\(\int \frac {6+e^x (8+8 x^2)+(-8+e^x (-8+8 x)) \log (x)}{x^2} \, dx\) [7586]

Optimal result

Integrand size = 31, antiderivative size = 23 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=2 \left (3+\frac {x+4 \left (x+e^x x\right ) (x+\log (x))}{x^2}\right ) \]

[Out]

2*(4*(x+ln(x))*(exp(x)*x+x)+x)/x^2+6

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {14, 2341, 2326} \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {8 e^x \left (x^2+x \log (x)\right )}{x^2}+\frac {8}{x}-\frac {2 (3-4 \log (x))}{x} \]

[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 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]

Rule 2341

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {2+8 e^x x+8 \left (1+e^x\right ) \log (x)}{x} \]

[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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
norman	\(\frac {2+8 \,{\mathrm e}^{x} x +8 \,{\mathrm e}^{x} \ln \left (x \right )+8 \ln \left (x \right )}{x}\)	\(22\)
parallelrisch	\(-\frac {-8 \,{\mathrm e}^{x} x -2-8 \,{\mathrm e}^{x} \ln \left (x \right )-8 \ln \left (x \right )}{x}\)	\(23\)
risch	\(\frac {8 \left ({\mathrm e}^{x}+1\right ) \ln \left (x \right )}{x}+\frac {8 \,{\mathrm e}^{x} x +2}{x}\)	\(25\)
default	\(\frac {8 \,{\mathrm e}^{x} x +8 \,{\mathrm e}^{x} \ln \left (x \right )}{x}+\frac {2}{x}+\frac {8 \ln \left (x \right )}{x}\)	\(30\)
parts	\(\frac {8 \,{\mathrm e}^{x} x +8 \,{\mathrm e}^{x} \ln \left (x \right )}{x}+\frac {2}{x}+\frac {8 \ln \left (x \right )}{x}\)	\(30\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {2 \, {\left (4 \, x e^{x} + 4 \, {\left (e^{x} + 1\right )} \log \left (x\right ) + 1\right )}}{x} \]

[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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {\left (8 x + 8 \log {\left (x \right )}\right ) e^{x}}{x} + \frac {8 \log {\left (x \right )}}{x} + \frac {2}{x} \]

[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

Maxima [F]

\[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\int { \frac {2 \, {\left (4 \, {\left (x^{2} + 1\right )} e^{x} + 4 \, {\left ({\left (x - 1\right )} e^{x} - 1\right )} \log \left (x\right ) + 3\right )}}{x^{2}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=\frac {2 \, {\left (4 \, x e^{x} + 4 \, e^{x} \log \left (x\right ) + 4 \, \log \left (x\right ) + 1\right )}}{x} \]

[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

Mupad [B] (verification not implemented)

Time = 13.52 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {6+e^x \left (8+8 x^2\right )+\left (-8+e^x (-8+8 x)\right ) \log (x)}{x^2} \, dx=8\,{\mathrm {e}}^x+\frac {8\,\ln \left (x\right )+8\,{\mathrm {e}}^x\,\ln \left (x\right )+2}{x} \]

[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