3.8.45 \(\int \frac {25+150 x+50 x^2+e^5 (-25 x-25 x^2)+25 x \log (x)}{4 x+2 x^2-e^5 x^2+x \log (x)} \, dx\)

Optimal. Leaf size=23 \[ 25 \left (4+x+\log \left (x+\frac {4+x+\log (x)}{1-e^5}\right )\right ) \]

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Rubi [A]  time = 0.32, antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 6741, 12, 6742, 6684} \begin {gather*} 25 x+25 \log \left (\left (2-e^5\right ) x+\log (x)+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + 150*x + 50*x^2 + E^5*(-25*x - 25*x^2) + 25*x*Log[x])/(4*x + 2*x^2 - E^5*x^2 + x*Log[x]),x]

[Out]

25*x + 25*Log[4 + (2 - E^5)*x + Log[x]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25+150 x+50 x^2+e^5 \left (-25 x-25 x^2\right )+25 x \log (x)}{4 x+\left (2-e^5\right ) x^2+x \log (x)} \, dx\\ &=\int \frac {25 \left (1+6 \left (1-\frac {e^5}{6}\right ) x+2 \left (1-\frac {e^5}{2}\right ) x^2+x \log (x)\right )}{4 x+\left (2-e^5\right ) x^2+x \log (x)} \, dx\\ &=25 \int \frac {1+6 \left (1-\frac {e^5}{6}\right ) x+2 \left (1-\frac {e^5}{2}\right ) x^2+x \log (x)}{4 x+\left (2-e^5\right ) x^2+x \log (x)} \, dx\\ &=25 \int \left (1+\frac {1+\left (2-e^5\right ) x}{x \left (4+2 \left (1-\frac {e^5}{2}\right ) x+\log (x)\right )}\right ) \, dx\\ &=25 x+25 \int \frac {1+\left (2-e^5\right ) x}{x \left (4+2 \left (1-\frac {e^5}{2}\right ) x+\log (x)\right )} \, dx\\ &=25 x+25 \log \left (4+\left (2-e^5\right ) x+\log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 18, normalized size = 0.78 \begin {gather*} 25 \left (x+\log \left (4+2 x-e^5 x+\log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + 150*x + 50*x^2 + E^5*(-25*x - 25*x^2) + 25*x*Log[x])/(4*x + 2*x^2 - E^5*x^2 + x*Log[x]),x]

[Out]

25*(x + Log[4 + 2*x - E^5*x + Log[x]])

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fricas [A]  time = 0.69, size = 19, normalized size = 0.83 \begin {gather*} 25 \, x + 25 \, \log \left (-x e^{5} + 2 \, x + \log \relax (x) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x*log(x)+(-25*x^2-25*x)*exp(5)+50*x^2+150*x+25)/(x*log(x)-x^2*exp(5)+2*x^2+4*x),x, algorithm="fr
icas")

[Out]

25*x + 25*log(-x*e^5 + 2*x + log(x) + 4)

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giac [A]  time = 0.40, size = 20, normalized size = 0.87 \begin {gather*} 25 \, x + 25 \, \log \left (x e^{5} - 2 \, x - \log \relax (x) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x*log(x)+(-25*x^2-25*x)*exp(5)+50*x^2+150*x+25)/(x*log(x)-x^2*exp(5)+2*x^2+4*x),x, algorithm="gi
ac")

[Out]

25*x + 25*log(x*e^5 - 2*x - log(x) - 4)

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maple [A]  time = 0.04, size = 20, normalized size = 0.87

method	result	size
risch	\(25 x +25 \ln \left (-x \,{\mathrm e}^{5}+2 x +\ln \relax (x )+4\right )\)	\(20\)
norman	\(25 x +25 \ln \left (x \,{\mathrm e}^{5}-\ln \relax (x )-2 x -4\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x*ln(x)+(-25*x^2-25*x)*exp(5)+50*x^2+150*x+25)/(x*ln(x)-x^2*exp(5)+2*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

25*x+25*ln(-x*exp(5)+2*x+ln(x)+4)

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maxima [A]  time = 0.49, size = 18, normalized size = 0.78 \begin {gather*} 25 \, x + 25 \, \log \left (-x {\left (e^{5} - 2\right )} + \log \relax (x) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x*log(x)+(-25*x^2-25*x)*exp(5)+50*x^2+150*x+25)/(x*log(x)-x^2*exp(5)+2*x^2+4*x),x, algorithm="ma
xima")

[Out]

25*x + 25*log(-x*(e^5 - 2) + log(x) + 4)

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mupad [B]  time = 0.92, size = 19, normalized size = 0.83 \begin {gather*} 25\,x+25\,\ln \left (2\,x+\ln \relax (x)-x\,{\mathrm {e}}^5+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((150*x - exp(5)*(25*x + 25*x^2) + 25*x*log(x) + 50*x^2 + 25)/(4*x - x^2*exp(5) + x*log(x) + 2*x^2),x)

[Out]

25*x + 25*log(2*x + log(x) - x*exp(5) + 4)

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sympy [A]  time = 0.15, size = 19, normalized size = 0.83 \begin {gather*} 25 x + 25 \log {\left (- x e^{5} + 2 x + \log {\relax (x )} + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x*ln(x)+(-25*x**2-25*x)*exp(5)+50*x**2+150*x+25)/(x*ln(x)-x**2*exp(5)+2*x**2+4*x),x)

[Out]

25*x + 25*log(-x*exp(5) + 2*x + log(x) + 4)

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