3.23.43 \(\int \frac {15+5 x^2+e^{5+x-x^2} (-3+3 x-6 x^2)-15 \log (x)}{5 x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {77}{16}+x+\frac {3 \left (\frac {1}{5} e^{5+x-x^2}+\log (x)\right )}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 40, normalized size of antiderivative = 1.48, number of steps used = 9, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2288, 2304} \begin {gather*} \frac {3 e^{-x^2+x+5} \left (x-2 x^2\right )}{5 (1-2 x) x^2}+x+\frac {3 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 + 5*x^2 + E^(5 + x - x^2)*(-3 + 3*x - 6*x^2) - 15*Log[x])/(5*x^2),x]

[Out]

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

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 2288

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]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {15+5 x^2+e^{5+x-x^2} \left (-3+3 x-6 x^2\right )-15 \log (x)}{x^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {3 e^{5+x-x^2} \left (1-x+2 x^2\right )}{x^2}+\frac {5 \left (3+x^2-3 \log (x)\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {3}{5} \int \frac {e^{5+x-x^2} \left (1-x+2 x^2\right )}{x^2} \, dx\right )+\int \frac {3+x^2-3 \log (x)}{x^2} \, dx\\ &=\frac {3 e^{5+x-x^2} \left (x-2 x^2\right )}{5 (1-2 x) x^2}+\int \left (\frac {3+x^2}{x^2}-\frac {3 \log (x)}{x^2}\right ) \, dx\\ &=\frac {3 e^{5+x-x^2} \left (x-2 x^2\right )}{5 (1-2 x) x^2}-3 \int \frac {\log (x)}{x^2} \, dx+\int \frac {3+x^2}{x^2} \, dx\\ &=\frac {3}{x}+\frac {3 e^{5+x-x^2} \left (x-2 x^2\right )}{5 (1-2 x) x^2}+\frac {3 \log (x)}{x}+\int \left (1+\frac {3}{x^2}\right ) \, dx\\ &=x+\frac {3 e^{5+x-x^2} \left (x-2 x^2\right )}{5 (1-2 x) x^2}+\frac {3 \log (x)}{x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 29, normalized size = 1.07 \begin {gather*} \frac {3 e^{5+x-x^2}+5 x^2+15 \log (x)}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 + 5*x^2 + E^(5 + x - x^2)*(-3 + 3*x - 6*x^2) - 15*Log[x])/(5*x^2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.84, size = 26, normalized size = 0.96 \begin {gather*} \frac {5 \, x^{2} + 3 \, e^{\left (-x^{2} + x + 5\right )} + 15 \, \log \relax (x)}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-15*log(x)+(-6*x^2+3*x-3)*exp(-x^2+x+5)+5*x^2+15)/x^2,x, algorithm="fricas")

[Out]

1/5*(5*x^2 + 3*e^(-x^2 + x + 5) + 15*log(x))/x

________________________________________________________________________________________

giac [A]  time = 0.28, size = 26, normalized size = 0.96 \begin {gather*} \frac {5 \, x^{2} + 3 \, e^{\left (-x^{2} + x + 5\right )} + 15 \, \log \relax (x)}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-15*log(x)+(-6*x^2+3*x-3)*exp(-x^2+x+5)+5*x^2+15)/x^2,x, algorithm="giac")

[Out]

1/5*(5*x^2 + 3*e^(-x^2 + x + 5) + 15*log(x))/x

________________________________________________________________________________________

maple [A]  time = 0.06, size = 24, normalized size = 0.89

method	result	size
default	\(x +\frac {3 \ln \relax (x )}{x}+\frac {3 \,{\mathrm e}^{-x^{2}+x +5}}{5 x}\)	\(24\)
norman	\(\frac {x^{2}+\frac {3 \,{\mathrm e}^{-x^{2}+x +5}}{5}+3 \ln \relax (x )}{x}\)	\(24\)
risch	\(\frac {3 \ln \relax (x )}{x}+\frac {5 x^{2}+3 \,{\mathrm e}^{-x^{2}+x +5}}{5 x}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-15*ln(x)+(-6*x^2+3*x-3)*exp(-x^2+x+5)+5*x^2+15)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {3}{5} \, \sqrt {\pi } \operatorname {erf}\left (x - \frac {1}{2}\right ) e^{\frac {21}{4}} + x + \frac {3 \, {\left (\log \relax (x) + 1\right )}}{x} - \frac {3}{x} + \frac {1}{5} \, \int \frac {3 \, {\left (x e^{5} - e^{5}\right )} e^{\left (-x^{2} + x\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-15*log(x)+(-6*x^2+3*x-3)*exp(-x^2+x+5)+5*x^2+15)/x^2,x, algorithm="maxima")

[Out]

-3/5*sqrt(pi)*erf(x - 1/2)*e^(21/4) + x + 3*(log(x) + 1)/x - 3/x + 1/5*integrate(3*(x*e^5 - e^5)*e^(-x^2 + x)/
x^2, x)

________________________________________________________________________________________

mupad [B]  time = 1.37, size = 24, normalized size = 0.89 \begin {gather*} x+\frac {3\,\ln \relax (x)}{x}+\frac {3\,{\mathrm {e}}^5\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^x}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*log(x) + (exp(x - x^2 + 5)*(6*x^2 - 3*x + 3))/5 - x^2 - 3)/x^2,x)

[Out]

x + (3*log(x))/x + (3*exp(5)*exp(-x^2)*exp(x))/(5*x)

________________________________________________________________________________________

sympy [A]  time = 0.30, size = 20, normalized size = 0.74 \begin {gather*} x + \frac {3 e^{- x^{2} + x + 5}}{5 x} + \frac {3 \log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-15*ln(x)+(-6*x**2+3*x-3)*exp(-x**2+x+5)+5*x**2+15)/x**2,x)

[Out]

x + 3*exp(-x**2 + x + 5)/(5*x) + 3*log(x)/x

________________________________________________________________________________________