3.73.56 $\int \frac{-2-25x+e^{x^2}(-27+4x^2)+27\log (x)+(-e^{x^2}-x+\log (x))\log (\frac{e^{2x^2}+2e^{x^2}x+x^2+(-2e^{x^2}-2x)\log (x)+{\log }^2(x)}{x^2})}{-e^{x^2}{x}^2-x^3+x^2\log (x)}dx$

Optimal. Leaf size=27 $$5-\frac{25+\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x}$$

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Rubi [A]  time = 1.97, antiderivative size = 29, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 5, integrand size = 103, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.049, Rules used = {6742, 14, 30, 2555, 12} $$\begin{array}{c}-\frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x}-\frac{25}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-2 - 25*x + E^x^2*(-27 + 4*x^2) + 27*Log[x] + (-E^x^2 - x + Log[x])*Log[(E^(2*x^2) + 2*E^x^2*x + x^2 + (-
2*E^x^2 - 2*x)*Log[x] + Log[x]^2)/x^2])/(-(E^x^2*x^2) - x^3 + x^2*Log[x]),x]

[Out]

-25/x - Log[(E^x^2 + x - Log[x])^2/x^2]/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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{2(1-x+2x^3-2x^2\log (x))}{x^2(e^{x^2}+x-\log (x))}+\frac{27-4x^2+\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x^2})dx\\ & =2\int \frac{1-x+2x^3-2x^2\log (x)}{x^2(e^{x^2}+x-\log (x))}dx+\int \frac{27-4x^2+\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x^2}dx\\ & =2\int (\frac{1}{x^2(e^{x^2}+x-\log (x))}-\frac{1}{x(e^{x^2}+x-\log (x))}+\frac{2x}{e^{x^2}+x-\log (x)}-\frac{2\log (x)}{e^{x^2}+x-\log (x)})dx+\int (\frac{27-4x^2}{x^2}+\frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x^2})dx\\ & =2\int \frac{1}{x^2(e^{x^2}+x-\log (x))}dx-2\int \frac{1}{x(e^{x^2}+x-\log (x))}dx+4\int \frac{x}{e^{x^2}+x-\log (x)}dx-4\int \frac{\log (x)}{e^{x^2}+x-\log (x)}dx+\int \frac{27-4x^2}{x^2}dx+\int \frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x^2}dx\\ & =-\frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x}+2\int \frac{1}{x^2(e^{x^2}+x-\log (x))}dx-2\int \frac{1}{x(e^{x^2}+x-\log (x))}dx+4\int \frac{x}{e^{x^2}+x-\log (x)}dx-4\int \frac{\log (x)}{e^{x^2}+x-\log (x)}dx+\int (-4+\frac{27}{x^2})dx-\int -\frac{2(-1+e^{x^2}(-1+2x^2)+\log (x))}{x^2(e^{x^2}+x-\log (x))}dx\\ & =-\frac{27}{x}-4x-\frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x}+2\int \frac{1}{x^2(e^{x^2}+x-\log (x))}dx-2\int \frac{1}{x(e^{x^2}+x-\log (x))}dx+2\int \frac{-1+e^{x^2}(-1+2x^2)+\log (x)}{x^2(e^{x^2}+x-\log (x))}dx+4\int \frac{x}{e^{x^2}+x-\log (x)}dx-4\int \frac{\log (x)}{e^{x^2}+x-\log (x)}dx\\ & =-\frac{27}{x}-4x-\frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x}+2\int \frac{1}{x^2(e^{x^2}+x-\log (x))}dx-2\int \frac{1}{x(e^{x^2}+x-\log (x))}dx+2\int (\frac{-1+2x^2}{x^2}-\frac{1-x+2x^3-2x^2\log (x)}{x^2(e^{x^2}+x-\log (x))})dx+4\int \frac{x}{e^{x^2}+x-\log (x)}dx-4\int \frac{\log (x)}{e^{x^2}+x-\log (x)}dx\\ & =-\frac{27}{x}-4x-\frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x}+2\int \frac{-1+2x^2}{x^2}dx+2\int \frac{1}{x^2(e^{x^2}+x-\log (x))}dx-2\int \frac{1}{x(e^{x^2}+x-\log (x))}dx-2\int \frac{1-x+2x^3-2x^2\log (x)}{x^2(e^{x^2}+x-\log (x))}dx+4\int \frac{x}{e^{x^2}+x-\log (x)}dx-4\int \frac{\log (x)}{e^{x^2}+x-\log (x)}dx\\ & =-\frac{27}{x}-4x-\frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x}+2\int (2-\frac{1}{x^2})dx+2\int \frac{1}{x^2(e^{x^2}+x-\log (x))}dx-2\int \frac{1}{x(e^{x^2}+x-\log (x))}dx-2\int (\frac{1}{x^2(e^{x^2}+x-\log (x))}-\frac{1}{x(e^{x^2}+x-\log (x))}+\frac{2x}{e^{x^2}+x-\log (x)}-\frac{2\log (x)}{e^{x^2}+x-\log (x)})dx+4\int \frac{x}{e^{x^2}+x-\log (x)}dx-4\int \frac{\log (x)}{e^{x^2}+x-\log (x)}dx\\ & =-\frac{25}{x}-\frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.11, size = 29, normalized size = 1.07 $$\begin{array}{c}-\frac{25}{x}-\frac{\log \left(\frac{{(e^{x^2}+x-\log (x))}^2}{x^2}\right)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 25*x + E^x^2*(-27 + 4*x^2) + 27*Log[x] + (-E^x^2 - x + Log[x])*Log[(E^(2*x^2) + 2*E^x^2*x + x^
2 + (-2*E^x^2 - 2*x)*Log[x] + Log[x]^2)/x^2])/(-(E^x^2*x^2) - x^3 + x^2*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.54, size = 43, normalized size = 1.59 $$\begin{array}{c}-\frac{\log \left(\frac{x^2+2x{e}^{\left(x^2\right)}-2(x+e^{\left(x^2\right)})\log (x)+\log (x{)}^2+e^{\left(2x^2\right)}}{x^2}\right)+25}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)-exp(x^2)-x)*log((log(x)^2+(-2*exp(x^2)-2*x)*log(x)+exp(x^2)^2+2*exp(x^2)*x+x^2)/x^2)+27*log
(x)+(4*x^2-27)*exp(x^2)-25*x-2)/(x^2*log(x)-x^2*exp(x^2)-x^3),x, algorithm="fricas")

