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

Optimal. Leaf size=83 \[ \frac {2 \sinh ^3(a+b x)}{27 b^3}-\frac {4 \sinh (a+b x)}{9 b^3}+\frac {4 x \cosh (a+b x)}{9 b^2}-\frac {2 x \sinh ^2(a+b x) \cosh (a+b x)}{9 b^2}+\frac {x^2 \sinh ^3(a+b x)}{3 b} \]

[Out]

4/9*x*cosh(b*x+a)/b^2-4/9*sinh(b*x+a)/b^3-2/9*x*cosh(b*x+a)*sinh(b*x+a)^2/b^2+2/27*sinh(b*x+a)^3/b^3+1/3*x^2*s
inh(b*x+a)^3/b

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Rubi [A]  time = 0.08, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5372, 3310, 3296, 2637} \[ \frac {2 \sinh ^3(a+b x)}{27 b^3}-\frac {4 \sinh (a+b x)}{9 b^3}+\frac {4 x \cosh (a+b x)}{9 b^2}-\frac {2 x \sinh ^2(a+b x) \cosh (a+b x)}{9 b^2}+\frac {x^2 \sinh ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x*Cosh[a + b*x])/(9*b^2) - (4*Sinh[a + b*x])/(9*b^3) - (2*x*Cosh[a + b*x]*Sinh[a + b*x]^2)/(9*b^2) + (2*Sin
h[a + b*x]^3)/(27*b^3) + (x^2*Sinh[a + b*x]^3)/(3*b)

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 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]

Rubi steps

\begin {align*} \int x^2 \cosh (a+b x) \sinh ^2(a+b x) \, dx &=\frac {x^2 \sinh ^3(a+b x)}{3 b}-\frac {2 \int x \sinh ^3(a+b x) \, dx}{3 b}\\ &=-\frac {2 x \cosh (a+b x) \sinh ^2(a+b x)}{9 b^2}+\frac {2 \sinh ^3(a+b x)}{27 b^3}+\frac {x^2 \sinh ^3(a+b x)}{3 b}+\frac {4 \int x \sinh (a+b x) \, dx}{9 b}\\ &=\frac {4 x \cosh (a+b x)}{9 b^2}-\frac {2 x \cosh (a+b x) \sinh ^2(a+b x)}{9 b^2}+\frac {2 \sinh ^3(a+b x)}{27 b^3}+\frac {x^2 \sinh ^3(a+b x)}{3 b}-\frac {4 \int \cosh (a+b x) \, dx}{9 b^2}\\ &=\frac {4 x \cosh (a+b x)}{9 b^2}-\frac {4 \sinh (a+b x)}{9 b^3}-\frac {2 x \cosh (a+b x) \sinh ^2(a+b x)}{9 b^2}+\frac {2 \sinh ^3(a+b x)}{27 b^3}+\frac {x^2 \sinh ^3(a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 66, normalized size = 0.80 \[ \frac {\sinh (a+b x) \left (\left (9 b^2 x^2+2\right ) \cosh (2 (a+b x))-9 b^2 x^2-26\right )+27 b x \cosh (a+b x)-3 b x \cosh (3 (a+b x))}{54 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(27*b*x*Cosh[a + b*x] - 3*b*x*Cosh[3*(a + b*x)] + (-26 - 9*b^2*x^2 + (2 + 9*b^2*x^2)*Cosh[2*(a + b*x)])*Sinh[a
+ b*x])/(54*b^3)

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fricas [A]  time = 0.46, size = 104, normalized size = 1.25 \[ -\frac {6 \, b x \cosh \left (b x + a\right )^{3} + 18 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - {\left (9 \, b^{2} x^{2} + 2\right )} \sinh \left (b x + a\right )^{3} - 54 \, b x \cosh \left (b x + a\right ) + 3 \, {\left (9 \, b^{2} x^{2} - {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} + 18\right )} \sinh \left (b x + a\right )}{108 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/108*(6*b*x*cosh(b*x + a)^3 + 18*b*x*cosh(b*x + a)*sinh(b*x + a)^2 - (9*b^2*x^2 + 2)*sinh(b*x + a)^3 - 54*b*
x*cosh(b*x + a) + 3*(9*b^2*x^2 - (9*b^2*x^2 + 2)*cosh(b*x + a)^2 + 18)*sinh(b*x + a))/b^3

