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

Optimal. Leaf size=202 \[ \frac{3 x^2 \sinh (a+b x)}{8 b^2}+\frac{x^2 \sinh (3 a+3 b x)}{48 b^2}-\frac{3 x^2 \sinh (5 a+5 b x)}{400 b^2}+\frac{3 \sinh (a+b x)}{4 b^4}+\frac{\sinh (3 a+3 b x)}{216 b^4}-\frac{3 \sinh (5 a+5 b x)}{5000 b^4}-\frac{3 x \cosh (a+b x)}{4 b^3}-\frac{x \cosh (3 a+3 b x)}{72 b^3}+\frac{3 x \cosh (5 a+5 b x)}{1000 b^3}-\frac{x^3 \cosh (a+b x)}{8 b}-\frac{x^3 \cosh (3 a+3 b x)}{48 b}+\frac{x^3 \cosh (5 a+5 b x)}{80 b} \]

[Out]

(-3*x*Cosh[a + b*x])/(4*b^3) - (x^3*Cosh[a + b*x])/(8*b) - (x*Cosh[3*a + 3*b*x])/(72*b^3) - (x^3*Cosh[3*a + 3*
b*x])/(48*b) + (3*x*Cosh[5*a + 5*b*x])/(1000*b^3) + (x^3*Cosh[5*a + 5*b*x])/(80*b) + (3*Sinh[a + b*x])/(4*b^4)
+ (3*x^2*Sinh[a + b*x])/(8*b^2) + Sinh[3*a + 3*b*x]/(216*b^4) + (x^2*Sinh[3*a + 3*b*x])/(48*b^2) - (3*Sinh[5*
a + 5*b*x])/(5000*b^4) - (3*x^2*Sinh[5*a + 5*b*x])/(400*b^2)

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Rubi [A]  time = 0.262065, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {5448, 3296, 2637} \[ \frac{3 x^2 \sinh (a+b x)}{8 b^2}+\frac{x^2 \sinh (3 a+3 b x)}{48 b^2}-\frac{3 x^2 \sinh (5 a+5 b x)}{400 b^2}+\frac{3 \sinh (a+b x)}{4 b^4}+\frac{\sinh (3 a+3 b x)}{216 b^4}-\frac{3 \sinh (5 a+5 b x)}{5000 b^4}-\frac{3 x \cosh (a+b x)}{4 b^3}-\frac{x \cosh (3 a+3 b x)}{72 b^3}+\frac{3 x \cosh (5 a+5 b x)}{1000 b^3}-\frac{x^3 \cosh (a+b x)}{8 b}-\frac{x^3 \cosh (3 a+3 b x)}{48 b}+\frac{x^3 \cosh (5 a+5 b x)}{80 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*Cosh[a + b*x])/(4*b^3) - (x^3*Cosh[a + b*x])/(8*b) - (x*Cosh[3*a + 3*b*x])/(72*b^3) - (x^3*Cosh[3*a + 3*
b*x])/(48*b) + (3*x*Cosh[5*a + 5*b*x])/(1000*b^3) + (x^3*Cosh[5*a + 5*b*x])/(80*b) + (3*Sinh[a + b*x])/(4*b^4)
+ (3*x^2*Sinh[a + b*x])/(8*b^2) + Sinh[3*a + 3*b*x]/(216*b^4) + (x^2*Sinh[3*a + 3*b*x])/(48*b^2) - (3*Sinh[5*
a + 5*b*x])/(5000*b^4) - (3*x^2*Sinh[5*a + 5*b*x])/(400*b^2)

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]

