Integrand size = 16, antiderivative size = 64 \[ \int x^2 \cosh (a+b x) \sinh (a+b x) \, dx=\frac {x^2}{4 b}-\frac {x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac {\sinh ^2(a+b x)}{4 b^3}+\frac {x^2 \sinh ^2(a+b x)}{2 b} \] Output:
1/4*x^2/b-1/2*x*cosh(b*x+a)*sinh(b*x+a)/b^2+1/4*sinh(b*x+a)^2/b^3+1/2*x^2* sinh(b*x+a)^2/b
Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.61 \[ \int x^2 \cosh (a+b x) \sinh (a+b x) \, dx=\frac {\left (1+2 b^2 x^2\right ) \cosh (2 (a+b x))-2 b x \sinh (2 (a+b x))}{8 b^3} \] Input:
Integrate[x^2*Cosh[a + b*x]*Sinh[a + b*x],x]
Output:
((1 + 2*b^2*x^2)*Cosh[2*(a + b*x)] - 2*b*x*Sinh[2*(a + b*x)])/(8*b^3)
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5895, 3042, 25, 3791, 15}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \sinh (a+b x) \cosh (a+b x) \, dx\) |
\(\Big \downarrow \) 5895 |
\(\displaystyle \frac {x^2 \sinh ^2(a+b x)}{2 b}-\frac {\int x \sinh ^2(a+b x)dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {x^2 \sinh ^2(a+b x)}{2 b}-\frac {\int -x \sin (i a+i b x)^2dx}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {x^2 \sinh ^2(a+b x)}{2 b}+\frac {\int x \sin (i a+i b x)^2dx}{b}\) |
\(\Big \downarrow \) 3791 |
\(\displaystyle \frac {\frac {\int xdx}{2}+\frac {\sinh ^2(a+b x)}{4 b^2}-\frac {x \sinh (a+b x) \cosh (a+b x)}{2 b}}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\frac {\sinh ^2(a+b x)}{4 b^2}-\frac {x \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {x^2}{4}}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}\) |
Input:
Int[x^2*Cosh[a + b*x]*Sinh[a + b*x],x]
Output:
(x^2*Sinh[a + b*x]^2)/(2*b) + (x^2/4 - (x*Cosh[a + b*x]*Sinh[a + b*x])/(2* b) + Sinh[a + b*x]^2/(4*b^2))/b
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x ]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n) Int[(c + d* x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]
Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.) ]^(p_.), x_Symbol] :> Simp[x^(m - n + 1)*(Sinh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Simp[(m - n + 1)/(b*n*(p + 1)) Int[x^(m - n)*Sinh[a + b*x^n]^ (p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]
Time = 0.87 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {\left (2 b^{2} x^{2}-2 b x +1\right ) {\mathrm e}^{2 b x +2 a}}{16 b^{3}}+\frac {\left (2 b^{2} x^{2}+2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{16 b^{3}}\) | \(58\) |
orering | \(-\frac {\left (4 b^{2} x^{2}+1\right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{4 b^{4} x}+\frac {\left (2 b^{2} x^{2}+1\right ) \left (2 x \cosh \left (b x +a \right ) \sinh \left (b x +a \right )+x^{2} b \sinh \left (b x +a \right )^{2}+x^{2} \cosh \left (b x +a \right )^{2} b \right )}{8 b^{4} x^{2}}\) | \(92\) |
derivativedivides | \(\frac {\frac {a^{2} \cosh \left (b x +a \right )^{2}}{2}-2 a \left (\frac {\left (b x +a \right ) \cosh \left (b x +a \right )^{2}}{2}-\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{4}-\frac {b x}{4}-\frac {a}{4}\right )+\frac {\left (b x +a \right )^{2} \cosh \left (b x +a \right )^{2}}{2}-\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}-\frac {\left (b x +a \right )^{2}}{4}+\frac {\cosh \left (b x +a \right )^{2}}{4}}{b^{3}}\) | \(114\) |
default | \(\frac {\frac {a^{2} \cosh \left (b x +a \right )^{2}}{2}-2 a \left (\frac {\left (b x +a \right ) \cosh \left (b x +a \right )^{2}}{2}-\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{4}-\frac {b x}{4}-\frac {a}{4}\right )+\frac {\left (b x +a \right )^{2} \cosh \left (b x +a \right )^{2}}{2}-\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}-\frac {\left (b x +a \right )^{2}}{4}+\frac {\cosh \left (b x +a \right )^{2}}{4}}{b^{3}}\) | \(114\) |
