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

Optimal. Leaf size=101 \[ \frac {\sinh ^4(a+b x)}{32 b^3}-\frac {3 \sinh ^2(a+b x)}{32 b^3}-\frac {x \sinh ^3(a+b x) \cosh (a+b x)}{8 b^2}+\frac {3 x \sinh (a+b x) \cosh (a+b x)}{16 b^2}+\frac {x^2 \sinh ^4(a+b x)}{4 b}-\frac {3 x^2}{32 b} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5372, 3310, 30} \[ \frac {\sinh ^4(a+b x)}{32 b^3}-\frac {3 \sinh ^2(a+b x)}{32 b^3}-\frac {x \sinh ^3(a+b x) \cosh (a+b x)}{8 b^2}+\frac {3 x \sinh (a+b x) \cosh (a+b x)}{16 b^2}+\frac {x^2 \sinh ^4(a+b x)}{4 b}-\frac {3 x^2}{32 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x^2)/(32*b) + (3*x*Cosh[a + b*x]*Sinh[a + b*x])/(16*b^2) - (3*Sinh[a + b*x]^2)/(32*b^3) - (x*Cosh[a + b*x]
*Sinh[a + b*x]^3)/(8*b^2) + Sinh[a + b*x]^4/(32*b^3) + (x^2*Sinh[a + b*x]^4)/(4*b)

Rule 30

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

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

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Mathematica [A]  time = 0.23, size = 72, normalized size = 0.71 \[ \frac {-16 \left (2 b^2 x^2+1\right ) \cosh (2 (a+b x))+\left (8 b^2 x^2+1\right ) \cosh (4 (a+b x))+4 b x (8 \sinh (2 (a+b x))-\sinh (4 (a+b x)))}{256 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*(1 + 2*b^2*x^2)*Cosh[2*(a + b*x)] + (1 + 8*b^2*x^2)*Cosh[4*(a + b*x)] + 4*b*x*(8*Sinh[2*(a + b*x)] - Sinh
[4*(a + b*x)]))/(256*b^3)

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fricas [A]  time = 0.48, size = 154, normalized size = 1.52 \[ -\frac {16 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} - {\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{4} - {\left (8 \, b^{2} x^{2} + 1\right )} \sinh \left (b x + a\right )^{4} + 16 \, {\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (16 \, b^{2} x^{2} - 3 \, {\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} + 8\right )} \sinh \left (b x + a\right )^{2} + 16 \, {\left (b x \cosh \left (b x + a\right )^{3} - 4 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{256 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/256*(16*b*x*cosh(b*x + a)*sinh(b*x + a)^3 - (8*b^2*x^2 + 1)*cosh(b*x + a)^4 - (8*b^2*x^2 + 1)*sinh(b*x + a)
^4 + 16*(2*b^2*x^2 + 1)*cosh(b*x + a)^2 + 2*(16*b^2*x^2 - 3*(8*b^2*x^2 + 1)*cosh(b*x + a)^2 + 8)*sinh(b*x + a)
^2 + 16*(b*x*cosh(b*x + a)^3 - 4*b*x*cosh(b*x + a))*sinh(b*x + a))/b^3

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giac [A]  time = 0.12, size = 113, normalized size = 1.12 \[ \frac {{\left (8 \, b^{2} x^{2} - 4 \, b x + 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{512 \, b^{3}} - \frac {{\left (2 \, b^{2} x^{2} - 2 \, b x + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b^{3}} - \frac {{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{3}} + \frac {{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/512*(8*b^2*x^2 - 4*b*x + 1)*e^(4*b*x + 4*a)/b^3 - 1/32*(2*b^2*x^2 - 2*b*x + 1)*e^(2*b*x + 2*a)/b^3 - 1/32*(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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maple [A]  time = 0.33, size = 161, normalized size = 1.59 \[ \frac {\frac {\left (b x +a \right )^{2} \left (\sinh ^{4}\left (b x +a \right )\right )}{4}-\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \left (\sinh ^{3}\left (b x +a \right )\right )}{8}+\frac {3 \left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{16}-\frac {3 \left (b x +a \right )^{2}}{32}+\frac {\left (\sinh ^{4}\left (b x +a \right )\right )}{32}-\frac {3 \left (\cosh ^{2}\left (b x +a \right )\right )}{32}-2 a \left (\frac {\left (b x +a \right ) \left (\sinh ^{4}\left (b x +a \right )\right )}{4}-\frac {\cosh \left (b x +a \right ) \left (\sinh ^{3}\left (b x +a \right )\right )}{16}+\frac {3 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{32}-\frac {3 b x}{32}-\frac {3 a}{32}\right )+\frac {a^{2} \left (\sinh ^{4}\left (b x +a \right )\right )}{4}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 127, normalized size = 1.26 \[ \frac {{\left (8 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 4 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{512 \, b^{3}} - \frac {{\left (2 \, b^{2} x^{2} e^{\left (2 \, a\right )} - 2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{32 \, b^{3}} - \frac {{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{3}} + \frac {{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 - 1/32*(2*b^2*x^2*e^(2*a) - 2*b*x*e^(2*a) +
e^(2*a))*e^(2*b*x)/b^3 - 1/32*(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 = 1.61, size = 89, normalized size = 0.88 \[ -\frac {\frac {\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{16}-\frac {\mathrm {cosh}\left (4\,a+4\,b\,x\right )}{256}+b^2\,\left (\frac {x^2\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{8}-\frac {x^2\,\mathrm {cosh}\left (4\,a+4\,b\,x\right )}{32}\right )-b\,\left (\frac {x\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{8}-\frac {x\,\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{64}\right )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cosh(2*a + 2*b*x)/16 - cosh(4*a + 4*b*x)/256 + b^2*((x^2*cosh(2*a + 2*b*x))/8 - (x^2*cosh(4*a + 4*b*x))/32)
- b*((x*sinh(2*a + 2*b*x))/8 - (x*sinh(4*a + 4*b*x))/64))/b^3

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sympy [A]  time = 2.90, size = 150, normalized size = 1.49 \[ \begin {cases} \frac {5 x^{2} \sinh ^{4}{\left (a + b x \right )}}{32 b} + \frac {3 x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b} - \frac {3 x^{2} \cosh ^{4}{\left (a + b x \right )}}{32 b} - \frac {5 x \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{16 b^{2}} + \frac {3 x \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{16 b^{2}} + \frac {5 \sinh ^{4}{\left (a + b x \right )}}{64 b^{3}} - \frac {3 \cosh ^{4}{\left (a + b x \right )}}{64 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \sinh ^{3}{\relax (a )} \cosh {\relax (a )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*x**2*sinh(a + b*x)**4/(32*b) + 3*x**2*sinh(a + b*x)**2*cosh(a + b*x)**2/(16*b) - 3*x**2*cosh(a +
b*x)**4/(32*b) - 5*x*sinh(a + b*x)**3*cosh(a + b*x)/(16*b**2) + 3*x*sinh(a + b*x)*cosh(a + b*x)**3/(16*b**2) +
5*sinh(a + b*x)**4/(64*b**3) - 3*cosh(a + b*x)**4/(64*b**3), Ne(b, 0)), (x**3*sinh(a)**3*cosh(a)/3, True))

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