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giac [A]  time = 0.12, size = 108, normalized size = 1.30 \[ \frac {{\left (9 \, b^{2} x^{2} - 6 \, b x + 2\right )} e^{\left (3 \, b x + 3 \, a\right )}}{216 \, b^{3}} - \frac {{\left (b^{2} x^{2} - 2 \, b x + 2\right )} e^{\left (b x + a\right )}}{8 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{8 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{216 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/216*(9*b^2*x^2 - 6*b*x + 2)*e^(3*b*x + 3*a)/b^3 - 1/8*(b^2*x^2 - 2*b*x + 2)*e^(b*x + a)/b^3 + 1/8*(b^2*x^2 +
2*b*x + 2)*e^(-b*x - a)/b^3 - 1/216*(9*b^2*x^2 + 6*b*x + 2)*e^(-3*b*x - 3*a)/b^3

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maple [A]  time = 0.33, size = 131, normalized size = 1.58 \[ \frac {\frac {\left (b x +a \right )^{2} \left (\sinh ^{3}\left (b x +a \right )\right )}{3}+\frac {4 \left (b x +a \right ) \cosh \left (b x +a \right )}{9}-\frac {2 \left (b x +a \right ) \cosh \left (b x +a \right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{9}-\frac {4 \sinh \left (b x +a \right )}{9}+\frac {2 \left (\sinh ^{3}\left (b x +a \right )\right )}{27}-2 a \left (\frac {\left (b x +a \right ) \left (\sinh ^{3}\left (b x +a \right )\right )}{3}+\frac {2 \cosh \left (b x +a \right )}{9}-\frac {\cosh \left (b x +a \right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{9}\right )+\frac {a^{2} \left (\sinh ^{3}\left (b x +a \right )\right )}{3}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/3*(b*x+a)^2*sinh(b*x+a)^3+4/9*(b*x+a)*cosh(b*x+a)-2/9*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)^2-4/9*sinh(b*x+
a)+2/27*sinh(b*x+a)^3-2*a*(1/3*(b*x+a)*sinh(b*x+a)^3+2/9*cosh(b*x+a)-1/9*cosh(b*x+a)*sinh(b*x+a)^2)+1/3*a^2*si
nh(b*x+a)^3)

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maxima [A]  time = 0.35, size = 122, normalized size = 1.47 \[ \frac {{\left (9 \, b^{2} x^{2} e^{\left (3 \, a\right )} - 6 \, b x e^{\left (3 \, a\right )} + 2 \, e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{216 \, b^{3}} - \frac {{\left (b^{2} x^{2} e^{a} - 2 \, b x e^{a} + 2 \, e^{a}\right )} e^{\left (b x\right )}}{8 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{8 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{216 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/216*(9*b^2*x^2*e^(3*a) - 6*b*x*e^(3*a) + 2*e^(3*a))*e^(3*b*x)/b^3 - 1/8*(b^2*x^2*e^a - 2*b*x*e^a + 2*e^a)*e^
(b*x)/b^3 + 1/8*(b^2*x^2 + 2*b*x + 2)*e^(-b*x - a)/b^3 - 1/216*(9*b^2*x^2 + 6*b*x + 2)*e^(-3*b*x - 3*a)/b^3

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mupad [B]  time = 1.68, size = 82, normalized size = 0.99 \[ \frac {\frac {4\,x\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{9}-\frac {2\,x\,\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{3}}{b^2}+\frac {14\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{27\,b^3}-\frac {4\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )}{9\,b^3}+\frac {x^2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((4*x*cosh(a + b*x)^3)/9 - (2*x*cosh(a + b*x)*sinh(a + b*x)^2)/3)/b^2 + (14*sinh(a + b*x)^3)/(27*b^3) - (4*cos
h(a + b*x)^2*sinh(a + b*x))/(9*b^3) + (x^2*sinh(a + b*x)^3)/(3*b)

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sympy [A]  time = 1.70, size = 105, normalized size = 1.27 \[ \begin {cases} \frac {x^{2} \sinh ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 x \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{3 b^{2}} + \frac {4 x \cosh ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {14 \sinh ^{3}{\left (a + b x \right )}}{27 b^{3}} - \frac {4 \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{9 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \sinh ^{2}{\relax (a )} \cosh {\relax (a )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*sinh(a + b*x)**3/(3*b) - 2*x*sinh(a + b*x)**2*cosh(a + b*x)/(3*b**2) + 4*x*cosh(a + b*x)**3/(9
*b**2) + 14*sinh(a + b*x)**3/(27*b**3) - 4*sinh(a + b*x)*cosh(a + b*x)**2/(9*b**3), Ne(b, 0)), (x**3*sinh(a)**
2*cosh(a)/3, True))

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