### 3.301 $$\int x \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0958022, 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, 2638} $\frac{\cosh (a+b x)}{8 b^2}-\frac{\cosh (3 a+3 b x)}{144 b^2}-\frac{\cosh (5 a+5 b x)}{400 b^2}-\frac{x \sinh (a+b x)}{8 b}+\frac{x \sinh (3 a+3 b x)}{48 b}+\frac{x \sinh (5 a+5 b x)}{80 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 2638

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

Rubi steps

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

Mathematica [A]  time = 0.192816, size = 70, normalized size = 0.74 $\frac{-450 b x \sinh (a+b x)+75 b x \sinh (3 (a+b x))+45 b x \sinh (5 (a+b x))+450 \cosh (a+b x)-25 \cosh (3 (a+b x))-9 \cosh (5 (a+b x))}{3600 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 143, normalized size = 1.5 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{ \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}{5}}-{\frac{ \left ( 2\,bx+2\,a \right ) \sinh \left ( bx+a \right ) }{15}}-{\frac{ \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{15}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{25}}-{\frac{4\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{225}}+{\frac{26\,\cosh \left ( bx+a \right ) }{225}}-a \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4}\sinh \left ( bx+a \right ) }{5}}-{\frac{\sinh \left ( bx+a \right ) }{5} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) } \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06624, 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)^3*sinh(b*x+a)^2,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 = 1.834, size = 425, normalized size = 4.52 \begin{align*} \frac{45 \, b x \sinh \left (b x + a\right )^{5} - 9 \, \cosh \left (b x + a\right )^{5} - 45 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 75 \,{\left (6 \, b x \cosh \left (b x + a\right )^{2} + b x\right )} \sinh \left (b x + a\right )^{3} - 25 \, \cosh \left (b x + a\right )^{3} - 15 \,{\left (6 \, \cosh \left (b x + a\right )^{3} + 5 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 225 \,{\left (b x \cosh \left (b x + a\right )^{4} + b x \cosh \left (b x + a\right )^{2} - 2 \, b x\right )} \sinh \left (b x + a\right ) + 450 \, \cosh \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)^3*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

1/3600*(45*b*x*sinh(b*x + a)^5 - 9*cosh(b*x + a)^5 - 45*cosh(b*x + a)*sinh(b*x + a)^4 + 75*(6*b*x*cosh(b*x + a
)^2 + b*x)*sinh(b*x + a)^3 - 25*cosh(b*x + a)^3 - 15*(6*cosh(b*x + a)^3 + 5*cosh(b*x + a))*sinh(b*x + a)^2 + 2
25*(b*x*cosh(b*x + a)^4 + b*x*cosh(b*x + a)^2 - 2*b*x)*sinh(b*x + a) + 450*cosh(b*x + a))/b^2

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Sympy [A]  time = 4.41347, size = 112, normalized size = 1.19 \begin{align*} \begin{cases} - \frac{2 x \sinh ^{5}{\left (a + b x \right )}}{15 b} + \frac{x \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b} + \frac{2 \sinh ^{4}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{15 b^{2}} - \frac{13 \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{45 b^{2}} + \frac{26 \cosh ^{5}{\left (a + b x \right )}}{225 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \sinh ^{2}{\left (a \right )} \cosh ^{3}{\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)**3*sinh(b*x+a)**2,x)

[Out]

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

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Giac [A]  time = 1.16221, 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)^3*sinh(b*x+a)^2,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