3.24 \(\int x^2 \cosh ^4(a+b x) \, dx\)

Optimal. Leaf size=134 \[ -\frac{x \cosh ^4(a+b x)}{8 b^2}-\frac{3 x \cosh ^2(a+b x)}{8 b^2}+\frac{\sinh (a+b x) \cosh ^3(a+b x)}{32 b^3}+\frac{15 \sinh (a+b x) \cosh (a+b x)}{64 b^3}+\frac{x^2 \sinh (a+b x) \cosh ^3(a+b x)}{4 b}+\frac{3 x^2 \sinh (a+b x) \cosh (a+b x)}{8 b}+\frac{15 x}{64 b^2}+\frac{x^3}{8} \]

[Out]

(15*x)/(64*b^2) + x^3/8 - (3*x*Cosh[a + b*x]^2)/(8*b^2) - (x*Cosh[a + b*x]^4)/(8*b^2) + (15*Cosh[a + b*x]*Sinh
[a + b*x])/(64*b^3) + (3*x^2*Cosh[a + b*x]*Sinh[a + b*x])/(8*b) + (Cosh[a + b*x]^3*Sinh[a + b*x])/(32*b^3) + (
x^2*Cosh[a + b*x]^3*Sinh[a + b*x])/(4*b)

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Rubi [A]  time = 0.106692, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3311, 30, 2635, 8} \[ -\frac{x \cosh ^4(a+b x)}{8 b^2}-\frac{3 x \cosh ^2(a+b x)}{8 b^2}+\frac{\sinh (a+b x) \cosh ^3(a+b x)}{32 b^3}+\frac{15 \sinh (a+b x) \cosh (a+b x)}{64 b^3}+\frac{x^2 \sinh (a+b x) \cosh ^3(a+b x)}{4 b}+\frac{3 x^2 \sinh (a+b x) \cosh (a+b x)}{8 b}+\frac{15 x}{64 b^2}+\frac{x^3}{8} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cosh[a + b*x]^4,x]

[Out]

(15*x)/(64*b^2) + x^3/8 - (3*x*Cosh[a + b*x]^2)/(8*b^2) - (x*Cosh[a + b*x]^4)/(8*b^2) + (15*Cosh[a + b*x]*Sinh
[a + b*x])/(64*b^3) + (3*x^2*Cosh[a + b*x]*Sinh[a + b*x])/(8*b) + (Cosh[a + b*x]^3*Sinh[a + b*x])/(32*b^3) + (
x^2*Cosh[a + b*x]^3*Sinh[a + b*x])/(4*b)

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^2 \cosh ^4(a+b x) \, dx &=-\frac{x \cosh ^4(a+b x)}{8 b^2}+\frac{x^2 \cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac{3}{4} \int x^2 \cosh ^2(a+b x) \, dx+\frac{\int \cosh ^4(a+b x) \, dx}{8 b^2}\\ &=-\frac{3 x \cosh ^2(a+b x)}{8 b^2}-\frac{x \cosh ^4(a+b x)}{8 b^2}+\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac{\cosh ^3(a+b x) \sinh (a+b x)}{32 b^3}+\frac{x^2 \cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac{3 \int x^2 \, dx}{8}+\frac{3 \int \cosh ^2(a+b x) \, dx}{32 b^2}+\frac{3 \int \cosh ^2(a+b x) \, dx}{8 b^2}\\ &=\frac{x^3}{8}-\frac{3 x \cosh ^2(a+b x)}{8 b^2}-\frac{x \cosh ^4(a+b x)}{8 b^2}+\frac{15 \cosh (a+b x) \sinh (a+b x)}{64 b^3}+\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac{\cosh ^3(a+b x) \sinh (a+b x)}{32 b^3}+\frac{x^2 \cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac{3 \int 1 \, dx}{64 b^2}+\frac{3 \int 1 \, dx}{16 b^2}\\ &=\frac{15 x}{64 b^2}+\frac{x^3}{8}-\frac{3 x \cosh ^2(a+b x)}{8 b^2}-\frac{x \cosh ^4(a+b x)}{8 b^2}+\frac{15 \cosh (a+b x) \sinh (a+b x)}{64 b^3}+\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac{\cosh ^3(a+b x) \sinh (a+b x)}{32 b^3}+\frac{x^2 \cosh ^3(a+b x) \sinh (a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.157266, size = 90, normalized size = 0.67 \[ \frac{64 b^2 x^2 \sinh (2 (a+b x))+8 b^2 x^2 \sinh (4 (a+b x))+32 \sinh (2 (a+b x))+\sinh (4 (a+b x))-64 b x \cosh (2 (a+b x))-4 b x \cosh (4 (a+b x))+32 b^3 x^3}{256 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cosh[a + b*x]^4,x]

[Out]

(32*b^3*x^3 - 64*b*x*Cosh[2*(a + b*x)] - 4*b*x*Cosh[4*(a + b*x)] + 32*Sinh[2*(a + b*x)] + 64*b^2*x^2*Sinh[2*(a
+ b*x)] + Sinh[4*(a + b*x)] + 8*b^2*x^2*Sinh[4*(a + b*x)])/(256*b^3)

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Maple [B]  time = 0.007, size = 253, normalized size = 1.9 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{4}}+{\frac{3\, \left ( bx+a \right ) ^{2}\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{8}}+{\frac{ \left ( bx+a \right ) ^{3}}{8}}-{\frac{ \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{8}}-{\frac{ \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2}}+{\frac{\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{32}}+{\frac{15\,\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{64}}+{\frac{15\,bx}{64}}+{\frac{15\,a}{64}}-2\,a \left ( 1/4\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}+3/8\, \left ( bx+a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) +3/16\, \left ( bx+a \right ) ^{2}-1/16\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}-1/4\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \right ) +{a}^{2} \left ( \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{4}}+{\frac{3\,\cosh \left ( bx+a \right ) }{8}} \right ) \sinh \left ( bx+a \right ) +{\frac{3\,bx}{8}}+{\frac{3\,a}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosh(b*x+a)^4,x)

