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

Optimal. Leaf size=155 \[ -\frac{3 x^2 \sinh ^3(a+b x) \cosh (a+b x)}{16 b^2}+\frac{9 x^2 \sinh (a+b x) \cosh (a+b x)}{32 b^2}+\frac{3 x \sinh ^4(a+b x)}{32 b^3}-\frac{9 x \sinh ^2(a+b x)}{32 b^3}-\frac{3 \sinh ^3(a+b x) \cosh (a+b x)}{128 b^4}+\frac{45 \sinh (a+b x) \cosh (a+b x)}{256 b^4}+\frac{x^3 \sinh ^4(a+b x)}{4 b}-\frac{45 x}{256 b^3}-\frac{3 x^3}{32 b} \]

[Out]

(-45*x)/(256*b^3) - (3*x^3)/(32*b) + (45*Cosh[a + b*x]*Sinh[a + b*x])/(256*b^4) + (9*x^2*Cosh[a + b*x]*Sinh[a
+ b*x])/(32*b^2) - (9*x*Sinh[a + b*x]^2)/(32*b^3) - (3*Cosh[a + b*x]*Sinh[a + b*x]^3)/(128*b^4) - (3*x^2*Cosh[
a + b*x]*Sinh[a + b*x]^3)/(16*b^2) + (3*x*Sinh[a + b*x]^4)/(32*b^3) + (x^3*Sinh[a + b*x]^4)/(4*b)

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Rubi [A]  time = 0.13935, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {5372, 3311, 30, 2635, 8} \[ -\frac{3 x^2 \sinh ^3(a+b x) \cosh (a+b x)}{16 b^2}+\frac{9 x^2 \sinh (a+b x) \cosh (a+b x)}{32 b^2}+\frac{3 x \sinh ^4(a+b x)}{32 b^3}-\frac{9 x \sinh ^2(a+b x)}{32 b^3}-\frac{3 \sinh ^3(a+b x) \cosh (a+b x)}{128 b^4}+\frac{45 \sinh (a+b x) \cosh (a+b x)}{256 b^4}+\frac{x^3 \sinh ^4(a+b x)}{4 b}-\frac{45 x}{256 b^3}-\frac{3 x^3}{32 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-45*x)/(256*b^3) - (3*x^3)/(32*b) + (45*Cosh[a + b*x]*Sinh[a + b*x])/(256*b^4) + (9*x^2*Cosh[a + b*x]*Sinh[a
+ b*x])/(32*b^2) - (9*x*Sinh[a + b*x]^2)/(32*b^3) - (3*Cosh[a + b*x]*Sinh[a + b*x]^3)/(128*b^4) - (3*x^2*Cosh[
a + b*x]*Sinh[a + b*x]^3)/(16*b^2) + (3*x*Sinh[a + b*x]^4)/(32*b^3) + (x^3*Sinh[a + b*x]^4)/(4*b)

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]

