Integrand size = 48, antiderivative size = 31 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=e^{2 x} \left (4-x-\frac {1}{4} \left (-3+e^5-x\right ) x\right ) \left (5-x^2\right ) \] Output:
(4-x-1/4*(exp(5)-x-3)*x)*exp(x)^2*(-x^2+5)
Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=-\frac {1}{4} e^{2 x} \left (-5+x^2\right ) \left (16-\left (1+e^5\right ) x+x^2\right ) \] Input:
Integrate[(E^(2*x)*(155 - 32*x - 19*x^2 - 2*x^3 - 2*x^4 + E^5*(-5 - 10*x + 3*x^2 + 2*x^3)))/4,x]
Output:
-1/4*(E^(2*x)*(-5 + x^2)*(16 - (1 + E^5)*x + x^2))
Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(31)=62\).
Time = 0.44 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.26, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {27, 2626, 2009}
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.
\(\displaystyle \int \frac {1}{4} e^{2 x} \left (-2 x^4-2 x^3-19 x^2+e^5 \left (2 x^3+3 x^2-10 x-5\right )-32 x+155\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int e^{2 x} \left (-2 x^4-2 x^3-19 x^2-32 x-e^5 \left (-2 x^3-3 x^2+10 x+5\right )+155\right )dx\) |
\(\Big \downarrow \) 2626 |
\(\displaystyle \frac {1}{4} \int \left (-2 e^{2 x} x^4-2 e^{2 x} x^3-19 e^{2 x} x^2-32 e^{2 x} x+155 e^{2 x}+e^{2 x+5} \left (2 x^3+3 x^2-10 x-5\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-e^{2 x} x^4+e^{2 x} x^3+e^{2 x+5} x^3-11 e^{2 x} x^2-5 e^{2 x} x-5 e^{2 x+5} x+80 e^{2 x}\right )\) |
Input:
Int[(E^(2*x)*(155 - 32*x - 19*x^2 - 2*x^3 - 2*x^4 + E^5*(-5 - 10*x + 3*x^2 + 2*x^3)))/4,x]
Output:
(80*E^(2*x) - 5*E^(2*x)*x - 5*E^(5 + 2*x)*x - 11*E^(2*x)*x^2 + E^(2*x)*x^3 + E^(5 + 2*x)*x^3 - E^(2*x)*x^4)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(F_)^(v_)*(Px_), x_Symbol] :> Int[ExpandIntegrand[F^v, Px, x], x] /; Fr eeQ[F, x] && PolynomialQ[Px, x] && LinearQ[v, x] && !TrueQ[$UseGamma]
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16
method | result | size |
gosper | \(\frac {{\mathrm e}^{2 x} \left (x^{3} {\mathrm e}^{5}-x^{4}+x^{3}-5 x \,{\mathrm e}^{5}-11 x^{2}-5 x +80\right )}{4}\) | \(36\) |
risch | \(\frac {{\mathrm e}^{2 x} \left (x^{3} {\mathrm e}^{5}-x^{4}+x^{3}-5 x \,{\mathrm e}^{5}-11 x^{2}-5 x +80\right )}{4}\) | \(36\) |
norman | \(\left (-\frac {5}{4}-\frac {5 \,{\mathrm e}^{5}}{4}\right ) x \,{\mathrm e}^{2 x}+\left (\frac {1}{4}+\frac {{\mathrm e}^{5}}{4}\right ) x^{3} {\mathrm e}^{2 x}+20 \,{\mathrm e}^{2 x}-\frac {11 \,{\mathrm e}^{2 x} x^{2}}{4}-\frac {{\mathrm e}^{2 x} x^{4}}{4}\) | \(52\) |
parallelrisch | \(\frac {{\mathrm e}^{5} {\mathrm e}^{2 x} x^{3}}{4}-\frac {{\mathrm e}^{2 x} x^{4}}{4}+\frac {{\mathrm e}^{2 x} x^{3}}{4}-\frac {5 \,{\mathrm e}^{5} {\mathrm e}^{2 x} x}{4}-\frac {11 \,{\mathrm e}^{2 x} x^{2}}{4}-\frac {5 x \,{\mathrm e}^{2 x}}{4}+20 \,{\mathrm e}^{2 x}\) | \(62\) |
