3.46.11 \(\int \frac {5 x-10 x^2+5 x^3+e^2 (4 x^2-2 x^3)+(2-4 x+2 x^2) \log (5)}{x-2 x^2+x^3} \, dx\)

Optimal. Leaf size=32 \[ x+2 \left (2 x+\frac {e^2 x^2}{1-x}-\log (5) (5-\log (x))\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 27, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 3, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {1594, 27, 1620} \begin {gather*} \left (5-2 e^2\right ) x+\frac {2 e^2}{1-x}+\log (25) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x - 10*x^2 + 5*x^3 + E^2*(4*x^2 - 2*x^3) + (2 - 4*x + 2*x^2)*Log[5])/(x - 2*x^2 + x^3),x]

[Out]

(2*E^2)/(1 - x) + (5 - 2*E^2)*x + Log[25]*Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x-10 x^2+5 x^3+e^2 \left (4 x^2-2 x^3\right )+\left (2-4 x+2 x^2\right ) \log (5)}{x \left (1-2 x+x^2\right )} \, dx\\ &=\int \frac {5 x-10 x^2+5 x^3+e^2 \left (4 x^2-2 x^3\right )+\left (2-4 x+2 x^2\right ) \log (5)}{(-1+x)^2 x} \, dx\\ &=\int \left (5 \left (1-\frac {2 e^2}{5}\right )+\frac {2 e^2}{(-1+x)^2}+\frac {\log (25)}{x}\right ) \, dx\\ &=\frac {2 e^2}{1-x}+\left (5-2 e^2\right ) x+\log (25) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 24, normalized size = 0.75 \begin {gather*} e^2 \left (-\frac {2}{-1+x}-2 x\right )+5 x+\log (25) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x - 10*x^2 + 5*x^3 + E^2*(4*x^2 - 2*x^3) + (2 - 4*x + 2*x^2)*Log[5])/(x - 2*x^2 + x^3),x]

[Out]

E^2*(-2/(-1 + x) - 2*x) + 5*x + Log[25]*Log[x]

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fricas [A]  time = 0.43, size = 36, normalized size = 1.12 \begin {gather*} \frac {2 \, {\left (x - 1\right )} \log \relax (5) \log \relax (x) + 5 \, x^{2} - 2 \, {\left (x^{2} - x + 1\right )} e^{2} - 5 \, x}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-4*x+2)*log(5)+(-2*x^3+4*x^2)*exp(1)^2+5*x^3-10*x^2+5*x)/(x^3-2*x^2+x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.13, size = 25, normalized size = 0.78 \begin {gather*} -2 \, x e^{2} + 2 \, \log \relax (5) \log \left ({\left | x \right |}\right ) + 5 \, x - \frac {2 \, e^{2}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-4*x+2)*log(5)+(-2*x^3+4*x^2)*exp(1)^2+5*x^3-10*x^2+5*x)/(x^3-2*x^2+x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 25, normalized size = 0.78

method	result	size
default	\(-2 \,{\mathrm e}^{2} x +5 x -\frac {2 \,{\mathrm e}^{2}}{x -1}+2 \ln \relax (5) \ln \relax (x )\)	\(25\)
risch	\(-2 \,{\mathrm e}^{2} x +5 x -\frac {2 \,{\mathrm e}^{2}}{x -1}+2 \ln \relax (5) \ln \left (-x \right )\)	\(27\)
norman	\(\frac {\left (-2 \,{\mathrm e}^{2}+5\right ) x^{2}-5}{x -1}+2 \ln \relax (5) \ln \relax (x )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-4*x+2)*ln(5)+(-2*x^3+4*x^2)*exp(1)^2+5*x^3-10*x^2+5*x)/(x^3-2*x^2+x),x,method=_RETURNVERBOSE)

[Out]

-2*exp(2)*x+5*x-2*exp(2)/(x-1)+2*ln(5)*ln(x)

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maxima [A]  time = 0.35, size = 25, normalized size = 0.78 \begin {gather*} -x {\left (2 \, e^{2} - 5\right )} + 2 \, \log \relax (5) \log \relax (x) - \frac {2 \, e^{2}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-4*x+2)*log(5)+(-2*x^3+4*x^2)*exp(1)^2+5*x^3-10*x^2+5*x)/(x^3-2*x^2+x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.12, size = 53, normalized size = 1.66 \begin {gather*} 2\,\ln \relax (5)\,\ln \relax (x)-\ln \left (x-1\right )\,\left (2\,\ln \relax (5)-\ln \left (25\right )\right )-\frac {2\,{\mathrm {e}}^2+2\,\ln \relax (5)+\ln \left (25\right )-\ln \left (625\right )}{x-1}-x\,\left (2\,{\mathrm {e}}^2-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + log(5)*(2*x^2 - 4*x + 2) + exp(2)*(4*x^2 - 2*x^3) - 10*x^2 + 5*x^3)/(x - 2*x^2 + x^3),x)

[Out]

2*log(5)*log(x) - log(x - 1)*(2*log(5) - log(25)) - (2*exp(2) + 2*log(5) + log(25) - log(625))/(x - 1) - x*(2*
exp(2) - 5)

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sympy [A]  time = 0.25, size = 24, normalized size = 0.75 \begin {gather*} x \left (5 - 2 e^{2}\right ) + 2 \log {\relax (5 )} \log {\relax (x )} - \frac {2 e^{2}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-4*x+2)*ln(5)+(-2*x**3+4*x**2)*exp(1)**2+5*x**3-10*x**2+5*x)/(x**3-2*x**2+x),x)

[Out]

x*(5 - 2*exp(2)) + 2*log(5)*log(x) - 2*exp(2)/(x - 1)

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