### 3.299 $$\int x^3 \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.267255, 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, 2638} $\frac{3 x^2 \cosh (a+b x)}{8 b^2}-\frac{x^2 \cosh (3 a+3 b x)}{48 b^2}-\frac{3 x^2 \cosh (5 a+5 b x)}{400 b^2}-\frac{3 x \sinh (a+b x)}{4 b^3}+\frac{x \sinh (3 a+3 b x)}{72 b^3}+\frac{3 x \sinh (5 a+5 b x)}{1000 b^3}+\frac{3 \cosh (a+b x)}{4 b^4}-\frac{\cosh (3 a+3 b x)}{216 b^4}-\frac{3 \cosh (5 a+5 b x)}{5000 b^4}-\frac{x^3 \sinh (a+b x)}{8 b}+\frac{x^3 \sinh (3 a+3 b x)}{48 b}+\frac{x^3 \sinh (5 a+5 b x)}{80 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 2638

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

Rubi steps

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

Mathematica [A]  time = 1.08862, size = 125, normalized size = 0.62 $\frac{101250 \left (b^2 x^2+2\right ) \cosh (a+b x)-625 \left (9 b^2 x^2+2\right ) \cosh (3 (a+b x))-81 \left (25 b^2 x^2+2\right ) \cosh (5 (a+b x))+30 b x \sinh (a+b x) \left (8 \left (75 b^2 x^2+38\right ) \cosh (2 (a+b x))+9 \left (25 b^2 x^2+6\right ) \cosh (4 (a+b x))-825 b^2 x^2-6598\right )}{270000 b^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(101250*(2 + b^2*x^2)*Cosh[a + b*x] - 625*(2 + 9*b^2*x^2)*Cosh[3*(a + b*x)] - 81*(2 + 25*b^2*x^2)*Cosh[5*(a +
b*x)] + 30*b*x*(-6598 - 825*b^2*x^2 + 8*(38 + 75*b^2*x^2)*Cosh[2*(a + b*x)] + 9*(6 + 25*b^2*x^2)*Cosh[4*(a + b
*x)])*Sinh[a + b*x])/(270000*b^4)

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Maple [B]  time = 0.012, size = 534, normalized size = 2.6 \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)^3*sinh(b*x+a)^2,x)

[Out]

1/b^4*(1/5*(b*x+a)^3*sinh(b*x+a)*cosh(b*x+a)^4-2/15*(b*x+a)^3*sinh(b*x+a)-1/15*(b*x+a)^3*sinh(b*x+a)*cosh(b*x+
a)^2-3/25*(b*x+a)^2*sinh(b*x+a)^2*cosh(b*x+a)^3-4/75*(b*x+a)^2*sinh(b*x+a)^2*cosh(b*x+a)+26/75*(b*x+a)^2*cosh(
b*x+a)+6/125*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^4-856/1125*(b*x+a)*sinh(b*x+a)+22/1125*(b*x+a)*sinh(b*x+a)*cosh(b
*x+a)^2-6/625*cosh(b*x+a)^3*sinh(b*x+a)^2-272/16875*cosh(b*x+a)*sinh(b*x+a)^2+12568/16875*cosh(b*x+a)-3*a*(1/5
*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^4-2/15*(b*x+a)^2*sinh(b*x+a)-1/15*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^2-2/25*
(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^3-8/225*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)+52/225*(b*x+a)*cosh(b*x+a)+2/125*c
osh(b*x+a)^4*sinh(b*x+a)-856/3375*sinh(b*x+a)+22/3375*sinh(b*x+a)*cosh(b*x+a)^2)+3*a^2*(1/5*(b*x+a)*sinh(b*x+a
)*cosh(b*x+a)^4-2/15*(b*x+a)*sinh(b*x+a)-1/15*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^2-1/25*cosh(b*x+a)^3*sinh(b*x+a)
^2-4/225*cosh(b*x+a)*sinh(b*x+a)^2+26/225*cosh(b*x+a))-a^3*(1/5*cosh(b*x+a)^4*sinh(b*x+a)-1/5*(2/3+1/3*cosh(b*
x+a)^2)*sinh(b*x+a)))

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Maxima [A]  time = 1.08415, 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)^3*sinh(b*x+a)^2,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 = 1.75494, size = 716, normalized size = 3.54 \begin{align*} -\frac{81 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{5} + 405 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 135 \,{\left (25 \, b^{3} x^{3} + 6 \, b x\right )} \sinh \left (b x + a\right )^{5} + 625 \,{\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{3} - 75 \,{\left (75 \, b^{3} x^{3} + 18 \,{\left (25 \, b^{3} x^{3} + 6 \, b x\right )} \cosh \left (b x + a\right )^{2} + 50 \, b x\right )} \sinh \left (b x + a\right )^{3} + 15 \,{\left (54 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{3} + 125 \,{\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 101250 \,{\left (b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) + 225 \,{\left (150 \, b^{3} x^{3} - 3 \,{\left (25 \, b^{3} x^{3} + 6 \, b x\right )} \cosh \left (b x + a\right )^{4} - 25 \,{\left (3 \, b^{3} x^{3} + 2 \, b x\right )} \cosh \left (b x + a\right )^{2} + 900 \, b x\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)^3*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/270000*(81*(25*b^2*x^2 + 2)*cosh(b*x + a)^5 + 405*(25*b^2*x^2 + 2)*cosh(b*x + a)*sinh(b*x + a)^4 - 135*(25*
b^3*x^3 + 6*b*x)*sinh(b*x + a)^5 + 625*(9*b^2*x^2 + 2)*cosh(b*x + a)^3 - 75*(75*b^3*x^3 + 18*(25*b^3*x^3 + 6*b
*x)*cosh(b*x + a)^2 + 50*b*x)*sinh(b*x + a)^3 + 15*(54*(25*b^2*x^2 + 2)*cosh(b*x + a)^3 + 125*(9*b^2*x^2 + 2)*
cosh(b*x + a))*sinh(b*x + a)^2 - 101250*(b^2*x^2 + 2)*cosh(b*x + a) + 225*(150*b^3*x^3 - 3*(25*b^3*x^3 + 6*b*x
)*cosh(b*x + a)^4 - 25*(3*b^3*x^3 + 2*b*x)*cosh(b*x + a)^2 + 900*b*x)*sinh(b*x + a))/b^4

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Sympy [A]  time = 13.5314, size = 253, normalized size = 1.25 \begin{align*} \begin{cases} - \frac{2 x^{3} \sinh ^{5}{\left (a + b x \right )}}{15 b} + \frac{x^{3} \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b} + \frac{2 x^{2} \sinh ^{4}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{5 b^{2}} - \frac{13 x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{15 b^{2}} + \frac{26 x^{2} \cosh ^{5}{\left (a + b x \right )}}{75 b^{2}} - \frac{856 x \sinh ^{5}{\left (a + b x \right )}}{1125 b^{3}} + \frac{338 x \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{225 b^{3}} - \frac{52 x \sinh{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{75 b^{3}} + \frac{856 \sinh ^{4}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{1125 b^{4}} - \frac{5114 \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3375 b^{4}} + \frac{12568 \cosh ^{5}{\left (a + b x \right )}}{16875 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \sinh ^{2}{\left (a \right )} \cosh ^{3}{\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)**3*sinh(b*x+a)**2,x)

[Out]

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

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Giac [A]  time = 1.1853, 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)^3*sinh(b*x+a)^2,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