[Out]

-(log((x^2 + 2*x*e^(x^2) - 2*(x + e^(x^2))*log(x) + log(x)^2 + e^(2*x^2))/x^2) + 25)/x

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Timed out}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)-exp(x^2)-x)*log((log(x)^2+(-2*exp(x^2)-2*x)*log(x)+exp(x^2)^2+2*exp(x^2)*x+x^2)/x^2)+27*log
(x)+(4*x^2-27)*exp(x^2)-25*x-2)/(x^2*log(x)-x^2*exp(x^2)-x^3),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.16, size = 338, normalized size = 12.52

method	result	size
risch	$-\frac{2\ln ({\mathrm{e}}^{x^2}-\ln (x)+x)}{x}+\frac{-i\pi \mathrm{csgn}{\left(ix\right)}^2\mathrm{csgn}\left(i{x}^2\right)+2i\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(i{x}^2\right)}^2-i\pi \mathrm{csgn}{\left(i{x}^2\right)}^3+i\pi \mathrm{csgn}\left(\frac{i}{x^2}\right)\mathrm{csgn}\left(i{(\ln (x)-{\mathrm{e}}^{x^2}-x)}^2\right)\mathrm{csgn}\left(\frac{i{(\ln (x)-{\mathrm{e}}^{x^2}-x)}^2}{x^2}\right)-i\pi \mathrm{csgn}\left(\frac{i}{x^2}\right)\mathrm{csgn}{\left(\frac{i{(\ln (x)-{\mathrm{e}}^{x^2}-x)}^2}{x^2}\right)}^2+i\pi \mathrm{csgn}{\left(i(\ln (x)-{\mathrm{e}}^{x^2}-x)\right)}^2\mathrm{csgn}\left(i{(\ln (x)-{\mathrm{e}}^{x^2}-x)}^2\right)+2i\pi \mathrm{csgn}\left(i(\ln (x)-{\mathrm{e}}^{x^2}-x)\right)\mathrm{csgn}{\left(i{(\ln (x)-{\mathrm{e}}^{x^2}-x)}^2\right)}^2+i\pi \mathrm{csgn}{\left(i{(\ln (x)-{\mathrm{e}}^{x^2}-x)}^2\right)}^3-i\pi \mathrm{csgn}\left(i{(\ln (x)-{\mathrm{e}}^{x^2}-x)}^2\right)\mathrm{csgn}{\left(\frac{i{(\ln (x)-{\mathrm{e}}^{x^2}-x)}^2}{x^2}\right)}^2+i\pi \mathrm{csgn}{\left(\frac{i{(\ln (x)-{\mathrm{e}}^{x^2}-x)}^2}{x^2}\right)}^3+4\ln (x)-50}{2x}$	$338$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x)-exp(x^2)-x)*ln((ln(x)^2+(-2*exp(x^2)-2*x)*ln(x)+exp(x^2)^2+2*exp(x^2)*x+x^2)/x^2)+27*ln(x)+(4*x^2-
27)*exp(x^2)-25*x-2)/(x^2*ln(x)-x^2*exp(x^2)-x^3),x,method=_RETURNVERBOSE)

