### 3.292 $$\int x \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=41 $-\frac{\cosh (4 a+4 b x)}{128 b^2}+\frac{x \sinh (4 a+4 b x)}{32 b}-\frac{x^2}{16}$

[Out]

-x^2/16 - Cosh[4*a + 4*b*x]/(128*b^2) + (x*Sinh[4*a + 4*b*x])/(32*b)

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Rubi [A]  time = 0.0504448, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {5448, 3296, 2638} $-\frac{\cosh (4 a+4 b x)}{128 b^2}+\frac{x \sinh (4 a+4 b x)}{32 b}-\frac{x^2}{16}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-x^2/16 - Cosh[4*a + 4*b*x]/(128*b^2) + (x*Sinh[4*a + 4*b*x])/(32*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 ^2(a+b x) \sinh ^2(a+b x) \, dx &=\int \left (-\frac{x}{8}+\frac{1}{8} x \cosh (4 a+4 b x)\right ) \, dx\\ &=-\frac{x^2}{16}+\frac{1}{8} \int x \cosh (4 a+4 b x) \, dx\\ &=-\frac{x^2}{16}+\frac{x \sinh (4 a+4 b x)}{32 b}-\frac{\int \sinh (4 a+4 b x) \, dx}{32 b}\\ &=-\frac{x^2}{16}-\frac{\cosh (4 a+4 b x)}{128 b^2}+\frac{x \sinh (4 a+4 b x)}{32 b}\\ \end{align*}

Mathematica [A]  time = 0.135183, size = 41, normalized size = 1. $-\frac{-8 a^2-4 b x \sinh (4 (a+b x))+\cosh (4 (a+b x))+8 b^2 x^2}{128 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.005, size = 114, normalized size = 2.8 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{ \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{4}}-{\frac{ \left ( bx+a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{8}}-{\frac{ \left ( bx+a \right ) ^{2}}{16}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{16}}-a \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}\sinh \left ( bx+a \right ) }{4}}-{\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{8}}-{\frac{bx}{8}}-{\frac{a}{8}} \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)^2,x)

[Out]

1/b^2*(1/4*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^3-1/8*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)-1/16*(b*x+a)^2-1/16*cosh(b*x+
a)^2*sinh(b*x+a)^2-a*(1/4*cosh(b*x+a)^3*sinh(b*x+a)-1/8*cosh(b*x+a)*sinh(b*x+a)-1/8*b*x-1/8*a))

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Maxima [A]  time = 1.10955, size = 69, normalized size = 1.68 \begin{align*} -\frac{1}{16} \, x^{2} + \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 (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)^2*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/16*x^2 + 1/256*(4*b*x*e^(4*a) - e^(4*a))*e^(4*b*x)/b^2 - 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2

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Fricas [B]  time = 1.83311, size = 234, normalized size = 5.71 \begin{align*} \frac{16 \, b x \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 16 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} - 8 \, b^{2} x^{2} - \cosh \left (b x + a\right )^{4} - 6 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} - \sinh \left (b x + a\right )^{4}}{128 \, 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)^2,x, algorithm="fricas")

[Out]

1/128*(16*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 16*b*x*cosh(b*x + a)*sinh(b*x + a)^3 - 8*b^2*x^2 - cosh(b*x + a)
^4 - 6*cosh(b*x + a)^2*sinh(b*x + a)^2 - sinh(b*x + a)^4)/b^2

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Sympy [A]  time = 2.43515, size = 131, normalized size = 3.2 \begin{align*} \begin{cases} - \frac{x^{2} \sinh ^{4}{\left (a + b x \right )}}{16} + \frac{x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8} - \frac{x^{2} \cosh ^{4}{\left (a + b x \right )}}{16} + \frac{x \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8 b} + \frac{x \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b} - \frac{\sinh ^{4}{\left (a + b x \right )}}{32 b^{2}} - \frac{\cosh ^{4}{\left (a + b x \right )}}{32 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \sinh ^{2}{\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)**2,x)

[Out]

Piecewise((-x**2*sinh(a + b*x)**4/16 + x**2*sinh(a + b*x)**2*cosh(a + b*x)**2/8 - x**2*cosh(a + b*x)**4/16 + x
*sinh(a + b*x)**3*cosh(a + b*x)/(8*b) + x*sinh(a + b*x)*cosh(a + b*x)**3/(8*b) - sinh(a + b*x)**4/(32*b**2) -
cosh(a + b*x)**4/(32*b**2), Ne(b, 0)), (x**2*sinh(a)**2*cosh(a)**2/2, True))

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Giac [A]  time = 1.22371, size = 62, normalized size = 1.51 \begin{align*} -\frac{1}{16} \, x^{2} + \frac{{\left (4 \, b x - 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{256 \, 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)^2*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

-1/16*x^2 + 1/256*(4*b*x - 1)*e^(4*b*x + 4*a)/b^2 - 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2