3.1.24 \(\int x^2 \cosh ^4(a+b x) \, dx\) [24]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 134, normalized size of antiderivative = 1.00, 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 = {3392, 30, 2715, 8} \begin {gather*} \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 \cosh ^4(a+b x)}{8 b^2}-\frac {3 x \cosh ^2(a+b x)}{8 b^2}+\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} \end {gather*}

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2715

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] && Integ
erQ[2*n]

Rule 3392

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]

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*}

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Mathematica [A]
time = 0.12, size = 90, normalized size = 0.67 \begin {gather*} \frac {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} \end {gather*}

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 [A]
time = 1.26, size = 213, normalized size = 1.59

method	result	size
risch	\(\frac {x^{3}}{8}+\frac {\left (8 b^{2} x^{2}-4 b x +1\right ) {\mathrm e}^{4 b x +4 a}}{512 b^{3}}+\frac {\left (2 b^{2} x^{2}-2 b x +1\right ) {\mathrm e}^{2 b x +2 a}}{16 b^{3}}-\frac {\left (2 b^{2} x^{2}+2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{16 b^{3}}-\frac {\left (8 b^{2} x^{2}+4 b x +1\right ) {\mathrm e}^{-4 b x -4 a}}{512 b^{3}}\)	\(119\)
default	\(\frac {x^{3}}{8}+\frac {\left (2 b x +2 a \right )^{2} \sinh \left (2 b x +2 a \right )-2 \left (2 b x +2 a \right ) \cosh \left (2 b x +2 a \right )+2 \sinh \left (2 b x +2 a \right )-4 a \left (\left (2 b x +2 a \right ) \sinh \left (2 b x +2 a \right )-\cosh \left (2 b x +2 a \right )\right )+4 a^{2} \sinh \left (2 b x +2 a \right )}{16 b^{3}}+\frac {\left (4 b x +4 a \right )^{2} \sinh \left (4 b x +4 a \right )-2 \left (4 b x +4 a \right ) \cosh \left (4 b x +4 a \right )+2 \sinh \left (4 b x +4 a \right )-8 a \left (\left (4 b x +4 a \right ) \sinh \left (4 b x +4 a \right )-\cosh \left (4 b x +4 a \right )\right )+16 a^{2} \sinh \left (4 b x +4 a \right )}{512 b^{3}}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 132, normalized size = 0.99 \begin {gather*} \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 {gather*}

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 = 0.34, size = 147, normalized size = 1.10 \begin {gather*} \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 {gather*}

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 = 0.45, size = 209, normalized size = 1.56 \begin {gather*} \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 {gather*}

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 = 0.41, size = 118, normalized size = 0.88 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.25, size = 94, normalized size = 0.70 \begin {gather*} \frac {\frac {\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{8}+\frac {\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{256}-b\,\left (\frac {x\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{4}+\frac {x\,\mathrm {cosh}\left (4\,a+4\,b\,x\right )}{64}\right )+b^2\,\left (\frac {x^2\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{4}+\frac {x^2\,\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{32}\right )}{b^3}+\frac {x^3}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosh(a + b*x)^4,x)

[Out]

(sinh(2*a + 2*b*x)/8 + sinh(4*a + 4*b*x)/256 - b*((x*cosh(2*a + 2*b*x))/4 + (x*cosh(4*a + 4*b*x))/64) + b^2*((
x^2*sinh(2*a + 2*b*x))/4 + (x^2*sinh(4*a + 4*b*x))/32))/b^3 + x^3/8

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