### 3.270 $$\int x^2 \cosh ^3(a+b x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=101 $\frac{\cosh ^4(a+b x)}{32 b^3}+\frac{3 \cosh ^2(a+b x)}{32 b^3}-\frac{x \sinh (a+b x) \cosh ^3(a+b x)}{8 b^2}-\frac{3 x \sinh (a+b x) \cosh (a+b x)}{16 b^2}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{3 x^2}{32 b}$

[Out]

(-3*x^2)/(32*b) + (3*Cosh[a + b*x]^2)/(32*b^3) + Cosh[a + b*x]^4/(32*b^3) + (x^2*Cosh[a + b*x]^4)/(4*b) - (3*x
*Cosh[a + b*x]*Sinh[a + b*x])/(16*b^2) - (x*Cosh[a + b*x]^3*Sinh[a + b*x])/(8*b^2)

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Rubi [A]  time = 0.0760245, antiderivative size = 101, 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 = {5373, 3310, 30} $\frac{\cosh ^4(a+b x)}{32 b^3}+\frac{3 \cosh ^2(a+b x)}{32 b^3}-\frac{x \sinh (a+b x) \cosh ^3(a+b x)}{8 b^2}-\frac{3 x \sinh (a+b x) \cosh (a+b x)}{16 b^2}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{3 x^2}{32 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3*x^2)/(32*b) + (3*Cosh[a + b*x]^2)/(32*b^3) + Cosh[a + b*x]^4/(32*b^3) + (x^2*Cosh[a + b*x]^4)/(4*b) - (3*x
*Cosh[a + b*x]*Sinh[a + b*x])/(16*b^2) - (x*Cosh[a + b*x]^3*Sinh[a + b*x])/(8*b^2)

Rule 5373

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[(x^(m -
n + 1)*Cosh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Cosh[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 ^3(a+b x) \sinh (a+b x) \, dx &=\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{\int x \cosh ^4(a+b x) \, dx}{2 b}\\ &=\frac{\cosh ^4(a+b x)}{32 b^3}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{x \cosh ^3(a+b x) \sinh (a+b x)}{8 b^2}-\frac{3 \int x \cosh ^2(a+b x) \, dx}{8 b}\\ &=\frac{3 \cosh ^2(a+b x)}{32 b^3}+\frac{\cosh ^4(a+b x)}{32 b^3}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{3 x \cosh (a+b x) \sinh (a+b x)}{16 b^2}-\frac{x \cosh ^3(a+b x) \sinh (a+b x)}{8 b^2}-\frac{3 \int x \, dx}{16 b}\\ &=-\frac{3 x^2}{32 b}+\frac{3 \cosh ^2(a+b x)}{32 b^3}+\frac{\cosh ^4(a+b x)}{32 b^3}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{3 x \cosh (a+b x) \sinh (a+b x)}{16 b^2}-\frac{x \cosh ^3(a+b x) \sinh (a+b x)}{8 b^2}\\ \end{align*}

Mathematica [A]  time = 0.231325, size = 70, normalized size = 0.69 $\frac{16 \left (2 b^2 x^2+1\right ) \cosh (2 (a+b x))+\left (8 b^2 x^2+1\right ) \cosh (4 (a+b x))-4 b x (8 \sinh (2 (a+b x))+\sinh (4 (a+b x)))}{256 b^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(16*(1 + 2*b^2*x^2)*Cosh[2*(a + b*x)] + (1 + 8*b^2*x^2)*Cosh[4*(a + b*x)] - 4*b*x*(8*Sinh[2*(a + b*x)] + Sinh[
4*(a + b*x)]))/(256*b^3)

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Maple [B]  time = 0.006, size = 237, normalized size = 2.4 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( bx+a \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}}+{\frac{ \left ( bx+a \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}}-{\frac{ \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{8}}-{\frac{ \left ( 3\,bx+3\,a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{16}}-{\frac{3\, \left ( bx+a \right ) ^{2}}{32}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{32}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{8}}-2\,a \left ( 1/4\, \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}+1/4\, \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}-1/16\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3}\sinh \left ( bx+a \right ) -{\frac{3\,\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{32}}-{\frac{3\,bx}{32}}-{\frac{3\,a}{32}} \right ) +{a}^{2} \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^2*cosh(b*x+a)^3*sinh(b*x+a),x)

[Out]

