3.168 \(\int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=104 \[ -\frac {a x \left (2 a^2-3 b^2\right )}{2 b^4}+\frac {\sinh (x) \left (2 \left (a^2-b^2\right )-a b \cosh (x)\right )}{2 b^3}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}+\frac {\sinh ^3(x)}{3 b} \]

[Out]

-1/2*a*(2*a^2-3*b^2)*x/b^4+2*(a-b)^(3/2)*(a+b)^(3/2)*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/b^4+1/2*(2*a
^2-2*b^2-a*b*cosh(x))*sinh(x)/b^3+1/3*sinh(x)^3/b

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Rubi [A]  time = 0.24, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2695, 2865, 2735, 2659, 208} \[ -\frac {a x \left (2 a^2-3 b^2\right )}{2 b^4}+\frac {\sinh (x) \left (2 \left (a^2-b^2\right )-a b \cosh (x)\right )}{2 b^3}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}+\frac {\sinh ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(a + b*Cosh[x]),x]

[Out]

-(a*(2*a^2 - 3*b^2)*x)/(2*b^4) + (2*(a - b)^(3/2)*(a + b)^(3/2)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/
b^4 + ((2*(a^2 - b^2) - a*b*Cosh[x])*Sinh[x])/(2*b^3) + Sinh[x]^3/(3*b)

Rule 208

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\sinh ^4(x)}{a+b \cosh (x)} \, dx &=\frac {\sinh ^3(x)}{3 b}+\frac {\int \frac {(-b-a \cosh (x)) \sinh ^2(x)}{a+b \cosh (x)} \, dx}{b}\\ &=\frac {\left (2 \left (a^2-b^2\right )-a b \cosh (x)\right ) \sinh (x)}{2 b^3}+\frac {\sinh ^3(x)}{3 b}-\frac {\int \frac {b \left (a^2-2 b^2\right )+a \left (2 a^2-3 b^2\right ) \cosh (x)}{a+b \cosh (x)} \, dx}{2 b^3}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {\left (2 \left (a^2-b^2\right )-a b \cosh (x)\right ) \sinh (x)}{2 b^3}+\frac {\sinh ^3(x)}{3 b}+\frac {\left (a^2-b^2\right )^2 \int \frac {1}{a+b \cosh (x)} \, dx}{b^4}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {\left (2 \left (a^2-b^2\right )-a b \cosh (x)\right ) \sinh (x)}{2 b^3}+\frac {\sinh ^3(x)}{3 b}+\frac {\left (2 \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{b^4}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}+\frac {\left (2 \left (a^2-b^2\right )-a b \cosh (x)\right ) \sinh (x)}{2 b^3}+\frac {\sinh ^3(x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 95, normalized size = 0.91 \[ \frac {-12 a^3 x-24 \left (b^2-a^2\right )^{3/2} \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )+12 a^2 b \sinh (x)+18 a b^2 x-3 a b^2 \sinh (2 x)-15 b^3 \sinh (x)+b^3 \sinh (3 x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4/(a + b*Cosh[x]),x]

[Out]

(-12*a^3*x + 18*a*b^2*x - 24*(-a^2 + b^2)^(3/2)*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]] + 12*a^2*b*Sinh[x
] - 15*b^3*Sinh[x] - 3*a*b^2*Sinh[2*x] + b^3*Sinh[3*x])/(12*b^4)

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fricas [B]  time = 2.78, size = 1099, normalized size = 10.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

[1/24*(b^3*cosh(x)^6 + b^3*sinh(x)^6 - 3*a*b^2*cosh(x)^5 + 3*(2*b^3*cosh(x) - a*b^2)*sinh(x)^5 - 12*(2*a^3 - 3
*a*b^2)*x*cosh(x)^3 + 3*(4*a^2*b - 5*b^3)*cosh(x)^4 + 3*(5*b^3*cosh(x)^2 - 5*a*b^2*cosh(x) + 4*a^2*b - 5*b^3)*
sinh(x)^4 + 3*a*b^2*cosh(x) + 2*(10*b^3*cosh(x)^3 - 15*a*b^2*cosh(x)^2 - 6*(2*a^3 - 3*a*b^2)*x + 6*(4*a^2*b -
5*b^3)*cosh(x))*sinh(x)^3 - b^3 - 3*(4*a^2*b - 5*b^3)*cosh(x)^2 + 3*(5*b^3*cosh(x)^4 - 10*a*b^2*cosh(x)^3 - 4*
a^2*b + 5*b^3 - 12*(2*a^3 - 3*a*b^2)*x*cosh(x) + 6*(4*a^2*b - 5*b^3)*cosh(x)^2)*sinh(x)^2 - 24*((a^2 - b^2)*co
sh(x)^3 + 3*(a^2 - b^2)*cosh(x)^2*sinh(x) + 3*(a^2 - b^2)*cosh(x)*sinh(x)^2 + (a^2 - b^2)*sinh(x)^3)*sqrt(a^2
- b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) + 2*sq
rt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x
) + b)) + 3*(2*b^3*cosh(x)^5 - 5*a*b^2*cosh(x)^4 - 12*(2*a^3 - 3*a*b^2)*x*cosh(x)^2 + 4*(4*a^2*b - 5*b^3)*cosh
(x)^3 + a*b^2 - 2*(4*a^2*b - 5*b^3)*cosh(x))*sinh(x))/(b^4*cosh(x)^3 + 3*b^4*cosh(x)^2*sinh(x) + 3*b^4*cosh(x)
*sinh(x)^2 + b^4*sinh(x)^3), 1/24*(b^3*cosh(x)^6 + b^3*sinh(x)^6 - 3*a*b^2*cosh(x)^5 + 3*(2*b^3*cosh(x) - a*b^
2)*sinh(x)^5 - 12*(2*a^3 - 3*a*b^2)*x*cosh(x)^3 + 3*(4*a^2*b - 5*b^3)*cosh(x)^4 + 3*(5*b^3*cosh(x)^2 - 5*a*b^2
*cosh(x) + 4*a^2*b - 5*b^3)*sinh(x)^4 + 3*a*b^2*cosh(x) + 2*(10*b^3*cosh(x)^3 - 15*a*b^2*cosh(x)^2 - 6*(2*a^3
- 3*a*b^2)*x + 6*(4*a^2*b - 5*b^3)*cosh(x))*sinh(x)^3 - b^3 - 3*(4*a^2*b - 5*b^3)*cosh(x)^2 + 3*(5*b^3*cosh(x)
^4 - 10*a*b^2*cosh(x)^3 - 4*a^2*b + 5*b^3 - 12*(2*a^3 - 3*a*b^2)*x*cosh(x) + 6*(4*a^2*b - 5*b^3)*cosh(x)^2)*si
nh(x)^2 - 48*((a^2 - b^2)*cosh(x)^3 + 3*(a^2 - b^2)*cosh(x)^2*sinh(x) + 3*(a^2 - b^2)*cosh(x)*sinh(x)^2 + (a^2
- b^2)*sinh(x)^3)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + 3*(2*b
^3*cosh(x)^5 - 5*a*b^2*cosh(x)^4 - 12*(2*a^3 - 3*a*b^2)*x*cosh(x)^2 + 4*(4*a^2*b - 5*b^3)*cosh(x)^3 + a*b^2 -
2*(4*a^2*b - 5*b^3)*cosh(x))*sinh(x))/(b^4*cosh(x)^3 + 3*b^4*cosh(x)^2*sinh(x) + 3*b^4*cosh(x)*sinh(x)^2 + b^4
*sinh(x)^3)]