[Out]

-2/x*ln(exp(x^2)-ln(x)+x)+1/2*(-I*Pi*csgn(I*x)^2*csgn(I*x^2)+2*I*Pi*csgn(I*x)*csgn(I*x^2)^2-I*Pi*csgn(I*x^2)^3
+I*Pi*csgn(I/x^2)*csgn(I*(ln(x)-exp(x^2)-x)^2)*csgn(I*(ln(x)-exp(x^2)-x)^2/x^2)-I*Pi*csgn(I/x^2)*csgn(I*(ln(x)
-exp(x^2)-x)^2/x^2)^2+I*Pi*csgn(I*(ln(x)-exp(x^2)-x))^2*csgn(I*(ln(x)-exp(x^2)-x)^2)+2*I*Pi*csgn(I*(ln(x)-exp(
x^2)-x))*csgn(I*(ln(x)-exp(x^2)-x)^2)^2+I*Pi*csgn(I*(ln(x)-exp(x^2)-x)^2)^3-I*Pi*csgn(I*(ln(x)-exp(x^2)-x)^2)*
csgn(I*(ln(x)-exp(x^2)-x)^2/x^2)^2+I*Pi*csgn(I*(ln(x)-exp(x^2)-x)^2/x^2)^3+4*ln(x)-50)/x

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maxima [A]  time = 0.39, size = 25, normalized size = 0.93 $$\begin{array}{c}\frac{2\log (x)-2\log (-x-e^{\left(x^2\right)}+\log (x))-25}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)-exp(x^2)-x)*log((log(x)^2+(-2*exp(x^2)-2*x)*log(x)+exp(x^2)^2+2*exp(x^2)*x+x^2)/x^2)+27*log
(x)+(4*x^2-27)*exp(x^2)-25*x-2)/(x^2*log(x)-x^2*exp(x^2)-x^3),x, algorithm="maxima")

[Out]

(2*log(x) - 2*log(-x - e^(x^2) + log(x)) - 25)/x

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mupad [B]  time = 4.57, size = 47, normalized size = 1.74 $$\begin{array}{c}-\frac{\ln \left(\frac{{\mathrm{e}}^{2x^2}+2x{\mathrm{e}}^{x^2}-\ln (x)(2x+2{\mathrm{e}}^{x^2})+{\ln (x)}^2+x^2}{x^2}\right)+25}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x - 27*log(x) - exp(x^2)*(4*x^2 - 27) + log((exp(2*x^2) + 2*x*exp(x^2) - log(x)*(2*x + 2*exp(x^2)) + l
og(x)^2 + x^2)/x^2)*(x + exp(x^2) - log(x)) + 2)/(x^2*exp(x^2) - x^2*log(x) + x^3),x)

[Out]

-(log((exp(2*x^2) + 2*x*exp(x^2) - log(x)*(2*x + 2*exp(x^2)) + log(x)^2 + x^2)/x^2) + 25)/x

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sympy [B]  time = 1.07, size = 49, normalized size = 1.81 $$\begin{array}{c}-\frac{\log \left(\frac{x^2+2x{e}^{x^2}+(-2x-2e^{x^2})\log (x)+e^{2x^2}+\log {(x)}^2}{x^2}\right)}{x}-\frac{25}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x)-exp(x**2)-x)*ln((ln(x)**2+(-2*exp(x**2)-2*x)*ln(x)+exp(x**2)**2+2*exp(x**2)*x+x**2)/x**2)+27
*ln(x)+(4*x**2-27)*exp(x**2)-25*x-2)/(x**2*ln(x)-x**2*exp(x**2)-x**3),x)

[Out]

-log((x**2 + 2*x*exp(x**2) + (-2*x - 2*exp(x**2))*log(x) + exp(2*x**2) + log(x)**2)/x**2)/x - 25/x

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