3.3.81 \(\int x^3 \cosh (a+b x) \sinh ^2(a+b x) \, dx\) [281]

Optimal. Leaf size=117 \[ \frac {14 \cosh (a+b x)}{9 b^4}+\frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {2 \cosh ^3(a+b x)}{27 b^4}-\frac {4 x \sinh (a+b x)}{3 b^3}-\frac {x^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}+\frac {x^3 \sinh ^3(a+b x)}{3 b} \]

[Out]

14/9*cosh(b*x+a)/b^4+2/3*x^2*cosh(b*x+a)/b^2-2/27*cosh(b*x+a)^3/b^4-4/3*x*sinh(b*x+a)/b^3-1/3*x^2*cosh(b*x+a)*
sinh(b*x+a)^2/b^2+2/9*x*sinh(b*x+a)^3/b^3+1/3*x^3*sinh(b*x+a)^3/b

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Rubi [A]
time = 0.10, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5480, 3392, 3377, 2718, 2713} \begin {gather*} -\frac {2 \cosh ^3(a+b x)}{27 b^4}+\frac {14 \cosh (a+b x)}{9 b^4}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}-\frac {4 x \sinh (a+b x)}{3 b^3}+\frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {x^2 \sinh ^2(a+b x) \cosh (a+b x)}{3 b^2}+\frac {x^3 \sinh ^3(a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(14*Cosh[a + b*x])/(9*b^4) + (2*x^2*Cosh[a + b*x])/(3*b^2) - (2*Cosh[a + b*x]^3)/(27*b^4) - (4*x*Sinh[a + b*x]
)/(3*b^3) - (x^2*Cosh[a + b*x]*Sinh[a + b*x]^2)/(3*b^2) + (2*x*Sinh[a + b*x]^3)/(9*b^3) + (x^3*Sinh[a + b*x]^3
)/(3*b)

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

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 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]

Rule 5480

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^3 \cosh (a+b x) \sinh ^2(a+b x) \, dx &=\frac {x^3 \sinh ^3(a+b x)}{3 b}-\frac {\int x^2 \sinh ^3(a+b x) \, dx}{b}\\ &=-\frac {x^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}+\frac {x^3 \sinh ^3(a+b x)}{3 b}-\frac {2 \int \sinh ^3(a+b x) \, dx}{9 b^3}+\frac {2 \int x^2 \sinh (a+b x) \, dx}{3 b}\\ &=\frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {x^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}+\frac {x^3 \sinh ^3(a+b x)}{3 b}+\frac {2 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (a+b x)\right )}{9 b^4}-\frac {4 \int x \cosh (a+b x) \, dx}{3 b^2}\\ &=\frac {2 \cosh (a+b x)}{9 b^4}+\frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {2 \cosh ^3(a+b x)}{27 b^4}-\frac {4 x \sinh (a+b x)}{3 b^3}-\frac {x^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}+\frac {x^3 \sinh ^3(a+b x)}{3 b}+\frac {4 \int \sinh (a+b x) \, dx}{3 b^3}\\ &=\frac {14 \cosh (a+b x)}{9 b^4}+\frac {2 x^2 \cosh (a+b x)}{3 b^2}-\frac {2 \cosh ^3(a+b x)}{27 b^4}-\frac {4 x \sinh (a+b x)}{3 b^3}-\frac {x^2 \cosh (a+b x) \sinh ^2(a+b x)}{3 b^2}+\frac {2 x \sinh ^3(a+b x)}{9 b^3}+\frac {x^3 \sinh ^3(a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 84, normalized size = 0.72 \begin {gather*} \frac {81 \left (2+b^2 x^2\right ) \cosh (a+b x)-\left (2+9 b^2 x^2\right ) \cosh (3 (a+b x))+6 b x \left (-26-3 b^2 x^2+\left (2+3 b^2 x^2\right ) \cosh (2 (a+b x))\right ) \sinh (a+b x)}{108 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(81*(2 + b^2*x^2)*Cosh[a + b*x] - (2 + 9*b^2*x^2)*Cosh[3*(a + b*x)] + 6*b*x*(-26 - 3*b^2*x^2 + (2 + 3*b^2*x^2)
*Cosh[2*(a + b*x)])*Sinh[a + b*x])/(108*b^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(313\) vs. \(2(103)=206\).
time = 1.63, size = 314, normalized size = 2.68