1/b^3*(1/4*(b*x+a)^2*sinh(b*x+a)^2*cosh(b*x+a)^2+1/4*(b*x+a)^2*cosh(b*x+a)^2-1/8*(b*x+a)*sinh(b*x+a)*cosh(b*x+
a)^3-3/16*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)-3/32*(b*x+a)^2+1/32*cosh(b*x+a)^2*sinh(b*x+a)^2+1/8*cosh(b*x+a)^2-2*
a*(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)-3/32*cosh(
b*x+a)*sinh(b*x+a)-3/32*b*x-3/32*a)+a^2*(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 = 1.20604, size = 171, normalized size = 1.69 \begin{align*} \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 (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 )}}{32 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, 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)^3*sinh(b*x+a),x, algorithm="maxima")

[Out]

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 + 1/32*(2*b^2*x^2*e^(2*a) - 2*b*x*e^(2*a) +
e^(2*a))*e^(2*b*x)/b^3 + 1/32*(2*b^2*x^2 + 2*b*x + 1)*e^(-2*b*x - 2*a)/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 [A]  time = 1.75687, size = 394, normalized size = 3.9 \begin{align*} -\frac{16 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} -{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{4} -{\left (8 \, b^{2} x^{2} + 1\right )} \sinh \left (b x + a\right )^{4} - 16 \,{\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} - 2 \,{\left (16 \, b^{2} x^{2} + 3 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} + 8\right )} \sinh \left (b x + a\right )^{2} + 16 \,{\left (b x \cosh \left (b x + a\right )^{3} + 4 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{256 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)^3*sinh(b*x+a),x, algorithm="fricas")

[Out]

-1/256*(16*b*x*cosh(b*x + a)*sinh(b*x + a)^3 - (8*b^2*x^2 + 1)*cosh(b*x + a)^4 - (8*b^2*x^2 + 1)*sinh(b*x + a)
^4 - 16*(2*b^2*x^2 + 1)*cosh(b*x + a)^2 - 2*(16*b^2*x^2 + 3*(8*b^2*x^2 + 1)*cosh(b*x + a)^2 + 8)*sinh(b*x + a)
^2 + 16*(b*x*cosh(b*x + a)^3 + 4*b*x*cosh(b*x + a))*sinh(b*x + a))/b^3

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Sympy [A]  time = 4.30484, size = 150, normalized size = 1.49 \begin{align*} \begin{cases} - \frac{3 x^{2} \sinh ^{4}{\left (a + b x \right )}}{32 b} + \frac{3 x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b} + \frac{5 x^{2} \cosh ^{4}{\left (a + b x \right )}}{32 b} + \frac{3 x \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{16 b^{2}} - \frac{5 x \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{16 b^{2}} - \frac{3 \sinh ^{4}{\left (a + b x \right )}}{64 b^{3}} + \frac{5 \cosh ^{4}{\left (a + b x \right )}}{64 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \sinh{\left (a \right )} \cosh ^{3}{\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)**3*sinh(b*x+a),x)

[Out]

Piecewise((-3*x**2*sinh(a + b*x)**4/(32*b) + 3*x**2*sinh(a + b*x)**2*cosh(a + b*x)**2/(16*b) + 5*x**2*cosh(a +
b*x)**4/(32*b) + 3*x*sinh(a + b*x)**3*cosh(a + b*x)/(16*b**2) - 5*x*sinh(a + b*x)*cosh(a + b*x)**3/(16*b**2)
- 3*sinh(a + b*x)**4/(64*b**3) + 5*cosh(a + b*x)**4/(64*b**3), Ne(b, 0)), (x**3*sinh(a)*cosh(a)**3/3, True))

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Giac [A]  time = 1.17006, size = 153, normalized size = 1.51 \begin{align*} \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 (2 \, b^{2} x^{2} - 2 \, b x + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, 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)^3*sinh(b*x+a),x, algorithm="giac")

[Out]

1/512*(8*b^2*x^2 - 4*b*x + 1)*e^(4*b*x + 4*a)/b^3 + 1/32*(2*b^2*x^2 - 2*b*x + 1)*e^(2*b*x + 2*a)/b^3 + 1/32*(2
*b^2*x^2 + 2*b*x + 1)*e^(-2*b*x - 2*a)/b^3 + 1/512*(8*b^2*x^2 + 4*b*x + 1)*e^(-4*b*x - 4*a)/b^3