### 3.291 $$\int x^2 \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=60 $\frac{\sinh (4 a+4 b x)}{256 b^3}-\frac{x \cosh (4 a+4 b x)}{64 b^2}+\frac{x^2 \sinh (4 a+4 b x)}{32 b}-\frac{x^3}{24}$

[Out]

-x^3/24 - (x*Cosh[4*a + 4*b*x])/(64*b^2) + Sinh[4*a + 4*b*x]/(256*b^3) + (x^2*Sinh[4*a + 4*b*x])/(32*b)

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Rubi [A]  time = 0.10117, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {5448, 3296, 2637} $\frac{\sinh (4 a+4 b x)}{256 b^3}-\frac{x \cosh (4 a+4 b x)}{64 b^2}+\frac{x^2 \sinh (4 a+4 b x)}{32 b}-\frac{x^3}{24}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-x^3/24 - (x*Cosh[4*a + 4*b*x])/(64*b^2) + Sinh[4*a + 4*b*x]/(256*b^3) + (x^2*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 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^2 \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx &=\int \left (-\frac{x^2}{8}+\frac{1}{8} x^2 \cosh (4 a+4 b x)\right ) \, dx\\ &=-\frac{x^3}{24}+\frac{1}{8} \int x^2 \cosh (4 a+4 b x) \, dx\\ &=-\frac{x^3}{24}+\frac{x^2 \sinh (4 a+4 b x)}{32 b}-\frac{\int x \sinh (4 a+4 b x) \, dx}{16 b}\\ &=-\frac{x^3}{24}-\frac{x \cosh (4 a+4 b x)}{64 b^2}+\frac{x^2 \sinh (4 a+4 b x)}{32 b}+\frac{\int \cosh (4 a+4 b x) \, dx}{64 b^2}\\ &=-\frac{x^3}{24}-\frac{x \cosh (4 a+4 b x)}{64 b^2}+\frac{\sinh (4 a+4 b x)}{256 b^3}+\frac{x^2 \sinh (4 a+4 b x)}{32 b}\\ \end{align*}

Mathematica [A]  time = 0.164927, size = 48, normalized size = 0.8 $\frac{3 \left (8 b^2 x^2+1\right ) \sinh (4 (a+b x))-12 b x \cosh (4 (a+b x))-32 b^3 x^3}{768 b^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-32*b^3*x^3 - 12*b*x*Cosh[4*(a + b*x)] + 3*(1 + 8*b^2*x^2)*Sinh[4*(a + b*x)])/(768*b^3)

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Maple [B]  time = 0.007, size = 232, normalized size = 3.9 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{4}}-{\frac{ \left ( bx+a \right ) ^{2}\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{8}}-{\frac{ \left ( bx+a \right ) ^{3}}{24}}-{\frac{ \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{8}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}\sinh \left ( bx+a \right ) }{32}}-{\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{64}}-{\frac{bx}{64}}-{\frac{a}{64}}-2\,a \left ( 1/4\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}-1/8\, \left ( bx+a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/16\, \left ( bx+a \right ) ^{2}-1/16\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \right ) +{a}^{2} \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^2*cosh(b*x+a)^2*sinh(b*x+a)^2,x)

[Out]

1/b^3*(1/4*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^3-1/8*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)-1/24*(b*x+a)^3-1/8*(b*x+a
)*sinh(b*x+a)^2*cosh(b*x+a)^2+1/32*cosh(b*x+a)^3*sinh(b*x+a)-1/64*cosh(b*x+a)*sinh(b*x+a)-1/64*b*x-1/64*a-2*a*
(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*s
inh(b*x+a)^2)+a^2*(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 = 1.16365, size = 93, normalized size = 1.55 \begin{align*} -\frac{1}{24} \, x^{3} + \frac{{\left (8 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 4 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{512 \, b^{3}} - \frac{{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*x^3 + 1/512*(8*b^2*x^2*e^(4*a) - 4*b*x*e^(4*a) + e^(4*a))*e^(4*b*x)/b^3 - 1/512*(8*b^2*x^2 + 4*b*x + 1)*
e^(-4*b*x - 4*a)/b^3

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Fricas [B]  time = 1.85527, size = 288, normalized size = 4.8 \begin{align*} -\frac{8 \, b^{3} x^{3} + 3 \, b x \cosh \left (b x + a\right )^{4} + 18 \, b x \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 3 \, b x \sinh \left (b x + a\right )^{4} - 3 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) - 3 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3}}{192 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(8*b^3*x^3 + 3*b*x*cosh(b*x + a)^4 + 18*b*x*cosh(b*x + a)^2*sinh(b*x + a)^2 + 3*b*x*sinh(b*x + a)^4 - 3
*(8*b^2*x^2 + 1)*cosh(b*x + a)^3*sinh(b*x + a) - 3*(8*b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3)/b^3

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Sympy [A]  time = 4.66419, size = 204, normalized size = 3.4 \begin{align*} \begin{cases} - \frac{x^{3} \sinh ^{4}{\left (a + b x \right )}}{24} + \frac{x^{3} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{12} - \frac{x^{3} \cosh ^{4}{\left (a + b x \right )}}{24} + \frac{x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8 b} + \frac{x^{2} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b} - \frac{x \sinh ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac{3 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{32 b^{2}} - \frac{x \cosh ^{4}{\left (a + b x \right )}}{64 b^{2}} + \frac{\sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{64 b^{3}} + \frac{\sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{64 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \sinh ^{2}{\left (a \right )} \cosh ^{2}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x**3*sinh(a + b*x)**4/24 + x**3*sinh(a + b*x)**2*cosh(a + b*x)**2/12 - x**3*cosh(a + b*x)**4/24 +
x**2*sinh(a + b*x)**3*cosh(a + b*x)/(8*b) + x**2*sinh(a + b*x)*cosh(a + b*x)**3/(8*b) - x*sinh(a + b*x)**4/(64
*b**2) - 3*x*sinh(a + b*x)**2*cosh(a + b*x)**2/(32*b**2) - x*cosh(a + b*x)**4/(64*b**2) + sinh(a + b*x)**3*cos
h(a + b*x)/(64*b**3) + sinh(a + b*x)*cosh(a + b*x)**3/(64*b**3), Ne(b, 0)), (x**3*sinh(a)**2*cosh(a)**2/3, Tru
e))

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Giac [A]  time = 1.29334, size = 84, normalized size = 1.4 \begin{align*} -\frac{1}{24} \, x^{3} + \frac{{\left (8 \, b^{2} x^{2} - 4 \, b x + 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{512 \, b^{3}} - \frac{{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*x^3 + 1/512*(8*b^2*x^2 - 4*b*x + 1)*e^(4*b*x + 4*a)/b^3 - 1/512*(8*b^2*x^2 + 4*b*x + 1)*e^(-4*b*x - 4*a)
/b^3