### 3.300 $$\int x^2 \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=148 $-\frac{\sinh (a+b x)}{4 b^3}+\frac{\sinh (3 a+3 b x)}{216 b^3}+\frac{\sinh (5 a+5 b x)}{1000 b^3}+\frac{x \cosh (a+b x)}{4 b^2}-\frac{x \cosh (3 a+3 b x)}{72 b^2}-\frac{x \cosh (5 a+5 b x)}{200 b^2}-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{x^2 \sinh (5 a+5 b x)}{80 b}$

[Out]

(x*Cosh[a + b*x])/(4*b^2) - (x*Cosh[3*a + 3*b*x])/(72*b^2) - (x*Cosh[5*a + 5*b*x])/(200*b^2) - Sinh[a + b*x]/(
4*b^3) - (x^2*Sinh[a + b*x])/(8*b) + Sinh[3*a + 3*b*x]/(216*b^3) + (x^2*Sinh[3*a + 3*b*x])/(48*b) + Sinh[5*a +
5*b*x]/(1000*b^3) + (x^2*Sinh[5*a + 5*b*x])/(80*b)

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Rubi [A]  time = 0.180373, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {5448, 3296, 2637} $-\frac{\sinh (a+b x)}{4 b^3}+\frac{\sinh (3 a+3 b x)}{216 b^3}+\frac{\sinh (5 a+5 b x)}{1000 b^3}+\frac{x \cosh (a+b x)}{4 b^2}-\frac{x \cosh (3 a+3 b x)}{72 b^2}-\frac{x \cosh (5 a+5 b x)}{200 b^2}-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{x^2 \sinh (5 a+5 b x)}{80 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x*Cosh[a + b*x])/(4*b^2) - (x*Cosh[3*a + 3*b*x])/(72*b^2) - (x*Cosh[5*a + 5*b*x])/(200*b^2) - Sinh[a + b*x]/(
4*b^3) - (x^2*Sinh[a + b*x])/(8*b) + Sinh[3*a + 3*b*x]/(216*b^3) + (x^2*Sinh[3*a + 3*b*x])/(48*b) + Sinh[5*a +
5*b*x]/(1000*b^3) + (x^2*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 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^2 \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx &=\int \left (-\frac{1}{8} x^2 \cosh (a+b x)+\frac{1}{16} x^2 \cosh (3 a+3 b x)+\frac{1}{16} x^2 \cosh (5 a+5 b x)\right ) \, dx\\ &=\frac{1}{16} \int x^2 \cosh (3 a+3 b x) \, dx+\frac{1}{16} \int x^2 \cosh (5 a+5 b x) \, dx-\frac{1}{8} \int x^2 \cosh (a+b x) \, dx\\ &=-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{x^2 \sinh (5 a+5 b x)}{80 b}-\frac{\int x \sinh (5 a+5 b x) \, dx}{40 b}-\frac{\int x \sinh (3 a+3 b x) \, dx}{24 b}+\frac{\int x \sinh (a+b x) \, dx}{4 b}\\ &=\frac{x \cosh (a+b x)}{4 b^2}-\frac{x \cosh (3 a+3 b x)}{72 b^2}-\frac{x \cosh (5 a+5 b x)}{200 b^2}-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{x^2 \sinh (5 a+5 b x)}{80 b}+\frac{\int \cosh (5 a+5 b x) \, dx}{200 b^2}+\frac{\int \cosh (3 a+3 b x) \, dx}{72 b^2}-\frac{\int \cosh (a+b x) \, dx}{4 b^2}\\ &=\frac{x \cosh (a+b x)}{4 b^2}-\frac{x \cosh (3 a+3 b x)}{72 b^2}-\frac{x \cosh (5 a+5 b x)}{200 b^2}-\frac{\sinh (a+b x)}{4 b^3}-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{\sinh (3 a+3 b x)}{216 b^3}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{\sinh (5 a+5 b x)}{1000 b^3}+\frac{x^2 \sinh (5 a+5 b x)}{80 b}\\ \end{align*}

