3.11.51 \(\int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+(-4 x-4 e^x x) \log (x)}{4 x} \, dx\)

Optimal. Leaf size=24 \[ -1+x \left (x+\frac {x^2}{4}\right )-\left (e^x+x\right ) (2+\log (x)) \]

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Rubi [A]  time = 0.07, antiderivative size = 35, normalized size of antiderivative = 1.46, number of steps used = 5, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {12, 14, 2295, 2288} \begin {gather*} \frac {x^3}{4}+x^2-2 x-x \log (x)-\frac {e^x (2 x+x \log (x))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-4 - 8*x) - 12*x + 8*x^2 + 3*x^3 + (-4*x - 4*E^x*x)*Log[x])/(4*x),x]

[Out]

-2*x + x^2 + x^3/4 - x*Log[x] - (E^x*(2*x + x*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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {e^x (-4-8 x)-12 x+8 x^2+3 x^3+\left (-4 x-4 e^x x\right ) \log (x)}{x} \, dx\\ &=\frac {1}{4} \int \left (-12+8 x+3 x^2-4 \log (x)-\frac {4 e^x (1+2 x+x \log (x))}{x}\right ) \, dx\\ &=-3 x+x^2+\frac {x^3}{4}-\int \log (x) \, dx-\int \frac {e^x (1+2 x+x \log (x))}{x} \, dx\\ &=-2 x+x^2+\frac {x^3}{4}-x \log (x)-\frac {e^x (2 x+x \log (x))}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 1.17 \begin {gather*} -2 e^x-2 x+x^2+\frac {x^3}{4}-\left (e^x+x\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-4 - 8*x) - 12*x + 8*x^2 + 3*x^3 + (-4*x - 4*E^x*x)*Log[x])/(4*x),x]

[Out]

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

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fricas [A]  time = 0.66, size = 24, normalized size = 1.00 \begin {gather*} \frac {1}{4} \, x^{3} + x^{2} - {\left (x + e^{x}\right )} \log \relax (x) - 2 \, x - 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-4*exp(x)*x-4*x)*log(x)+(-8*x-4)*exp(x)+3*x^3+8*x^2-12*x)/x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.45, size = 27, normalized size = 1.12 \begin {gather*} \frac {1}{4} \, x^{3} + x^{2} - x \log \relax (x) - e^{x} \log \relax (x) - 2 \, x - 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-4*exp(x)*x-4*x)*log(x)+(-8*x-4)*exp(x)+3*x^3+8*x^2-12*x)/x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 28, normalized size = 1.17

method	result	size
default	\(-2 x -{\mathrm e}^{x} \ln \relax (x )-2 \,{\mathrm e}^{x}+x^{2}+\frac {x^{3}}{4}-x \ln \relax (x )\)	\(28\)
norman	\(-2 x -{\mathrm e}^{x} \ln \relax (x )-2 \,{\mathrm e}^{x}+x^{2}+\frac {x^{3}}{4}-x \ln \relax (x )\)	\(28\)
risch	\(\frac {\left (-4 x -4 \,{\mathrm e}^{x}\right ) \ln \relax (x )}{4}+\frac {x^{3}}{4}+x^{2}-2 x -2 \,{\mathrm e}^{x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-4*exp(x)*x-4*x)*ln(x)+(-8*x-4)*exp(x)+3*x^3+8*x^2-12*x)/x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.46, size = 27, normalized size = 1.12 \begin {gather*} \frac {1}{4} \, x^{3} + x^{2} - x \log \relax (x) - e^{x} \log \relax (x) - 2 \, x - 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-4*exp(x)*x-4*x)*log(x)+(-8*x-4)*exp(x)+3*x^3+8*x^2-12*x)/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.73, size = 27, normalized size = 1.12 \begin {gather*} x^2-2\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\ln \relax (x)-x\,\ln \relax (x)-2\,x+\frac {x^3}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + (log(x)*(4*x + 4*x*exp(x)))/4 + (exp(x)*(8*x + 4))/4 - 2*x^2 - (3*x^3)/4)/x,x)

[Out]

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

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sympy [A]  time = 0.33, size = 26, normalized size = 1.08 \begin {gather*} \frac {x^{3}}{4} + x^{2} - x \log {\relax (x )} - 2 x + \left (- \log {\relax (x )} - 2\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-4*exp(x)*x-4*x)*ln(x)+(-8*x-4)*exp(x)+3*x**3+8*x**2-12*x)/x,x)

[Out]

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

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