3.3.50 \(\int \frac {4 x+e^x (3-4 x+3 x^2)+e^x (3 x+6 x^2+3 x^3) \log (x)}{4 x} \, dx\) [250]

Optimal. Leaf size=25 \[ \frac {1}{10}+x+\frac {1}{4} e^x \left (-4+3 \left (\log (x)+x^2 \log (x)\right )\right ) \]

[Out]

1/4*exp(x)*(3*ln(x)+3*x^2*ln(x)-4)+x+1/10

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 14, 2326} \begin {gather*} x-\frac {e^x \left (-3 x^3 \log (x)+4 x-3 x \log (x)\right )}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x + E^x*(3 - 4*x + 3*x^2) + E^x*(3*x + 6*x^2 + 3*x^3)*Log[x])/(4*x),x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.07, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{4} \left (-4 e^x+4 x+3 e^x \left (1+x^2\right ) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x + E^x*(3 - 4*x + 3*x^2) + E^x*(3*x + 6*x^2 + 3*x^3)*Log[x])/(4*x),x]

[Out]

(-4*E^x + 4*x + 3*E^x*(1 + x^2)*Log[x])/4

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 22, normalized size = 0.88

method	result	size
risch	\(\frac {3 \left (x^{2}+1\right ) {\mathrm e}^{x} \ln \left (x \right )}{4}+x -{\mathrm e}^{x}\)	\(18\)
default	\(x +\frac {3 \,{\mathrm e}^{x} \ln \left (x \right )}{4}+\frac {3 x^{2} {\mathrm e}^{x} \ln \left (x \right )}{4}-{\mathrm e}^{x}\)	\(22\)
norman	\(x +\frac {3 \,{\mathrm e}^{x} \ln \left (x \right )}{4}+\frac {3 x^{2} {\mathrm e}^{x} \ln \left (x \right )}{4}-{\mathrm e}^{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((3*x^3+6*x^2+3*x)*exp(x)*ln(x)+(3*x^2-4*x+3)*exp(x)+4*x)/x,x,method=_RETURNVERBOSE)

[Out]

x+3/4*exp(x)*ln(x)+3/4*x^2*exp(x)*ln(x)-exp(x)

________________________________________________________________________________________

Maxima [A]
time = 0.31, size = 34, normalized size = 1.36 \begin {gather*} \frac {3}{4} \, {\left (x^{2} \log \left (x\right ) - x + 1\right )} e^{x} + \frac {3}{4} \, {\left (x - 1\right )} e^{x} + \frac {3}{4} \, e^{x} \log \left (x\right ) + x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((3*x^3+6*x^2+3*x)*exp(x)*log(x)+(3*x^2-4*x+3)*exp(x)+4*x)/x,x, algorithm="maxima")

[Out]

3/4*(x^2*log(x) - x + 1)*e^x + 3/4*(x - 1)*e^x + 3/4*e^x*log(x) + x - e^x

________________________________________________________________________________________

Fricas [A]
time = 0.30, size = 17, normalized size = 0.68 \begin {gather*} \frac {3}{4} \, {\left (x^{2} + 1\right )} e^{x} \log \left (x\right ) + x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((3*x^3+6*x^2+3*x)*exp(x)*log(x)+(3*x^2-4*x+3)*exp(x)+4*x)/x,x, algorithm="fricas")

[Out]

3/4*(x^2 + 1)*e^x*log(x) + x - e^x

________________________________________________________________________________________

Sympy [A]
time = 0.12, size = 20, normalized size = 0.80 \begin {gather*} x + \frac {\left (3 x^{2} \log {\left (x \right )} + 3 \log {\left (x \right )} - 4\right ) e^{x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((3*x**3+6*x**2+3*x)*exp(x)*ln(x)+(3*x**2-4*x+3)*exp(x)+4*x)/x,x)

[Out]

x + (3*x**2*log(x) + 3*log(x) - 4)*exp(x)/4

________________________________________________________________________________________

Giac [A]
time = 0.40, size = 21, normalized size = 0.84 \begin {gather*} \frac {3}{4} \, x^{2} e^{x} \log \left (x\right ) + \frac {3}{4} \, e^{x} \log \left (x\right ) + x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((3*x^3+6*x^2+3*x)*exp(x)*log(x)+(3*x^2-4*x+3)*exp(x)+4*x)/x,x, algorithm="giac")

[Out]

3/4*x^2*e^x*log(x) + 3/4*e^x*log(x) + x - e^x

________________________________________________________________________________________

Mupad [B]
time = 0.46, size = 21, normalized size = 0.84 \begin {gather*} x-{\mathrm {e}}^x+\frac {3\,{\mathrm {e}}^x\,\ln \left (x\right )}{4}+\frac {3\,x^2\,{\mathrm {e}}^x\,\ln \left (x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + (exp(x)*(3*x^2 - 4*x + 3))/4 + (exp(x)*log(x)*(3*x + 6*x^2 + 3*x^3))/4)/x,x)

[Out]

x - exp(x) + (3*exp(x)*log(x))/4 + (3*x^2*exp(x)*log(x))/4

________________________________________________________________________________________