3.54.25 \(\int \frac {-8+406 x+78 x^2-6 x^3+(8 x+2 x^2) \log (e^{x^2} x)+(2-102 x+6 x^2-2 x \log (e^{x^2} x)) \log (-50+x-\log (e^{x^2} x))}{50 x-x^2+x \log (e^{x^2} x)} \, dx\)

Optimal. Leaf size=21 \[ \left (-4-x+\log \left (-50+x-\log \left (e^{x^2} x\right )\right )\right )^2 \]

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Rubi [A]  time = 0.21, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 91, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6688, 12, 6686} \begin {gather*} \left (-\log \left (-\log \left (e^{x^2} x\right )+x-50\right )+x+4\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8 + 406*x + 78*x^2 - 6*x^3 + (8*x + 2*x^2)*Log[E^x^2*x] + (2 - 102*x + 6*x^2 - 2*x*Log[E^x^2*x])*Log[-50
+ x - Log[E^x^2*x]])/(50*x - x^2 + x*Log[E^x^2*x]),x]

[Out]

(4 + x - Log[-50 + x - Log[E^x^2*x]])^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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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 \frac {2 \left (1-51 x+3 x^2-x \log \left (e^{x^2} x\right )\right ) \left (-4-x+\log \left (-50+x-\log \left (e^{x^2} x\right )\right )\right )}{x \left (50-x+\log \left (e^{x^2} x\right )\right )} \, dx\\ &=2 \int \frac {\left (1-51 x+3 x^2-x \log \left (e^{x^2} x\right )\right ) \left (-4-x+\log \left (-50+x-\log \left (e^{x^2} x\right )\right )\right )}{x \left (50-x+\log \left (e^{x^2} x\right )\right )} \, dx\\ &=\left (4+x-\log \left (-50+x-\log \left (e^{x^2} x\right )\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 21, normalized size = 1.00 \begin {gather*} \left (4+x-\log \left (-50+x-\log \left (e^{x^2} x\right )\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 406*x + 78*x^2 - 6*x^3 + (8*x + 2*x^2)*Log[E^x^2*x] + (2 - 102*x + 6*x^2 - 2*x*Log[E^x^2*x])*L
og[-50 + x - Log[E^x^2*x]])/(50*x - x^2 + x*Log[E^x^2*x]),x]

[Out]

(4 + x - Log[-50 + x - Log[E^x^2*x]])^2

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fricas [A]  time = 0.59, size = 40, normalized size = 1.90 \begin {gather*} x^{2} - 2 \, {\left (x + 4\right )} \log \left (x - \log \left (x e^{\left (x^{2}\right )}\right ) - 50\right ) + \log \left (x - \log \left (x e^{\left (x^{2}\right )}\right ) - 50\right )^{2} + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(exp(x^2)*x)+6*x^2-102*x+2)*log(-log(exp(x^2)*x)+x-50)+(2*x^2+8*x)*log(exp(x^2)*x)-6*x^3+7
8*x^2+406*x-8)/(x*log(exp(x^2)*x)-x^2+50*x),x, algorithm="fricas")

[Out]

x^2 - 2*(x + 4)*log(x - log(x*e^(x^2)) - 50) + log(x - log(x*e^(x^2)) - 50)^2 + 8*x

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giac [B]  time = 0.15, size = 53, normalized size = 2.52 \begin {gather*} x^{2} - 2 \, x \log \left (-x^{2} + x - \log \relax (x) - 50\right ) + \log \left (-x^{2} + x - \log \relax (x) - 50\right )^{2} + 8 \, x - 8 \, \log \left (-x^{2} + x - \log \relax (x) - 50\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(exp(x^2)*x)+6*x^2-102*x+2)*log(-log(exp(x^2)*x)+x-50)+(2*x^2+8*x)*log(exp(x^2)*x)-6*x^3+7
8*x^2+406*x-8)/(x*log(exp(x^2)*x)-x^2+50*x),x, algorithm="giac")

[Out]

x^2 - 2*x*log(-x^2 + x - log(x) - 50) + log(-x^2 + x - log(x) - 50)^2 + 8*x - 8*log(-x^2 + x - log(x) - 50)

