Integrand size = 61, antiderivative size = 32 \[ \int \frac {-18+e^{x^2} \left (10 x^4-3 x^5+10 x^6-2 x^7+e^x \left (-5 x^3-3 x^4-9 x^5+2 x^6\right )\right )}{3 x^3} \, dx=\frac {3}{x^2}+\frac {1}{3} \left (1-e^{x^2} (5-x) \left (e^x-x\right ) x\right ) \]
Time = 2.41 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {-18+e^{x^2} \left (10 x^4-3 x^5+10 x^6-2 x^7+e^x \left (-5 x^3-3 x^4-9 x^5+2 x^6\right )\right )}{3 x^3} \, dx=\frac {3}{x^2}+\frac {1}{3} e^{x^2} \left (5 x^2-x^3+e^x \left (-5 x+x^2\right )\right ) \]
Integrate[(-18 + E^x^2*(10*x^4 - 3*x^5 + 10*x^6 - 2*x^7 + E^x*(-5*x^3 - 3* x^4 - 9*x^5 + 2*x^6)))/(3*x^3),x]
Time = 0.89 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.59, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {27, 25, 2010, 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.
\(\displaystyle \int \frac {e^{x^2} \left (-2 x^7+10 x^6-3 x^5+10 x^4+e^x \left (2 x^6-9 x^5-3 x^4-5 x^3\right )\right )-18}{3 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {18-e^{x^2} \left (-2 x^7+10 x^6-3 x^5+10 x^4-e^x \left (-2 x^6+9 x^5+3 x^4+5 x^3\right )\right )}{x^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} \int \frac {18-e^{x^2} \left (-2 x^7+10 x^6-3 x^5+10 x^4-e^x \left (-2 x^6+9 x^5+3 x^4+5 x^3\right )\right )}{x^3}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{3} \int \left (\frac {18}{x^3}-e^{x^2} \left (-2 x^4+2 e^x x^3+10 x^3-9 e^x x^2-3 x^2-3 e^x x+10 x-5 e^x\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (5 e^{x^2} x^2+e^{x^2+x} x^2-5 e^{x^2+x} x+\frac {9}{x^2}-e^{x^2} x^3\right )\) |
Int[(-18 + E^x^2*(10*x^4 - 3*x^5 + 10*x^6 - 2*x^7 + E^x*(-5*x^3 - 3*x^4 - 9*x^5 + 2*x^6)))/(3*x^3),x]
3.11.71.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
risch | \(\frac {3}{x^{2}}+\frac {\left (-x^{3}+{\mathrm e}^{x} x^{2}+5 x^{2}-5 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{x^{2}}}{3}\) | \(35\) |
default | \(\frac {3}{x^{2}}+\frac {5 x^{2} {\mathrm e}^{x^{2}}}{3}-\frac {x^{3} {\mathrm e}^{x^{2}}}{3}-\frac {5 x \,{\mathrm e}^{x^{2}+x}}{3}+\frac {x^{2} {\mathrm e}^{x^{2}+x}}{3}\) | \(45\) |
parts | \(\frac {3}{x^{2}}+\frac {5 x^{2} {\mathrm e}^{x^{2}}}{3}-\frac {x^{3} {\mathrm e}^{x^{2}}}{3}-\frac {5 x \,{\mathrm e}^{x^{2}+x}}{3}+\frac {x^{2} {\mathrm e}^{x^{2}+x}}{3}\) | \(45\) |
norman | \(\frac {3+\frac {5 x^{4} {\mathrm e}^{x^{2}}}{3}-\frac {{\mathrm e}^{x^{2}} x^{5}}{3}-\frac {5 \,{\mathrm e}^{x} {\mathrm e}^{x^{2}} x^{3}}{3}+\frac {{\mathrm e}^{x} {\mathrm e}^{x^{2}} x^{4}}{3}}{x^{2}}\) | \(47\) |
parallelrisch | \(-\frac {{\mathrm e}^{x^{2}} x^{5}-{\mathrm e}^{x} {\mathrm e}^{x^{2}} x^{4}-5 x^{4} {\mathrm e}^{x^{2}}+5 \,{\mathrm e}^{x} {\mathrm e}^{x^{2}} x^{3}-9}{3 x^{2}}\) | \(47\) |
int(1/3*(((2*x^6-9*x^5-3*x^4-5*x^3)*exp(x)-2*x^7+10*x^6-3*x^5+10*x^4)*exp( x^2)-18)/x^3,x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-18+e^{x^2} \left (10 x^4-3 x^5+10 x^6-2 x^7+e^x \left (-5 x^3-3 x^4-9 x^5+2 x^6\right )\right )}{3 x^3} \, dx=-\frac {{\left (x^{5} - 5 \, x^{4} - {\left (x^{4} - 5 \, x^{3}\right )} e^{x}\right )} e^{\left (x^{2}\right )} - 9}{3 \, x^{2}} \]
integrate(1/3*(((2*x^6-9*x^5-3*x^4-5*x^3)*exp(x)-2*x^7+10*x^6-3*x^5+10*x^4 )*exp(x^2)-18)/x^3,x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {-18+e^{x^2} \left (10 x^4-3 x^5+10 x^6-2 x^7+e^x \left (-5 x^3-3 x^4-9 x^5+2 x^6\right )\right )}{3 x^3} \, dx=\frac {\left (- x^{3} + x^{2} e^{x} + 5 x^{2} - 5 x e^{x}\right ) e^{x^{2}}}{3} + \frac {3}{x^{2}} \]