Mathematica [A]  time = 0.308318, size = 105, normalized size = 0.71 $\frac{-6750 \left (\left (b^2 x^2+2\right ) \sinh (a+b x)-2 b x \cosh (a+b x)\right )+125 \left (\left (9 b^2 x^2+2\right ) \sinh (3 (a+b x))-6 b x \cosh (3 (a+b x))\right )+27 \left (\left (25 b^2 x^2+2\right ) \sinh (5 (a+b x))-10 b x \cosh (5 (a+b x))\right )}{54000 b^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-6750*(-2*b*x*Cosh[a + b*x] + (2 + b^2*x^2)*Sinh[a + b*x]) + 125*(-6*b*x*Cosh[3*(a + b*x)] + (2 + 9*b^2*x^2)*
Sinh[3*(a + b*x)]) + 27*(-10*b*x*Cosh[5*(a + b*x)] + (2 + 25*b^2*x^2)*Sinh[5*(a + b*x)]))/(54000*b^3)

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Maple [B]  time = 0.008, size = 306, normalized size = 2.1 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}{5}}-{\frac{2\, \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) }{15}}-{\frac{ \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{15}}-{\frac{ \left ( 2\,bx+2\,a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{25}}-{\frac{ \left ( 8\,bx+8\,a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\cosh \left ( bx+a \right ) }{225}}+{\frac{ \left ( 52\,bx+52\,a \right ) \cosh \left ( bx+a \right ) }{225}}+{\frac{2\, \left ( \cosh \left ( bx+a \right ) \right ) ^{4}\sinh \left ( bx+a \right ) }{125}}-{\frac{856\,\sinh \left ( bx+a \right ) }{3375}}+{\frac{22\,\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3375}}-2\,a \left ( 1/5\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{4}-2/15\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/15\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}-1/25\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}-{\frac{4\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{225}}+{\frac{26\,\cosh \left ( bx+a \right ) }{225}} \right ) +{a}^{2} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4}\sinh \left ( bx+a \right ) }{5}}-{\frac{\sinh \left ( bx+a \right ) }{5} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) } \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(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*cosh(b*x+a)^4*sinh(b*x+a)-856/3375*sinh(b*x+a)+22/3375*sinh(b*x+a)*cosh(b*x+a)^2-2*a*(1/5*(b*x+a)*sin
h(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^2*(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 = 1.06848, size = 252, normalized size = 1.7 \begin{align*} \frac{{\left (25 \, b^{2} x^{2} e^{\left (5 \, a\right )} - 10 \, b x e^{\left (5 \, a\right )} + 2 \, e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )}}{4000 \, b^{3}} + \frac{{\left (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^{3}} - \frac{{\left (b^{2} x^{2} e^{a} - 2 \, b x e^{a} + 2 \, e^{a}\right )} e^{\left (b x\right )}}{16 \, b^{3}} + \frac{{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{16 \, b^{3}} - \frac{{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{864 \, b^{3}} - \frac{{\left (25 \, b^{2} x^{2} + 10 \, b x + 2\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{4000 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4000*(25*b^2*x^2*e^(5*a) - 10*b*x*e^(5*a) + 2*e^(5*a))*e^(5*b*x)/b^3 + 1/864*(9*b^2*x^2*e^(3*a) - 6*b*x*e^(3
*a) + 2*e^(3*a))*e^(3*b*x)/b^3 - 1/16*(b^2*x^2*e^a - 2*b*x*e^a + 2*e^a)*e^(b*x)/b^3 + 1/16*(b^2*x^2 + 2*b*x +
2)*e^(-b*x - a)/b^3 - 1/864*(9*b^2*x^2 + 6*b*x + 2)*e^(-3*b*x - 3*a)/b^3 - 1/4000*(25*b^2*x^2 + 10*b*x + 2)*e^
(-5*b*x - 5*a)/b^3

