### 3.253 $$\int x \cosh (a+b x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=44 $-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}$

[Out]

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

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Rubi [A]  time = 0.0234004, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.214, Rules used = {5372, 2635, 8} $-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Cosh[a + b*x]*Sinh[a + b*x],x]

[Out]

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

Mathematica [A]  time = 0.0549291, size = 28, normalized size = 0.64 $-\frac{\sinh (2 (a+b x))-2 b x \cosh (2 (a+b x))}{8 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 53, normalized size = 1.2 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{ \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2}}-{\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{4}}-{\frac{bx}{4}}-{\frac{a}{4}}-{\frac{a \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*cosh(b*x+a)*sinh(b*x+a),x)

[Out]

1/b^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*cosh(b*x+a)^2)

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Maxima [A]  time = 1.19771, size = 62, normalized size = 1.41 \begin{align*} \frac{{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{16 \, b^{2}} + \frac{{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.99045, size = 112, normalized size = 2.55 \begin{align*} \frac{b x \cosh \left (b x + a\right )^{2} + b x \sinh \left (b x + a\right )^{2} - \cosh \left (b x + a\right ) \sinh \left (b x + a\right )}{4 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.610406, size = 56, normalized size = 1.27 \begin{align*} \begin{cases} \frac{x \sinh ^{2}{\left (a + b x \right )}}{4 b} + \frac{x \cosh ^{2}{\left (a + b x \right )}}{4 b} - \frac{\sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{4 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \sinh{\left (a \right )} \cosh{\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)*sinh(b*x+a),x)

[Out]

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

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Giac [A]  time = 1.16262, size = 55, normalized size = 1.25 \begin{align*} \frac{{\left (2 \, b x - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{2}} + \frac{{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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