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

Optimal. Leaf size=40 \[ \frac {\left (a^2-b^2\right ) \log (a+b \cosh (x))}{b^3}-\frac {a \cosh (x)}{b^2}+\frac {\cosh ^2(x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2668, 697} \[ \frac {\left (a^2-b^2\right ) \log (a+b \cosh (x))}{b^3}-\frac {a \cosh (x)}{b^2}+\frac {\cosh ^2(x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(a + b*Cosh[x]),x]

[Out]

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

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2668

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 40, normalized size = 1.00 \[ \frac {\left (a^2-b^2\right ) \log (a+b \cosh (x))}{b^3}-\frac {a \cosh (x)}{b^2}+\frac {\cosh (2 x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.42, size = 234, normalized size = 5.85 \[ \frac {b^{2} \cosh \relax (x)^{4} + b^{2} \sinh \relax (x)^{4} - 4 \, a b \cosh \relax (x)^{3} - 8 \, {\left (a^{2} - b^{2}\right )} x \cosh \relax (x)^{2} + 4 \, {\left (b^{2} \cosh \relax (x) - a b\right )} \sinh \relax (x)^{3} - 4 \, a b \cosh \relax (x) + 2 \, {\left (3 \, b^{2} \cosh \relax (x)^{2} - 6 \, a b \cosh \relax (x) - 4 \, {\left (a^{2} - b^{2}\right )} x\right )} \sinh \relax (x)^{2} + b^{2} + 8 \, {\left ({\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{2} - b^{2}\right )} \sinh \relax (x)^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 4 \, {\left (b^{2} \cosh \relax (x)^{3} - 3 \, a b \cosh \relax (x)^{2} - 4 \, {\left (a^{2} - b^{2}\right )} x \cosh \relax (x) - a b\right )} \sinh \relax (x)}{8 \, {\left (b^{3} \cosh \relax (x)^{2} + 2 \, b^{3} \cosh \relax (x) \sinh \relax (x) + b^{3} \sinh \relax (x)^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^2*cosh(x)^4 + b^2*sinh(x)^4 - 4*a*b*cosh(x)^3 - 8*(a^2 - b^2)*x*cosh(x)^2 + 4*(b^2*cosh(x) - a*b)*sinh(
x)^3 - 4*a*b*cosh(x) + 2*(3*b^2*cosh(x)^2 - 6*a*b*cosh(x) - 4*(a^2 - b^2)*x)*sinh(x)^2 + b^2 + 8*((a^2 - b^2)*
cosh(x)^2 + 2*(a^2 - b^2)*cosh(x)*sinh(x) + (a^2 - b^2)*sinh(x)^2)*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x)))
+ 4*(b^2*cosh(x)^3 - 3*a*b*cosh(x)^2 - 4*(a^2 - b^2)*x*cosh(x) - a*b)*sinh(x))/(b^3*cosh(x)^2 + 2*b^3*cosh(x)*
sinh(x) + b^3*sinh(x)^2)

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giac [A]  time = 0.13, size = 56, normalized size = 1.40 \[ \frac {b {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4 \, a {\left (e^{\left (-x\right )} + e^{x}\right )}}{8 \, b^{2}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 283, normalized size = 7.08 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{2}}{b^{3}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a}{b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a}{b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{2}}{b^{3}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) a^{3}}{b^{3} \left (a -b \right )}-\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) a^{2}}{b^{2} \left (a -b \right )}-\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) a}{b \left (a -b \right )}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}{a -b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.31, size = 84, normalized size = 2.10 \[ -\frac {{\left (4 \, a e^{\left (-x\right )} - b\right )} e^{\left (2 \, x\right )}}{8 \, b^{2}} - \frac {4 \, a e^{\left (-x\right )} - b e^{\left (-2 \, x\right )}}{8 \, b^{2}} + \frac {{\left (a^{2} - b^{2}\right )} x}{b^{3}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.04, size = 79, normalized size = 1.98 \[ \frac {{\mathrm {e}}^{-2\,x}}{8\,b}+\frac {{\mathrm {e}}^{2\,x}}{8\,b}+\frac {\ln \left (b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}\right )\,\left (a^2-b^2\right )}{b^3}-\frac {a\,{\mathrm {e}}^x}{2\,b^2}-\frac {a\,{\mathrm {e}}^{-x}}{2\,b^2}-\frac {x\,\left (a^2-b^2\right )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a + b*cosh(x)),x)

[Out]

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

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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+b*cosh(x)),x)

[Out]

Timed out

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