\(\int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} (-5-5 x-5 x^2-5 x^2 \log (x))}{2 x^3} \, dx\) [3808]

Optimal result

Integrand size = 47, antiderivative size = 21 \[ \int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} \left (-5-5 x-5 x^2-5 x^2 \log (x)\right )}{2 x^3} \, dx=\frac {5 \left (-3+e^{2+\frac {1}{x}-x \log (x)}\right )}{2 x} \]

[Out]

5/x*(1/2*exp(2+1/x-x*ln(x))-3/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {12, 14, 2326} \[ \int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} \left (-5-5 x-5 x^2-5 x^2 \log (x)\right )}{2 x^3} \, dx=\frac {5}{2} e^{\frac {1}{x}+2} x^{-x-1}-\frac {15}{2 x} \]

[In]

Int[(15*x + E^((1 + 2*x - x^2*Log[x])/x)*(-5 - 5*x - 5*x^2 - 5*x^2*Log[x]))/(2*x^3),x]

[Out]

-15/(2*x) + (5*E^(2 + x^(-1))*x^(-1 - x))/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

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]]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} \left (-5-5 x-5 x^2-5 x^2 \log (x)\right )}{x^3} \, dx \\ & = \frac {1}{2} \int \left (\frac {15}{x^2}-5 e^{2+\frac {1}{x}} x^{-3-x} \left (1+x+x^2+x^2 \log (x)\right )\right ) \, dx \\ & = -\frac {15}{2 x}-\frac {5}{2} \int e^{2+\frac {1}{x}} x^{-3-x} \left (1+x+x^2+x^2 \log (x)\right ) \, dx \\ & = -\frac {15}{2 x}+\frac {5}{2} e^{2+\frac {1}{x}} x^{-1-x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} \left (-5-5 x-5 x^2-5 x^2 \log (x)\right )}{2 x^3} \, dx=-\frac {5}{2} \left (\frac {3}{x}-e^{2+\frac {1}{x}} x^{-1-x}\right ) \]

[In]

Integrate[(15*x + E^((1 + 2*x - x^2*Log[x])/x)*(-5 - 5*x - 5*x^2 - 5*x^2*Log[x]))/(2*x^3),x]

[Out]

(-5*(3/x - E^(2 + x^(-1))*x^(-1 - x)))/2

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29

method	result	size
risch	\(\frac {5 x^{-x} {\mathrm e}^{\frac {1+2 x}{x}}}{2 x}-\frac {15}{2 x}\)	\(27\)
parallelrisch	\(-\frac {15-5 \,{\mathrm e}^{-\frac {x^{2} \ln \left (x \right )-2 x -1}{x}}}{2 x}\)	\(27\)
default	\(\frac {5 \,{\mathrm e}^{\frac {-x^{2} \ln \left (x \right )+2 x +1}{x}}}{2 x}-\frac {15}{2 x}\)	\(29\)
norman	\(\frac {-\frac {15 x}{2}+\frac {5 x \,{\mathrm e}^{\frac {-x^{2} \ln \left (x \right )+2 x +1}{x}}}{2}}{x^{2}}\)	\(29\)
parts	\(\frac {5 \,{\mathrm e}^{\frac {-x^{2} \ln \left (x \right )+2 x +1}{x}}}{2 x}-\frac {15}{2 x}\)	\(29\)

[In]

int(1/2*((-5*x^2*ln(x)-5*x^2-5*x-5)*exp((-x^2*ln(x)+2*x+1)/x)+15*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

5/2/x*x^(-x)*exp((1+2*x)/x)-15/2/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} \left (-5-5 x-5 x^2-5 x^2 \log (x)\right )}{2 x^3} \, dx=\frac {5 \, {\left (e^{\left (-\frac {x^{2} \log \left (x\right ) - 2 \, x - 1}{x}\right )} - 3\right )}}{2 \, x} \]

[In]

integrate(1/2*((-5*x^2*log(x)-5*x^2-5*x-5)*exp((-x^2*log(x)+2*x+1)/x)+15*x)/x^3,x, algorithm="fricas")

[Out]

5/2*(e^(-(x^2*log(x) - 2*x - 1)/x) - 3)/x

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} \left (-5-5 x-5 x^2-5 x^2 \log (x)\right )}{2 x^3} \, dx=\frac {5 e^{\frac {- x^{2} \log {\left (x \right )} + 2 x + 1}{x}}}{2 x} - \frac {15}{2 x} \]

[In]

integrate(1/2*((-5*x**2*ln(x)-5*x**2-5*x-5)*exp((-x**2*ln(x)+2*x+1)/x)+15*x)/x**3,x)

[Out]

5*exp((-x**2*log(x) + 2*x + 1)/x)/(2*x) - 15/(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} \left (-5-5 x-5 x^2-5 x^2 \log (x)\right )}{2 x^3} \, dx=\frac {5 \, e^{\left (-x \log \left (x\right ) + \frac {1}{x} + 2\right )}}{2 \, x} - \frac {15}{2 \, x} \]

[In]

integrate(1/2*((-5*x^2*log(x)-5*x^2-5*x-5)*exp((-x^2*log(x)+2*x+1)/x)+15*x)/x^3,x, algorithm="maxima")

[Out]

5/2*e^(-x*log(x) + 1/x + 2)/x - 15/2/x

Giac [F]

\[ \int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} \left (-5-5 x-5 x^2-5 x^2 \log (x)\right )}{2 x^3} \, dx=\int { -\frac {5 \, {\left ({\left (x^{2} \log \left (x\right ) + x^{2} + x + 1\right )} e^{\left (-\frac {x^{2} \log \left (x\right ) - 2 \, x - 1}{x}\right )} - 3 \, x\right )}}{2 \, x^{3}} \,d x } \]

[In]

integrate(1/2*((-5*x^2*log(x)-5*x^2-5*x-5)*exp((-x^2*log(x)+2*x+1)/x)+15*x)/x^3,x, algorithm="giac")

[Out]

integrate(-5/2*((x^2*log(x) + x^2 + x + 1)*e^(-(x^2*log(x) - 2*x - 1)/x) - 3*x)/x^3, x)

Mupad [B] (verification not implemented)

Time = 9.45 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {15 x+e^{\frac {1+2 x-x^2 \log (x)}{x}} \left (-5-5 x-5 x^2-5 x^2 \log (x)\right )}{2 x^3} \, dx=\frac {5\,\left (\frac {{\mathrm {e}}^{\frac {1}{x}+2}}{x^x}-3\right )}{2\,x} \]

[In]

int(((15*x)/2 - (exp((2*x - x^2*log(x) + 1)/x)*(5*x + 5*x^2*log(x) + 5*x^2 + 5))/2)/x^3,x)

[Out]

(5*(exp(1/x + 2)/x^x - 3))/(2*x)