3.281 \(\int x^3 \cosh (a+b x) \sinh ^2(a+b x) \, dx\)

Optimal. Leaf size=117 \[ -\frac {2 \cosh ^3(a+b x)}{27 b^4}+\frac {14 \cosh (a+b x)}{9 b^4}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}-\frac {4 x \sinh (a+b x)}{3 b^3}+\frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {x^2 \sinh ^2(a+b x) \cosh (a+b x)}{3 b^2}+\frac {x^3 \sinh ^3(a+b x)}{3 b} \]

[Out]

14/9*cosh(b*x+a)/b^4+2/3*x^2*cosh(b*x+a)/b^2-2/27*cosh(b*x+a)^3/b^4-4/3*x*sinh(b*x+a)/b^3-1/3*x^2*cosh(b*x+a)*
sinh(b*x+a)^2/b^2+2/9*x*sinh(b*x+a)^3/b^3+1/3*x^3*sinh(b*x+a)^3/b

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Rubi [A]  time = 0.14, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5372, 3311, 3296, 2638, 2633} \[ \frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {x^2 \sinh ^2(a+b x) \cosh (a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}-\frac {4 x \sinh (a+b x)}{3 b^3}-\frac {2 \cosh ^3(a+b x)}{27 b^4}+\frac {14 \cosh (a+b x)}{9 b^4}+\frac {x^3 \sinh ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 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 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^3 \cosh (a+b x) \sinh ^2(a+b x) \, dx &=\frac {x^3 \sinh ^3(a+b x)}{3 b}-\frac {\int x^2 \sinh ^3(a+b x) \, dx}{b}\\ &=-\frac {x^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}+\frac {x^3 \sinh ^3(a+b x)}{3 b}-\frac {2 \int \sinh ^3(a+b x) \, dx}{9 b^3}+\frac {2 \int x^2 \sinh (a+b x) \, dx}{3 b}\\ &=\frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {x^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}+\frac {x^3 \sinh ^3(a+b x)}{3 b}+\frac {2 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (a+b x)\right )}{9 b^4}-\frac {4 \int x \cosh (a+b x) \, dx}{3 b^2}\\ &=\frac {2 \cosh (a+b x)}{9 b^4}+\frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {2 \cosh ^3(a+b x)}{27 b^4}-\frac {4 x \sinh (a+b x)}{3 b^3}-\frac {x^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}+\frac {x^3 \sinh ^3(a+b x)}{3 b}+\frac {4 \int \sinh (a+b x) \, dx}{3 b^3}\\ &=\frac {14 \cosh (a+b x)}{9 b^4}+\frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {2 \cosh ^3(a+b x)}{27 b^4}-\frac {4 x \sinh (a+b x)}{3 b^3}-\frac {x^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}+\frac {x^3 \sinh ^3(a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 84, normalized size = 0.72 \[ \frac {81 \left (b^2 x^2+2\right ) \cosh (a+b x)-\left (9 b^2 x^2+2\right ) \cosh (3 (a+b x))+6 b x \sinh (a+b x) \left (\left (3 b^2 x^2+2\right ) \cosh (2 (a+b x))-3 b^2 x^2-26\right )}{108 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(81*(2 + b^2*x^2)*Cosh[a + b*x] - (2 + 9*b^2*x^2)*Cosh[3*(a + b*x)] + 6*b*x*(-26 - 3*b^2*x^2 + (2 + 3*b^2*x^2)
*Cosh[2*(a + b*x)])*Sinh[a + b*x])/(108*b^4)

