3.95.31 \(\int \frac {3+6 x^2+e^x x^2+(12 x^2+e^x (2 x^2+x^3)) \log (x)}{x} \, dx\)

Optimal. Leaf size=16 \[ \left (6+e^x+\frac {3}{x^2}\right ) x^2 \log (x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 21, normalized size of antiderivative = 1.31, number of steps used = 8, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {14, 2288, 2304} \begin {gather*} e^x x^2 \log (x)+6 x^2 \log (x)+3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 6*x^2 + E^x*x^2 + (12*x^2 + E^x*(2*x^2 + x^3))*Log[x])/x,x]

[Out]

3*Log[x] + 6*x^2*Log[x] + E^x*x^2*Log[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} &=\int \left (e^x x (1+2 \log (x)+x \log (x))+\frac {3 \left (1+2 x^2+4 x^2 \log (x)\right )}{x}\right ) \, dx\\ &=3 \int \frac {1+2 x^2+4 x^2 \log (x)}{x} \, dx+\int e^x x (1+2 \log (x)+x \log (x)) \, dx\\ &=e^x x^2 \log (x)+3 \int \left (\frac {1+2 x^2}{x}+4 x \log (x)\right ) \, dx\\ &=e^x x^2 \log (x)+3 \int \frac {1+2 x^2}{x} \, dx+12 \int x \log (x) \, dx\\ &=-3 x^2+6 x^2 \log (x)+e^x x^2 \log (x)+3 \int \left (\frac {1}{x}+2 x\right ) \, dx\\ &=3 \log (x)+6 x^2 \log (x)+e^x x^2 \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 14, normalized size = 0.88 \begin {gather*} \left (3+\left (6+e^x\right ) x^2\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 6*x^2 + E^x*x^2 + (12*x^2 + E^x*(2*x^2 + x^3))*Log[x])/x,x]

[Out]

(3 + (6 + E^x)*x^2)*Log[x]

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 16, normalized size = 1.00 \begin {gather*} {\left (x^{2} e^{x} + 6 \, x^{2} + 3\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3+2*x^2)*exp(x)+12*x^2)*log(x)+exp(x)*x^2+6*x^2+3)/x,x, algorithm="fricas")

[Out]

(x^2*e^x + 6*x^2 + 3)*log(x)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 20, normalized size = 1.25 \begin {gather*} x^{2} e^{x} \log \relax (x) + 6 \, x^{2} \log \relax (x) + 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3+2*x^2)*exp(x)+12*x^2)*log(x)+exp(x)*x^2+6*x^2+3)/x,x, algorithm="giac")

[Out]

x^2*e^x*log(x) + 6*x^2*log(x) + 3*log(x)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 21, normalized size = 1.31

method	result	size
default	\(x^{2} {\mathrm e}^{x} \ln \relax (x )+6 x^{2} \ln \relax (x )+3 \ln \relax (x )\)	\(21\)
norman	\(x^{2} {\mathrm e}^{x} \ln \relax (x )+6 x^{2} \ln \relax (x )+3 \ln \relax (x )\)	\(21\)
risch	\(\left ({\mathrm e}^{x} x^{2}+6 x^{2}\right ) \ln \relax (x )+3 \ln \relax (x )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^3+2*x^2)*exp(x)+12*x^2)*ln(x)+exp(x)*x^2+6*x^2+3)/x,x,method=_RETURNVERBOSE)

[Out]

x^2*exp(x)*ln(x)+6*x^2*ln(x)+3*ln(x)

________________________________________________________________________________________

maxima [B]  time = 0.39, size = 32, normalized size = 2.00 \begin {gather*} 6 \, x^{2} \log \relax (x) + {\left (x^{2} \log \relax (x) - x + 1\right )} e^{x} + {\left (x - 1\right )} e^{x} + 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3+2*x^2)*exp(x)+12*x^2)*log(x)+exp(x)*x^2+6*x^2+3)/x,x, algorithm="maxima")

[Out]

6*x^2*log(x) + (x^2*log(x) - x + 1)*e^x + (x - 1)*e^x + 3*log(x)

________________________________________________________________________________________

mupad [B]  time = 7.22, size = 16, normalized size = 1.00 \begin {gather*} \ln \relax (x)\,\left (x^2\,{\mathrm {e}}^x+6\,x^2+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(x) + log(x)*(exp(x)*(2*x^2 + x^3) + 12*x^2) + 6*x^2 + 3)/x,x)

[Out]

log(x)*(x^2*exp(x) + 6*x^2 + 3)

________________________________________________________________________________________

sympy [A]  time = 0.33, size = 22, normalized size = 1.38 \begin {gather*} x^{2} e^{x} \log {\relax (x )} + 6 x^{2} \log {\relax (x )} + 3 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**3+2*x**2)*exp(x)+12*x**2)*ln(x)+exp(x)*x**2+6*x**2+3)/x,x)

[Out]

x**2*exp(x)*log(x) + 6*x**2*log(x) + 3*log(x)

________________________________________________________________________________________