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x^3 \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx &=\int \left (-\frac{1}{8} x^3 \sinh (a+b x)-\frac{1}{16} x^3 \sinh (3 a+3 b x)+\frac{1}{16} x^3 \sinh (5 a+5 b x)\right ) \, dx\\ &=-\left (\frac{1}{16} \int x^3 \sinh (3 a+3 b x) \, dx\right )+\frac{1}{16} \int x^3 \sinh (5 a+5 b x) \, dx-\frac{1}{8} \int x^3 \sinh (a+b x) \, dx\\ &=-\frac{x^3 \cosh (a+b x)}{8 b}-\frac{x^3 \cosh (3 a+3 b x)}{48 b}+\frac{x^3 \cosh (5 a+5 b x)}{80 b}-\frac{3 \int x^2 \cosh (5 a+5 b x) \, dx}{80 b}+\frac{\int x^2 \cosh (3 a+3 b x) \, dx}{16 b}+\frac{3 \int x^2 \cosh (a+b x) \, dx}{8 b}\\ &=-\frac{x^3 \cosh (a+b x)}{8 b}-\frac{x^3 \cosh (3 a+3 b x)}{48 b}+\frac{x^3 \cosh (5 a+5 b x)}{80 b}+\frac{3 x^2 \sinh (a+b x)}{8 b^2}+\frac{x^2 \sinh (3 a+3 b x)}{48 b^2}-\frac{3 x^2 \sinh (5 a+5 b x)}{400 b^2}+\frac{3 \int x \sinh (5 a+5 b x) \, dx}{200 b^2}-\frac{\int x \sinh (3 a+3 b x) \, dx}{24 b^2}-\frac{3 \int x \sinh (a+b x) \, dx}{4 b^2}\\ &=-\frac{3 x \cosh (a+b x)}{4 b^3}-\frac{x^3 \cosh (a+b x)}{8 b}-\frac{x \cosh (3 a+3 b x)}{72 b^3}-\frac{x^3 \cosh (3 a+3 b x)}{48 b}+\frac{3 x \cosh (5 a+5 b x)}{1000 b^3}+\frac{x^3 \cosh (5 a+5 b x)}{80 b}+\frac{3 x^2 \sinh (a+b x)}{8 b^2}+\frac{x^2 \sinh (3 a+3 b x)}{48 b^2}-\frac{3 x^2 \sinh (5 a+5 b x)}{400 b^2}-\frac{3 \int \cosh (5 a+5 b x) \, dx}{1000 b^3}+\frac{\int \cosh (3 a+3 b x) \, dx}{72 b^3}+\frac{3 \int \cosh (a+b x) \, dx}{4 b^3}\\ &=-\frac{3 x \cosh (a+b x)}{4 b^3}-\frac{x^3 \cosh (a+b x)}{8 b}-\frac{x \cosh (3 a+3 b x)}{72 b^3}-\frac{x^3 \cosh (3 a+3 b x)}{48 b}+\frac{3 x \cosh (5 a+5 b x)}{1000 b^3}+\frac{x^3 \cosh (5 a+5 b x)}{80 b}+\frac{3 \sinh (a+b x)}{4 b^4}+\frac{3 x^2 \sinh (a+b x)}{8 b^2}+\frac{\sinh (3 a+3 b x)}{216 b^4}+\frac{x^2 \sinh (3 a+3 b x)}{48 b^2}-\frac{3 \sinh (5 a+5 b x)}{5000 b^4}-\frac{3 x^2 \sinh (5 a+5 b x)}{400 b^2}\\ \end{align*}

Mathematica [A]  time = 0.490951, size = 136, normalized size = 0.67 \[ \frac{-33750 \left (b x \left (b^2 x^2+6\right ) \cosh (a+b x)-3 \left (b^2 x^2+2\right ) \sinh (a+b x)\right )-625 \left (\left (9 b^3 x^3+6 b x\right ) \cosh (3 (a+b x))-\left (9 b^2 x^2+2\right ) \sinh (3 (a+b x))\right )+27 \left (5 b x \left (25 b^2 x^2+6\right ) \cosh (5 (a+b x))-3 \left (25 b^2 x^2+2\right ) \sinh (5 (a+b x))\right )}{270000 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-33750*(b*x*(6 + b^2*x^2)*Cosh[a + b*x] - 3*(2 + b^2*x^2)*Sinh[a + b*x]) - 625*((6*b*x + 9*b^3*x^3)*Cosh[3*(a
+ b*x)] - (2 + 9*b^2*x^2)*Sinh[3*(a + b*x)]) + 27*(5*b*x*(6 + 25*b^2*x^2)*Cosh[5*(a + b*x)] - 3*(2 + 25*b^2*x
^2)*Sinh[5*(a + b*x)]))/(270000*b^4)

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Maple [B]  time = 0.01, size = 542, 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)^2*sinh(b*x+a)^3,x)

[Out]