orering | \(\frac {\left (x^{3} {\mathrm e}^{5}-x^{4}+x^{3}-5 x \,{\mathrm e}^{5}-11 x^{2}-5 x +80\right ) \left (\left (2 x^{3}+3 x^{2}-10 x -5\right ) {\mathrm e}^{5}-2 x^{4}-2 x^{3}-19 x^{2}-32 x +155\right ) {\mathrm e}^{2 x}}{8 x^{3} {\mathrm e}^{5}-8 x^{4}+12 x^{2} {\mathrm e}^{5}-8 x^{3}-40 x \,{\mathrm e}^{5}-76 x^{2}-20 \,{\mathrm e}^{5}-128 x +620}\) | \(119\) |
meijerg | \(-19+\frac {155 \,{\mathrm e}^{2 x}}{8}-\frac {\left (80 x^{4}-160 x^{3}+240 x^{2}-240 x +120\right ) {\mathrm e}^{2 x}}{320}+\frac {\left (\frac {{\mathrm e}^{5}}{2}-\frac {1}{2}\right ) \left (6-\frac {\left (-32 x^{3}+48 x^{2}-48 x +24\right ) {\mathrm e}^{2 x}}{4}\right )}{16}-\frac {\left (\frac {3 \,{\mathrm e}^{5}}{4}-\frac {19}{4}\right ) \left (2-\frac {\left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{3}\right )}{8}+\frac {\left (-\frac {5 \,{\mathrm e}^{5}}{2}-8\right ) \left (1-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\right )}{4}+\frac {5 \,{\mathrm e}^{5} \left (1-{\mathrm e}^{2 x}\right )}{8}\) | \(125\) |
default | \(20 \,{\mathrm e}^{2 x}-\frac {5 x \,{\mathrm e}^{2 x}}{4}-\frac {5 \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{8}-\frac {11 \,{\mathrm e}^{2 x} x^{2}}{4}+\frac {{\mathrm e}^{2 x} x^{3}}{4}-\frac {{\mathrm e}^{2 x} x^{4}}{4}-\frac {5 \,{\mathrm e}^{5} \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{2 x}}{4}\right )}{2}+\frac {3 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{2 x} x^{2}}{2}-\frac {x \,{\mathrm e}^{2 x}}{2}+\frac {{\mathrm e}^{2 x}}{4}\right )}{4}+\frac {{\mathrm e}^{5} \left (\frac {{\mathrm e}^{2 x} x^{3}}{2}-\frac {3 \,{\mathrm e}^{2 x} x^{2}}{4}+\frac {3 x \,{\mathrm e}^{2 x}}{4}-\frac {3 \,{\mathrm e}^{2 x}}{8}\right )}{2}\) | \(131\) |
Input:
int(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2 ,x,method=_RETURNVERBOSE)
Output:
1/4*exp(x)^2*(x^3*exp(5)-x^4+x^3-5*x*exp(5)-11*x^2-5*x+80)
Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=-\frac {1}{4} \, {\left (x^{4} - x^{3} + 11 \, x^{2} - {\left (x^{3} - 5 \, x\right )} e^{5} + 5 \, x - 80\right )} e^{\left (2 \, x\right )} \] Input:
integrate(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*ex p(x)^2,x, algorithm="fricas")
Output:
-1/4*(x^4 - x^3 + 11*x^2 - (x^3 - 5*x)*e^5 + 5*x - 80)*e^(2*x)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=\frac {\left (- x^{4} + x^{3} + x^{3} e^{5} - 11 x^{2} - 5 x e^{5} - 5 x + 80\right ) e^{2 x}}{4} \] Input:
integrate(1/4*((2*x**3+3*x**2-10*x-5)*exp(5)-2*x**4-2*x**3-19*x**2-32*x+15 5)*exp(x)**2,x)
Output:
(-x**4 + x**3 + x**3*exp(5) - 11*x**2 - 5*x*exp(5) - 5*x + 80)*exp(2*x)/4
Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (25) = 50\).
