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

Optimal. Leaf size=101 \[ \frac {a^2 b x}{\left (a^2-b^2\right )^2}+\frac {b x}{2 \left (a^2-b^2\right )}+\frac {a \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {b \sinh (x) \cosh (x)}{2 \left (a^2-b^2\right )}-\frac {a^3 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2} \]

[Out]

a^2*b*x/(a^2-b^2)^2+1/2*b*x/(a^2-b^2)-a^3*ln(a*cosh(x)+b*sinh(x))/(a^2-b^2)^2-1/2*b*cosh(x)*sinh(x)/(a^2-b^2)+
1/2*a*sinh(x)^2/(a^2-b^2)

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Rubi [A]  time = 0.13, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3099, 3097, 3133, 2635, 8} \[ \frac {a^2 b x}{\left (a^2-b^2\right )^2}+\frac {b x}{2 \left (a^2-b^2\right )}+\frac {a \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {b \sinh (x) \cosh (x)}{2 \left (a^2-b^2\right )}-\frac {a^3 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*b*x)/(a^2 - b^2)^2 + (b*x)/(2*(a^2 - b^2)) - (a^3*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 - b^2)^2 - (b*Cosh[x]*
Sinh[x])/(2*(a^2 - b^2)) + (a*Sinh[x]^2)/(2*(a^2 - b^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3097

Int[sin[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(b*x)/(a^2 + b^2), x] - Dist[a/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3099

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

Rule 3133

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + Simp[((c*B - b*C)*L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

\begin {align*} \int \frac {\sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx &=\frac {a \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {a^2 \int \frac {\sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}-\frac {b \int \sinh ^2(x) \, dx}{a^2-b^2}\\ &=\frac {a^2 b x}{\left (a^2-b^2\right )^2}-\frac {b \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )}+\frac {a \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {\left (i a^3\right ) \int \frac {-i b \cosh (x)-i a \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {b \int 1 \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac {a^2 b x}{\left (a^2-b^2\right )^2}+\frac {b x}{2 \left (a^2-b^2\right )}-\frac {a^3 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2}-\frac {b \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )}+\frac {a \sinh ^2(x)}{2 \left (a^2-b^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 75, normalized size = 0.74 \[ \frac {-4 a^3 \log (a \cosh (x)+b \sinh (x))+\left (b^3-a^2 b\right ) \sinh (2 x)+a \left (a^2-b^2\right ) \cosh (2 x)+6 a^2 b x-2 b^3 x}{4 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^2*b*x - 2*b^3*x + a*(a^2 - b^2)*Cosh[2*x] - 4*a^3*Log[a*Cosh[x] + b*Sinh[x]] + (-(a^2*b) + b^3)*Sinh[2*x]
)/(4*(a - b)^2*(a + b)^2)

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fricas [B]  time = 0.43, size = 337, normalized size = 3.34 \[ \frac {{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \relax (x)^{4} + 4 \, {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} x \cosh \relax (x)^{2} + a^{3} + a^{2} b - a b^{2} - b^{3} + 2 \, {\left (3 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{2} + 2 \, {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} x\right )} \sinh \relax (x)^{2} - 8 \, {\left (a^{3} \cosh \relax (x)^{2} + 2 \, a^{3} \cosh \relax (x) \sinh \relax (x) + a^{3} \sinh \relax (x)^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 4 \, {\left ({\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{3} + 2 \, {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} x \cosh \relax (x)\right )} \sinh \relax (x)}{8 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^4 + 4*(a^3 - a^2*b - a*b^2 + b^3)*cosh(x)*sinh(x)^3 + (a^3 - a^2*b -
a*b^2 + b^3)*sinh(x)^4 + 4*(2*a^3 + 3*a^2*b - b^3)*x*cosh(x)^2 + a^3 + a^2*b - a*b^2 - b^3 + 2*(3*(a^3 - a^2*b
- a*b^2 + b^3)*cosh(x)^2 + 2*(2*a^3 + 3*a^2*b - b^3)*x)*sinh(x)^2 - 8*(a^3*cosh(x)^2 + 2*a^3*cosh(x)*sinh(x)
+ a^3*sinh(x)^2)*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x))) + 4*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^3
+ 2*(2*a^3 + 3*a^2*b - b^3)*x*cosh(x))*sinh(x))/((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 + 2*(a^4 - 2*a^2*b^2 + b^4
)*cosh(x)*sinh(x) + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^2)

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giac [A]  time = 0.14, size = 114, normalized size = 1.13 \[ -\frac {a^{3} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (2 \, a - b\right )} x}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (4 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} - a + b\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^4 - 2*a^2*b^2 + b^4) + 1/2*(2*a - b)*x/(a^2 - 2*a*b + b^2) - 1
/8*(4*a*e^(2*x) - 2*b*e^(2*x) - a + b)*e^(-2*x)/(a^2 - 2*a*b + b^2) + 1/8*e^(2*x)/(a + b)

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maple [A]  time = 0.22, size = 175, normalized size = 1.73 \[ \frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {16}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a}{\left (a +b \right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b}{2 \left (a +b \right )^{2}}-\frac {a^{3} \ln \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {16}{\left (32 a -32 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {8}{\left (16 a -16 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a}{\left (a -b \right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b}{2 \left (a -b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a*cosh(x)+b*sinh(x)),x)

[Out]

8/(16*a+16*b)/(tanh(1/2*x)-1)^2+16/(32*a+32*b)/(tanh(1/2*x)-1)+1/(a+b)^2*ln(tanh(1/2*x)-1)*a+1/2/(a+b)^2*ln(ta
nh(1/2*x)-1)*b-a^3/(a-b)^2/(a+b)^2*ln(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)-16/(32*a-32*b)/(tanh(1/2*x)+1)+8/(16*
a-16*b)/(tanh(1/2*x)+1)^2+1/(a-b)^2*ln(tanh(1/2*x)+1)*a-1/2/(a-b)^2*ln(tanh(1/2*x)+1)*b

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maxima [A]  time = 0.56, size = 87, normalized size = 0.86 \[ -\frac {a^{3} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (2 \, a + b\right )} x}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} + \frac {e^{\left (-2 \, x\right )}}{8 \, {\left (a - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*log(-(a - b)*e^(-2*x) - a - b)/(a^4 - 2*a^2*b^2 + b^4) - 1/2*(2*a + b)*x/(a^2 + 2*a*b + b^2) + 1/8*e^(2*x
)/(a + b) + 1/8*e^(-2*x)/(a - b)

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mupad [B]  time = 1.96, size = 86, normalized size = 0.85 \[ \frac {{\mathrm {e}}^{-2\,x}}{8\,a-8\,b}+\frac {{\mathrm {e}}^{2\,x}}{8\,a+8\,b}-\frac {a^3\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {x\,\left (2\,a-b\right )}{2\,{\left (a-b\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a*cosh(x) + b*sinh(x)),x)

[Out]

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

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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)**3/(a*cosh(x)+b*sinh(x)),x)

[Out]

Timed out

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