1/b^4*(1/5*(b*x+a)^3*sinh(b*x+a)^2*cosh(b*x+a)^3-2/15*(b*x+a)^3*cosh(b*x+a)*sinh(b*x+a)^2-2/15*(b*x+a)^3*cosh(
b*x+a)-3/25*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^4+26/75*(b*x+a)^2*sinh(b*x+a)+13/75*(b*x+a)^2*sinh(b*x+a)*cosh(b
*x+a)^2+6/125*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^3-76/1125*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)-856/1125*(b*x+a)*c
osh(b*x+a)-6/625*cosh(b*x+a)^4*sinh(b*x+a)+12568/16875*sinh(b*x+a)+434/16875*sinh(b*x+a)*cosh(b*x+a)^2-3*a*(1/
5*(b*x+a)^2*sinh(b*x+a)^2*cosh(b*x+a)^3-2/15*(b*x+a)^2*sinh(b*x+a)^2*cosh(b*x+a)-2/15*(b*x+a)^2*cosh(b*x+a)-2/
25*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^4+52/225*(b*x+a)*sinh(b*x+a)+26/225*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^2+2/125
*cosh(b*x+a)^3*sinh(b*x+a)^2-76/3375*cosh(b*x+a)*sinh(b*x+a)^2-856/3375*cosh(b*x+a))+3*a^2*(1/5*(b*x+a)*sinh(b
*x+a)^2*cosh(b*x+a)^3-2/15*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)-2/15*(b*x+a)*cosh(b*x+a)-1/25*cosh(b*x+a)^4*sinh(
b*x+a)+26/225*sinh(b*x+a)+13/225*sinh(b*x+a)*cosh(b*x+a)^2)-a^3*(1/5*cosh(b*x+a)^3*sinh(b*x+a)^2-2/15*cosh(b*x
+a)*sinh(b*x+a)^2-2/15*cosh(b*x+a)))

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Maxima [A]  time = 1.09067, size = 331, normalized size = 1.64 \begin{align*} \frac{{\left (125 \, b^{3} x^{3} e^{\left (5 \, a\right )} - 75 \, b^{2} x^{2} e^{\left (5 \, a\right )} + 30 \, b x e^{\left (5 \, a\right )} - 6 \, e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )}}{20000 \, b^{4}} - \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 )}}{864 \, 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 )}}{16 \, b^{4}} - \frac{{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x - a\right )}}{16 \, 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 )}}{864 \, b^{4}} + \frac{{\left (125 \, b^{3} x^{3} + 75 \, b^{2} x^{2} + 30 \, b x + 6\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{20000 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20000*(125*b^3*x^3*e^(5*a) - 75*b^2*x^2*e^(5*a) + 30*b*x*e^(5*a) - 6*e^(5*a))*e^(5*b*x)/b^4 - 1/864*(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 - 1/16*(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 - 1/16*(b^3*x^3 + 3*b^2*x^2 + 6*b*x + 6)*e^(-b*x - a)/b^4 - 1/864*(9*b^3*x^3
+ 9*b^2*x^2 + 6*b*x + 2)*e^(-3*b*x - 3*a)/b^4 + 1/20000*(125*b^3*x^3 + 75*b^2*x^2 + 30*b*x + 6)*e^(-5*b*x - 5*
a)/b^4

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Fricas [A]  time = 1.77431, size = 722, normalized size = 3.57 \begin{align*} \frac{135 \,{\left (25 \, b^{3} x^{3} + 6 \, b x\right )} \cosh \left (b x + a\right )^{5} + 675 \,{\left (25 \, b^{3} x^{3} + 6 \, b x\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 81 \,{\left (25 \, b^{2} x^{2} + 2\right )} \sinh \left (b x + a\right )^{5} - 1875 \,{\left (3 \, b^{3} x^{3} + 2 \, b x\right )} \cosh \left (b x + a\right )^{3} + 5 \,{\left (1125 \, b^{2} x^{2} - 162 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} + 250\right )} \sinh \left (b x + a\right )^{3} + 225 \,{\left (6 \,{\left (25 \, b^{3} x^{3} + 6 \, b x\right )} \cosh \left (b x + a\right )^{3} - 25 \,{\left (3 \, b^{3} x^{3} + 2 \, b x\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 33750 \,{\left (b^{3} x^{3} + 6 \, b x\right )} \cosh \left (b x + a\right ) - 15 \,{\left (27 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{4} - 6750 \, b^{2} x^{2} - 125 \,{\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} - 13500\right )} \sinh \left (b x + a\right )}{270000 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/270000*(135*(25*b^3*x^3 + 6*b*x)*cosh(b*x + a)^5 + 675*(25*b^3*x^3 + 6*b*x)*cosh(b*x + a)*sinh(b*x + a)^4 -
81*(25*b^2*x^2 + 2)*sinh(b*x + a)^5 - 1875*(3*b^3*x^3 + 2*b*x)*cosh(b*x + a)^3 + 5*(1125*b^2*x^2 - 162*(25*b^2
*x^2 + 2)*cosh(b*x + a)^2 + 250)*sinh(b*x + a)^3 + 225*(6*(25*b^3*x^3 + 6*b*x)*cosh(b*x + a)^3 - 25*(3*b^3*x^3
+ 2*b*x)*cosh(b*x + a))*sinh(b*x + a)^2 - 33750*(b^3*x^3 + 6*b*x)*cosh(b*x + a) - 15*(27*(25*b^2*x^2 + 2)*cos
h(b*x + a)^4 - 6750*b^2*x^2 - 125*(9*b^2*x^2 + 2)*cosh(b*x + a)^2 - 13500)*sinh(b*x + a))/b^4