Input:
int(x^2*cosh(b*x+a)*sinh(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/16*(2*b^2*x^2-2*b*x+1)/b^3*exp(2*b*x+2*a)+1/16*(2*b^2*x^2+2*b*x+1)/b^3*e xp(-2*b*x-2*a)
Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97 \[ \int x^2 \cosh (a+b x) \sinh (a+b x) \, dx=-\frac {4 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} - {\left (2 \, b^{2} x^{2} + 1\right )} \sinh \left (b x + a\right )^{2}}{8 \, b^{3}} \] Input:
integrate(x^2*cosh(b*x+a)*sinh(b*x+a),x, algorithm="fricas")
Output:
-1/8*(4*b*x*cosh(b*x + a)*sinh(b*x + a) - (2*b^2*x^2 + 1)*cosh(b*x + a)^2 - (2*b^2*x^2 + 1)*sinh(b*x + a)^2)/b^3
Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int x^2 \cosh (a+b x) \sinh (a+b x) \, dx=\begin {cases} \frac {x^{2} \sinh ^{2}{\left (a + b x \right )}}{4 b} + \frac {x^{2} \cosh ^{2}{\left (a + b x \right )}}{4 b} - \frac {x \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{2 b^{2}} + \frac {\cosh ^{2}{\left (a + b x \right )}}{4 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \sinh {\left (a \right )} \cosh {\left (a \right )}}{3} & \text {otherwise} \end {cases} \] Input:
integrate(x**2*cosh(b*x+a)*sinh(b*x+a),x)
Output:
Piecewise((x**2*sinh(a + b*x)**2/(4*b) + x**2*cosh(a + b*x)**2/(4*b) - x*s inh(a + b*x)*cosh(a + b*x)/(2*b**2) + cosh(a + b*x)**2/(4*b**3), Ne(b, 0)) , (x**3*sinh(a)*cosh(a)/3, True))
Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00 \[ \int x^2 \cosh (a+b x) \sinh (a+b x) \, dx=\frac {{\left (2 \, b^{2} x^{2} e^{\left (2 \, a\right )} - 2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{16 \, b^{3}} + \frac {{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{3}} \] Input:
integrate(x^2*cosh(b*x+a)*sinh(b*x+a),x, algorithm="maxima")
Output:
1/16*(2*b^2*x^2*e^(2*a) - 2*b*x*e^(2*a) + e^(2*a))*e^(2*b*x)/b^3 + 1/16*(2 *b^2*x^2 + 2*b*x + 1)*e^(-2*b*x - 2*a)/b^3
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89 \[ \int x^2 \cosh (a+b x) \sinh (a+b x) \, dx=\frac {{\left (2 \, b^{2} x^{2} - 2 \, b x + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{3}} + \frac {{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{3}} \] Input:
integrate(x^2*cosh(b*x+a)*sinh(b*x+a),x, algorithm="giac")
Output:
1/16*(2*b^2*x^2 - 2*b*x + 1)*e^(2*b*x + 2*a)/b^3 + 1/16*(2*b^2*x^2 + 2*b*x + 1)*e^(-2*b*x - 2*a)/b^3
Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.72 \[ \int x^2 \cosh (a+b x) \sinh (a+b x) \, dx=\frac {\frac {\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{2}-b\,x\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )+b^2\,x^2\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{4\,b^3} \] Input:
int(x^2*cosh(a + b*x)*sinh(a + b*x),x)
Output:
(cosh(2*a + 2*b*x)/2 - b*x*sinh(2*a + 2*b*x) + b^2*x^2*cosh(2*a + 2*b*x))/ (4*b^3)
Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int x^2 \cosh (a+b x) \sinh (a+b x) \, dx=\frac {\cosh \left (b x +a \right )^{2} b^{2} x^{2}+\cosh \left (b x +a \right )^{2}-2 \cosh \left (b x +a \right ) \sinh \left (b x +a \right ) b x +\sinh \left (b x +a \right )^{2} b^{2} x^{2}}{4 b^{3}} \] Input:
int(x^2*cosh(b*x+a)*sinh(b*x+a),x)
Output:
(cosh(a + b*x)**2*b**2*x**2 + cosh(a + b*x)**2 - 2*cosh(a + b*x)*sinh(a + b*x)*b*x + sinh(a + b*x)**2*b**2*x**2)/(4*b**3)