### 3.326 $$\int x^3 \cosh ^3(a+b x) \sinh ^3(a+b x) \, dx$$

Optimal. Leaf size=143 $\frac{9 x^2 \sinh (2 a+2 b x)}{128 b^2}-\frac{x^2 \sinh (6 a+6 b x)}{384 b^2}+\frac{9 \sinh (2 a+2 b x)}{256 b^4}-\frac{\sinh (6 a+6 b x)}{6912 b^4}-\frac{9 x \cosh (2 a+2 b x)}{128 b^3}+\frac{x \cosh (6 a+6 b x)}{1152 b^3}-\frac{3 x^3 \cosh (2 a+2 b x)}{64 b}+\frac{x^3 \cosh (6 a+6 b x)}{192 b}$

[Out]

(-9*x*Cosh[2*a + 2*b*x])/(128*b^3) - (3*x^3*Cosh[2*a + 2*b*x])/(64*b) + (x*Cosh[6*a + 6*b*x])/(1152*b^3) + (x^
3*Cosh[6*a + 6*b*x])/(192*b) + (9*Sinh[2*a + 2*b*x])/(256*b^4) + (9*x^2*Sinh[2*a + 2*b*x])/(128*b^2) - Sinh[6*
a + 6*b*x]/(6912*b^4) - (x^2*Sinh[6*a + 6*b*x])/(384*b^2)

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Rubi [A]  time = 0.197204, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {5448, 3296, 2637} $\frac{9 x^2 \sinh (2 a+2 b x)}{128 b^2}-\frac{x^2 \sinh (6 a+6 b x)}{384 b^2}+\frac{9 \sinh (2 a+2 b x)}{256 b^4}-\frac{\sinh (6 a+6 b x)}{6912 b^4}-\frac{9 x \cosh (2 a+2 b x)}{128 b^3}+\frac{x \cosh (6 a+6 b x)}{1152 b^3}-\frac{3 x^3 \cosh (2 a+2 b x)}{64 b}+\frac{x^3 \cosh (6 a+6 b x)}{192 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-9*x*Cosh[2*a + 2*b*x])/(128*b^3) - (3*x^3*Cosh[2*a + 2*b*x])/(64*b) + (x*Cosh[6*a + 6*b*x])/(1152*b^3) + (x^
3*Cosh[6*a + 6*b*x])/(192*b) + (9*Sinh[2*a + 2*b*x])/(256*b^4) + (9*x^2*Sinh[2*a + 2*b*x])/(128*b^2) - Sinh[6*
a + 6*b*x]/(6912*b^4) - (x^2*Sinh[6*a + 6*b*x])/(384*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 ^3(a+b x) \sinh ^3(a+b x) \, dx &=\int \left (-\frac{3}{32} x^3 \sinh (2 a+2 b x)+\frac{1}{32} x^3 \sinh (6 a+6 b x)\right ) \, dx\\ &=\frac{1}{32} \int x^3 \sinh (6 a+6 b x) \, dx-\frac{3}{32} \int x^3 \sinh (2 a+2 b x) \, dx\\ &=-\frac{3 x^3 \cosh (2 a+2 b x)}{64 b}+\frac{x^3 \cosh (6 a+6 b x)}{192 b}-\frac{\int x^2 \cosh (6 a+6 b x) \, dx}{64 b}+\frac{9 \int x^2 \cosh (2 a+2 b x) \, dx}{64 b}\\ &=-\frac{3 x^3 \cosh (2 a+2 b x)}{64 b}+\frac{x^3 \cosh (6 a+6 b x)}{192 b}+\frac{9 x^2 \sinh (2 a+2 b x)}{128 b^2}-\frac{x^2 \sinh (6 a+6 b x)}{384 b^2}+\frac{\int x \sinh (6 a+6 b x) \, dx}{192 b^2}-\frac{9 \int x \sinh (2 a+2 b x) \, dx}{64 b^2}\\ &=-\frac{9 x \cosh (2 a+2 b x)}{128 b^3}-\frac{3 x^3 \cosh (2 a+2 b x)}{64 b}+\frac{x \cosh (6 a+6 b x)}{1152 b^3}+\frac{x^3 \cosh (6 a+6 b x)}{192 b}+\frac{9 x^2 \sinh (2 a+2 b x)}{128 b^2}-\frac{x^2 \sinh (6 a+6 b x)}{384 b^2}-\frac{\int \cosh (6 a+6 b x) \, dx}{1152 b^3}+\frac{9 \int \cosh (2 a+2 b x) \, dx}{128 b^3}\\ &=-\frac{9 x \cosh (2 a+2 b x)}{128 b^3}-\frac{3 x^3 \cosh (2 a+2 b x)}{64 b}+\frac{x \cosh (6 a+6 b x)}{1152 b^3}+\frac{x^3 \cosh (6 a+6 b x)}{192 b}+\frac{9 \sinh (2 a+2 b x)}{256 b^4}+\frac{9 x^2 \sinh (2 a+2 b x)}{128 b^2}-\frac{\sinh (6 a+6 b x)}{6912 b^4}-\frac{x^2 \sinh (6 a+6 b x)}{384 b^2}\\ \end{align*}

