3.70.94 \(\int \frac {2-15 x-15 e^4 x+30 x^2+(e^4-2 x) \log (x^2)+(-15 x+\log (x^2)) \log (-15 x+\log (x^2))}{(15 e^4 x^2-15 x^3) \log ^2(5)+(-e^4 x+x^2) \log ^2(5) \log (x^2)+(15 x^2 \log ^2(5)-x \log ^2(5) \log (x^2)) \log (-15 x+\log (x^2))+(-15 e^4 x^2+15 x^3+(e^4 x-x^2) \log (x^2)+(-15 x^2+x \log (x^2)) \log (-15 x+\log (x^2))) \log (e^4 x-x^2+x \log (-15 x+\log (x^2)))} \, dx\)

Optimal. Leaf size=27 \[ \log \left (\log ^2(5)-\log \left (x \left (e^4-x+\log \left (-15 x+\log \left (x^2\right )\right )\right )\right )\right ) \]

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Rubi [A]  time = 0.44, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 195, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6, 6688, 6684} \begin {gather*} \log \left (\log ^2(5)-\log \left (x \left (\log \left (\log \left (x^2\right )-15 x\right )-x+e^4\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 15*x - 15*E^4*x + 30*x^2 + (E^4 - 2*x)*Log[x^2] + (-15*x + Log[x^2])*Log[-15*x + Log[x^2]])/((15*E^4*
x^2 - 15*x^3)*Log[5]^2 + (-(E^4*x) + x^2)*Log[5]^2*Log[x^2] + (15*x^2*Log[5]^2 - x*Log[5]^2*Log[x^2])*Log[-15*
x + Log[x^2]] + (-15*E^4*x^2 + 15*x^3 + (E^4*x - x^2)*Log[x^2] + (-15*x^2 + x*Log[x^2])*Log[-15*x + Log[x^2]])
*Log[E^4*x - x^2 + x*Log[-15*x + Log[x^2]]]),x]

[Out]

Log[Log[5]^2 - Log[x*(E^4 - x + Log[-15*x + Log[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 {2+\left (-15-15 e^4\right ) x+30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (-15 x+\log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )}{\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (-e^4 x+x^2\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )+\left (-15 e^4 x^2+15 x^3+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (-15 x^2+x \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )\right ) \log \left (e^4 x-x^2+x \log \left (-15 x+\log \left (x^2\right )\right )\right )} \, dx\\ &=\int \frac {2-15 \left (1+e^4\right ) x+30 x^2-15 x \log \left (-15 x+\log \left (x^2\right )\right )+\log \left (x^2\right ) \left (e^4-2 x+\log \left (-15 x+\log \left (x^2\right )\right )\right )}{x \left (15 x-\log \left (x^2\right )\right ) \left (e^4-x+\log \left (-15 x+\log \left (x^2\right )\right )\right ) \left (\log ^2(5)-\log \left (x \left (e^4-x+\log \left (-15 x+\log \left (x^2\right )\right )\right )\right )\right )} \, dx\\ &=\log \left (\log ^2(5)-\log \left (x \left (e^4-x+\log \left (-15 x+\log \left (x^2\right )\right )\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 27, normalized size = 1.00 \begin {gather*} \log \left (\log ^2(5)-\log \left (x \left (e^4-x+\log \left (-15 x+\log \left (x^2\right )\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 15*x - 15*E^4*x + 30*x^2 + (E^4 - 2*x)*Log[x^2] + (-15*x + Log[x^2])*Log[-15*x + Log[x^2]])/((1
5*E^4*x^2 - 15*x^3)*Log[5]^2 + (-(E^4*x) + x^2)*Log[5]^2*Log[x^2] + (15*x^2*Log[5]^2 - x*Log[5]^2*Log[x^2])*Lo
g[-15*x + Log[x^2]] + (-15*E^4*x^2 + 15*x^3 + (E^4*x - x^2)*Log[x^2] + (-15*x^2 + x*Log[x^2])*Log[-15*x + Log[
x^2]])*Log[E^4*x - x^2 + x*Log[-15*x + Log[x^2]]]),x]

[Out]

Log[Log[5]^2 - Log[x*(E^4 - x + Log[-15*x + Log[x^2]])]]

