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

Optimal. Leaf size=137 \[ -\frac{b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}+\frac{a \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac{a \cosh (x)}{a^2-b^2}-\frac{a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac{a^3 b \tan ^{-1}\left (\frac{a \sinh (x)+b \cosh (x)}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]

[Out]

-((a^3*b*ArcTan[(b*Cosh[x] + a*Sinh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2)) - (a*b^2*Cosh[x])/(a^2 - b^2)^2 -
(a*Cosh[x])/(a^2 - b^2) + (a*Cosh[x]^3)/(3*(a^2 - b^2)) + (a^2*b*Sinh[x])/(a^2 - b^2)^2 - (b*Sinh[x]^3)/(3*(a
^2 - b^2))

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Rubi [A]  time = 0.202991, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3109, 2564, 30, 2633, 3099, 3074, 206, 2638} \[ -\frac{b \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}+\frac{a \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac{a \cosh (x)}{a^2-b^2}-\frac{a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac{a^3 b \tan ^{-1}\left (\frac{a \sinh (x)+b \cosh (x)}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*b*ArcTan[(b*Cosh[x] + a*Sinh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2)) - (a*b^2*Cosh[x])/(a^2 - b^2)^2 -
(a*Cosh[x])/(a^2 - b^2) + (a*Cosh[x]^3)/(3*(a^2 - b^2)) + (a^2*b*Sinh[x])/(a^2 - b^2)^2 - (b*Sinh[x]^3)/(3*(a
^2 - b^2))

Rule 3109

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 2564

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 30

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/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 3074

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\cosh (x) \sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx &=\frac{a \int \sinh ^3(x) \, dx}{a^2-b^2}-\frac{b \int \cosh (x) \sinh ^2(x) \, dx}{a^2-b^2}+\frac{(a b) \int \frac{\sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}\\ &=\frac{a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}-\frac{\left (a^3 b\right ) \int \frac{1}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}-\frac{\left (a b^2\right ) \int \sinh (x) \, dx}{\left (a^2-b^2\right )^2}-\frac{a \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (x)\right )}{a^2-b^2}-\frac{(i b) \operatorname{Subst}\left (\int x^2 \, dx,x,i \sinh (x)\right )}{a^2-b^2}\\ &=-\frac{a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac{a \cosh (x)}{a^2-b^2}+\frac{a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}-\frac{b \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac{\left (i a^3 b\right ) \operatorname{Subst}\left (\int \frac{1}{a^2-b^2-x^2} \, dx,x,-i b \cosh (x)-i a \sinh (x)\right )}{\left (a^2-b^2\right )^2}\\ &=-\frac{a^3 b \tan ^{-1}\left (\frac{b \cosh (x)+a \sinh (x)}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{a b^2 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac{a \cosh (x)}{a^2-b^2}+\frac{a \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a^2 b \sinh (x)}{\left (a^2-b^2\right )^2}-\frac{b \sinh ^3(x)}{3 \left (a^2-b^2\right )}\\ \end{align*}

Mathematica [A]  time = 1.10541, size = 180, normalized size = 1.31 \[ \frac{-3 a \sqrt{a-b} \sqrt{a+b} \left (3 a^2+b^2\right ) \cosh (x)+a \sqrt{a-b} \sqrt{a+b} \left (a^2-b^2\right ) \cosh (3 x)+b \left (3 \sqrt{a-b} \sqrt{a+b} \left (5 a^2-b^2\right ) \sinh (x)-\sqrt{a-b} \sqrt{a+b} \left (a^2-b^2\right ) \sinh (3 x)-24 a^3 \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a-b} \sqrt{a+b}}\right )\right )}{12 (a-b)^{5/2} (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*Sqrt[a - b]*Sqrt[a + b]*(3*a^2 + b^2)*Cosh[x] + a*Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*Cosh[3*x] + b*(-24
*a^3*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])] + 3*Sqrt[a - b]*Sqrt[a + b]*(5*a^2 - b^2)*Sinh[x] - S
qrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*Sinh[3*x]))/(12*(a - b)^(5/2)*(a + b)^(5/2))