method	result	size
risch	\(\frac {\left (9 b^{3} x^{3}-9 b^{2} x^{2}+6 b x -2\right ) {\mathrm e}^{3 b x +3 a}}{216 b^{4}}-\frac {\left (b^{3} x^{3}-3 b^{2} x^{2}+6 b x -6\right ) {\mathrm e}^{b x +a}}{8 b^{4}}+\frac {\left (b^{3} x^{3}+3 b^{2} x^{2}+6 b x +6\right ) {\mathrm e}^{-b x -a}}{8 b^{4}}-\frac {\left (9 b^{3} x^{3}+9 b^{2} x^{2}+6 b x +2\right ) {\mathrm e}^{-3 b x -3 a}}{216 b^{4}}\)	\(141\)
default	\(-\frac {\left (b x +a \right )^{3} \sinh \left (b x +a \right )-3 \left (b x +a \right )^{2} \cosh \left (b x +a \right )+6 \left (b x +a \right ) \sinh \left (b x +a \right )-6 \cosh \left (b x +a \right )-3 a \left (\left (b x +a \right )^{2} \sinh \left (b x +a \right )-2 \left (b x +a \right ) \cosh \left (b x +a \right )+2 \sinh \left (b x +a \right )\right )+3 a^{2} \left (\left (b x +a \right ) \sinh \left (b x +a \right )-\cosh \left (b x +a \right )\right )-a^{3} \sinh \left (b x +a \right )}{4 b^{4}}+\frac {\left (3 b x +3 a \right )^{3} \sinh \left (3 b x +3 a \right )-3 \left (3 b x +3 a \right )^{2} \cosh \left (3 b x +3 a \right )+6 \left (3 b x +3 a \right ) \sinh \left (3 b x +3 a \right )-6 \cosh \left (3 b x +3 a \right )-9 a \left (\left (3 b x +3 a \right )^{2} \sinh \left (3 b x +3 a \right )-2 \left (3 b x +3 a \right ) \cosh \left (3 b x +3 a \right )+2 \sinh \left (3 b x +3 a \right )\right )+27 a^{2} \left (\left (3 b x +3 a \right ) \sinh \left (3 b x +3 a \right )-\cosh \left (3 b x +3 a \right )\right )-27 a^{3} \sinh \left (3 b x +3 a \right )}{324 b^{4}}\)	\(314\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/b^4*((b*x+a)^3*sinh(b*x+a)-3*(b*x+a)^2*cosh(b*x+a)+6*(b*x+a)*sinh(b*x+a)-6*cosh(b*x+a)-3*a*((b*x+a)^2*sin
h(b*x+a)-2*(b*x+a)*cosh(b*x+a)+2*sinh(b*x+a))+3*a^2*((b*x+a)*sinh(b*x+a)-cosh(b*x+a))-a^3*sinh(b*x+a))+1/324/b
^4*((3*b*x+3*a)^3*sinh(3*b*x+3*a)-3*(3*b*x+3*a)^2*cosh(3*b*x+3*a)+6*(3*b*x+3*a)*sinh(3*b*x+3*a)-6*cosh(3*b*x+3
*a)-9*a*((3*b*x+3*a)^2*sinh(3*b*x+3*a)-2*(3*b*x+3*a)*cosh(3*b*x+3*a)+2*sinh(3*b*x+3*a))+27*a^2*((3*b*x+3*a)*si
nh(3*b*x+3*a)-cosh(3*b*x+3*a))-27*a^3*sinh(3*b*x+3*a))

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Maxima [A]
time = 0.28, size = 160, normalized size = 1.37 \begin {gather*} \frac {{\left (9 \, b^{3} x^{3} e^{\left (3 \, a\right )} - 9 \, b^{2} x^{2} e^{\left (3 \, a\right )} + 6 \, b x e^{\left (3 \, a\right )} - 2 \, e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{216 \, b^{4}} - \frac {{\left (b^{3} x^{3} e^{a} - 3 \, b^{2} x^{2} e^{a} + 6 \, b x e^{a} - 6 \, e^{a}\right )} e^{\left (b x\right )}}{8 \, b^{4}} + \frac {{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x - a\right )}}{8 \, b^{4}} - \frac {{\left (9 \, b^{3} x^{3} + 9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{216 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/216*(9*b^3*x^3*e^(3*a) - 9*b^2*x^2*e^(3*a) + 6*b*x*e^(3*a) - 2*e^(3*a))*e^(3*b*x)/b^4 - 1/8*(b^3*x^3*e^a - 3
*b^2*x^2*e^a + 6*b*x*e^a - 6*e^a)*e^(b*x)/b^4 + 1/8*(b^3*x^3 + 3*b^2*x^2 + 6*b*x + 6)*e^(-b*x - a)/b^4 - 1/216
*(9*b^3*x^3 + 9*b^2*x^2 + 6*b*x + 2)*e^(-3*b*x - 3*a)/b^4