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^3 \cosh (a+b x) \sinh ^3(a+b x) \, dx &=\frac{x^3 \sinh ^4(a+b x)}{4 b}-\frac{3 \int x^2 \sinh ^4(a+b x) \, dx}{4 b}\\ &=-\frac{3 x^2 \cosh (a+b x) \sinh ^3(a+b x)}{16 b^2}+\frac{3 x \sinh ^4(a+b x)}{32 b^3}+\frac{x^3 \sinh ^4(a+b x)}{4 b}-\frac{3 \int \sinh ^4(a+b x) \, dx}{32 b^3}+\frac{9 \int x^2 \sinh ^2(a+b x) \, dx}{16 b}\\ &=\frac{9 x^2 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac{9 x \sinh ^2(a+b x)}{32 b^3}-\frac{3 \cosh (a+b x) \sinh ^3(a+b x)}{128 b^4}-\frac{3 x^2 \cosh (a+b x) \sinh ^3(a+b x)}{16 b^2}+\frac{3 x \sinh ^4(a+b x)}{32 b^3}+\frac{x^3 \sinh ^4(a+b x)}{4 b}+\frac{9 \int \sinh ^2(a+b x) \, dx}{128 b^3}+\frac{9 \int \sinh ^2(a+b x) \, dx}{32 b^3}-\frac{9 \int x^2 \, dx}{32 b}\\ &=-\frac{3 x^3}{32 b}+\frac{45 \cosh (a+b x) \sinh (a+b x)}{256 b^4}+\frac{9 x^2 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac{9 x \sinh ^2(a+b x)}{32 b^3}-\frac{3 \cosh (a+b x) \sinh ^3(a+b x)}{128 b^4}-\frac{3 x^2 \cosh (a+b x) \sinh ^3(a+b x)}{16 b^2}+\frac{3 x \sinh ^4(a+b x)}{32 b^3}+\frac{x^3 \sinh ^4(a+b x)}{4 b}-\frac{9 \int 1 \, dx}{256 b^3}-\frac{9 \int 1 \, dx}{64 b^3}\\ &=-\frac{45 x}{256 b^3}-\frac{3 x^3}{32 b}+\frac{45 \cosh (a+b x) \sinh (a+b x)}{256 b^4}+\frac{9 x^2 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac{9 x \sinh ^2(a+b x)}{32 b^3}-\frac{3 \cosh (a+b x) \sinh ^3(a+b x)}{128 b^4}-\frac{3 x^2 \cosh (a+b x) \sinh ^3(a+b x)}{16 b^2}+\frac{3 x \sinh ^4(a+b x)}{32 b^3}+\frac{x^3 \sinh ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.635343, size = 95, normalized size = 0.61 \[ \frac{48 \left (2 b^2 x^2+1\right ) \sinh (2 (a+b x))+2 b x \left (8 b^2 x^2+3\right ) \cosh (4 (a+b x))-\left (3 \left (8 b^2 x^2+1\right ) \sinh (2 (a+b x))+32 b x \left (2 b^2 x^2+3\right )\right ) \cosh (2 (a+b x))}{512 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*x*(3 + 8*b^2*x^2)*Cosh[4*(a + b*x)] + 48*(1 + 2*b^2*x^2)*Sinh[2*(a + b*x)] - Cosh[2*(a + b*x)]*(32*b*x*(3
+ 2*b^2*x^2) + 3*(1 + 8*b^2*x^2)*Sinh[2*(a + b*x)]))/(512*b^4)

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Maple [B]  time = 0.006, size = 414, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^4*(1/4*(b*x+a)^3*sinh(b*x+a)^2*cosh(b*x+a)^2-1/4*(b*x+a)^3*cosh(b*x+a)^2-3/16*(b*x+a)^2*sinh(b*x+a)*cosh(b
*x+a)^3+15/32*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)+5/32*(b*x+a)^3+3/32*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^2-3/128*
cosh(b*x+a)^3*sinh(b*x+a)+51/256*cosh(b*x+a)*sinh(b*x+a)+51/256*b*x+51/256*a-3/8*(b*x+a)*cosh(b*x+a)^2-3*a*(1/
4*(b*x+a)^2*sinh(b*x+a)^2*cosh(b*x+a)^2-1/4*(b*x+a)^2*cosh(b*x+a)^2-1/8*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^3+5/16
*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)+5/32*(b*x+a)^2+1/32*cosh(b*x+a)^2*sinh(b*x+a)^2-1/8*cosh(b*x+a)^2)+3*a^2*(1/4
*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^2-1/4*(b*x+a)*cosh(b*x+a)^2-1/16*cosh(b*x+a)^3*sinh(b*x+a)+5/32*cosh(b*x+a)
*sinh(b*x+a)+5/32*b*x+5/32*a)-a^3*(1/4*cosh(b*x+a)^2*sinh(b*x+a)^2-1/4*cosh(b*x+a)^2))

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Maxima [A]  time = 1.07176, size = 231, normalized size = 1.49 \begin{align*} \frac{{\left (32 \, b^{3} x^{3} e^{\left (4 \, a\right )} - 24 \, b^{2} x^{2} e^{\left (4 \, a\right )} + 12 \, b x e^{\left (4 \, a\right )} - 3 \, e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{2048 \, b^{4}} - \frac{{\left (4 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 6 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 6 \, b x e^{\left (2 \, a\right )} - 3 \, e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{64 \, b^{4}} - \frac{{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{64 \, b^{4}} + \frac{{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2048*(32*b^3*x^3*e^(4*a) - 24*b^2*x^2*e^(4*a) + 12*b*x*e^(4*a) - 3*e^(4*a))*e^(4*b*x)/b^4 - 1/64*(4*b^3*x^3*
e^(2*a) - 6*b^2*x^2*e^(2*a) + 6*b*x*e^(2*a) - 3*e^(2*a))*e^(2*b*x)/b^4 - 1/64*(4*b^3*x^3 + 6*b^2*x^2 + 6*b*x +
3)*e^(-2*b*x - 2*a)/b^4 + 1/2048*(32*b^3*x^3 + 24*b^2*x^2 + 12*b*x + 3)*e^(-4*b*x - 4*a)/b^4

