + b^3)*cosh(x)^2)*sin
h(x)^4 + 4*(5*(a*b^2 + b^3)*cosh(x)^3 - 3*(4*a^2*b + a*b^2 - 3*b^3)*cosh(x))*sinh(x)^3 - a*b^2 - b^3 + 3*(4*a^
2*b + a*b^2 - 3*b^3)*cosh(x)^2 + 3*(5*(a*b^2 + b^3)*cosh(x)^4 + 4*a^2*b + a*b^2 - 3*b^3 - 6*(4*a^2*b + a*b^2 -
3*b^3)*cosh(x)^2)*sinh(x)^2 + 24*(a^2*cosh(x)^3 + 3*a^2*cosh(x)^2*sinh(x) + 3*a^2*cosh(x)*sinh(x)^2 + a^2*sin
h(x)^3)*sqrt(a*b + b^2)*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)*sinh(x))/sqrt(a*b + b^2)) + 24*(a^2*cosh(x)^3 + 3*a^2*cosh(x)^2*sinh(x) + 3*a^2*co
sh(x)*sinh(x)^2 + a^2*sinh(x)^3)*sqrt(a*b + b^2)*arctan(1/2*sqrt(a*b + b^2)*(cosh(x) + sinh(x))/(a + b)) + 6*(
(a*b^2 + b^3)*cosh(x)^5 - 2*(4*a^2*b + a*b^2 - 3*b^3)*cosh(x)^3 + (4*a^2*b + a*b^2 - 3*b^3)*cosh(x))*sinh(x))/
((a*b^3 + b^4)*cosh(x)^3 + 3*(a*b^3 + b^4)*cosh(x)^2*sinh(x) + 3*(a*b^3 + b^4)*cosh(x)*sinh(x)^2 + (a*b^3 + b^
4)*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(cosh(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(cosh(x)^5/(a+b*cosh(x)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError