### 3.252 $$\int x^2 \cosh (a+b x) \sinh (a+b x) \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0427648, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.188, Rules used = {5372, 3310, 30} $\frac{\sinh ^2(a+b x)}{4 b^3}-\frac{x \sinh (a+b x) \cosh (a+b x)}{2 b^2}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{x^2}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x^2/(4*b) - (x*Cosh[a + b*x]*Sinh[a + b*x])/(2*b^2) + Sinh[a + b*x]^2/(4*b^3) + (x^2*Sinh[a + b*x]^2)/(2*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 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0673341, size = 39, normalized size = 0.61 $\frac{\left (2 b^2 x^2+1\right ) \cosh (2 (a+b x))-2 b x \sinh (2 (a+b x))}{8 b^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.004, size = 114, normalized size = 1.8 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( bx+a \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2}}-{\frac{ \left ( bx+a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{2}}-{\frac{ \left ( bx+a \right ) ^{2}}{4}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}}-2\,a \left ( 1/2\, \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}-1/4\,\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/4\,bx-a/4 \right ) +{\frac{{a}^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.08568, size = 86, normalized size = 1.34 \begin{align*} \frac{{\left (2 \, b^{2} x^{2} e^{\left (2 \, a\right )} - 2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{16 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.123, size = 154, normalized size = 2.41 \begin{align*} -\frac{4 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) -{\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} -{\left (2 \, b^{2} x^{2} + 1\right )} \sinh \left (b x + a\right )^{2}}{8 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.21291, size = 75, normalized size = 1.17 \begin{align*} \begin{cases} \frac{x^{2} \sinh ^{2}{\left (a + b x \right )}}{4 b} + \frac{x^{2} \cosh ^{2}{\left (a + b x \right )}}{4 b} - \frac{x \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{2 b^{2}} + \frac{\sinh ^{2}{\left (a + b x \right )}}{4 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \sinh{\left (a \right )} \cosh{\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)*sinh(b*x+a),x)

[Out]

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

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Giac [A]  time = 1.14793, size = 77, normalized size = 1.2 \begin{align*} \frac{{\left (2 \, b^{2} x^{2} - 2 \, b x + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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