### 3.251 $$\int x^3 \cosh (a+b x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=94 $-\frac{3 x^2 \sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}-\frac{3 \sinh (a+b x) \cosh (a+b x)}{8 b^4}+\frac{x^3 \sinh ^2(a+b x)}{2 b}+\frac{3 x}{8 b^3}+\frac{x^3}{4 b}$

[Out]

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

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Rubi [A]  time = 0.0691723, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.312, Rules used = {5372, 3311, 30, 2635, 8} $-\frac{3 x^2 \sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}-\frac{3 \sinh (a+b x) \cosh (a+b x)}{8 b^4}+\frac{x^3 \sinh ^2(a+b x)}{2 b}+\frac{3 x}{8 b^3}+\frac{x^3}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*x)/(8*b^3) + x^3/(4*b) - (3*Cosh[a + b*x]*Sinh[a + b*x])/(8*b^4) - (3*x^2*Cosh[a + b*x]*Sinh[a + b*x])/(4*b
^2) + (3*x*Sinh[a + b*x]^2)/(4*b^3) + (x^3*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 3311

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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^3 \cosh (a+b x) \sinh (a+b x) \, dx &=\frac{x^3 \sinh ^2(a+b x)}{2 b}-\frac{3 \int x^2 \sinh ^2(a+b x) \, dx}{2 b}\\ &=-\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}+\frac{x^3 \sinh ^2(a+b x)}{2 b}-\frac{3 \int \sinh ^2(a+b x) \, dx}{4 b^3}+\frac{3 \int x^2 \, dx}{4 b}\\ &=\frac{x^3}{4 b}-\frac{3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}+\frac{x^3 \sinh ^2(a+b x)}{2 b}+\frac{3 \int 1 \, dx}{8 b^3}\\ &=\frac{3 x}{8 b^3}+\frac{x^3}{4 b}-\frac{3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}+\frac{x^3 \sinh ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.103852, size = 50, normalized size = 0.53 $\frac{\left (4 b^3 x^3+6 b x\right ) \cosh (2 (a+b x))-3 \left (2 b^2 x^2+1\right ) \sinh (2 (a+b x))}{16 b^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.004, size = 203, normalized size = 2.2 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{ \left ( bx+a \right ) ^{3} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2}}-{\frac{3\, \left ( bx+a \right ) ^{2}\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{4}}-{\frac{ \left ( bx+a \right ) ^{3}}{4}}+{\frac{ \left ( 3\,bx+3\,a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}}-{\frac{3\,\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{8}}-{\frac{3\,bx}{8}}-{\frac{3\,a}{8}}-3\,a \left ( 1/2\, \left ( bx+a \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}-1/2\, \left ( bx+a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/4\, \left ( bx+a \right ) ^{2}+1/4\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \right ) +3\,{a}^{2} \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}^{3} \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^3*cosh(b*x+a)*sinh(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.10756, size = 116, normalized size = 1.23 \begin{align*} \frac{{\left (4 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 6 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 6 \, b x e^{\left (2 \, a\right )} - 3 \, e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{32 \, b^{4}} + \frac{{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(4*b^3*x^3*e^(2*a) - 6*b^2*x^2*e^(2*a) + 6*b*x*e^(2*a) - 3*e^(2*a))*e^(2*b*x)/b^4 + 1/32*(4*b^3*x^3 + 6*b
^2*x^2 + 6*b*x + 3)*e^(-2*b*x - 2*a)/b^4

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Fricas [A]  time = 2.05681, size = 180, normalized size = 1.91 \begin{align*} \frac{{\left (2 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{2} - 3 \,{\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (2 \, b^{3} x^{3} + 3 \, b x\right )} \sinh \left (b x + a\right )^{2}}{8 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.38946, size = 119, normalized size = 1.27 \begin{align*} \begin{cases} \frac{x^{3} \sinh ^{2}{\left (a + b x \right )}}{4 b} + \frac{x^{3} \cosh ^{2}{\left (a + b x \right )}}{4 b} - \frac{3 x^{2} \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{4 b^{2}} + \frac{3 x \sinh ^{2}{\left (a + b x \right )}}{8 b^{3}} + \frac{3 x \cosh ^{2}{\left (a + b x \right )}}{8 b^{3}} - \frac{3 \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \sinh{\left (a \right )} \cosh{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cosh(b*x+a)*sinh(b*x+a),x)

[Out]

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

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Giac [A]  time = 1.16116, size = 99, normalized size = 1.05 \begin{align*} \frac{{\left (4 \, b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, b x - 3\right )} e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b^{4}} + \frac{{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(4*b^3*x^3 - 6*b^2*x^2 + 6*b*x - 3)*e^(2*b*x + 2*a)/b^4 + 1/32*(4*b^3*x^3 + 6*b^2*x^2 + 6*b*x + 3)*e^(-2*
b*x - 2*a)/b^4