### 3.310 $$\int x \cosh (a+b x) \sinh ^3(a+b x) \, dx$$

Optimal. Leaf size=65 $-\frac{\sinh ^3(a+b x) \cosh (a+b x)}{16 b^2}+\frac{3 \sinh (a+b x) \cosh (a+b x)}{32 b^2}+\frac{x \sinh ^4(a+b x)}{4 b}-\frac{3 x}{32 b}$

[Out]

(-3*x)/(32*b) + (3*Cosh[a + b*x]*Sinh[a + b*x])/(32*b^2) - (Cosh[a + b*x]*Sinh[a + b*x]^3)/(16*b^2) + (x*Sinh[
a + b*x]^4)/(4*b)

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Rubi [A]  time = 0.0430793, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.188, Rules used = {5372, 2635, 8} $-\frac{\sinh ^3(a+b x) \cosh (a+b x)}{16 b^2}+\frac{3 \sinh (a+b x) \cosh (a+b x)}{32 b^2}+\frac{x \sinh ^4(a+b x)}{4 b}-\frac{3 x}{32 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3*x)/(32*b) + (3*Cosh[a + b*x]*Sinh[a + b*x])/(32*b^2) - (Cosh[a + b*x]*Sinh[a + b*x]^3)/(16*b^2) + (x*Sinh[
a + b*x]^4)/(4*b)

Rule 5372

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m -
n + 1)*Sinh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 2635

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] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int x \cosh (a+b x) \sinh ^3(a+b x) \, dx &=\frac{x \sinh ^4(a+b x)}{4 b}-\frac{\int \sinh ^4(a+b x) \, dx}{4 b}\\ &=-\frac{\cosh (a+b x) \sinh ^3(a+b x)}{16 b^2}+\frac{x \sinh ^4(a+b x)}{4 b}+\frac{3 \int \sinh ^2(a+b x) \, dx}{16 b}\\ &=\frac{3 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac{\cosh (a+b x) \sinh ^3(a+b x)}{16 b^2}+\frac{x \sinh ^4(a+b x)}{4 b}-\frac{3 \int 1 \, dx}{32 b}\\ &=-\frac{3 x}{32 b}+\frac{3 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac{\cosh (a+b x) \sinh ^3(a+b x)}{16 b^2}+\frac{x \sinh ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.156814, size = 50, normalized size = 0.77 $-\frac{-8 \sinh (2 (a+b x))+\sinh (4 (a+b x))+16 b x \cosh (2 (a+b x))-4 b x \cosh (4 (a+b x))}{128 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(16*b*x*Cosh[2*(a + b*x)] - 4*b*x*Cosh[4*(a + b*x)] - 8*Sinh[2*(a + b*x)] + Sinh[4*(a + b*x)])/(128*b^2)

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Maple [A]  time = 0.005, size = 113, normalized size = 1.7 \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 ) ^{2}}{4}}-{\frac{ \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}\sinh \left ( bx+a \right ) }{16}}+{\frac{5\,\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{32}}+{\frac{5\,bx}{32}}+{\frac{5\,a}{32}}-a \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{4}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(1/4*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^2-1/4*(b*x+a)*cosh(b*x+a)^2-1/16*cosh(b*x+a)^3*sinh(b*x+a)+5/32*c
osh(b*x+a)*sinh(b*x+a)+5/32*b*x+5/32*a-a*(1/4*cosh(b*x+a)^2*sinh(b*x+a)^2-1/4*cosh(b*x+a)^2))

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Maxima [A]  time = 1.12009, size = 123, normalized size = 1.89 \begin{align*} \frac{{\left (4 \, b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{256 \, b^{2}} - \frac{{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{32 \, b^{2}} - \frac{{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{2}} + \frac{{\left (4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(4*b*x*e^(4*a) - e^(4*a))*e^(4*b*x)/b^2 - 1/32*(2*b*x*e^(2*a) - e^(2*a))*e^(2*b*x)/b^2 - 1/32*(2*b*x + 1
)*e^(-2*b*x - 2*a)/b^2 + 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2

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Fricas [A]  time = 1.67987, size = 289, normalized size = 4.45 \begin{align*} \frac{b x \cosh \left (b x + a\right )^{4} + b x \sinh \left (b x + a\right )^{4} - 4 \, b x \cosh \left (b x + a\right )^{2} - \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 2 \,{\left (3 \, b x \cosh \left (b x + a\right )^{2} - 2 \, b x\right )} \sinh \left (b x + a\right )^{2} -{\left (\cosh \left (b x + a\right )^{3} - 4 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{32 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(b*x*cosh(b*x + a)^4 + b*x*sinh(b*x + a)^4 - 4*b*x*cosh(b*x + a)^2 - cosh(b*x + a)*sinh(b*x + a)^3 + 2*(3
*b*x*cosh(b*x + a)^2 - 2*b*x)*sinh(b*x + a)^2 - (cosh(b*x + a)^3 - 4*cosh(b*x + a))*sinh(b*x + a))/b^2

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Sympy [A]  time = 2.32101, size = 110, normalized size = 1.69 \begin{align*} \begin{cases} \frac{5 x \sinh ^{4}{\left (a + b x \right )}}{32 b} + \frac{3 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b} - \frac{3 x \cosh ^{4}{\left (a + b x \right )}}{32 b} - \frac{5 \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{32 b^{2}} + \frac{3 \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{32 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \sinh ^{3}{\left (a \right )} \cosh{\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)*sinh(b*x+a)**3,x)

[Out]

Piecewise((5*x*sinh(a + b*x)**4/(32*b) + 3*x*sinh(a + b*x)**2*cosh(a + b*x)**2/(16*b) - 3*x*cosh(a + b*x)**4/(
32*b) - 5*sinh(a + b*x)**3*cosh(a + b*x)/(32*b**2) + 3*sinh(a + b*x)*cosh(a + b*x)**3/(32*b**2), Ne(b, 0)), (x
**2*sinh(a)**3*cosh(a)/2, True))

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Giac [A]  time = 1.17258, size = 109, normalized size = 1.68 \begin{align*} \frac{{\left (4 \, b x - 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{256 \, b^{2}} - \frac{{\left (2 \, b x - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b^{2}} - \frac{{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{2}} + \frac{{\left (4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(4*b*x - 1)*e^(4*b*x + 4*a)/b^2 - 1/32*(2*b*x - 1)*e^(2*b*x + 2*a)/b^2 - 1/32*(2*b*x + 1)*e^(-2*b*x - 2*
a)/b^2 + 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2