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Sympy [A]  time = 13.8801, size = 253, normalized size = 1.25 \begin{align*} \begin{cases} \frac{x^{3} \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac{2 x^{3} \cosh ^{5}{\left (a + b x \right )}}{15 b} + \frac{26 x^{2} \sinh ^{5}{\left (a + b x \right )}}{75 b^{2}} - \frac{13 x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{15 b^{2}} + \frac{2 x^{2} \sinh{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{5 b^{2}} - \frac{52 x \sinh ^{4}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{75 b^{3}} + \frac{338 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{225 b^{3}} - \frac{856 x \cosh ^{5}{\left (a + b x \right )}}{1125 b^{3}} + \frac{12568 \sinh ^{5}{\left (a + b x \right )}}{16875 b^{4}} - \frac{5114 \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3375 b^{4}} + \frac{856 \sinh{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{1125 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \sinh ^{3}{\left (a \right )} \cosh ^{2}{\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)**2*sinh(b*x+a)**3,x)

[Out]

Piecewise((x**3*sinh(a + b*x)**2*cosh(a + b*x)**3/(3*b) - 2*x**3*cosh(a + b*x)**5/(15*b) + 26*x**2*sinh(a + b*
x)**5/(75*b**2) - 13*x**2*sinh(a + b*x)**3*cosh(a + b*x)**2/(15*b**2) + 2*x**2*sinh(a + b*x)*cosh(a + b*x)**4/
(5*b**2) - 52*x*sinh(a + b*x)**4*cosh(a + b*x)/(75*b**3) + 338*x*sinh(a + b*x)**2*cosh(a + b*x)**3/(225*b**3)
- 856*x*cosh(a + b*x)**5/(1125*b**3) + 12568*sinh(a + b*x)**5/(16875*b**4) - 5114*sinh(a + b*x)**3*cosh(a + b*
x)**2/(3375*b**4) + 856*sinh(a + b*x)*cosh(a + b*x)**4/(1125*b**4), Ne(b, 0)), (x**4*sinh(a)**3*cosh(a)**2/4,
True))

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Giac [A]  time = 1.17057, size = 286, normalized size = 1.42 \begin{align*} \frac{{\left (125 \, b^{3} x^{3} - 75 \, b^{2} x^{2} + 30 \, b x - 6\right )} e^{\left (5 \, b x + 5 \, a\right )}}{20000 \, 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 )}}{864 \, b^{4}} - \frac{{\left (b^{3} x^{3} - 3 \, b^{2} x^{2} + 6 \, b x - 6\right )} e^{\left (b x + a\right )}}{16 \, b^{4}} - \frac{{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x - a\right )}}{16 \, 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 )}}{864 \, b^{4}} + \frac{{\left (125 \, b^{3} x^{3} + 75 \, b^{2} x^{2} + 30 \, b x + 6\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{20000 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20000*(125*b^3*x^3 - 75*b^2*x^2 + 30*b*x - 6)*e^(5*b*x + 5*a)/b^4 - 1/864*(9*b^3*x^3 - 9*b^2*x^2 + 6*b*x - 2
)*e^(3*b*x + 3*a)/b^4 - 1/16*(b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*e^(b*x + a)/b^4 - 1/16*(b^3*x^3 + 3*b^2*x^2 + 6
*b*x + 6)*e^(-b*x - a)/b^4 - 1/864*(9*b^3*x^3 + 9*b^2*x^2 + 6*b*x + 2)*e^(-3*b*x - 3*a)/b^4 + 1/20000*(125*b^3
*x^3 + 75*b^2*x^2 + 30*b*x + 6)*e^(-5*b*x - 5*a)/b^4