3.262 \(\int x \cosh ^2(a+b x) \sinh (a+b x) \, dx\)

Optimal. Leaf size=45 \[ -\frac {\sinh ^3(a+b x)}{9 b^2}-\frac {\sinh (a+b x)}{3 b^2}+\frac {x \cosh ^3(a+b x)}{3 b} \]

[Out]

1/3*x*cosh(b*x+a)^3/b-1/3*sinh(b*x+a)/b^2-1/9*sinh(b*x+a)^3/b^2

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5373, 2633} \[ -\frac {\sinh ^3(a+b x)}{9 b^2}-\frac {\sinh (a+b x)}{3 b^2}+\frac {x \cosh ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Cosh[a + b*x]^3)/(3*b) - Sinh[a + b*x]/(3*b^2) - Sinh[a + b*x]^3/(9*b^2)

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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]

Rubi steps

\begin {align*} \int x \cosh ^2(a+b x) \sinh (a+b x) \, dx &=\frac {x \cosh ^3(a+b x)}{3 b}-\frac {\int \cosh ^3(a+b x) \, dx}{3 b}\\ &=\frac {x \cosh ^3(a+b x)}{3 b}-\frac {i \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh (a+b x)\right )}{3 b^2}\\ &=\frac {x \cosh ^3(a+b x)}{3 b}-\frac {\sinh (a+b x)}{3 b^2}-\frac {\sinh ^3(a+b x)}{9 b^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 46, normalized size = 1.02 \[ -\frac {9 \sinh (a+b x)+\sinh (3 (a+b x))-9 b x \cosh (a+b x)-3 b x \cosh (3 (a+b x))}{36 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/36*(-9*b*x*Cosh[a + b*x] - 3*b*x*Cosh[3*(a + b*x)] + 9*Sinh[a + b*x] + Sinh[3*(a + b*x)])/b^2

________________________________________________________________________________________

fricas [A]  time = 0.72, size = 74, normalized size = 1.64 \[ \frac {3 \, b x \cosh \left (b x + a\right )^{3} + 9 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 9 \, b x \cosh \left (b x + a\right ) - \sinh \left (b x + a\right )^{3} - 3 \, {\left (\cosh \left (b x + a\right )^{2} + 3\right )} \sinh \left (b x + a\right )}{36 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(3*b*x*cosh(b*x + a)^3 + 9*b*x*cosh(b*x + a)*sinh(b*x + a)^2 + 9*b*x*cosh(b*x + a) - sinh(b*x + a)^3 - 3*
(cosh(b*x + a)^2 + 3)*sinh(b*x + a))/b^2

________________________________________________________________________________________

giac [A]  time = 0.12, size = 76, normalized size = 1.69 \[ \frac {{\left (3 \, b x - 1\right )} e^{\left (3 \, b x + 3 \, a\right )}}{72 \, b^{2}} + \frac {{\left (b x - 1\right )} e^{\left (b x + a\right )}}{8 \, b^{2}} + \frac {{\left (b x + 1\right )} e^{\left (-b x - a\right )}}{8 \, b^{2}} + \frac {{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{72 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(3*b*x - 1)*e^(3*b*x + 3*a)/b^2 + 1/8*(b*x - 1)*e^(b*x + a)/b^2 + 1/8*(b*x + 1)*e^(-b*x - a)/b^2 + 1/72*(
3*b*x + 1)*e^(-3*b*x - 3*a)/b^2

________________________________________________________________________________________

maple [A]  time = 0.32, size = 56, normalized size = 1.24 \[ \frac {\frac {\left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{3}-\frac {2 \sinh \left (b x +a \right )}{9}-\frac {\left (\cosh ^{2}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{9}-\frac {a \left (\cosh ^{3}\left (b x +a \right )\right )}{3}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(1/3*(b*x+a)*cosh(b*x+a)^3-2/9*sinh(b*x+a)-1/9*cosh(b*x+a)^2*sinh(b*x+a)-1/3*a*cosh(b*x+a)^3)

________________________________________________________________________________________

maxima [B]  time = 0.32, size = 84, normalized size = 1.87 \[ \frac {{\left (3 \, b x e^{\left (3 \, a\right )} - e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{72 \, b^{2}} + \frac {{\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{8 \, b^{2}} + \frac {{\left (b x + 1\right )} e^{\left (-b x - a\right )}}{8 \, b^{2}} + \frac {{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{72 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(3*b*x*e^(3*a) - e^(3*a))*e^(3*b*x)/b^2 + 1/8*(b*x*e^a - e^a)*e^(b*x)/b^2 + 1/8*(b*x + 1)*e^(-b*x - a)/b^
2 + 1/72*(3*b*x + 1)*e^(-3*b*x - 3*a)/b^2

________________________________________________________________________________________

mupad [B]  time = 0.08, size = 41, normalized size = 0.91 \[ -\frac {-3\,b\,x\,{\mathrm {cosh}\left (a+b\,x\right )}^3+\mathrm {sinh}\left (a+b\,x\right )\,{\mathrm {cosh}\left (a+b\,x\right )}^2+2\,\mathrm {sinh}\left (a+b\,x\right )}{9\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*sinh(a + b*x) + cosh(a + b*x)^2*sinh(a + b*x) - 3*b*x*cosh(a + b*x)^3)/(9*b^2)

________________________________________________________________________________________

sympy [A]  time = 0.81, size = 61, normalized size = 1.36 \[ \begin {cases} \frac {x \cosh ^{3}{\left (a + b x \right )}}{3 b} + \frac {2 \sinh ^{3}{\left (a + b x \right )}}{9 b^{2}} - \frac {\sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \sinh {\relax (a )} \cosh ^{2}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*cosh(a + b*x)**3/(3*b) + 2*sinh(a + b*x)**3/(9*b**2) - sinh(a + b*x)*cosh(a + b*x)**2/(3*b**2), N
e(b, 0)), (x**2*sinh(a)*cosh(a)**2/2, True))

________________________________________________________________________________________