Mathematica [A]  time = 0.88507, size = 90, normalized size = 0.63 $-\frac{81 b x \left (2 b^2 x^2+3\right ) \cosh (2 (a+b x))-3 \left (6 b^3 x^3+b x\right ) \cosh (6 (a+b x))+\sinh (2 (a+b x)) \left (\left (18 b^2 x^2+1\right ) \cosh (4 (a+b x))-234 b^2 x^2-121\right )}{3456 b^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(81*b*x*(3 + 2*b^2*x^2)*Cosh[2*(a + b*x)] - 3*(b*x + 6*b^3*x^3)*Cosh[6*(a + b*x)] + (-121 - 234*b^2*x^2 + (1
+ 18*b^2*x^2)*Cosh[4*(a + b*x)])*Sinh[2*(a + b*x)])/(3456*b^4)

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Maple [B]  time = 0.01, size = 622, normalized size = 4.4 \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)^3,x)

[Out]

1/b^4*(1/6*(b*x+a)^3*sinh(b*x+a)^2*cosh(b*x+a)^4-1/12*(b*x+a)^3*sinh(b*x+a)^2*cosh(b*x+a)^2-1/12*(b*x+a)^3*cos
h(b*x+a)^2-1/12*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^5+1/12*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^3+1/36*(b*x+a)*sinh
(b*x+a)^2*cosh(b*x+a)^4-1/72*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^2-5/36*(b*x+a)*cosh(b*x+a)^2-1/216*sinh(b*x+a)*
cosh(b*x+a)^5+1/216*cosh(b*x+a)^3*sinh(b*x+a)+5/72*cosh(b*x+a)*sinh(b*x+a)+5/72*b*x+5/72*a+1/8*(b*x+a)^2*cosh(
b*x+a)*sinh(b*x+a)+1/24*(b*x+a)^3-3*a*(1/6*(b*x+a)^2*sinh(b*x+a)^2*cosh(b*x+a)^4-1/12*(b*x+a)^2*sinh(b*x+a)^2*
cosh(b*x+a)^2-1/12*(b*x+a)^2*cosh(b*x+a)^2-1/18*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^5+1/18*(b*x+a)*sinh(b*x+a)*cos
h(b*x+a)^3+1/108*cosh(b*x+a)^4*sinh(b*x+a)^2-1/216*cosh(b*x+a)^2*sinh(b*x+a)^2-5/108*cosh(b*x+a)^2+1/12*(b*x+a
)*cosh(b*x+a)*sinh(b*x+a)+1/24*(b*x+a)^2)+3*a^2*(1/6*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^4-1/12*(b*x+a)*sinh(b*x
+a)^2*cosh(b*x+a)^2-1/12*(b*x+a)*cosh(b*x+a)^2-1/36*sinh(b*x+a)*cosh(b*x+a)^5+1/36*cosh(b*x+a)^3*sinh(b*x+a)+1
/24*cosh(b*x+a)*sinh(b*x+a)+1/24*b*x+1/24*a)-a^3*(1/6*cosh(b*x+a)^4*sinh(b*x+a)^2-1/12*cosh(b*x+a)^2*sinh(b*x+
a)^2-1/12*cosh(b*x+a)^2))

