\(\int \frac {\cosh ^3(x)}{a+b \sinh (x)} \, dx\) [191]

Optimal result

Integrand size = 13, antiderivative size = 38 \[ \int \frac {\cosh ^3(x)}{a+b \sinh (x)} \, dx=\frac {\left (a^2+b^2\right ) \log (a+b \sinh (x))}{b^3}-\frac {a \sinh (x)}{b^2}+\frac {\sinh ^2(x)}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2747, 711} \[ \int \frac {\cosh ^3(x)}{a+b \sinh (x)} \, dx=\frac {\left (a^2+b^2\right ) \log (a+b \sinh (x))}{b^3}-\frac {a \sinh (x)}{b^2}+\frac {\sinh ^2(x)}{2 b} \]

[In]

Int[Cosh[x]^3/(a + b*Sinh[x]),x]

[Out]

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

Rule 711

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 2747

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {\cosh ^3(x)}{a+b \sinh (x)} \, dx=-\frac {-\left (\left (a^2+b^2\right ) \log (a+b \sinh (x))\right )+a b \sinh (x)-\frac {1}{2} b^2 \sinh ^2(x)}{b^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.99 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(-\frac {-\frac {b \sinh \left (x \right )^{2}}{2}+a \sinh \left (x \right )}{b^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (x \right )\right )}{b^{3}}\)	\(37\)
default	\(-\frac {-\frac {b \sinh \left (x \right )^{2}}{2}+a \sinh \left (x \right )}{b^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (a +b \sinh \left (x \right )\right )}{b^{3}}\)	\(37\)
risch	\(-\frac {x \,a^{2}}{b^{3}}-\frac {x}{b}+\frac {{\mathrm e}^{2 x}}{8 b}-\frac {a \,{\mathrm e}^{x}}{2 b^{2}}+\frac {a \,{\mathrm e}^{-x}}{2 b^{2}}+\frac {{\mathrm e}^{-2 x}}{8 b}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right ) a^{2}}{b^{3}}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right )}{b}\)	\(94\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (36) = 72\).

Time = 0.29 (sec) , antiderivative size = 221, normalized size of antiderivative = 5.82 \[ \int \frac {\cosh ^3(x)}{a+b \sinh (x)} \, dx=\frac {b^{2} \cosh \left (x\right )^{4} + b^{2} \sinh \left (x\right )^{4} - 4 \, a b \cosh \left (x\right )^{3} - 8 \, {\left (a^{2} + b^{2}\right )} x \cosh \left (x\right )^{2} + 4 \, {\left (b^{2} \cosh \left (x\right ) - a b\right )} \sinh \left (x\right )^{3} + 4 \, a b \cosh \left (x\right ) + 2 \, {\left (3 \, b^{2} \cosh \left (x\right )^{2} - 6 \, a b \cosh \left (x\right ) - 4 \, {\left (a^{2} + b^{2}\right )} x\right )} \sinh \left (x\right )^{2} + b^{2} + 8 \, {\left ({\left (a^{2} + b^{2}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{2} + b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{2} + b^{2}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \, {\left (b^{2} \cosh \left (x\right )^{3} - 3 \, a b \cosh \left (x\right )^{2} - 4 \, {\left (a^{2} + b^{2}\right )} x \cosh \left (x\right ) + a b\right )} \sinh \left (x\right )}{8 \, {\left (b^{3} \cosh \left (x\right )^{2} + 2 \, b^{3} \cosh \left (x\right ) \sinh \left (x\right ) + b^{3} \sinh \left (x\right )^{2}\right )}} \]

[In]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cosh ^3(x)}{a+b \sinh (x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (36) = 72\).

Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.13 \[ \int \frac {\cosh ^3(x)}{a+b \sinh (x)} \, dx=-\frac {{\left (4 \, a e^{\left (-x\right )} - b\right )} e^{\left (2 \, x\right )}}{8 \, b^{2}} + \frac {4 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )}}{8 \, b^{2}} + \frac {{\left (a^{2} + b^{2}\right )} x}{b^{3}} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{b^{3}} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.61 \[ \int \frac {\cosh ^3(x)}{a+b \sinh (x)} \, dx=\frac {b {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4 \, a {\left (e^{\left (-x\right )} - e^{x}\right )}}{8 \, b^{2}} + \frac {{\left (a^{2} + b^{2}\right )} \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{b^{3}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 1.39 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.03 \[ \int \frac {\cosh ^3(x)}{a+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^{-2\,x}}{8\,b}+\frac {{\mathrm {e}}^{2\,x}}{8\,b}+\frac {\ln \left (2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}\right )\,\left (a^2+b^2\right )}{b^3}-\frac {a\,{\mathrm {e}}^x}{2\,b^2}-\frac {x\,\left (a^2+b^2\right )}{b^3}+\frac {a\,{\mathrm {e}}^{-x}}{2\,b^2} \]

[In]

int(cosh(x)^3/(a + b*sinh(x)),x)

[Out]

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