3.54.63 \(\int \frac {-4 x^2+(5-e^5+3 x) \log (15)}{(10 x^2-2 e^5 x^2+2 x^3+(-5 x+e^5 x-x^2) \log (15)) \log (\frac {-2 x+\log (15)}{e^{15} x+e^{10} (-10 x-2 x^2)+e^5 (25 x+10 x^2+x^3)})} \, dx\)

Optimal. Leaf size=24 \[ \log \left (\log \left (\frac {-2+\frac {\log (15)}{x}}{e^5 \left (-5+e^5-x\right )^2}\right )\right ) \]

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Rubi [A]  time = 0.42, antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, integrand size = 105, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6, 6688, 6684} \begin {gather*} \log \left (\log \left (-\frac {2 x-\log (15)}{e^5 x \left (x-e^5+5\right )^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*x^2 + (5 - E^5 + 3*x)*Log[15])/((10*x^2 - 2*E^5*x^2 + 2*x^3 + (-5*x + E^5*x - x^2)*Log[15])*Log[(-2*x
+ Log[15])/(E^15*x + E^10*(-10*x - 2*x^2) + E^5*(25*x + 10*x^2 + x^3))]),x]

[Out]

Log[Log[-((2*x - Log[15])/(E^5*x*(5 - E^5 + x)^2))]]

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 6684

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

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 {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (\left (10-2 e^5\right ) x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx\\ &=\int \frac {-4 x^2+\left (5-e^5\right ) \log (15)+3 x \log (15)}{x \left (5-e^5+x\right ) (2 x-\log (15)) \log \left (\frac {-2 x+\log (15)}{e^5 x \left (5-e^5+x\right )^2}\right )} \, dx\\ &=\log \left (\log \left (-\frac {2 x-\log (15)}{e^5 x \left (5-e^5+x\right )^2}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 25, normalized size = 1.04 \begin {gather*} \log \left (\log \left (\frac {-2 x+\log (15)}{e^5 x \left (5-e^5+x\right )^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x^2 + (5 - E^5 + 3*x)*Log[15])/((10*x^2 - 2*E^5*x^2 + 2*x^3 + (-5*x + E^5*x - x^2)*Log[15])*Log[
(-2*x + Log[15])/(E^15*x + E^10*(-10*x - 2*x^2) + E^5*(25*x + 10*x^2 + x^3))]),x]

[Out]

Log[Log[(-2*x + Log[15])/(E^5*x*(5 - E^5 + x)^2)]]

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fricas [B]  time = 0.59, size = 45, normalized size = 1.88 \begin {gather*} \log \left (\log \left (-\frac {2 \, x - \log \left (15\right )}{x e^{15} - 2 \, {\left (x^{2} + 5 \, x\right )} e^{10} + {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{5}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^2*exp(5)+2*x^3+10*x^2)/log((log(15)-
2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3+10*x^2+25*x)*exp(5))),x, algorithm="fricas")

[Out]

log(log(-(2*x - log(15))/(x*e^15 - 2*(x^2 + 5*x)*e^10 + (x^3 + 10*x^2 + 25*x)*e^5)))

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giac [B]  time = 0.31, size = 47, normalized size = 1.96 \begin {gather*} \log \left (-\log \left (x^{3} e^{5} - 2 \, x^{2} e^{10} + 10 \, x^{2} e^{5} + x e^{15} - 10 \, x e^{10} + 25 \, x e^{5}\right ) + \log \left (-2 \, x + \log \left (15\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^2*exp(5)+2*x^3+10*x^2)/log((log(15)-
2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3+10*x^2+25*x)*exp(5))),x, algorithm="giac")

[Out]

log(-log(x^3*e^5 - 2*x^2*e^10 + 10*x^2*e^5 + x*e^15 - 10*x*e^10 + 25*x*e^5) + log(-2*x + log(15)))

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maple [A]  time = 0.37, size = 46, normalized size = 1.92

method	result	size
risch	\(\ln \left (\ln \left (\frac {\ln \relax (3)+\ln \relax (5)-2 x}{x \,{\mathrm e}^{15}+\left (-2 x^{2}-10 x \right ) {\mathrm e}^{10}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{5}}\right )\right )\)	\(46\)
norman	\(\ln \left (\ln \left (\frac {\ln \left (15\right )-2 x}{x \,{\mathrm e}^{15}+\left (-2 x^{2}-10 x \right ) {\mathrm e}^{10}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{5}}\right )\right )\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(5)+3*x+5)*ln(15)-4*x^2)/((x*exp(5)-x^2-5*x)*ln(15)-2*x^2*exp(5)+2*x^3+10*x^2)/ln((ln(15)-2*x)/(x*ex
p(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3+10*x^2+25*x)*exp(5))),x,method=_RETURNVERBOSE)

[Out]

ln(ln((ln(3)+ln(5)-2*x)/(x*exp(15)+(-2*x^2-10*x)*exp(10)+(x^3+10*x^2+25*x)*exp(5))))

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maxima [A]  time = 0.58, size = 26, normalized size = 1.08 \begin {gather*} \log \left (-2 \, \log \left (x - e^{5} + 5\right ) - \log \relax (x) + \log \left (-2 \, x + \log \relax (5) + \log \relax (3)\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^2*exp(5)+2*x^3+10*x^2)/log((log(15)-
2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3+10*x^2+25*x)*exp(5))),x, algorithm="maxima")

[Out]

log(-2*log(x - e^5 + 5) - log(x) + log(-2*x + log(5) + log(3)) - 5)

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mupad [B]  time = 6.19, size = 47, normalized size = 1.96 \begin {gather*} \ln \left (\ln \left (-\frac {2\,x-\ln \left (15\right )}{{\mathrm {e}}^5\,\left (x^3+10\,x^2+25\,x\right )-{\mathrm {e}}^{10}\,\left (2\,x^2+10\,x\right )+x\,{\mathrm {e}}^{15}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(15)*(3*x - exp(5) + 5) - 4*x^2)/(log(-(2*x - log(15))/(exp(5)*(25*x + 10*x^2 + x^3) - exp(10)*(10*x
+ 2*x^2) + x*exp(15)))*(log(15)*(5*x - x*exp(5) + x^2) + 2*x^2*exp(5) - 10*x^2 - 2*x^3)),x)

[Out]

log(log(-(2*x - log(15))/(exp(5)*(25*x + 10*x^2 + x^3) - exp(10)*(10*x + 2*x^2) + x*exp(15))))

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sympy [B]  time = 0.66, size = 42, normalized size = 1.75 \begin {gather*} \log {\left (\log {\left (\frac {- 2 x + \log {\left (15 \right )}}{x e^{15} + \left (- 2 x^{2} - 10 x\right ) e^{10} + \left (x^{3} + 10 x^{2} + 25 x\right ) e^{5}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(5)+3*x+5)*ln(15)-4*x**2)/((x*exp(5)-x**2-5*x)*ln(15)-2*x**2*exp(5)+2*x**3+10*x**2)/ln((ln(15)
-2*x)/(x*exp(5)**3+(-2*x**2-10*x)*exp(5)**2+(x**3+10*x**2+25*x)*exp(5))),x)

[Out]

log(log((-2*x + log(15))/(x*exp(15) + (-2*x**2 - 10*x)*exp(10) + (x**3 + 10*x**2 + 25*x)*exp(5))))

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