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maple [C]  time = 0.17, size = 245, normalized size = 11.67

method	result	size
risch	\(\ln \left (-\ln \relax (x )-\ln \left ({\mathrm e}^{x^{2}}\right )+\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )+\mathrm {csgn}\left (i {\mathrm e}^{x^{2}}\right )\right )}{2}+x -50\right )^{2}-2 x \ln \left (-\ln \relax (x )-\ln \left ({\mathrm e}^{x^{2}}\right )+\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )+\mathrm {csgn}\left (i {\mathrm e}^{x^{2}}\right )\right )}{2}+x -50\right )+x^{2}+8 x -8 \ln \left (\ln \left ({\mathrm e}^{x^{2}}\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{x^{2}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{x^{2}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )^{2}+\pi \mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2}}\right )^{3}-2 i x +2 i \ln \relax (x )+100 i\right )}{2}\right )\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*ln(exp(x^2)*x)+6*x^2-102*x+2)*ln(-ln(exp(x^2)*x)+x-50)+(2*x^2+8*x)*ln(exp(x^2)*x)-6*x^3+78*x^2+406*
x-8)/(x*ln(exp(x^2)*x)-x^2+50*x),x,method=_RETURNVERBOSE)

[Out]

ln(-ln(x)-ln(exp(x^2))+1/2*I*Pi*csgn(I*x*exp(x^2))*(-csgn(I*x*exp(x^2))+csgn(I*x))*(-csgn(I*x*exp(x^2))+csgn(I
*exp(x^2)))+x-50)^2-2*x*ln(-ln(x)-ln(exp(x^2))+1/2*I*Pi*csgn(I*x*exp(x^2))*(-csgn(I*x*exp(x^2))+csgn(I*x))*(-c
sgn(I*x*exp(x^2))+csgn(I*exp(x^2)))+x-50)+x^2+8*x-8*ln(ln(exp(x^2))-1/2*I*(Pi*csgn(I*x)*csgn(I*exp(x^2))*csgn(
I*x*exp(x^2))-Pi*csgn(I*x)*csgn(I*x*exp(x^2))^2-Pi*csgn(I*exp(x^2))*csgn(I*x*exp(x^2))^2+Pi*csgn(I*x*exp(x^2))
^3-2*I*x+2*I*ln(x)+100*I))

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maxima [A]  time = 0.41, size = 40, normalized size = 1.90 \begin {gather*} x^{2} - 2 \, {\left (x + 4\right )} \log \left (-x^{2} + x - \log \relax (x) - 50\right ) + \log \left (-x^{2} + x - \log \relax (x) - 50\right )^{2} + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(exp(x^2)*x)+6*x^2-102*x+2)*log(-log(exp(x^2)*x)+x-50)+(2*x^2+8*x)*log(exp(x^2)*x)-6*x^3+7
8*x^2+406*x-8)/(x*log(exp(x^2)*x)-x^2+50*x),x, algorithm="maxima")

[Out]

x^2 - 2*(x + 4)*log(-x^2 + x - log(x) - 50) + log(-x^2 + x - log(x) - 50)^2 + 8*x

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mupad [B]  time = 3.71, size = 51, normalized size = 2.43 \begin {gather*} 8\,x-8\,\ln \left (\ln \relax (x)-x+x^2+50\right )+{\ln \left (x-\ln \relax (x)-x^2-50\right )}^2-2\,x\,\ln \left (x-\ln \relax (x)-x^2-50\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((406*x - log(x - log(x*exp(x^2)) - 50)*(102*x + 2*x*log(x*exp(x^2)) - 6*x^2 - 2) + log(x*exp(x^2))*(8*x +
2*x^2) + 78*x^2 - 6*x^3 - 8)/(50*x + x*log(x*exp(x^2)) - x^2),x)

[Out]

8*x - 8*log(log(x) - x + x^2 + 50) + log(x - log(x) - x^2 - 50)^2 - 2*x*log(x - log(x) - x^2 - 50) + x^2

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sympy [B]  time = 0.74, size = 53, normalized size = 2.52 \begin {gather*} x^{2} - 2 x \log {\left (x - \log {\left (x e^{x^{2}} \right )} - 50 \right )} + 8 x - 8 \log {\left (- x + \log {\left (x e^{x^{2}} \right )} + 50 \right )} + \log {\left (x - \log {\left (x e^{x^{2}} \right )} - 50 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*ln(exp(x**2)*x)+6*x**2-102*x+2)*ln(-ln(exp(x**2)*x)+x-50)+(2*x**2+8*x)*ln(exp(x**2)*x)-6*x**3
+78*x**2+406*x-8)/(x*ln(exp(x**2)*x)-x**2+50*x),x)

[Out]

x**2 - 2*x*log(x - log(x*exp(x**2)) - 50) + 8*x - 8*log(-x + log(x*exp(x**2)) + 50) + log(x - log(x*exp(x**2))
- 50)**2

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