3.35.79 \(\int \frac {2+2 x+2 x^2+(1+x) \log (\frac {e^{2 x} x^2}{100+200 x+100 x^2+e^3 (-40-80 x-40 x^2)+e^6 (4+8 x+4 x^2)})}{2+2 x} \, dx\)

Optimal. Leaf size=30 \[ \frac {1}{2} x \log \left (\frac {e^{2 x} x^2}{4 \left (-5+e^3\right )^2 (1+x)^2}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 32, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 4, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6688, 698, 2548, 12} \begin {gather*} \frac {1}{2} x \log \left (\frac {e^{2 x} x^2}{4 \left (5-e^3\right )^2 (x+1)^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 2*x + 2*x^2 + (1 + x)*Log[(E^(2*x)*x^2)/(100 + 200*x + 100*x^2 + E^3*(-40 - 80*x - 40*x^2) + E^6*(4 +
8*x + 4*x^2))])/(2 + 2*x),x]

[Out]

(x*Log[(E^(2*x)*x^2)/(4*(5 - E^3)^2*(1 + x)^2)])/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1+x+x^2}{1+x}+\frac {1}{2} \log \left (\frac {e^{2 x} x^2}{4 \left (-5+e^3\right )^2 (1+x)^2}\right )\right ) \, dx\\ &=\frac {1}{2} \int \log \left (\frac {e^{2 x} x^2}{4 \left (-5+e^3\right )^2 (1+x)^2}\right ) \, dx+\int \frac {1+x+x^2}{1+x} \, dx\\ &=\frac {1}{2} x \log \left (\frac {e^{2 x} x^2}{4 \left (5-e^3\right )^2 (1+x)^2}\right )-\frac {1}{2} \int \frac {2 \left (1+x+x^2\right )}{1+x} \, dx+\int \left (x+\frac {1}{1+x}\right ) \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} x \log \left (\frac {e^{2 x} x^2}{4 \left (5-e^3\right )^2 (1+x)^2}\right )+\log (1+x)-\int \frac {1+x+x^2}{1+x} \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} x \log \left (\frac {e^{2 x} x^2}{4 \left (5-e^3\right )^2 (1+x)^2}\right )+\log (1+x)-\int \left (x+\frac {1}{1+x}\right ) \, dx\\ &=\frac {1}{2} x \log \left (\frac {e^{2 x} x^2}{4 \left (5-e^3\right )^2 (1+x)^2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 33, normalized size = 1.10 \begin {gather*} \frac {1}{2} \left (1+x \log \left (\frac {e^{2 x} x^2}{4 \left (-5+e^3\right )^2 (1+x)^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 2*x + 2*x^2 + (1 + x)*Log[(E^(2*x)*x^2)/(100 + 200*x + 100*x^2 + E^3*(-40 - 80*x - 40*x^2) + E^
6*(4 + 8*x + 4*x^2))])/(2 + 2*x),x]

[Out]

(1 + x*Log[(E^(2*x)*x^2)/(4*(-5 + E^3)^2*(1 + x)^2)])/2

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fricas [A]  time = 0.61, size = 48, normalized size = 1.60 \begin {gather*} \frac {1}{2} \, x \log \left (\frac {x^{2} e^{\left (2 \, x\right )}}{4 \, {\left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{6} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{3} + 50 \, x + 25\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*log(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40)*exp(3)+100*x^2+200*x+100))+2*x^2+2
*x+2)/(2*x+2),x, algorithm="fricas")

[Out]

1/2*x*log(1/4*x^2*e^(2*x)/(25*x^2 + (x^2 + 2*x + 1)*e^6 - 10*(x^2 + 2*x + 1)*e^3 + 50*x + 25))

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giac [B]  time = 0.46, size = 54, normalized size = 1.80 \begin {gather*} x^{2} + \frac {1}{2} \, x \log \left (\frac {x^{2}}{4 \, {\left (x^{2} e^{6} - 10 \, x^{2} e^{3} + 25 \, x^{2} + 2 \, x e^{6} - 20 \, x e^{3} + 50 \, x + e^{6} - 10 \, e^{3} + 25\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*log(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40)*exp(3)+100*x^2+200*x+100))+2*x^2+2
*x+2)/(2*x+2),x, algorithm="giac")

[Out]

x^2 + 1/2*x*log(1/4*x^2/(x^2*e^6 - 10*x^2*e^3 + 25*x^2 + 2*x*e^6 - 20*x*e^3 + 50*x + e^6 - 10*e^3 + 25))

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maple [B]  time = 0.36, size = 53, normalized size = 1.77