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Fricas [A]  time = 1.69249, size = 468, normalized size = 3.02 \begin{align*} \frac{{\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{4} - 3 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} +{\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \sinh \left (b x + a\right )^{4} - 16 \,{\left (2 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{2} - 2 \,{\left (16 \, b^{3} x^{3} - 3 \,{\left (8 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{2} + 24 \, b x\right )} \sinh \left (b x + a\right )^{2} - 3 \,{\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 )}{256 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*((8*b^3*x^3 + 3*b*x)*cosh(b*x + a)^4 - 3*(8*b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (8*b^3*x^3 + 3*
b*x)*sinh(b*x + a)^4 - 16*(2*b^3*x^3 + 3*b*x)*cosh(b*x + a)^2 - 2*(16*b^3*x^3 - 3*(8*b^3*x^3 + 3*b*x)*cosh(b*x
+ a)^2 + 24*b*x)*sinh(b*x + a)^2 - 3*((8*b^2*x^2 + 1)*cosh(b*x + a)^3 - 16*(2*b^2*x^2 + 1)*cosh(b*x + a))*sin
h(b*x + a))/b^4

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Sympy [A]  time = 7.94228, size = 226, normalized size = 1.46 \begin{align*} \begin{cases} \frac{5 x^{3} \sinh ^{4}{\left (a + b x \right )}}{32 b} + \frac{3 x^{3} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b} - \frac{3 x^{3} \cosh ^{4}{\left (a + b x \right )}}{32 b} - \frac{15 x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{32 b^{2}} + \frac{9 x^{2} \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{32 b^{2}} + \frac{51 x \sinh ^{4}{\left (a + b x \right )}}{256 b^{3}} + \frac{9 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{128 b^{3}} - \frac{45 x \cosh ^{4}{\left (a + b x \right )}}{256 b^{3}} - \frac{51 \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{256 b^{4}} + \frac{45 \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{256 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \sinh ^{3}{\left (a \right )} \cosh{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*x**3*sinh(a + b*x)**4/(32*b) + 3*x**3*sinh(a + b*x)**2*cosh(a + b*x)**2/(16*b) - 3*x**3*cosh(a +
b*x)**4/(32*b) - 15*x**2*sinh(a + b*x)**3*cosh(a + b*x)/(32*b**2) + 9*x**2*sinh(a + b*x)*cosh(a + b*x)**3/(32*
b**2) + 51*x*sinh(a + b*x)**4/(256*b**3) + 9*x*sinh(a + b*x)**2*cosh(a + b*x)**2/(128*b**3) - 45*x*cosh(a + b*
x)**4/(256*b**3) - 51*sinh(a + b*x)**3*cosh(a + b*x)/(256*b**4) + 45*sinh(a + b*x)*cosh(a + b*x)**3/(256*b**4)
, Ne(b, 0)), (x**4*sinh(a)**3*cosh(a)/4, True))

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Giac [A]  time = 1.2151, size = 196, normalized size = 1.26 \begin{align*} \frac{{\left (32 \, b^{3} x^{3} - 24 \, b^{2} x^{2} + 12 \, b x - 3\right )} e^{\left (4 \, b x + 4 \, a\right )}}{2048 \, b^{4}} - \frac{{\left (4 \, b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, b x - 3\right )} e^{\left (2 \, b x + 2 \, a\right )}}{64 \, b^{4}} - \frac{{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{64 \, b^{4}} + \frac{{\left (32 \, b^{3} x^{3} + 24 \, b^{2} x^{2} + 12 \, b x + 3\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{2048 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2048*(32*b^3*x^3 - 24*b^2*x^2 + 12*b*x - 3)*e^(4*b*x + 4*a)/b^4 - 1/64*(4*b^3*x^3 - 6*b^2*x^2 + 6*b*x - 3)*e
^(2*b*x + 2*a)/b^4 - 1/64*(4*b^3*x^3 + 6*b^2*x^2 + 6*b*x + 3)*e^(-2*b*x - 2*a)/b^4 + 1/2048*(32*b^3*x^3 + 24*b
^2*x^2 + 12*b*x + 3)*e^(-4*b*x - 4*a)/b^4