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

Optimal. Leaf size=60 \[ \frac {\sinh (4 a+4 b x)}{256 b^3}-\frac {x \cosh (4 a+4 b x)}{64 b^2}+\frac {x^2 \sinh (4 a+4 b x)}{32 b}-\frac {x^3}{24} \]

[Out]

-1/24*x^3-1/64*x*cosh(4*b*x+4*a)/b^2+1/256*sinh(4*b*x+4*a)/b^3+1/32*x^2*sinh(4*b*x+4*a)/b

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Rubi [A]  time = 0.10, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5448, 3296, 2637} \[ \frac {\sinh (4 a+4 b x)}{256 b^3}-\frac {x \cosh (4 a+4 b x)}{64 b^2}+\frac {x^2 \sinh (4 a+4 b x)}{32 b}-\frac {x^3}{24} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^3/24 - (x*Cosh[4*a + 4*b*x])/(64*b^2) + Sinh[4*a + 4*b*x]/(256*b^3) + (x^2*Sinh[4*a + 4*b*x])/(32*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 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^2 \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx &=\int \left (-\frac {x^2}{8}+\frac {1}{8} x^2 \cosh (4 a+4 b x)\right ) \, dx\\ &=-\frac {x^3}{24}+\frac {1}{8} \int x^2 \cosh (4 a+4 b x) \, dx\\ &=-\frac {x^3}{24}+\frac {x^2 \sinh (4 a+4 b x)}{32 b}-\frac {\int x \sinh (4 a+4 b x) \, dx}{16 b}\\ &=-\frac {x^3}{24}-\frac {x \cosh (4 a+4 b x)}{64 b^2}+\frac {x^2 \sinh (4 a+4 b x)}{32 b}+\frac {\int \cosh (4 a+4 b x) \, dx}{64 b^2}\\ &=-\frac {x^3}{24}-\frac {x \cosh (4 a+4 b x)}{64 b^2}+\frac {\sinh (4 a+4 b x)}{256 b^3}+\frac {x^2 \sinh (4 a+4 b x)}{32 b}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 48, normalized size = 0.80 \[ \frac {3 \left (8 b^2 x^2+1\right ) \sinh (4 (a+b x))-12 b x \cosh (4 (a+b x))-32 b^3 x^3}{768 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*b^3*x^3 - 12*b*x*Cosh[4*(a + b*x)] + 3*(1 + 8*b^2*x^2)*Sinh[4*(a + b*x)])/(768*b^3)

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fricas [B]  time = 0.77, size = 110, normalized size = 1.83 \[ -\frac {8 \, b^{3} x^{3} + 3 \, b x \cosh \left (b x + a\right )^{4} + 18 \, b x \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 3 \, b x \sinh \left (b x + a\right )^{4} - 3 \, {\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) - 3 \, {\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3}}{192 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(8*b^3*x^3 + 3*b*x*cosh(b*x + a)^4 + 18*b*x*cosh(b*x + a)^2*sinh(b*x + a)^2 + 3*b*x*sinh(b*x + a)^4 - 3
*(8*b^2*x^2 + 1)*cosh(b*x + a)^3*sinh(b*x + a) - 3*(8*b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3)/b^3

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giac [A]  time = 0.12, size = 62, normalized size = 1.03 \[ -\frac {1}{24} \, x^{3} + \frac {{\left (8 \, b^{2} x^{2} - 4 \, b x + 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{512 \, b^{3}} - \frac {{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*x^3 + 1/512*(8*b^2*x^2 - 4*b*x + 1)*e^(4*b*x + 4*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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maple [B]  time = 0.32, size = 241, normalized size = 4.02 \[ \frac {\frac {\left (b x +a \right )^{2} \sinh \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{4}-\frac {\left (b x +a \right )^{2} \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}-\frac {\left (b x +a \right )^{3}}{24}-\frac {\left (b x +a \right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{8}+\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{32}-\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{64}-\frac {b x}{64}-\frac {a}{64}+\frac {\left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{8}-2 a \left (\frac {\left (b x +a \right ) \sinh \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{4}-\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}-\frac {\left (b x +a \right )^{2}}{16}-\frac {\left (\cosh ^{4}\left (b x +a \right )\right )}{16}+\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{16}\right )+a^{2} \left (\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{4}-\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}-\frac {b x}{8}-\frac {a}{8}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/4*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^3-1/8*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)-1/24*(b*x+a)^3-1/8*(b*x+a
)*cosh(b*x+a)^4+1/32*cosh(b*x+a)^3*sinh(b*x+a)-1/64*cosh(b*x+a)*sinh(b*x+a)-1/64*b*x-1/64*a+1/8*(b*x+a)*cosh(b
*x+a)^2-2*a*(1/4*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^3-1/8*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)-1/16*(b*x+a)^2-1/16*cos
h(b*x+a)^4+1/16*cosh(b*x+a)^2)+a^2*(1/4*cosh(b*x+a)^3*sinh(b*x+a)-1/8*cosh(b*x+a)*sinh(b*x+a)-1/8*b*x-1/8*a))

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maxima [A]  time = 0.34, size = 69, normalized size = 1.15 \[ -\frac {1}{24} \, x^{3} + \frac {{\left (8 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 4 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{512 \, b^{3}} - \frac {{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*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 - 1/512*(8*b^2*x^2 + 4*b*x + 1)*
e^(-4*b*x - 4*a)/b^3

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mupad [B]  time = 0.14, size = 52, normalized size = 0.87 \[ \frac {\frac {\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{256}+\frac {b^2\,x^2\,\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{32}-\frac {b\,x\,\mathrm {cosh}\left (4\,a+4\,b\,x\right )}{64}}{b^3}-\frac {x^3}{24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sinh(4*a + 4*b*x)/256 + (b^2*x^2*sinh(4*a + 4*b*x))/32 - (b*x*cosh(4*a + 4*b*x))/64)/b^3 - x^3/24

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sympy [A]  time = 3.05, size = 204, normalized size = 3.40 \[ \begin {cases} - \frac {x^{3} \sinh ^{4}{\left (a + b x \right )}}{24} + \frac {x^{3} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{12} - \frac {x^{3} \cosh ^{4}{\left (a + b x \right )}}{24} + \frac {x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{8 b} + \frac {x^{2} \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b} - \frac {x \sinh ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac {3 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{32 b^{2}} - \frac {x \cosh ^{4}{\left (a + b x \right )}}{64 b^{2}} + \frac {\sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{64 b^{3}} + \frac {\sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{64 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \sinh ^{2}{\relax (a )} \cosh ^{2}{\relax (a )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x**3*sinh(a + b*x)**4/24 + x**3*sinh(a + b*x)**2*cosh(a + b*x)**2/12 - x**3*cosh(a + b*x)**4/24 +
x**2*sinh(a + b*x)**3*cosh(a + b*x)/(8*b) + x**2*sinh(a + b*x)*cosh(a + b*x)**3/(8*b) - x*sinh(a + b*x)**4/(64
*b**2) - 3*x*sinh(a + b*x)**2*cosh(a + b*x)**2/(32*b**2) - x*cosh(a + b*x)**4/(64*b**2) + sinh(a + b*x)**3*cos
h(a + b*x)/(64*b**3) + sinh(a + b*x)*cosh(a + b*x)**3/(64*b**3), Ne(b, 0)), (x**3*sinh(a)**2*cosh(a)**2/3, Tru
e))

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