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fricas [A]  time = 0.95, size = 30, normalized size = 1.11 \begin {gather*} \log \left (-\log \relax (5)^{2} + \log \left (-x^{2} + x e^{4} + x \log \left (-15 \, x + \log \left (x^{2}\right )\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x^2)-15*x)*log(log(x^2)-15*x)+(exp(4)-2*x)*log(x^2)-15*x*exp(4)+30*x^2-15*x+2)/(((x*log(x^2)-1
5*x^2)*log(log(x^2)-15*x)+(x*exp(4)-x^2)*log(x^2)-15*x^2*exp(4)+15*x^3)*log(x*log(log(x^2)-15*x)+x*exp(4)-x^2)
+(-x*log(5)^2*log(x^2)+15*x^2*log(5)^2)*log(log(x^2)-15*x)+(-x*exp(4)+x^2)*log(5)^2*log(x^2)+(15*x^2*exp(4)-15
*x^3)*log(5)^2),x, algorithm="fricas")

[Out]

log(-log(5)^2 + log(-x^2 + x*e^4 + x*log(-15*x + log(x^2))))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {30 \, x^{2} - 15 \, x e^{4} - {\left (2 \, x - e^{4}\right )} \log \left (x^{2}\right ) - {\left (15 \, x - \log \left (x^{2}\right )\right )} \log \left (-15 \, x + \log \left (x^{2}\right )\right ) - 15 \, x + 2}{{\left (x^{2} - x e^{4}\right )} \log \relax (5)^{2} \log \left (x^{2}\right ) - 15 \, {\left (x^{3} - x^{2} e^{4}\right )} \log \relax (5)^{2} + {\left (15 \, x^{3} - 15 \, x^{2} e^{4} - {\left (x^{2} - x e^{4}\right )} \log \left (x^{2}\right ) - {\left (15 \, x^{2} - x \log \left (x^{2}\right )\right )} \log \left (-15 \, x + \log \left (x^{2}\right )\right )\right )} \log \left (-x^{2} + x e^{4} + x \log \left (-15 \, x + \log \left (x^{2}\right )\right )\right ) + {\left (15 \, x^{2} \log \relax (5)^{2} - x \log \relax (5)^{2} \log \left (x^{2}\right )\right )} \log \left (-15 \, x + \log \left (x^{2}\right )\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x^2)-15*x)*log(log(x^2)-15*x)+(exp(4)-2*x)*log(x^2)-15*x*exp(4)+30*x^2-15*x+2)/(((x*log(x^2)-1
5*x^2)*log(log(x^2)-15*x)+(x*exp(4)-x^2)*log(x^2)-15*x^2*exp(4)+15*x^3)*log(x*log(log(x^2)-15*x)+x*exp(4)-x^2)
+(-x*log(5)^2*log(x^2)+15*x^2*log(5)^2)*log(log(x^2)-15*x)+(-x*exp(4)+x^2)*log(5)^2*log(x^2)+(15*x^2*exp(4)-15
*x^3)*log(5)^2),x, algorithm="giac")

[Out]

