3.20 \(\int \frac{\cosh ^7(x)}{a+b \cosh ^2(x)} \, dx\)

Optimal. Leaf size=78 \[ \frac{\left (a^2-a b+b^2\right ) \sinh (x)}{b^3}-\frac{a^3 \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{b^{7/2} \sqrt{a+b}}-\frac{(a-2 b) \sinh ^3(x)}{3 b^2}+\frac{\sinh ^5(x)}{5 b} \]

[Out]

-((a^3*ArcTan[(Sqrt[b]*Sinh[x])/Sqrt[a + b]])/(b^(7/2)*Sqrt[a + b])) + ((a^2 - a*b + b^2)*Sinh[x])/b^3 - ((a -
 2*b)*Sinh[x]^3)/(3*b^2) + Sinh[x]^5/(5*b)

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Rubi [A]  time = 0.0902578, antiderivative size = 78, 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} \[ \frac{\left (a^2-a b+b^2\right ) \sinh (x)}{b^3}-\frac{a^3 \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{b^{7/2} \sqrt{a+b}}-\frac{(a-2 b) \sinh ^3(x)}{3 b^2}+\frac{\sinh ^5(x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^7/(a + b*Cosh[x]^2),x]

[Out]

-((a^3*ArcTan[(Sqrt[b]*Sinh[x])/Sqrt[a + b]])/(b^(7/2)*Sqrt[a + b])) + ((a^2 - a*b + b^2)*Sinh[x])/b^3 - ((a -
 2*b)*Sinh[x]^3)/(3*b^2) + Sinh[x]^5/(5*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 ^7(x)}{a+b \cosh ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{a+b+b x^2} \, dx,x,\sinh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{a^2-a b+b^2}{b^3}-\frac{(a-2 b) x^2}{b^2}+\frac{x^4}{b}-\frac{a^3}{b^3 \left (a+b+b x^2\right )}\right ) \, dx,x,\sinh (x)\right )\\ &=\frac{\left (a^2-a b+b^2\right ) \sinh (x)}{b^3}-\frac{(a-2 b) \sinh ^3(x)}{3 b^2}+\frac{\sinh ^5(x)}{5 b}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\sinh (x)\right )}{b^3}\\ &=-\frac{a^3 \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{b^{7/2} \sqrt{a+b}}+\frac{\left (a^2-a b+b^2\right ) \sinh (x)}{b^3}-\frac{(a-2 b) \sinh ^3(x)}{3 b^2}+\frac{\sinh ^5(x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.273588, size = 86, normalized size = 1.1 \[ \frac{\left (8 a^2-6 a b+5 b^2\right ) \sinh (x)}{8 b^3}+\frac{a^3 \tan ^{-1}\left (\frac{\sqrt{a+b} \text{csch}(x)}{\sqrt{b}}\right )}{b^{7/2} \sqrt{a+b}}-\frac{(4 a-5 b) \sinh (3 x)}{48 b^2}+\frac{\sinh (5 x)}{80 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^7/(a + b*Cosh[x]^2),x]

[Out]

(a^3*ArcTan[(Sqrt[a + b]*Csch[x])/Sqrt[b]])/(b^(7/2)*Sqrt[a + b]) + ((8*a^2 - 6*a*b + 5*b^2)*Sinh[x])/(8*b^3)
- ((4*a - 5*b)*Sinh[3*x])/(48*b^2) + Sinh[5*x]/(80*b)

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Maple [B]  time = 0.046, size = 317, normalized size = 4.1 \begin{align*} -{\frac{1}{5\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-5}}+{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-4}}+{\frac{7}{8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{a}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{11}{12\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+{\frac{a}{3\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}-{\frac{{a}^{2}}{{b}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\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}^{3}\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{7}{2}}}{\frac{1}{\sqrt{a+b}}}}+{{a}^{3}\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{7}{2}}}{\frac{1}{\sqrt{a+b}}}}-{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-4}}-{\frac{1}{5\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-5}}-{\frac{7}{8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{a}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{11}{12\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}+{\frac{a}{3\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-{\frac{{a}^{2}}{{b}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\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)^7/(a+b*cosh(x)^2),x)

[Out]