method	result	size
norman	\(\frac {x \ln \left (\frac {x^{2} {\mathrm e}^{2 x}}{\left (4 x^{2}+8 x +4\right ) {\mathrm e}^{6}+\left (-40 x^{2}-80 x -40\right ) {\mathrm e}^{3}+100 x^{2}+200 x +100}\right )}{2}\)	\(53\)
default	\(\frac {x^{2}}{2}+\ln \left (x +1\right )+\frac {625 x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {25 x^{2}}{2 \left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )}-\frac {625 \ln \left (x +1\right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {25 x}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}+\frac {x \ln \left (\frac {x^{2} {\mathrm e}^{2 x}}{\left (4 x^{2}+8 x +4\right ) {\mathrm e}^{6}+\left (-40 x^{2}-80 x -40\right ) {\mathrm e}^{3}+100 x^{2}+200 x +100}\right )}{2}+\frac {10 \,{\mathrm e}^{3} x}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}-\frac {{\mathrm e}^{6} x}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}+\frac {500 \,{\mathrm e}^{3} \ln \left (x +1\right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {150 \,{\mathrm e}^{6} \ln \left (x +1\right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {20 \,{\mathrm e}^{9} \ln \left (x +1\right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {{\mathrm e}^{12} \ln \left (x +1\right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {5 \,{\mathrm e}^{3} x^{2}}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}-\frac {{\mathrm e}^{6} x^{2}}{2 \left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )}-\frac {500 \,{\mathrm e}^{3} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {150 \,{\mathrm e}^{6} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {20 \,{\mathrm e}^{9} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {{\mathrm e}^{12} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}\)	\(363\)
risch	\(-x \ln \relax (2)+x \ln \relax (x )-x \ln \left ({\mathrm e}^{3}-5\right )+x \ln \left ({\mathrm e}^{x}\right )-x \ln \left (x +1\right )+\frac {i x \pi \,\mathrm {csgn}\left (\frac {i}{\left (x +1\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right )^{2}}{4}+\frac {i x \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right )^{2}}{4}-\frac {i x \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right )^{3}}{4}+\frac {i x \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i x^{2} {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right )^{2}}{4}+\frac {i x \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i x^{2} {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right )^{2}}{4}+\frac {i x \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}}{2}+\frac {i x \pi \mathrm {csgn}\left (i \left (x +1\right )^{2}\right )^{3}}{4}-\frac {i x \pi \mathrm {csgn}\left (\frac {i x^{2} {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right )^{3}}{4}-\frac {i x \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )}{4}-\frac {i x \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}}{4}-\frac {i x \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i x^{2} {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right )}{4}-\frac {i x \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{4}-\frac {i x \pi \,\mathrm {csgn}\left (\frac {i}{\left (x +1\right )^{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (x +1\right )^{2}}\right )}{4}-\frac {i x \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{4}-\frac {i x \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (i \left (x +1\right )^{2}\right )^{2}}{2}+\frac {i x \pi \mathrm {csgn}\left (i \left (x +1\right )\right )^{2} \mathrm {csgn}\left (i \left (x +1\right )^{2}\right )}{4}+\frac {i x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}}{2}\)	\(444\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+1)*ln(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40)*exp(3)+100*x^2+200*x+100))+2*x^2+2*x+2)/(
2*x+2),x,method=_RETURNVERBOSE)

[Out]

1/2*x*ln(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40)*exp(3)+100*x^2+200*x+100))

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maxima [A]  time = 1.01, size = 32, normalized size = 1.07 \begin {gather*} x^{2} - x {\left (\log \relax (2) + \log \left (e^{3} - 5\right )\right )} - {\left (x + 1\right )} \log \left (x + 1\right ) + x \log \relax (x) + \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*log(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40)*exp(3)+100*x^2+200*x+100))+2*x^2+2
*x+2)/(2*x+2),x, algorithm="maxima")

[Out]

x^2 - x*(log(2) + log(e^3 - 5)) - (x + 1)*log(x + 1) + x*log(x) + log(x + 1)

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mupad [B]  time = 2.36, size = 56, normalized size = 1.87 \begin {gather*} \frac {x\,\left (2\,x-\ln \left (200\,x-40\,{\mathrm {e}}^3+4\,{\mathrm {e}}^6-80\,x\,{\mathrm {e}}^3+8\,x\,{\mathrm {e}}^6-40\,x^2\,{\mathrm {e}}^3+4\,x^2\,{\mathrm {e}}^6+100\,x^2+100\right )+\ln \left (x^2\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + log((x^2*exp(2*x))/(200*x + exp(6)*(8*x + 4*x^2 + 4) - exp(3)*(80*x + 40*x^2 + 40) + 100*x^2 + 100)
)*(x + 1) + 2*x^2 + 2)/(2*x + 2),x)

[Out]

(x*(2*x - log(200*x - 40*exp(3) + 4*exp(6) - 80*x*exp(3) + 8*x*exp(6) - 40*x^2*exp(3) + 4*x^2*exp(6) + 100*x^2
+ 100) + log(x^2)))/2

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sympy [B]  time = 0.45, size = 70, normalized size = 2.33 \begin {gather*} - \frac {x}{6} + \left (\frac {x}{2} + \frac {1}{12}\right ) \log {\left (\frac {x^{2} e^{2 x}}{100 x^{2} + 200 x + \left (- 40 x^{2} - 80 x - 40\right ) e^{3} + \left (4 x^{2} + 8 x + 4\right ) e^{6} + 100} \right )} - \frac {\log {\relax (x )}}{6} + \frac {\log {\left (x + 1 \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*ln(x**2*exp(x)**2/((4*x**2+8*x+4)*exp(3)**2+(-40*x**2-80*x-40)*exp(3)+100*x**2+200*x+100))+2*
x**2+2*x+2)/(2*x+2),x)

[Out]

-x/6 + (x/2 + 1/12)*log(x**2*exp(2*x)/(100*x**2 + 200*x + (-40*x**2 - 80*x - 40)*exp(3) + (4*x**2 + 8*x + 4)*e
xp(6) + 100)) - log(x)/6 + log(x + 1)/6

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