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Maxima [A]  time = 1.08534, size = 231, normalized size = 1.62 \begin{align*} \frac{{\left (36 \, b^{3} x^{3} e^{\left (6 \, a\right )} - 18 \, b^{2} x^{2} e^{\left (6 \, a\right )} + 6 \, b x e^{\left (6 \, a\right )} - e^{\left (6 \, a\right )}\right )} e^{\left (6 \, b x\right )}}{13824 \, b^{4}} - \frac{3 \,{\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 )}}{512 \, b^{4}} - \frac{3 \,{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{512 \, b^{4}} + \frac{{\left (36 \, b^{3} x^{3} + 18 \, b^{2} x^{2} + 6 \, b x + 1\right )} e^{\left (-6 \, b x - 6 \, a\right )}}{13824 \, 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)^3,x, algorithm="maxima")

[Out]

1/13824*(36*b^3*x^3*e^(6*a) - 18*b^2*x^2*e^(6*a) + 6*b*x*e^(6*a) - e^(6*a))*e^(6*b*x)/b^4 - 3/512*(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 - 3/512*(4*b^3*x^3 + 6*b^2*x^2 + 6*b*x +
3)*e^(-2*b*x - 2*a)/b^4 + 1/13824*(36*b^3*x^3 + 18*b^2*x^2 + 6*b*x + 1)*e^(-6*b*x - 6*a)/b^4

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Fricas [A]  time = 1.76797, size = 617, normalized size = 4.31 \begin{align*} \frac{3 \,{\left (6 \, b^{3} x^{3} + b x\right )} \cosh \left (b x + a\right )^{6} - 10 \,{\left (18 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right )^{3} + 45 \,{\left (6 \, b^{3} x^{3} + b x\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} - 3 \,{\left (18 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + 3 \,{\left (6 \, b^{3} x^{3} + b x\right )} \sinh \left (b x + a\right )^{6} - 81 \,{\left (2 \, b^{3} x^{3} + 3 \, b x\right )} \cosh \left (b x + a\right )^{2} - 9 \,{\left (18 \, b^{3} x^{3} - 5 \,{\left (6 \, b^{3} x^{3} + b x\right )} \cosh \left (b x + a\right )^{4} + 27 \, b x\right )} \sinh \left (b x + a\right )^{2} - 3 \,{\left ({\left (18 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{5} - 81 \,{\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{3456 \, 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)^3,x, algorithm="fricas")

[Out]

1/3456*(3*(6*b^3*x^3 + b*x)*cosh(b*x + a)^6 - 10*(18*b^2*x^2 + 1)*cosh(b*x + a)^3*sinh(b*x + a)^3 + 45*(6*b^3*
x^3 + b*x)*cosh(b*x + a)^2*sinh(b*x + a)^4 - 3*(18*b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^5 + 3*(6*b^3*x^3 +
b*x)*sinh(b*x + a)^6 - 81*(2*b^3*x^3 + 3*b*x)*cosh(b*x + a)^2 - 9*(18*b^3*x^3 - 5*(6*b^3*x^3 + b*x)*cosh(b*x
+ a)^4 + 27*b*x)*sinh(b*x + a)^2 - 3*((18*b^2*x^2 + 1)*cosh(b*x + a)^5 - 81*(2*b^2*x^2 + 1)*cosh(b*x + a))*sin
h(b*x + a))/b^4