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Fricas [A]
time = 0.36, size = 135, normalized size = 1.15 \begin {gather*} -\frac {{\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{3} + 3 \, {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 3 \, {\left (3 \, b^{3} x^{3} + 2 \, b x\right )} \sinh \left (b x + a\right )^{3} - 81 \, {\left (b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) + 9 \, {\left (3 \, b^{3} x^{3} - {\left (3 \, b^{3} x^{3} + 2 \, b x\right )} \cosh \left (b x + a\right )^{2} + 18 \, b x\right )} \sinh \left (b x + a\right )}{108 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/108*((9*b^2*x^2 + 2)*cosh(b*x + a)^3 + 3*(9*b^2*x^2 + 2)*cosh(b*x + a)*sinh(b*x + a)^2 - 3*(3*b^3*x^3 + 2*b
*x)*sinh(b*x + a)^3 - 81*(b^2*x^2 + 2)*cosh(b*x + a) + 9*(3*b^3*x^3 - (3*b^3*x^3 + 2*b*x)*cosh(b*x + a)^2 + 18
*b*x)*sinh(b*x + a))/b^4

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Sympy [A]
time = 0.43, size = 146, normalized size = 1.25 \begin {gather*} \begin {cases} \frac {x^{3} \sinh ^{3}{\left (a + b x \right )}}{3 b} - \frac {x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{b^{2}} + \frac {2 x^{2} \cosh ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac {14 x \sinh ^{3}{\left (a + b x \right )}}{9 b^{3}} - \frac {4 x \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b^{3}} - \frac {14 \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{9 b^{4}} + \frac {40 \cosh ^{3}{\left (a + b x \right )}}{27 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \sinh ^{2}{\left (a \right )} \cosh {\left (a \right )}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**3*sinh(a + b*x)**3/(3*b) - x**2*sinh(a + b*x)**2*cosh(a + b*x)/b**2 + 2*x**2*cosh(a + b*x)**3/(3
*b**2) + 14*x*sinh(a + b*x)**3/(9*b**3) - 4*x*sinh(a + b*x)*cosh(a + b*x)**2/(3*b**3) - 14*sinh(a + b*x)**2*co
sh(a + b*x)/(9*b**4) + 40*cosh(a + b*x)**3/(27*b**4), Ne(b, 0)), (x**4*sinh(a)**2*cosh(a)/4, True))

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Giac [A]
time = 0.41, size = 140, normalized size = 1.20 \begin {gather*} \frac {{\left (9 \, b^{3} x^{3} - 9 \, b^{2} x^{2} + 6 \, b x - 2\right )} e^{\left (3 \, b x + 3 \, a\right )}}{216 \, b^{4}} - \frac {{\left (b^{3} x^{3} - 3 \, b^{2} x^{2} + 6 \, b x - 6\right )} e^{\left (b x + a\right )}}{8 \, b^{4}} + \frac {{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x - a\right )}}{8 \, b^{4}} - \frac {{\left (9 \, b^{3} x^{3} + 9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{216 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/216*(9*b^3*x^3 - 9*b^2*x^2 + 6*b*x - 2)*e^(3*b*x + 3*a)/b^4 - 1/8*(b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*e^(b*x +
a)/b^4 + 1/8*(b^3*x^3 + 3*b^2*x^2 + 6*b*x + 6)*e^(-b*x - a)/b^4 - 1/216*(9*b^3*x^3 + 9*b^2*x^2 + 6*b*x + 2)*e
^(-3*b*x - 3*a)/b^4

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Mupad [B]
time = 0.17, size = 119, normalized size = 1.02 \begin {gather*} \frac {\frac {14\,x\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{9}-\frac {4\,x\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )}{3}}{b^3}+\frac {\frac {2\,x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{3}-x^2\,\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{b^2}+\frac {40\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{27\,b^4}-\frac {14\,\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{9\,b^4}+\frac {x^3\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((14*x*sinh(a + b*x)^3)/9 - (4*x*cosh(a + b*x)^2*sinh(a + b*x))/3)/b^3 + ((2*x^2*cosh(a + b*x)^3)/3 - x^2*cosh
(a + b*x)*sinh(a + b*x)^2)/b^2 + (40*cosh(a + b*x)^3)/(27*b^4) - (14*cosh(a + b*x)*sinh(a + b*x)^2)/(9*b^4) +
(x^3*sinh(a + b*x)^3)/(3*b)

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