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

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

[Out]

-1/4*cosh(b*x+a)/b^3-1/8*x^2*cosh(b*x+a)/b-1/216*cosh(3*b*x+3*a)/b^3-1/48*x^2*cosh(3*b*x+3*a)/b+1/1000*cosh(5*
b*x+5*a)/b^3+1/80*x^2*cosh(5*b*x+5*a)/b+1/4*x*sinh(b*x+a)/b^2+1/72*x*sinh(3*b*x+3*a)/b^2-1/200*x*sinh(5*b*x+5*
a)/b^2

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Rubi [A]  time = 0.18, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5448, 3296, 2638} \[ \frac {x \sinh (a+b x)}{4 b^2}+\frac {x \sinh (3 a+3 b x)}{72 b^2}-\frac {x \sinh (5 a+5 b x)}{200 b^2}-\frac {\cosh (a+b x)}{4 b^3}-\frac {\cosh (3 a+3 b x)}{216 b^3}+\frac {\cosh (5 a+5 b x)}{1000 b^3}-\frac {x^2 \cosh (a+b x)}{8 b}-\frac {x^2 \cosh (3 a+3 b x)}{48 b}+\frac {x^2 \cosh (5 a+5 b x)}{80 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2638

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.46, size = 98, normalized size = 0.66 \[ \frac {-6750 \left (b^2 x^2+2\right ) \cosh (a+b x)-125 \left (9 b^2 x^2+2\right ) \cosh (3 (a+b x))+27 \left (25 b^2 x^2+2\right ) \cosh (5 (a+b x))+30 b x (450 \sinh (a+b x)+25 \sinh (3 (a+b x))-9 \sinh (5 (a+b x)))}{54000 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6750*(2 + b^2*x^2)*Cosh[a + b*x] - 125*(2 + 9*b^2*x^2)*Cosh[3*(a + b*x)] + 27*(2 + 25*b^2*x^2)*Cosh[5*(a + b
*x)] + 30*b*x*(450*Sinh[a + b*x] + 25*Sinh[3*(a + b*x)] - 9*Sinh[5*(a + b*x)]))/(54000*b^3)

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fricas [A]  time = 0.66, size = 214, normalized size = 1.45 \[ -\frac {270 \, b x \sinh \left (b x + a\right )^{5} - 27 \, {\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{5} - 135 \, {\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 125 \, {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{3} + 150 \, {\left (18 \, b x \cosh \left (b x + a\right )^{2} - 5 \, b x\right )} \sinh \left (b x + a\right )^{3} - 15 \, {\left (18 \, {\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{3} - 25 \, {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 6750 \, {\left (b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) + 450 \, {\left (3 \, b x \cosh \left (b x + a\right )^{4} - 5 \, b x \cosh \left (b x + a\right )^{2} - 30 \, b x\right )} \sinh \left (b x + a\right )}{54000 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54000*(270*b*x*sinh(b*x + a)^5 - 27*(25*b^2*x^2 + 2)*cosh(b*x + a)^5 - 135*(25*b^2*x^2 + 2)*cosh(b*x + a)*s
inh(b*x + a)^4 + 125*(9*b^2*x^2 + 2)*cosh(b*x + a)^3 + 150*(18*b*x*cosh(b*x + a)^2 - 5*b*x)*sinh(b*x + a)^3 -
15*(18*(25*b^2*x^2 + 2)*cosh(b*x + a)^3 - 25*(9*b^2*x^2 + 2)*cosh(b*x + a))*sinh(b*x + a)^2 + 6750*(b^2*x^2 +
2)*cosh(b*x + a) + 450*(3*b*x*cosh(b*x + a)^4 - 5*b*x*cosh(b*x + a)^2 - 30*b*x)*sinh(b*x + a))/b^3