[Out]

1/b^3*(1/4*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^3+3/8*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)+1/8*(b*x+a)^3-1/8*(b*x+a)
*sinh(b*x+a)^2*cosh(b*x+a)^2-1/2*(b*x+a)*cosh(b*x+a)^2+1/32*sinh(b*x+a)*cosh(b*x+a)^3+15/64*cosh(b*x+a)*sinh(b
*x+a)+15/64*b*x+15/64*a-2*a*(1/4*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^3+3/8*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)+3/16*(b
*x+a)^2-1/16*sinh(b*x+a)^2*cosh(b*x+a)^2-1/4*cosh(b*x+a)^2)+a^2*((1/4*cosh(b*x+a)^3+3/8*cosh(b*x+a))*sinh(b*x+
a)+3/8*b*x+3/8*a))

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Maxima [A]  time = 1.04815, size = 178, normalized size = 1.33 \begin{align*} \frac{1}{8} \, x^{3} + \frac{{\left (8 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 4 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{512 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} e^{\left (2 \, a\right )} - 2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{16 \, b^{3}} - \frac{{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{3}} - \frac{{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)^4,x, algorithm="maxima")

[Out]

1/8*x^3 + 1/512*(8*b^2*x^2*e^(4*a) - 4*b*x*e^(4*a) + e^(4*a))*e^(4*b*x)/b^3 + 1/16*(2*b^2*x^2*e^(2*a) - 2*b*x*
e^(2*a) + e^(2*a))*e^(2*b*x)/b^3 - 1/16*(2*b^2*x^2 + 2*b*x + 1)*e^(-2*b*x - 2*a)/b^3 - 1/512*(8*b^2*x^2 + 4*b*
x + 1)*e^(-4*b*x - 4*a)/b^3

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Fricas [A]  time = 1.98625, size = 373, normalized size = 2.78 \begin{align*} \frac{8 \, b^{3} x^{3} - b x \cosh \left (b x + a\right )^{4} - b x \sinh \left (b x + a\right )^{4} +{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} - 16 \, b x \cosh \left (b x + a\right )^{2} - 2 \,{\left (3 \, b x \cosh \left (b x + a\right )^{2} + 8 \, b x\right )} \sinh \left (b x + a\right )^{2} +{\left ({\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} + 16 \,{\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{64 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)^4,x, algorithm="fricas")

[Out]

1/64*(8*b^3*x^3 - b*x*cosh(b*x + a)^4 - b*x*sinh(b*x + a)^4 + (8*b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3 -
16*b*x*cosh(b*x + a)^2 - 2*(3*b*x*cosh(b*x + a)^2 + 8*b*x)*sinh(b*x + a)^2 + ((8*b^2*x^2 + 1)*cosh(b*x + a)^3
+ 16*(2*b^2*x^2 + 1)*cosh(b*x + a))*sinh(b*x + a))/b^3

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Sympy [A]  time = 4.3555, size = 209, normalized size = 1.56 \begin{align*} \begin{cases} \frac{x^{3} \sinh ^{4}{\left (a + b x \right )}}{8} - \frac{x^{3} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{4} + \frac{x^{3} \cosh ^{4}{\left (a + b x \right )}}{8} - \frac{3 x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8 b} + \frac{5 x^{2} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b} + \frac{15 x \sinh ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac{3 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{32 b^{2}} - \frac{17 x \cosh ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac{15 \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{64 b^{3}} + \frac{17 \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{64 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \cosh ^{4}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cosh(b*x+a)**4,x)

[Out]

Piecewise((x**3*sinh(a + b*x)**4/8 - x**3*sinh(a + b*x)**2*cosh(a + b*x)**2/4 + x**3*cosh(a + b*x)**4/8 - 3*x*
*2*sinh(a + b*x)**3*cosh(a + b*x)/(8*b) + 5*x**2*sinh(a + b*x)*cosh(a + b*x)**3/(8*b) + 15*x*sinh(a + b*x)**4/
(64*b**2) - 3*x*sinh(a + b*x)**2*cosh(a + b*x)**2/(32*b**2) - 17*x*cosh(a + b*x)**4/(64*b**2) - 15*sinh(a + b*
x)**3*cosh(a + b*x)/(64*b**3) + 17*sinh(a + b*x)*cosh(a + b*x)**3/(64*b**3), Ne(b, 0)), (x**3*cosh(a)**4/3, Tr
ue))

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Giac [A]  time = 1.30187, size = 159, normalized size = 1.19 \begin{align*} \frac{1}{8} \, x^{3} + \frac{{\left (8 \, b^{2} x^{2} - 4 \, b x + 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{512 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} - 2 \, b x + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{3}} - \frac{{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{3}} - \frac{{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)^4,x, algorithm="giac")

[Out]

1/8*x^3 + 1/512*(8*b^2*x^2 - 4*b*x + 1)*e^(4*b*x + 4*a)/b^3 + 1/16*(2*b^2*x^2 - 2*b*x + 1)*e^(2*b*x + 2*a)/b^3
- 1/16*(2*b^2*x^2 + 2*b*x + 1)*e^(-2*b*x - 2*a)/b^3 - 1/512*(8*b^2*x^2 + 4*b*x + 1)*e^(-4*b*x - 4*a)/b^3