3.319 \(\int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx\)

Optimal. Leaf size=94 \[ \frac{\sinh (a+b x)}{8 b^2}+\frac{\sinh (3 a+3 b x)}{144 b^2}-\frac{\sinh (5 a+5 b x)}{400 b^2}-\frac{x \cosh (a+b x)}{8 b}-\frac{x \cosh (3 a+3 b x)}{48 b}+\frac{x \cosh (5 a+5 b x)}{80 b} \]

[Out]

-(x*Cosh[a + b*x])/(8*b) - (x*Cosh[3*a + 3*b*x])/(48*b) + (x*Cosh[5*a + 5*b*x])/(80*b) + Sinh[a + b*x]/(8*b^2)
+ Sinh[3*a + 3*b*x]/(144*b^2) - Sinh[5*a + 5*b*x]/(400*b^2)

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Rubi [A]  time = 0.0903493, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5448, 3296, 2637} \[ \frac{\sinh (a+b x)}{8 b^2}+\frac{\sinh (3 a+3 b x)}{144 b^2}-\frac{\sinh (5 a+5 b x)}{400 b^2}-\frac{x \cosh (a+b x)}{8 b}-\frac{x \cosh (3 a+3 b x)}{48 b}+\frac{x \cosh (5 a+5 b x)}{80 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Cosh[a + b*x]^2*Sinh[a + b*x]^3,x]

[Out]

-(x*Cosh[a + b*x])/(8*b) - (x*Cosh[3*a + 3*b*x])/(48*b) + (x*Cosh[5*a + 5*b*x])/(80*b) + Sinh[a + b*x]/(8*b^2)
+ Sinh[3*a + 3*b*x]/(144*b^2) - Sinh[5*a + 5*b*x]/(400*b^2)

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx &=\int \left (-\frac{1}{8} x \sinh (a+b x)-\frac{1}{16} x \sinh (3 a+3 b x)+\frac{1}{16} x \sinh (5 a+5 b x)\right ) \, dx\\ &=-\left (\frac{1}{16} \int x \sinh (3 a+3 b x) \, dx\right )+\frac{1}{16} \int x \sinh (5 a+5 b x) \, dx-\frac{1}{8} \int x \sinh (a+b x) \, dx\\ &=-\frac{x \cosh (a+b x)}{8 b}-\frac{x \cosh (3 a+3 b x)}{48 b}+\frac{x \cosh (5 a+5 b x)}{80 b}-\frac{\int \cosh (5 a+5 b x) \, dx}{80 b}+\frac{\int \cosh (3 a+3 b x) \, dx}{48 b}+\frac{\int \cosh (a+b x) \, dx}{8 b}\\ &=-\frac{x \cosh (a+b x)}{8 b}-\frac{x \cosh (3 a+3 b x)}{48 b}+\frac{x \cosh (5 a+5 b x)}{80 b}+\frac{\sinh (a+b x)}{8 b^2}+\frac{\sinh (3 a+3 b x)}{144 b^2}-\frac{\sinh (5 a+5 b x)}{400 b^2}\\ \end{align*}

Mathematica [A]  time = 0.188888, size = 70, normalized size = 0.74 \[ \frac{450 \sinh (a+b x)+25 \sinh (3 (a+b x))-9 \sinh (5 (a+b x))-450 b x \cosh (a+b x)-75 b x \cosh (3 (a+b x))+45 b x \cosh (5 (a+b x))}{3600 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cosh[a + b*x]^2*Sinh[a + b*x]^3,x]

[Out]

(-450*b*x*Cosh[a + b*x] - 75*b*x*Cosh[3*(a + b*x)] + 45*b*x*Cosh[5*(a + b*x)] + 450*Sinh[a + b*x] + 25*Sinh[3*
(a + b*x)] - 9*Sinh[5*(a + b*x)])/(3600*b^2)

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Maple [A]  time = 0.007, size = 149, normalized size = 1.6 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{ \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{5}}-{\frac{ \left ( 2\,bx+2\,a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\cosh \left ( bx+a \right ) }{15}}-{\frac{ \left ( 2\,bx+2\,a \right ) \cosh \left ( bx+a \right ) }{15}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4}\sinh \left ( bx+a \right ) }{25}}+{\frac{26\,\sinh \left ( bx+a \right ) }{225}}+{\frac{13\,\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{225}}-a \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{5}}-{\frac{2\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{15}}-{\frac{2\,\cosh \left ( bx+a \right ) }{15}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(b*x+a)^2*sinh(b*x+a)^3,x)

[Out]

1/b^2*(1/5*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^3-2/15*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)-2/15*(b*x+a)*cosh(b*x+a)
-1/25*cosh(b*x+a)^4*sinh(b*x+a)+26/225*sinh(b*x+a)+13/225*sinh(b*x+a)*cosh(b*x+a)^2-a*(1/5*cosh(b*x+a)^3*sinh(
b*x+a)^2-2/15*cosh(b*x+a)*sinh(b*x+a)^2-2/15*cosh(b*x+a)))

