3.69.57 $\int \frac{e^{{\log }^2(\frac{900x^4+60x^2\log (x)+{\log }^2(x)}{100x^4})}(-30e{x}^3-ex\log (x)+(e(16-4x)+e(-32+8x)\log (x))\log (\frac{900x^4+60x^2\log (x)+{\log }^2(x)}{100x^4}))}{30x^3+x\log (x)}dx$

Optimal. Leaf size=30 $$e^{1+{\log }^2\left(\frac{1}{4}{(6+\frac{\log (x)}{5x^2})}^2\right)}(4-x)$$

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Rubi [B]  time = 0.12, antiderivative size = 129, normalized size of antiderivative = 4.30, number of steps used = 1, number of rules used = 1, integrand size = 98, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.010, Rules used = {2288} $$\begin{array}{c}\frac{e^{{\log }^2\left(\frac{900x^4+60x^2\log (x)+{\log }^2(x)}{100x^4}\right)}(e(4-x)-2e(4-x)\log (x))(900x^4+60x^2\log (x)+{\log }^2(x))}{x^4(30x^3+x\log (x))(\frac{1800x^3+30x+60x\log (x)+\frac{\log (x)}{x}}{x^4}-\frac{2(900x^4+60x^2\log (x)+{\log }^2(x))}{x^5})}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]^2*(-30*E*x^3 - E*x*Log[x] + (E*(16 - 4*x) + E*(-32
+ 8*x)*Log[x])*Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]))/(30*x^3 + x*Log[x]),x]

[Out]

(E^Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]^2*(E*(4 - x) - 2*E*(4 - x)*Log[x])*(900*x^4 + 60*x^2*Lo
g[x] + Log[x]^2))/(x^4*(30*x^3 + x*Log[x])*((30*x + 1800*x^3 + Log[x]/x + 60*x*Log[x])/x^4 - (2*(900*x^4 + 60*
x^2*Log[x] + Log[x]^2))/x^5))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{e^{{\log }^2\left(\frac{900x^4+60x^2\log (x)+{\log }^2(x)}{100x^4}\right)}(e(4-x)-2e(4-x)\log (x))(900x^4+60x^2\log (x)+{\log }^2(x))}{x^4(30x^3+x\log (x))(\frac{30x+1800x^3+\frac{\log (x)}{x}+60x\log (x)}{x^4}-\frac{2(900x^4+60x^2\log (x)+{\log }^2(x))}{x^5})}\end{array}\end{array}$$

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Mathematica [A]  time = 0.10, size = 29, normalized size = 0.97 $$\begin{array}{c}-e^{1+{\log }^2\left(\frac{{(30x^2+\log (x))}^2}{100x^4}\right)}(-4+x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]^2*(-30*E*x^3 - E*x*Log[x] + (E*(16 - 4*x) + E
*(-32 + 8*x)*Log[x])*Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]))/(30*x^3 + x*Log[x]),x]

[Out]

-(E^(1 + Log[(30*x^2 + Log[x])^2/(100*x^4)]^2)*(-4 + x))

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fricas [A]  time = 0.53, size = 33, normalized size = 1.10 $$\begin{array}{c}-(x-4)e^{(\log {\left(\frac{900x^4+60x^2\log (x)+\log (x{)}^2}{100x^4}\right)}^2+1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-32)*exp(1)*log(x)+(-4*x+16)*exp(1))*log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)-x*exp(1)*
log(x)-30*x^3*exp(1))*exp(log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)^2)/(x*log(x)+30*x^3),x, algorithm="f
ricas")

[Out]

-(x - 4)*e^(log(1/100*(900*x^4 + 60*x^2*log(x) + log(x)^2)/x^4)^2 + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int -\frac{(30x^{3}e+xe\log (x)-4(2(x-4)e\log (x)-(x-4)e)\log \left(\frac{900x^4+60x^2\log (x)+\log (x{)}^2}{100x^4}\right))e^{(\log {\left(\frac{900x^4+60x^2\log (x)+\log (x{)}^2}{100x^4}\right)}^2)}}{30x^3+x\log (x)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-32)*exp(1)*log(x)+(-4*x+16)*exp(1))*log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)-x*exp(1)*
log(x)-30*x^3*exp(1))*exp(log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)^2)/(x*log(x)+30*x^3),x, algorithm="g
iac")