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giac [A]  time = 0.12, size = 146, normalized size = 1.40 \[ \frac {b^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 12 \, a^{2} e^{x} - 15 \, b^{2} e^{x}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} x}{2 \, b^{4}} + \frac {{\left (3 \, a b^{2} e^{x} - b^{3} - 3 \, {\left (4 \, a^{2} b - 5 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, b^{4}} + \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*cosh(x)),x, algorithm="giac")

[Out]

1/24*(b^2*e^(3*x) - 3*a*b*e^(2*x) + 12*a^2*e^x - 15*b^2*e^x)/b^3 - 1/2*(2*a^3 - 3*a*b^2)*x/b^4 + 1/24*(3*a*b^2
*e^x - b^3 - 3*(4*a^2*b - 5*b^3)*e^(2*x))*e^(-3*x)/b^4 + 2*(a^4 - 2*a^2*b^2 + b^4)*arctan((b*e^x + a)/sqrt(-a^
2 + b^2))/(sqrt(-a^2 + b^2)*b^4)

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maple [B]  time = 0.08, size = 338, normalized size = 3.25 \[ -\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {a^{2}}{b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a^{3} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{4}}-\frac {3 a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{2}}-\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a^{2}}{b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a^{3} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{4}}+\frac {3 a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{2}}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) a^{4}}{b^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {4 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) a^{2}}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a+b*cosh(x)),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.31, size = 222, normalized size = 2.13 \[ \frac {{\mathrm {e}}^{3\,x}}{24\,b}-\frac {{\mathrm {e}}^{-3\,x}}{24\,b}+\frac {x\,\left (3\,a\,b^2-2\,a^3\right )}{2\,b^4}+\frac {{\mathrm {e}}^x\,\left (4\,a^2-5\,b^2\right )}{8\,b^3}+\frac {a\,{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {a\,{\mathrm {e}}^{2\,x}}{8\,b^2}-\frac {{\mathrm {e}}^{-x}\,\left (4\,a^2-5\,b^2\right )}{8\,b^3}+\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x\,\left (a^4-2\,a^2\,b^2+b^4\right )}{b^5}-\frac {2\,{\left (a+b\right )}^{3/2}\,\left (b+a\,{\mathrm {e}}^x\right )\,{\left (a-b\right )}^{3/2}}{b^5}\right )\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}{b^4}-\frac {\ln \left (\frac {2\,{\left (a+b\right )}^{3/2}\,\left (b+a\,{\mathrm {e}}^x\right )\,{\left (a-b\right )}^{3/2}}{b^5}-\frac {2\,{\mathrm {e}}^x\,\left (a^4-2\,a^2\,b^2+b^4\right )}{b^5}\right )\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}{b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a + b*cosh(x)),x)

[Out]

exp(3*x)/(24*b) - exp(-3*x)/(24*b) + (x*(3*a*b^2 - 2*a^3))/(2*b^4) + (exp(x)*(4*a^2 - 5*b^2))/(8*b^3) + (a*exp
(-2*x))/(8*b^2) - (a*exp(2*x))/(8*b^2) - (exp(-x)*(4*a^2 - 5*b^2))/(8*b^3) + (log(- (2*exp(x)*(a^4 + b^4 - 2*a
^2*b^2))/b^5 - (2*(a + b)^(3/2)*(b + a*exp(x))*(a - b)^(3/2))/b^5)*(a + b)^(3/2)*(a - b)^(3/2))/b^4 - (log((2*
(a + b)^(3/2)*(b + a*exp(x))*(a - b)^(3/2))/b^5 - (2*exp(x)*(a^4 + b^4 - 2*a^2*b^2))/b^5)*(a + b)^(3/2)*(a - b
)^(3/2))/b^4

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**4/(a+b*cosh(x)),x)

[Out]

Timed out

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