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Maple [A]  time = 0.049, size = 166, normalized size = 1.2 \begin{align*} -8\,{\frac{1}{ \left ( 16\,a-16\,b \right ) \left ( \tanh \left ( x/2 \right ) +1 \right ) ^{2}}}+{\frac{16}{48\,a-48\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}-{\frac{a}{2\, \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{{a}^{3}b}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{16}{48\,a+48\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-8\,{\frac{1}{ \left ( 16\,a+16\,b \right ) \left ( \tanh \left ( x/2 \right ) -1 \right ) ^{2}}}+{\frac{a}{2\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/(16*a-16*b)/(tanh(1/2*x)+1)^2+16/3/(tanh(1/2*x)+1)^3/(16*a-16*b)-1/2*a/(a-b)^2/(tanh(1/2*x)+1)-2*a^3*b/(a-b
)^2/(a+b)^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tanh(1/2*x)+2*b)/(a^2-b^2)^(1/2))-16/3/(tanh(1/2*x)-1)^3/(16*a+16*
b)-8/(16*a+16*b)/(tanh(1/2*x)-1)^2+1/2*a/(a+b)^2/(tanh(1/2*x)-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.13573, size = 4208, normalized size = 30.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3
+ a*b^4 - b^5)*cosh(x)*sinh(x)^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^6 + a^5 + a^4*b
- 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 - 3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^4 -
3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b
^5)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(3*a^5 - 5*a
^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 -
a*b^4 + b^5)*cosh(x)^2 - 3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2
+ 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 + 6*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^2)*s
inh(x)^2 - 24*(a^3*b*cosh(x)^3 + 3*a^3*b*cosh(x)^2*sinh(x) + 3*a^3*b*cosh(x)*sinh(x)^2 + a^3*b*sinh(x)^3)*sqrt
(-a^2 + b^2)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + 2*sqrt(-a^2 + b^2)*(cosh
(x) + sinh(x)) - a + b)/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a - b)) + 6*((a^5
- a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^5 - 2*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4
- b^5)*cosh(x)^3 - (3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^
2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^6 - 3*a^4*b^2
+ 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^3), 1/24*((a^5 - a^4*b - 2*
a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)*s
inh(x)^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^6 + a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3
+ a*b^4 + b^5 - 3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^4 - 3*(3*a^5 - 5*a^4*b - 2*
a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^4
+ 4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2
*b^3 - a*b^4 - b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5)*cosh(x)^2 -
3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b
^5)*cosh(x)^4 + 6*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^2)*sinh(x)^2 + 48*(a^3*b*cos
h(x)^3 + 3*a^3*b*cosh(x)^2*sinh(x) + 3*a^3*b*cosh(x)*sinh(x)^2 + a^3*b*sinh(x)^3)*sqrt(a^2 - b^2)*arctan(sqrt(
a^2 - b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) + 6*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(
x)^5 - 2*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^3 - (3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*
a^2*b^3 - a*b^4 + b^5)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 +
3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^6 - 3*a^4*b
^2 + 3*a^2*b^4 - b^6)*sinh(x)^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.15681, size = 220, normalized size = 1.61 \begin{align*} -\frac{2 \, a^{3} b \arctan \left (\frac{a e^{x} + b e^{x}}{\sqrt{a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} - \frac{{\left (9 \, a e^{\left (2 \, x\right )} - 3 \, b e^{\left (2 \, x\right )} - a + b\right )} e^{\left (-3 \, x\right )}}{24 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{a^{2} e^{\left (3 \, x\right )} + 2 \, a b e^{\left (3 \, x\right )} + b^{2} e^{\left (3 \, x\right )} - 9 \, a^{2} e^{x} - 12 \, a b e^{x} - 3 \, b^{2} e^{x}}{24 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^3*b*arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) - 1/24*(9*a*e^(2*x)
- 3*b*e^(2*x) - a + b)*e^(-3*x)/(a^2 - 2*a*b + b^2) + 1/24*(a^2*e^(3*x) + 2*a*b*e^(3*x) + b^2*e^(3*x) - 9*a^2
*e^x - 12*a*b*e^x - 3*b^2*e^x)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3)