[Out]

integrate(-(30*x^3*e + x*e*log(x) - 4*(2*(x - 4)*e*log(x) - (x - 4)*e)*log(1/100*(900*x^4 + 60*x^2*log(x) + lo
g(x)^2)/x^4))*e^(log(1/100*(900*x^4 + 60*x^2*log(x) + log(x)^2)/x^4)^2)/(30*x^3 + x*log(x)), x)

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maple [C]  time = 0.86, size = 7474, normalized size = 249.13

method	result	size
risch	$\text{Expression too large to display}$	$7474$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x-32)*exp(1)*ln(x)+(-4*x+16)*exp(1))*ln(1/100*(ln(x)^2+60*x^2*ln(x)+900*x^4)/x^4)-x*exp(1)*ln(x)-30*x
^3*exp(1))*exp(ln(1/100*(ln(x)^2+60*x^2*ln(x)+900*x^4)/x^4)^2)/(x*ln(x)+30*x^3),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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maxima [B]  time = 0.67, size = 123, normalized size = 4.10 $$\begin{array}{c}-(2^{8\log (5)}x{e}^{(4\log (5{)}^2+4\log (2{)}^2+1)}-2^{8\log (5)+2}{e}^{(4\log (5{)}^2+4\log (2{)}^2+1)})e^{(-8\log (5)\log (30x^2+\log (x))-8\log (2)\log (30x^2+\log (x))+4\log {(30x^2+\log (x))}^2+16\log (5)\log (x)+16\log (2)\log (x)-16\log (30x^2+\log (x))\log (x)+16\log (x{)}^2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-32)*exp(1)*log(x)+(-4*x+16)*exp(1))*log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)-x*exp(1)*
log(x)-30*x^3*exp(1))*exp(log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)^2)/(x*log(x)+30*x^3),x, algorithm="m
axima")

[Out]

-(2^(8*log(5))*x*e^(4*log(5)^2 + 4*log(2)^2 + 1) - 2^(8*log(5) + 2)*e^(4*log(5)^2 + 4*log(2)^2 + 1))*e^(-8*log
(5)*log(30*x^2 + log(x)) - 8*log(2)*log(30*x^2 + log(x)) + 4*log(30*x^2 + log(x))^2 + 16*log(5)*log(x) + 16*lo
g(2)*log(x) - 16*log(30*x^2 + log(x))*log(x) + 16*log(x)^2)

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mupad [B]  time = 4.55, size = 103, normalized size = 3.43 $$\begin{array}{c}\frac{{\mathrm{e}}^{{\ln \left(\frac{1}{x^4}\right)}^2+{\ln (900x^4+60x^2\ln (x)+{\ln (x)}^2)}^2+4{\ln \left(10\right)}^2}(4\mathrm{e}-x\mathrm{e}){\left(\frac{1}{x^4}\right)}^{2\ln (900x^4+60x^2\ln (x)+{\ln (x)}^2)}}{{\left(\frac{1}{x^4}\right)}^{4\ln \left(10\right)}{(900x^4+60x^2\ln (x)+{\ln (x)}^2)}^{4\ln \left(10\right)}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(((3*x^2*log(x))/5 + log(x)^2/100 + 9*x^4)/x^4)^2)*(log(((3*x^2*log(x))/5 + log(x)^2/100 + 9*x^4)
/x^4)*(exp(1)*(4*x - 16) - exp(1)*log(x)*(8*x - 32)) + 30*x^3*exp(1) + x*exp(1)*log(x)))/(x*log(x) + 30*x^3),x
)

[Out]

(exp(log(1/x^4)^2 + log(60*x^2*log(x) + log(x)^2 + 900*x^4)^2 + 4*log(10)^2)*(4*exp(1) - x*exp(1))*(1/x^4)^(2*
log(60*x^2*log(x) + log(x)^2 + 900*x^4)))/((1/x^4)^(4*log(10))*(60*x^2*log(x) + log(x)^2 + 900*x^4)^(4*log(10)
))

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sympy [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((((8*x-32)*exp(1)*ln(x)+(-4*x+16)*exp(1))*ln(1/100*(ln(x)**2+60*x**2*ln(x)+900*x**4)/x**4)-x*exp(1)*
ln(x)-30*x**3*exp(1))*exp(ln(1/100*(ln(x)**2+60*x**2*ln(x)+900*x**4)/x**4)**2)/(x*ln(x)+30*x**3),x)

[Out]

Timed out

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