integrate(1/3*(((2*x**6-9*x**5-3*x**4-5*x**3)*exp(x)-2*x**7+10*x**6-3*x**5 +10*x**4)*exp(x**2)-18)/x**3,x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.25 (sec) , antiderivative size = 294, normalized size of antiderivative = 9.19 \[ \int \frac {-18+e^{x^2} \left (10 x^4-3 x^5+10 x^6-2 x^7+e^x \left (-5 x^3-3 x^4-9 x^5+2 x^6\right )\right )}{3 x^3} \, dx=\frac {5}{6} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x + \frac {1}{2} i\right ) e^{\left (-\frac {1}{4}\right )} + \frac {1}{24} \, {\left (\frac {12 \, {\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{\left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {\sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} + 6 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} - 8 \, \Gamma \left (2, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )\right )} e^{\left (-\frac {1}{4}\right )} + \frac {3}{8} \, {\left (\frac {4 \, {\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{\left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {\sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} + 4 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{4}\right )} + \frac {1}{4} \, {\left (\frac {\sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} - 2 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{4}\right )} - \frac {1}{6} \, {\left (2 \, x^{3} - 3 \, x\right )} e^{\left (x^{2}\right )} + \frac {5}{3} \, {\left (x^{2} - 1\right )} e^{\left (x^{2}\right )} - \frac {1}{2} \, x e^{\left (x^{2}\right )} + \frac {3}{x^{2}} + \frac {5}{3} \, e^{\left (x^{2}\right )} \]
integrate(1/3*(((2*x^6-9*x^5-3*x^4-5*x^3)*exp(x)-2*x^7+10*x^6-3*x^5+10*x^4 )*exp(x^2)-18)/x^3,x, algorithm=\
5/6*I*sqrt(pi)*erf(I*x + 1/2*I)*e^(-1/4) + 1/24*(12*(2*x + 1)^3*gamma(3/2, -1/4*(2*x + 1)^2)/(-(2*x + 1)^2)^(3/2) - sqrt(pi)*(2*x + 1)*(erf(1/2*sqrt (-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) + 6*e^(1/4*(2*x + 1)^2) - 8*gamma( 2, -1/4*(2*x + 1)^2))*e^(-1/4) + 3/8*(4*(2*x + 1)^3*gamma(3/2, -1/4*(2*x + 1)^2)/(-(2*x + 1)^2)^(3/2) - sqrt(pi)*(2*x + 1)*(erf(1/2*sqrt(-(2*x + 1)^ 2)) - 1)/sqrt(-(2*x + 1)^2) + 4*e^(1/4*(2*x + 1)^2))*e^(-1/4) + 1/4*(sqrt( pi)*(2*x + 1)*(erf(1/2*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) - 2*e^( 1/4*(2*x + 1)^2))*e^(-1/4) - 1/6*(2*x^3 - 3*x)*e^(x^2) + 5/3*(x^2 - 1)*e^( x^2) - 1/2*x*e^(x^2) + 3/x^2 + 5/3*e^(x^2)
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {-18+e^{x^2} \left (10 x^4-3 x^5+10 x^6-2 x^7+e^x \left (-5 x^3-3 x^4-9 x^5+2 x^6\right )\right )}{3 x^3} \, dx=-\frac {x^{5} e^{\left (x^{2}\right )} - x^{4} e^{\left (x^{2} + x\right )} - 5 \, x^{4} e^{\left (x^{2}\right )} + 5 \, x^{3} e^{\left (x^{2} + x\right )} - 9}{3 \, x^{2}} \]
integrate(1/3*(((2*x^6-9*x^5-3*x^4-5*x^3)*exp(x)-2*x^7+10*x^6-3*x^5+10*x^4 )*exp(x^2)-18)/x^3,x, algorithm=\
Time = 13.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-18+e^{x^2} \left (10 x^4-3 x^5+10 x^6-2 x^7+e^x \left (-5 x^3-3 x^4-9 x^5+2 x^6\right )\right )}{3 x^3} \, dx={\mathrm {e}}^{x^2}\,\left (\frac {x^2\,{\mathrm {e}}^x}{3}-\frac {5\,x\,{\mathrm {e}}^x}{3}+\frac {5\,x^2}{3}-\frac {x^3}{3}\right )+\frac {3}{x^2} \]