Time = 0.04 (sec) , antiderivative size = 156, normalized size of antiderivative = 5.03 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=-\frac {1}{8} \, {\left (2 \, x^{4} - 4 \, x^{3} + 6 \, x^{2} - 6 \, x + 3\right )} e^{\left (2 \, x\right )} + \frac {1}{16} \, {\left (4 \, x^{3} e^{5} - 6 \, x^{2} e^{5} + 6 \, x e^{5} - 3 \, e^{5}\right )} e^{\left (2 \, x\right )} - \frac {1}{16} \, {\left (4 \, x^{3} - 6 \, x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x\right )} + \frac {3}{16} \, {\left (2 \, x^{2} e^{5} - 2 \, x e^{5} + e^{5}\right )} e^{\left (2 \, x\right )} - \frac {19}{16} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - \frac {5}{8} \, {\left (2 \, x e^{5} - e^{5}\right )} e^{\left (2 \, x\right )} - 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + \frac {155}{8} \, e^{\left (2 \, x\right )} - \frac {5}{8} \, e^{\left (2 \, x + 5\right )} \] Input:
integrate(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*ex p(x)^2,x, algorithm="maxima")
Output:
-1/8*(2*x^4 - 4*x^3 + 6*x^2 - 6*x + 3)*e^(2*x) + 1/16*(4*x^3*e^5 - 6*x^2*e ^5 + 6*x*e^5 - 3*e^5)*e^(2*x) - 1/16*(4*x^3 - 6*x^2 + 6*x - 3)*e^(2*x) + 3 /16*(2*x^2*e^5 - 2*x*e^5 + e^5)*e^(2*x) - 19/16*(2*x^2 - 2*x + 1)*e^(2*x) - 5/8*(2*x*e^5 - e^5)*e^(2*x) - 2*(2*x - 1)*e^(2*x) + 155/8*e^(2*x) - 5/8* e^(2*x + 5)
Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=-\frac {1}{4} \, {\left (x^{4} - x^{3} + 11 \, x^{2} + 5 \, x - 80\right )} e^{\left (2 \, x\right )} + \frac {1}{4} \, {\left (x^{3} - 5 \, x\right )} e^{\left (2 \, x + 5\right )} \] Input:
integrate(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*ex p(x)^2,x, algorithm="giac")
Output:
-1/4*(x^4 - x^3 + 11*x^2 + 5*x - 80)*e^(2*x) + 1/4*(x^3 - 5*x)*e^(2*x + 5)
Time = 2.95 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=\frac {{\mathrm {e}}^{2\,x}\,\left (x^2-5\right )\,\left (x+x\,{\mathrm {e}}^5-x^2-16\right )}{4} \] Input:
int(-(exp(2*x)*(32*x + exp(5)*(10*x - 3*x^2 - 2*x^3 + 5) + 19*x^2 + 2*x^3 + 2*x^4 - 155))/4,x)
Output:
(exp(2*x)*(x^2 - 5)*(x + x*exp(5) - x^2 - 16))/4
Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=\frac {e^{2 x} \left (e^{5} x^{3}-5 e^{5} x -x^{4}+x^{3}-11 x^{2}-5 x +80\right )}{4} \] Input:
int(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2 ,x)
Output:
(e**(2*x)*(e**5*x**3 - 5*e**5*x - x**4 + x**3 - 11*x**2 - 5*x + 80))/4