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Fricas [A]  time = 1.75437, size = 579, normalized size = 3.91 \begin{align*} -\frac{270 \, b x \cosh \left (b x + a\right )^{5} + 1350 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 27 \,{\left (25 \, b^{2} x^{2} + 2\right )} \sinh \left (b x + a\right )^{5} + 750 \, b x \cosh \left (b x + a\right )^{3} - 5 \,{\left (225 \, b^{2} x^{2} + 54 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} + 50\right )} \sinh \left (b x + a\right )^{3} - 13500 \, b x \cosh \left (b x + a\right ) + 450 \,{\left (6 \, b x \cosh \left (b x + a\right )^{3} + 5 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 15 \,{\left (9 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{4} - 450 \, b^{2} x^{2} + 25 \,{\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} - 900\right )} \sinh \left (b x + a\right )}{54000 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54000*(270*b*x*cosh(b*x + a)^5 + 1350*b*x*cosh(b*x + a)*sinh(b*x + a)^4 - 27*(25*b^2*x^2 + 2)*sinh(b*x + a)
^5 + 750*b*x*cosh(b*x + a)^3 - 5*(225*b^2*x^2 + 54*(25*b^2*x^2 + 2)*cosh(b*x + a)^2 + 50)*sinh(b*x + a)^3 - 13
500*b*x*cosh(b*x + a) + 450*(6*b*x*cosh(b*x + a)^3 + 5*b*x*cosh(b*x + a))*sinh(b*x + a)^2 - 15*(9*(25*b^2*x^2
+ 2)*cosh(b*x + a)^4 - 450*b^2*x^2 + 25*(9*b^2*x^2 + 2)*cosh(b*x + a)^2 - 900)*sinh(b*x + a))/b^3

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Sympy [A]  time = 7.97986, size = 182, normalized size = 1.23 \begin{align*} \begin{cases} - \frac{2 x^{2} \sinh ^{5}{\left (a + b x \right )}}{15 b} + \frac{x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b} + \frac{4 x \sinh ^{4}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{15 b^{2}} - \frac{26 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{45 b^{2}} + \frac{52 x \cosh ^{5}{\left (a + b x \right )}}{225 b^{2}} - \frac{856 \sinh ^{5}{\left (a + b x \right )}}{3375 b^{3}} + \frac{338 \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{675 b^{3}} - \frac{52 \sinh{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{225 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \sinh ^{2}{\left (a \right )} \cosh ^{3}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*x**2*sinh(a + b*x)**5/(15*b) + x**2*sinh(a + b*x)**3*cosh(a + b*x)**2/(3*b) + 4*x*sinh(a + b*x)*
*4*cosh(a + b*x)/(15*b**2) - 26*x*sinh(a + b*x)**2*cosh(a + b*x)**3/(45*b**2) + 52*x*cosh(a + b*x)**5/(225*b**
2) - 856*sinh(a + b*x)**5/(3375*b**3) + 338*sinh(a + b*x)**3*cosh(a + b*x)**2/(675*b**3) - 52*sinh(a + b*x)*co
sh(a + b*x)**4/(225*b**3), Ne(b, 0)), (x**3*sinh(a)**2*cosh(a)**3/3, True))

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Giac [A]  time = 1.18424, size = 221, normalized size = 1.49 \begin{align*} \frac{{\left (25 \, b^{2} x^{2} - 10 \, b x + 2\right )} e^{\left (5 \, b x + 5 \, a\right )}}{4000 \, b^{3}} + \frac{{\left (9 \, b^{2} x^{2} - 6 \, b x + 2\right )} e^{\left (3 \, b x + 3 \, a\right )}}{864 \, b^{3}} - \frac{{\left (b^{2} x^{2} - 2 \, b x + 2\right )} e^{\left (b x + a\right )}}{16 \, b^{3}} + \frac{{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{16 \, b^{3}} - \frac{{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{864 \, b^{3}} - \frac{{\left (25 \, b^{2} x^{2} + 10 \, b x + 2\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{4000 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4000*(25*b^2*x^2 - 10*b*x + 2)*e^(5*b*x + 5*a)/b^3 + 1/864*(9*b^2*x^2 - 6*b*x + 2)*e^(3*b*x + 3*a)/b^3 - 1/1
6*(b^2*x^2 - 2*b*x + 2)*e^(b*x + a)/b^3 + 1/16*(b^2*x^2 + 2*b*x + 2)*e^(-b*x - a)/b^3 - 1/864*(9*b^2*x^2 + 6*b
*x + 2)*e^(-3*b*x - 3*a)/b^3 - 1/4000*(25*b^2*x^2 + 10*b*x + 2)*e^(-5*b*x - 5*a)/b^3