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Maxima [A]  time = 1.20229, size = 174, normalized size = 1.85 \begin{align*} \frac{{\left (5 \, b x e^{\left (5 \, a\right )} - e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )}}{800 \, b^{2}} - \frac{{\left (3 \, b x e^{\left (3 \, a\right )} - e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{288 \, b^{2}} - \frac{{\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{16 \, b^{2}} - \frac{{\left (b x + 1\right )} e^{\left (-b x - a\right )}}{16 \, b^{2}} - \frac{{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{288 \, b^{2}} + \frac{{\left (5 \, b x + 1\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{800 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

1/800*(5*b*x*e^(5*a) - e^(5*a))*e^(5*b*x)/b^2 - 1/288*(3*b*x*e^(3*a) - e^(3*a))*e^(3*b*x)/b^2 - 1/16*(b*x*e^a
- e^a)*e^(b*x)/b^2 - 1/16*(b*x + 1)*e^(-b*x - a)/b^2 - 1/288*(3*b*x + 1)*e^(-3*b*x - 3*a)/b^2 + 1/800*(5*b*x +
1)*e^(-5*b*x - 5*a)/b^2

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Fricas [A]  time = 1.69211, size = 433, normalized size = 4.61 \begin{align*} \frac{45 \, b x \cosh \left (b x + a\right )^{5} + 225 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 75 \, b x \cosh \left (b x + a\right )^{3} - 9 \, \sinh \left (b x + a\right )^{5} - 5 \,{\left (18 \, \cosh \left (b x + a\right )^{2} - 5\right )} \sinh \left (b x + a\right )^{3} - 450 \, b x \cosh \left (b x + a\right ) + 225 \,{\left (2 \, b x \cosh \left (b x + a\right )^{3} - b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 15 \,{\left (3 \, \cosh \left (b x + a\right )^{4} - 5 \, \cosh \left (b x + a\right )^{2} - 30\right )} \sinh \left (b x + a\right )}{3600 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/3600*(45*b*x*cosh(b*x + a)^5 + 225*b*x*cosh(b*x + a)*sinh(b*x + a)^4 - 75*b*x*cosh(b*x + a)^3 - 9*sinh(b*x +
a)^5 - 5*(18*cosh(b*x + a)^2 - 5)*sinh(b*x + a)^3 - 450*b*x*cosh(b*x + a) + 225*(2*b*x*cosh(b*x + a)^3 - b*x*
cosh(b*x + a))*sinh(b*x + a)^2 - 15*(3*cosh(b*x + a)^4 - 5*cosh(b*x + a)^2 - 30)*sinh(b*x + a))/b^2

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Sympy [A]  time = 4.18373, size = 112, normalized size = 1.19 \begin{align*} \begin{cases} \frac{x \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac{2 x \cosh ^{5}{\left (a + b x \right )}}{15 b} + \frac{26 \sinh ^{5}{\left (a + b x \right )}}{225 b^{2}} - \frac{13 \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{45 b^{2}} + \frac{2 \sinh{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{15 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \sinh ^{3}{\left (a \right )} \cosh ^{2}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)**2*sinh(b*x+a)**3,x)

[Out]

Piecewise((x*sinh(a + b*x)**2*cosh(a + b*x)**3/(3*b) - 2*x*cosh(a + b*x)**5/(15*b) + 26*sinh(a + b*x)**5/(225*
b**2) - 13*sinh(a + b*x)**3*cosh(a + b*x)**2/(45*b**2) + 2*sinh(a + b*x)*cosh(a + b*x)**4/(15*b**2), Ne(b, 0))
, (x**2*sinh(a)**3*cosh(a)**2/2, True))

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Giac [A]  time = 1.1768, size = 157, normalized size = 1.67 \begin{align*} \frac{{\left (5 \, b x - 1\right )} e^{\left (5 \, b x + 5 \, a\right )}}{800 \, b^{2}} - \frac{{\left (3 \, b x - 1\right )} e^{\left (3 \, b x + 3 \, a\right )}}{288 \, b^{2}} - \frac{{\left (b x - 1\right )} e^{\left (b x + a\right )}}{16 \, b^{2}} - \frac{{\left (b x + 1\right )} e^{\left (-b x - a\right )}}{16 \, b^{2}} - \frac{{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{288 \, b^{2}} + \frac{{\left (5 \, b x + 1\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{800 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

1/800*(5*b*x - 1)*e^(5*b*x + 5*a)/b^2 - 1/288*(3*b*x - 1)*e^(3*b*x + 3*a)/b^2 - 1/16*(b*x - 1)*e^(b*x + a)/b^2
- 1/16*(b*x + 1)*e^(-b*x - a)/b^2 - 1/288*(3*b*x + 1)*e^(-3*b*x - 3*a)/b^2 + 1/800*(5*b*x + 1)*e^(-5*b*x - 5*
a)/b^2