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giac [A]  time = 0.14, size = 164, normalized size = 1.11 \[ \frac {{\left (25 \, b^{2} x^{2} - 10 \, b x + 2\right )} e^{\left (5 \, b x + 5 \, a\right )}}{4000 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} - 6 \, b x + 2\right )} e^{\left (3 \, b x + 3 \, a\right )}}{864 \, b^{3}} - \frac {{\left (b^{2} x^{2} - 2 \, b x + 2\right )} e^{\left (b x + a\right )}}{16 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{16 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{864 \, b^{3}} + \frac {{\left (25 \, b^{2} x^{2} + 10 \, b x + 2\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{4000 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.30, size = 246, normalized size = 1.66 \[ \frac {\frac {\left (b x +a \right )^{2} \left (\sinh ^{2}\left (b x +a \right )\right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{5}-\frac {2 \left (b x +a \right )^{2} \left (\cosh ^{3}\left (b x +a \right )\right )}{15}-\frac {2 \left (b x +a \right ) \sinh \left (b x +a \right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{25}+\frac {52 \left (b x +a \right ) \sinh \left (b x +a \right )}{225}+\frac {26 \left (b x +a \right ) \sinh \left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{225}+\frac {2 \left (\cosh ^{5}\left (b x +a \right )\right )}{125}-\frac {52 \cosh \left (b x +a \right )}{225}-\frac {26 \left (\cosh ^{3}\left (b x +a \right )\right )}{675}-2 a \left (\frac {\left (b x +a \right ) \left (\sinh ^{2}\left (b x +a \right )\right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{5}-\frac {2 \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{15}-\frac {\sinh \left (b x +a \right ) \left (\cosh ^{4}\left (b x +a \right )\right )}{25}+\frac {26 \sinh \left (b x +a \right )}{225}+\frac {13 \left (\cosh ^{2}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{225}\right )+a^{2} \left (\frac {\left (\cosh ^{3}\left (b x +a \right )\right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{5}-\frac {2 \left (\cosh ^{3}\left (b x +a \right )\right )}{15}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/5*(b*x+a)^2*sinh(b*x+a)^2*cosh(b*x+a)^3-2/15*(b*x+a)^2*cosh(b*x+a)^3-2/25*(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)^5-52/225*cosh(b*x
+a)-26/675*cosh(b*x+a)^3-2*a*(1/5*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^3-2/15*(b*x+a)*cosh(b*x+a)^3-1/25*sinh(b*x
+a)*cosh(b*x+a)^4+26/225*sinh(b*x+a)+13/225*cosh(b*x+a)^2*sinh(b*x+a))+a^2*(1/5*cosh(b*x+a)^3*sinh(b*x+a)^2-2/
15*cosh(b*x+a)^3))

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maxima [A]  time = 0.34, size = 187, normalized size = 1.26 \[ \frac {{\left (25 \, b^{2} x^{2} e^{\left (5 \, a\right )} - 10 \, b x e^{\left (5 \, a\right )} + 2 \, e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )}}{4000 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} e^{\left (3 \, a\right )} - 6 \, b x e^{\left (3 \, a\right )} + 2 \, e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{864 \, b^{3}} - \frac {{\left (b^{2} x^{2} e^{a} - 2 \, b x e^{a} + 2 \, e^{a}\right )} e^{\left (b x\right )}}{16 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{16 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{864 \, b^{3}} + \frac {{\left (25 \, b^{2} x^{2} + 10 \, b x + 2\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{4000 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)^2*sinh(b*x+a)^3,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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mupad [B]  time = 0.23, size = 112, normalized size = 0.76 \[ -\frac {780\,\mathrm {cosh}\left (a+b\,x\right )+130\,{\mathrm {cosh}\left (a+b\,x\right )}^3-54\,{\mathrm {cosh}\left (a+b\,x\right )}^5-780\,b\,x\,\mathrm {sinh}\left (a+b\,x\right )+1125\,b^2\,x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^3-675\,b^2\,x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^5-390\,b\,x\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )+270\,b\,x\,{\mathrm {cosh}\left (a+b\,x\right )}^4\,\mathrm {sinh}\left (a+b\,x\right )}{3375\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(780*cosh(a + b*x) + 130*cosh(a + b*x)^3 - 54*cosh(a + b*x)^5 - 780*b*x*sinh(a + b*x) + 1125*b^2*x^2*cosh(a +
b*x)^3 - 675*b^2*x^2*cosh(a + b*x)^5 - 390*b*x*cosh(a + b*x)^2*sinh(a + b*x) + 270*b*x*cosh(a + b*x)^4*sinh(a
+ b*x))/(3375*b^3)

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sympy [A]  time = 5.06, size = 182, normalized size = 1.23 \[ \begin {cases} \frac {x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 x^{2} \cosh ^{5}{\left (a + b x \right )}}{15 b} + \frac {52 x \sinh ^{5}{\left (a + b x \right )}}{225 b^{2}} - \frac {26 x \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{45 b^{2}} + \frac {4 x \sinh {\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{15 b^{2}} - \frac {52 \sinh ^{4}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{225 b^{3}} + \frac {338 \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{675 b^{3}} - \frac {856 \cosh ^{5}{\left (a + b x \right )}}{3375 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \sinh ^{3}{\relax (a )} \cosh ^{2}{\relax (a )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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