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fricas [A]  time = 0.54, size = 135, normalized size = 1.15 \[ -\frac {{\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{3} + 3 \, {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 3 \, {\left (3 \, b^{3} x^{3} + 2 \, b x\right )} \sinh \left (b x + a\right )^{3} - 81 \, {\left (b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) + 9 \, {\left (3 \, b^{3} x^{3} - {\left (3 \, b^{3} x^{3} + 2 \, b x\right )} \cosh \left (b x + a\right )^{2} + 18 \, b x\right )} \sinh \left (b x + a\right )}{108 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/108*((9*b^2*x^2 + 2)*cosh(b*x + a)^3 + 3*(9*b^2*x^2 + 2)*cosh(b*x + a)*sinh(b*x + a)^2 - 3*(3*b^3*x^3 + 2*b
*x)*sinh(b*x + a)^3 - 81*(b^2*x^2 + 2)*cosh(b*x + a) + 9*(3*b^3*x^3 - (3*b^3*x^3 + 2*b*x)*cosh(b*x + a)^2 + 18
*b*x)*sinh(b*x + a))/b^4

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giac [A]  time = 0.12, size = 140, normalized size = 1.20 \[ \frac {{\left (9 \, b^{3} x^{3} - 9 \, b^{2} x^{2} + 6 \, b x - 2\right )} e^{\left (3 \, b x + 3 \, a\right )}}{216 \, b^{4}} - \frac {{\left (b^{3} x^{3} - 3 \, b^{2} x^{2} + 6 \, b x - 6\right )} e^{\left (b x + a\right )}}{8 \, b^{4}} + \frac {{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x - a\right )}}{8 \, b^{4}} - \frac {{\left (9 \, b^{3} x^{3} + 9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{216 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/216*(9*b^3*x^3 - 9*b^2*x^2 + 6*b*x - 2)*e^(3*b*x + 3*a)/b^4 - 1/8*(b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*e^(b*x +
a)/b^4 + 1/8*(b^3*x^3 + 3*b^2*x^2 + 6*b*x + 6)*e^(-b*x - a)/b^4 - 1/216*(9*b^3*x^3 + 9*b^2*x^2 + 6*b*x + 2)*e
^(-3*b*x - 3*a)/b^4

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maple [B]  time = 0.30, size = 244, normalized size = 2.09 \[ \frac {\frac {\left (b x +a \right )^{3} \left (\sinh ^{3}\left (b x +a \right )\right )}{3}+\frac {2 \left (b x +a \right )^{2} \cosh \left (b x +a \right )}{3}-\frac {\left (b x +a \right )^{2} \cosh \left (b x +a \right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{3}-\frac {4 \left (b x +a \right ) \sinh \left (b x +a \right )}{3}+\frac {40 \cosh \left (b x +a \right )}{27}+\frac {2 \left (b x +a \right ) \left (\sinh ^{3}\left (b x +a \right )\right )}{9}-\frac {2 \cosh \left (b x +a \right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{27}-3 a \left (\frac {\left (b x +a \right )^{2} \left (\sinh ^{3}\left (b x +a \right )\right )}{3}+\frac {4 \left (b x +a \right ) \cosh \left (b x +a \right )}{9}-\frac {2 \left (b x +a \right ) \cosh \left (b x +a \right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{9}-\frac {4 \sinh \left (b x +a \right )}{9}+\frac {2 \left (\sinh ^{3}\left (b x +a \right )\right )}{27}\right )+3 a^{2} \left (\frac {\left (b x +a \right ) \left (\sinh ^{3}\left (b x +a \right )\right )}{3}+\frac {2 \cosh \left (b x +a \right )}{9}-\frac {\cosh \left (b x +a \right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{9}\right )-\frac {a^{3} \left (\sinh ^{3}\left (b x +a \right )\right )}{3}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosh(b*x+a)*sinh(b*x+a)^2,x)

[Out]

1/b^4*(1/3*(b*x+a)^3*sinh(b*x+a)^3+2/3*(b*x+a)^2*cosh(b*x+a)-1/3*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)^2-4/3*(b*x+
a)*sinh(b*x+a)+40/27*cosh(b*x+a)+2/9*(b*x+a)*sinh(b*x+a)^3-2/27*cosh(b*x+a)*sinh(b*x+a)^2-3*a*(1/3*(b*x+a)^2*s
inh(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)
+3*a^2*(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^3*sinh(b*x+a)^3)