-1/5/b/(tanh(1/2*x)+1)^5+1/2/b/(tanh(1/2*x)+1)^4+7/8/b/(tanh(1/2*x)+1)^2-1/2/b^2/(tanh(1/2*x)+1)^2*a-11/12/b/(
tanh(1/2*x)+1)^3+1/3/b^2/(tanh(1/2*x)+1)^3*a-1/b^3/(tanh(1/2*x)+1)*a^2+1/b^2/(tanh(1/2*x)+1)*a-1/b/(tanh(1/2*x
)+1)-a^3/b^(7/2)/(a+b)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*(a+b)^(1/2)+2*a^(1/2))/b^(1/2))+a^3/b^(7/2)/(a+b)^(1/2)
*arctan(1/2*(-2*tanh(1/2*x)*(a+b)^(1/2)+2*a^(1/2))/b^(1/2))-1/2/b/(tanh(1/2*x)-1)^4-1/5/b/(tanh(1/2*x)-1)^5-7/
8/b/(tanh(1/2*x)-1)^2+1/2/b^2/(tanh(1/2*x)-1)^2*a-11/12/b/(tanh(1/2*x)-1)^3+1/3/b^2/(tanh(1/2*x)-1)^3*a-1/b^3/
(tanh(1/2*x)-1)*a^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 (3 \, b^{2} e^{\left (10 \, x\right )} - 3 \, b^{2} - 5 \,{\left (4 \, a b - 5 \, b^{2}\right )} e^{\left (8 \, x\right )} + 30 \,{\left (8 \, a^{2} - 6 \, a b + 5 \, b^{2}\right )} e^{\left (6 \, x\right )} - 30 \,{\left (8 \, a^{2} - 6 \, a b + 5 \, b^{2}\right )} e^{\left (4 \, x\right )} + 5 \,{\left (4 \, a b - 5 \, b^{2}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-5 \, x\right )}}{480 \, b^{3}} - \frac{1}{128} \, \int \frac{256 \,{\left (a^{3} e^{\left (3 \, x\right )} + a^{3} e^{x}\right )}}{b^{4} e^{\left (4 \, x\right )} + b^{4} + 2 \,{\left (2 \, a b^{3} + b^{4}\right )} e^{\left (2 \, x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^7/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

1/480*(3*b^2*e^(10*x) - 3*b^2 - 5*(4*a*b - 5*b^2)*e^(8*x) + 30*(8*a^2 - 6*a*b + 5*b^2)*e^(6*x) - 30*(8*a^2 - 6
*a*b + 5*b^2)*e^(4*x) + 5*(4*a*b - 5*b^2)*e^(2*x))*e^(-5*x)/b^3 - 1/128*integrate(256*(a^3*e^(3*x) + a^3*e^x)/
(b^4*e^(4*x) + b^4 + 2*(2*a*b^3 + b^4)*e^(2*x)), x)

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Fricas [B]  time = 2.12411, size = 6260, normalized size = 80.26 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^7/(a+b*cosh(x)^2),x, algorithm="fricas")

[Out]