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Sympy [A]  time = 23.1727, size = 314, normalized size = 2.2 \begin{align*} \begin{cases} - \frac{x^{3} \sinh ^{6}{\left (a + b x \right )}}{24 b} + \frac{x^{3} \sinh ^{4}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8 b} + \frac{x^{3} \sinh ^{2}{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{8 b} - \frac{x^{3} \cosh ^{6}{\left (a + b x \right )}}{24 b} + \frac{x^{2} \sinh ^{5}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{8 b^{2}} - \frac{x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b^{2}} + \frac{x^{2} \sinh{\left (a + b x \right )} \cosh ^{5}{\left (a + b x \right )}}{8 b^{2}} - \frac{5 x \sinh ^{6}{\left (a + b x \right )}}{72 b^{3}} + \frac{x \sinh ^{4}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{12 b^{3}} + \frac{x \sinh ^{2}{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{12 b^{3}} - \frac{5 x \cosh ^{6}{\left (a + b x \right )}}{72 b^{3}} + \frac{5 \sinh ^{5}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{72 b^{4}} - \frac{31 \sinh ^{3}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{216 b^{4}} + \frac{5 \sinh{\left (a + b x \right )} \cosh ^{5}{\left (a + b x \right )}}{72 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \sinh ^{3}{\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)**3,x)

[Out]

Piecewise((-x**3*sinh(a + b*x)**6/(24*b) + x**3*sinh(a + b*x)**4*cosh(a + b*x)**2/(8*b) + x**3*sinh(a + b*x)**
2*cosh(a + b*x)**4/(8*b) - x**3*cosh(a + b*x)**6/(24*b) + x**2*sinh(a + b*x)**5*cosh(a + b*x)/(8*b**2) - x**2*
sinh(a + b*x)**3*cosh(a + b*x)**3/(3*b**2) + x**2*sinh(a + b*x)*cosh(a + b*x)**5/(8*b**2) - 5*x*sinh(a + b*x)*
*6/(72*b**3) + x*sinh(a + b*x)**4*cosh(a + b*x)**2/(12*b**3) + x*sinh(a + b*x)**2*cosh(a + b*x)**4/(12*b**3) -
5*x*cosh(a + b*x)**6/(72*b**3) + 5*sinh(a + b*x)**5*cosh(a + b*x)/(72*b**4) - 31*sinh(a + b*x)**3*cosh(a + b*
x)**3/(216*b**4) + 5*sinh(a + b*x)*cosh(a + b*x)**5/(72*b**4), Ne(b, 0)), (x**4*sinh(a)**3*cosh(a)**3/4, True)
)

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Giac [A]  time = 1.30538, size = 196, normalized size = 1.37 \begin{align*} \frac{{\left (36 \, b^{3} x^{3} - 18 \, b^{2} x^{2} + 6 \, b x - 1\right )} e^{\left (6 \, b x + 6 \, a\right )}}{13824 \, b^{4}} - \frac{3 \,{\left (4 \, b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, b x - 3\right )} e^{\left (2 \, b x + 2 \, a\right )}}{512 \, b^{4}} - \frac{3 \,{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{512 \, b^{4}} + \frac{{\left (36 \, b^{3} x^{3} + 18 \, b^{2} x^{2} + 6 \, b x + 1\right )} e^{\left (-6 \, b x - 6 \, a\right )}}{13824 \, 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)^3,x, algorithm="giac")

[Out]

1/13824*(36*b^3*x^3 - 18*b^2*x^2 + 6*b*x - 1)*e^(6*b*x + 6*a)/b^4 - 3/512*(4*b^3*x^3 - 6*b^2*x^2 + 6*b*x - 3)*
e^(2*b*x + 2*a)/b^4 - 3/512*(4*b^3*x^3 + 6*b^2*x^2 + 6*b*x + 3)*e^(-2*b*x - 2*a)/b^4 + 1/13824*(36*b^3*x^3 + 1
8*b^2*x^2 + 6*b*x + 1)*e^(-6*b*x - 6*a)/b^4