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maxima [A]  time = 0.33, size = 160, normalized size = 1.37 \[ \frac {{\left (9 \, b^{3} x^{3} e^{\left (3 \, a\right )} - 9 \, b^{2} x^{2} e^{\left (3 \, a\right )} + 6 \, b x e^{\left (3 \, a\right )} - 2 \, e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{216 \, b^{4}} - \frac {{\left (b^{3} x^{3} e^{a} - 3 \, b^{2} x^{2} e^{a} + 6 \, b x e^{a} - 6 \, e^{a}\right )} e^{\left (b x\right )}}{8 \, b^{4}} + \frac {{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x - a\right )}}{8 \, b^{4}} - \frac {{\left (9 \, b^{3} x^{3} + 9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{216 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/216*(9*b^3*x^3*e^(3*a) - 9*b^2*x^2*e^(3*a) + 6*b*x*e^(3*a) - 2*e^(3*a))*e^(3*b*x)/b^4 - 1/8*(b^3*x^3*e^a - 3
*b^2*x^2*e^a + 6*b*x*e^a - 6*e^a)*e^(b*x)/b^4 + 1/8*(b^3*x^3 + 3*b^2*x^2 + 6*b*x + 6)*e^(-b*x - a)/b^4 - 1/216
*(9*b^3*x^3 + 9*b^2*x^2 + 6*b*x + 2)*e^(-3*b*x - 3*a)/b^4

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mupad [B]  time = 0.17, size = 119, normalized size = 1.02 \[ \frac {\frac {14\,x\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{9}-\frac {4\,x\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )}{3}}{b^3}+\frac {\frac {2\,x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{3}-x^2\,\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{b^2}+\frac {40\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{27\,b^4}-\frac {14\,\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{9\,b^4}+\frac {x^3\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosh(a + b*x)*sinh(a + b*x)^2,x)

[Out]

((14*x*sinh(a + b*x)^3)/9 - (4*x*cosh(a + b*x)^2*sinh(a + b*x))/3)/b^3 + ((2*x^2*cosh(a + b*x)^3)/3 - x^2*cosh
(a + b*x)*sinh(a + b*x)^2)/b^2 + (40*cosh(a + b*x)^3)/(27*b^4) - (14*cosh(a + b*x)*sinh(a + b*x)^2)/(9*b^4) +
(x^3*sinh(a + b*x)^3)/(3*b)

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sympy [A]  time = 2.93, size = 146, normalized size = 1.25 \[ \begin {cases} \frac {x^{3} \sinh ^{3}{\left (a + b x \right )}}{3 b} - \frac {x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{b^{2}} + \frac {2 x^{2} \cosh ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac {14 x \sinh ^{3}{\left (a + b x \right )}}{9 b^{3}} - \frac {4 x \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b^{3}} - \frac {14 \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{9 b^{4}} + \frac {40 \cosh ^{3}{\left (a + b x \right )}}{27 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \sinh ^{2}{\relax (a )} \cosh {\relax (a )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cosh(b*x+a)*sinh(b*x+a)**2,x)

[Out]

Piecewise((x**3*sinh(a + b*x)**3/(3*b) - x**2*sinh(a + b*x)**2*cosh(a + b*x)/b**2 + 2*x**2*cosh(a + b*x)**3/(3
*b**2) + 14*x*sinh(a + b*x)**3/(9*b**3) - 4*x*sinh(a + b*x)*cosh(a + b*x)**2/(3*b**3) - 14*sinh(a + b*x)**2*co
sh(a + b*x)/(9*b**4) + 40*cosh(a + b*x)**3/(27*b**4), Ne(b, 0)), (x**4*sinh(a)**2*cosh(a)/4, True))

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