3.8.9 \(\int \frac {\cosh ^2(x) \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx\) [709]

Optimal. Leaf size=102 \[ \frac {a^2 b x}{\left (a^2-b^2\right )^2}-\frac {b x}{2 \left (a^2-b^2\right )}-\frac {a b^2 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2}-\frac {b \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )}+\frac {a \sinh ^2(x)}{2 \left (a^2-b^2\right )} \]

[Out]

a^2*b*x/(a^2-b^2)^2-1/2*b*x/(a^2-b^2)-a*b^2*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 = 0.12, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3188, 2715, 8, 2644, 30, 3177, 3212} \begin {gather*} \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 b^2 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cosh[x]^2*Sinh[x])/(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*b^2*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2715

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] && Integ
erQ[2*n]

Rule 3177

Int[cos[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[a*(x/(a^2 + b^2)), x] + Dist[b/(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 3188

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

Rule 3212

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 {\cosh ^2(x) \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx &=\frac {a \int \cosh (x) \sinh (x) \, dx}{a^2-b^2}-\frac {b \int \cosh ^2(x) \, dx}{a^2-b^2}+\frac {(a b) \int \frac {\cosh (x)}{a \cosh (x)+b \sinh (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 {\left (i a b^2\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 {a \text {Subst}(\int x \, dx,x,i \sinh (x))}{a^2-b^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 b^2 \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 = 0.17, size = 73, normalized size = 0.72 \begin {gather*} \frac {a \left (a^2-b^2\right ) \cosh (2 x)+b \left (2 \left (a^2+b^2\right ) x-4 a b \log (a \cosh (x)+b \sinh (x))+\left (-a^2+b^2\right ) \sinh (2 x)\right )}{4 (a-b)^2 (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.00, size = 146, normalized size = 1.43

method	result	size
risch	\(\frac {b x}{2 \left (a +b \right )^{2}}+\frac {{\mathrm e}^{2 x}}{8 a +8 b}+\frac {{\mathrm e}^{-2 x}}{8 a -8 b}+\frac {2 a \,b^{2} x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {a \,b^{2} \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\)	\(99\)
default	\(-\frac {4}{\left (8 a -8 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {2}{\left (4 a -4 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 \left (a -b \right )^{2}}-\frac {a \,b^{2} \ln \left (a +2 b \tanh \left (\frac {x}{2}\right )+a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}+\frac {2}{\left (4 a +4 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 \left (a +b \right )^{2}}\)	\(146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2*sinh(x)/(a*cosh(x)+b*sinh(x)),x,method=_RETURNVERBOSE)

[Out]

-4/(8*a-8*b)/(tanh(1/2*x)+1)+2/(4*a-4*b)/(tanh(1/2*x)+1)^2+1/2*b/(a-b)^2*ln(tanh(1/2*x)+1)-a*b^2/(a-b)^2/(a+b)
^2*ln(a+2*b*tanh(1/2*x)+a*tanh(1/2*x)^2)+2/(4*a+4*b)/(tanh(1/2*x)-1)^2+4/(8*a+8*b)/(tanh(1/2*x)-1)-1/2*b/(a+b)
^2*ln(tanh(1/2*x)-1)

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Maxima [A]
time = 0.28, size = 84, normalized size = 0.82 \begin {gather*} -\frac {a b^{2} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {b 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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*b^2*log(-(a - b)*e^(-2*x) - a - b)/(a^4 - 2*a^2*b^2 + b^4) + 1/2*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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (96) = 192\).
time = 0.37, size = 334, normalized size = 3.27 \begin {gather*} \frac {{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \left (x\right )^{4} + 4 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} x \cosh \left (x\right )^{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 \left (x\right )^{2} + 2 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} x\right )} \sinh \left (x\right )^{2} - 8 \, {\left (a b^{2} \cosh \left (x\right )^{2} + 2 \, a b^{2} \cosh \left (x\right ) \sinh \left (x\right ) + a b^{2} \sinh \left (x\right )^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \, {\left ({\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{3} + 2 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} x \cosh \left (x\right )\right )} \sinh \left (x\right )}{8 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2*sinh(x)/(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*(a^2*b + 2*a*b^2 + 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*(a^2*b + 2*a*b^2 + b^3)*x)*sinh(x)^2 - 8*(a*b^2*cosh(x)^2 + 2*a*b^2*cosh(x)*sinh
(x) + a*b^2*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)*cos
h(x)^3 + 2*(a^2*b + 2*a*b^2 + 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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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**2*sinh(x)/(a*cosh(x)+b*sinh(x)),x)

[Out]

Timed out

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Giac [A]
time = 0.41, size = 102, normalized size = 1.00 \begin {gather*} -\frac {a b^{2} \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 {b x}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*b^2*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^4 - 2*a^2*b^2 + b^4) + 1/2*b*x/(a^2 - 2*a*b + b^2) - 1/8*(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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Mupad [B]
time = 1.91, size = 81, normalized size = 0.79 \begin {gather*} \frac {{\mathrm {e}}^{-2\,x}}{8\,a-8\,b}+\frac {{\mathrm {e}}^{2\,x}}{8\,a+8\,b}+\frac {b\,x}{2\,{\left (a-b\right )}^2}-\frac {a\,b^2\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^4-2\,a^2\,b^2+b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(x)^2*sinh(x))/(a*cosh(x) + b*sinh(x)),x)

[Out]

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

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