[1/480*(3*(a*b^3 + b^4)*cosh(x)^10 + 30*(a*b^3 + b^4)*cosh(x)*sinh(x)^9 + 3*(a*b^3 + b^4)*sinh(x)^10 - 5*(4*a^
2*b^2 - a*b^3 - 5*b^4)*cosh(x)^8 - 5*(4*a^2*b^2 - a*b^3 - 5*b^4 - 27*(a*b^3 + b^4)*cosh(x)^2)*sinh(x)^8 + 40*(
9*(a*b^3 + b^4)*cosh(x)^3 - (4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x))*sinh(x)^7 + 30*(8*a^3*b + 2*a^2*b^2 - a*b^3 +
 5*b^4)*cosh(x)^6 + 10*(63*(a*b^3 + b^4)*cosh(x)^4 + 24*a^3*b + 6*a^2*b^2 - 3*a*b^3 + 15*b^4 - 14*(4*a^2*b^2 -
 a*b^3 - 5*b^4)*cosh(x)^2)*sinh(x)^6 + 4*(189*(a*b^3 + b^4)*cosh(x)^5 - 70*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)
^3 + 45*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x))*sinh(x)^5 - 30*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*co
sh(x)^4 + 10*(63*(a*b^3 + b^4)*cosh(x)^6 - 35*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^4 - 24*a^3*b - 6*a^2*b^2 + 3
*a*b^3 - 15*b^4 + 45*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^2)*sinh(x)^4 - 3*a*b^3 - 3*b^4 + 40*(9*(a*b
^3 + b^4)*cosh(x)^7 - 7*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^5 + 15*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(
x)^3 - 3*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x))*sinh(x)^3 + 5*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^2 +
5*(27*(a*b^3 + b^4)*cosh(x)^8 - 28*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^6 + 90*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5
*b^4)*cosh(x)^4 + 4*a^2*b^2 - a*b^3 - 5*b^4 - 36*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^2)*sinh(x)^2 -
240*(a^3*cosh(x)^5 + 5*a^3*cosh(x)^4*sinh(x) + 10*a^3*cosh(x)^3*sinh(x)^2 + 10*a^3*cosh(x)^2*sinh(x)^3 + 5*a^3
*cosh(x)*sinh(x)^4 + a^3*sinh(x)^5)*sqrt(-a*b - b^2)*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*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)*s
inh(x)^2 + 4*(b*cosh(x)^3 + (2*a + b)*cosh(x))*sinh(x) + b)) + 10*(3*(a*b^3 + b^4)*cosh(x)^9 - 4*(4*a^2*b^2 -
a*b^3 - 5*b^4)*cosh(x)^7 + 18*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^5 - 12*(8*a^3*b + 2*a^2*b^2 - a*b^
3 + 5*b^4)*cosh(x)^3 + (4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x))*sinh(x))/((a*b^4 + b^5)*cosh(x)^5 + 5*(a*b^4 + b^5
)*cosh(x)^4*sinh(x) + 10*(a*b^4 + b^5)*cosh(x)^3*sinh(x)^2 + 10*(a*b^4 + b^5)*cosh(x)^2*sinh(x)^3 + 5*(a*b^4 +
 b^5)*cosh(x)*sinh(x)^4 + (a*b^4 + b^5)*sinh(x)^5), 1/480*(3*(a*b^3 + b^4)*cosh(x)^10 + 30*(a*b^3 + b^4)*cosh(
x)*sinh(x)^9 + 3*(a*b^3 + b^4)*sinh(x)^10 - 5*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^8 - 5*(4*a^2*b^2 - a*b^3 - 5
*b^4 - 27*(a*b^3 + b^4)*cosh(x)^2)*sinh(x)^8 + 40*(9*(a*b^3 + b^4)*cosh(x)^3 - (4*a^2*b^2 - a*b^3 - 5*b^4)*cos
h(x))*sinh(x)^7 + 30*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^6 + 10*(63*(a*b^3 + b^4)*cosh(x)^4 + 24*a^3
*b + 6*a^2*b^2 - 3*a*b^3 + 15*b^4 - 14*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^2)*sinh(x)^6 + 4*(189*(a*b^3 + b^4)
*cosh(x)^5 - 70*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^3 + 45*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x))*sinh
(x)^5 - 30*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^4 + 10*(63*(a*b^3 + b^4)*cosh(x)^6 - 35*(4*a^2*b^2 -
a*b^3 - 5*b^4)*cosh(x)^4 - 24*a^3*b - 6*a^2*b^2 + 3*a*b^3 - 15*b^4 + 45*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*
cosh(x)^2)*sinh(x)^4 - 3*a*b^3 - 3*b^4 + 40*(9*(a*b^3 + b^4)*cosh(x)^7 - 7*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)
^5 + 15*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^3 - 3*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x))*sin
h(x)^3 + 5*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^2 + 5*(27*(a*b^3 + b^4)*cosh(x)^8 - 28*(4*a^2*b^2 - a*b^3 - 5*b
^4)*cosh(x)^6 + 90*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^4 + 4*a^2*b^2 - a*b^3 - 5*b^4 - 36*(8*a^3*b +
 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^2)*sinh(x)^2 - 480*(a^3*cosh(x)^5 + 5*a^3*cosh(x)^4*sinh(x) + 10*a^3*cosh(
x)^3*sinh(x)^2 + 10*a^3*cosh(x)^2*sinh(x)^3 + 5*a^3*cosh(x)*sinh(x)^4 + a^3*sinh(x)^5)*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)*sin
h(x))/sqrt(a*b + b^2)) - 480*(a^3*cosh(x)^5 + 5*a^3*cosh(x)^4*sinh(x) + 10*a^3*cosh(x)^3*sinh(x)^2 + 10*a^3*co
sh(x)^2*sinh(x)^3 + 5*a^3*cosh(x)*sinh(x)^4 + a^3*sinh(x)^5)*sqrt(a*b + b^2)*arctan(1/2*sqrt(a*b + b^2)*(cosh(
x) + sinh(x))/(a + b)) + 10*(3*(a*b^3 + b^4)*cosh(x)^9 - 4*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^7 + 18*(8*a^3*b
 + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^5 - 12*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^3 + (4*a^2*b^2 - a*
b^3 - 5*b^4)*cosh(x))*sinh(x))/((a*b^4 + b^5)*cosh(x)^5 + 5*(a*b^4 + b^5)*cosh(x)^4*sinh(x) + 10*(a*b^4 + b^5)
*cosh(x)^3*sinh(x)^2 + 10*(a*b^4 + b^5)*cosh(x)^2*sinh(x)^3 + 5*(a*b^4 + b^5)*cosh(x)*sinh(x)^4 + (a*b^4 + b^5
)*sinh(x)^5)]

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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)**7/(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)^7/(a+b*cosh(x)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError