3.76.86 \(\int \frac {6+e^x (8+8 x^2)+(-8+e^x (-8+8 x)) \log (x)}{x^2} \, dx\)

Optimal. Leaf size=23 \[ 2 \left (3+\frac {x+4 \left (x+e^x x\right ) (x+\log (x))}{x^2}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 33, normalized size of antiderivative = 1.43, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {14, 2304, 2288} \begin {gather*} \frac {8 e^x \left (x^2+x \log (x)\right )}{x^2}+\frac {8}{x}-\frac {2 (3-4 \log (x))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + E^x*(8 + 8*x^2) + (-8 + E^x*(-8 + 8*x))*Log[x])/x^2,x]

[Out]

8/x - (2*(3 - 4*Log[x]))/x + (8*E^x*(x^2 + x*Log[x]))/x^2

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} &=\int \left (-\frac {2 (-3+4 \log (x))}{x^2}+\frac {8 e^x \left (1+x^2-\log (x)+x \log (x)\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {-3+4 \log (x)}{x^2} \, dx\right )+8 \int \frac {e^x \left (1+x^2-\log (x)+x \log (x)\right )}{x^2} \, dx\\ &=\frac {8}{x}-\frac {2 (3-4 \log (x))}{x}+\frac {8 e^x \left (x^2+x \log (x)\right )}{x^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 21, normalized size = 0.91 \begin {gather*} \frac {2+8 e^x x+8 \left (1+e^x\right ) \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + E^x*(8 + 8*x^2) + (-8 + E^x*(-8 + 8*x))*Log[x])/x^2,x]

[Out]

(2 + 8*E^x*x + 8*(1 + E^x)*Log[x])/x

________________________________________________________________________________________

fricas [A]  time = 0.93, size = 20, normalized size = 0.87 \begin {gather*} \frac {2 \, {\left (4 \, x e^{x} + 4 \, {\left (e^{x} + 1\right )} \log \relax (x) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-8)*exp(x)-8)*log(x)+(8*x^2+8)*exp(x)+6)/x^2,x, algorithm="fricas")

[Out]

2*(4*x*e^x + 4*(e^x + 1)*log(x) + 1)/x

________________________________________________________________________________________

giac [A]  time = 0.14, size = 22, normalized size = 0.96 \begin {gather*} \frac {2 \, {\left (4 \, x e^{x} + 4 \, e^{x} \log \relax (x) + 4 \, \log \relax (x) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-8)*exp(x)-8)*log(x)+(8*x^2+8)*exp(x)+6)/x^2,x, algorithm="giac")

[Out]

2*(4*x*e^x + 4*e^x*log(x) + 4*log(x) + 1)/x

________________________________________________________________________________________

maple [A]  time = 0.04, size = 22, normalized size = 0.96

method	result	size
norman	\(\frac {2+8 \,{\mathrm e}^{x} x +8 \,{\mathrm e}^{x} \ln \relax (x )+8 \ln \relax (x )}{x}\)	\(22\)
risch	\(\frac {8 \left ({\mathrm e}^{x}+1\right ) \ln \relax (x )}{x}+\frac {8 \,{\mathrm e}^{x} x +2}{x}\)	\(25\)
default	\(\frac {8 \,{\mathrm e}^{x} x +8 \,{\mathrm e}^{x} \ln \relax (x )}{x}+\frac {2}{x}+\frac {8 \ln \relax (x )}{x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x-8)*exp(x)-8)*ln(x)+(8*x^2+8)*exp(x)+6)/x^2,x,method=_RETURNVERBOSE)

[Out]

(2+8*exp(x)*x+8*exp(x)*ln(x)+8*ln(x))/x

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {8 \, e^{x} \log \relax (x)}{x} + \frac {8 \, \log \relax (x)}{x} + \frac {2}{x} + 8 \, e^{x} + 8 \, \Gamma \left (-1, -x\right ) - 8 \, \int \frac {e^{x}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-8)*exp(x)-8)*log(x)+(8*x^2+8)*exp(x)+6)/x^2,x, algorithm="maxima")

[Out]

8*e^x*log(x)/x + 8*log(x)/x + 2/x + 8*e^x + 8*gamma(-1, -x) - 8*integrate(e^x/x^2, x)

________________________________________________________________________________________

mupad [B]  time = 5.37, size = 21, normalized size = 0.91 \begin {gather*} 8\,{\mathrm {e}}^x+\frac {8\,\ln \relax (x)+8\,{\mathrm {e}}^x\,\ln \relax (x)+2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(exp(x)*(8*x - 8) - 8) + exp(x)*(8*x^2 + 8) + 6)/x^2,x)

[Out]

8*exp(x) + (8*log(x) + 8*exp(x)*log(x) + 2)/x

________________________________________________________________________________________

sympy [A]  time = 0.33, size = 22, normalized size = 0.96 \begin {gather*} \frac {\left (8 x + 8 \log {\relax (x )}\right ) e^{x}}{x} + \frac {8 \log {\relax (x )}}{x} + \frac {2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-8)*exp(x)-8)*ln(x)+(8*x**2+8)*exp(x)+6)/x**2,x)

[Out]

(8*x + 8*log(x))*exp(x)/x + 8*log(x)/x + 2/x

________________________________________________________________________________________