integrate((30*x^2 - 15*x*e^4 - (2*x - e^4)*log(x^2) - (15*x - log(x^2))*log(-15*x + log(x^2)) - 15*x + 2)/((x^
2 - x*e^4)*log(5)^2*log(x^2) - 15*(x^3 - x^2*e^4)*log(5)^2 + (15*x^3 - 15*x^2*e^4 - (x^2 - x*e^4)*log(x^2) - (
15*x^2 - x*log(x^2))*log(-15*x + log(x^2)))*log(-x^2 + x*e^4 + x*log(-15*x + log(x^2))) + (15*x^2*log(5)^2 - x
*log(5)^2*log(x^2))*log(-15*x + log(x^2))), x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (\ln \left (x^{2}\right )-15 x \right ) \ln \left (\ln \left (x^{2}\right )-15 x \right )+\left ({\mathrm e}^{4}-2 x \right ) \ln \left (x^{2}\right )-15 x \,{\mathrm e}^{4}+30 x^{2}-15 x +2}{\left (\left (x \ln \left (x^{2}\right )-15 x^{2}\right ) \ln \left (\ln \left (x^{2}\right )-15 x \right )+\left (x \,{\mathrm e}^{4}-x^{2}\right ) \ln \left (x^{2}\right )-15 x^{2} {\mathrm e}^{4}+15 x^{3}\right ) \ln \left (x \ln \left (\ln \left (x^{2}\right )-15 x \right )+x \,{\mathrm e}^{4}-x^{2}\right )+\left (-x \ln \relax (5)^{2} \ln \left (x^{2}\right )+15 x^{2} \ln \relax (5)^{2}\right ) \ln \left (\ln \left (x^{2}\right )-15 x \right )+\left (-x \,{\mathrm e}^{4}+x^{2}\right ) \ln \relax (5)^{2} \ln \left (x^{2}\right )+\left (15 x^{2} {\mathrm e}^{4}-15 x^{3}\right ) \ln \relax (5)^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x^2)-15*x)*ln(ln(x^2)-15*x)+(exp(4)-2*x)*ln(x^2)-15*x*exp(4)+30*x^2-15*x+2)/(((x*ln(x^2)-15*x^2)*ln(l
n(x^2)-15*x)+(x*exp(4)-x^2)*ln(x^2)-15*x^2*exp(4)+15*x^3)*ln(x*ln(ln(x^2)-15*x)+x*exp(4)-x^2)+(-x*ln(5)^2*ln(x
^2)+15*x^2*ln(5)^2)*ln(ln(x^2)-15*x)+(-x*exp(4)+x^2)*ln(5)^2*ln(x^2)+(15*x^2*exp(4)-15*x^3)*ln(5)^2),x)

[Out]

int(((ln(x^2)-15*x)*ln(ln(x^2)-15*x)+(exp(4)-2*x)*ln(x^2)-15*x*exp(4)+30*x^2-15*x+2)/(((x*ln(x^2)-15*x^2)*ln(l
n(x^2)-15*x)+(x*exp(4)-x^2)*ln(x^2)-15*x^2*exp(4)+15*x^3)*ln(x*ln(ln(x^2)-15*x)+x*exp(4)-x^2)+(-x*ln(5)^2*ln(x
^2)+15*x^2*ln(5)^2)*ln(ln(x^2)-15*x)+(-x*exp(4)+x^2)*ln(5)^2*ln(x^2)+(15*x^2*exp(4)-15*x^3)*ln(5)^2),x)

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maxima [A]  time = 0.55, size = 26, normalized size = 0.96 \begin {gather*} \log \left (-\log \relax (5)^{2} + \log \relax (x) + \log \left (-x + e^{4} + \log \left (-15 \, x + 2 \, \log \relax (x)\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x^2)-15*x)*log(log(x^2)-15*x)+(exp(4)-2*x)*log(x^2)-15*x*exp(4)+30*x^2-15*x+2)/(((x*log(x^2)-1
5*x^2)*log(log(x^2)-15*x)+(x*exp(4)-x^2)*log(x^2)-15*x^2*exp(4)+15*x^3)*log(x*log(log(x^2)-15*x)+x*exp(4)-x^2)
+(-x*log(5)^2*log(x^2)+15*x^2*log(5)^2)*log(log(x^2)-15*x)+(-x*exp(4)+x^2)*log(5)^2*log(x^2)+(15*x^2*exp(4)-15
*x^3)*log(5)^2),x, algorithm="maxima")

[Out]

log(-log(5)^2 + log(x) + log(-x + e^4 + log(-15*x + 2*log(x))))

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mupad [B]  time = 4.50, size = 26, normalized size = 0.96 \begin {gather*} \ln \left (\ln \left (x\,\left ({\mathrm {e}}^4-x+\ln \left (\ln \left (x^2\right )-15\,x\right )\right )\right )-{\ln \relax (5)}^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x + log(log(x^2) - 15*x)*(15*x - log(x^2)) + 15*x*exp(4) + log(x^2)*(2*x - exp(4)) - 30*x^2 - 2)/(log
(5)^2*(15*x^2*exp(4) - 15*x^3) + log(log(x^2) - 15*x)*(15*x^2*log(5)^2 - x*log(x^2)*log(5)^2) + log(x*exp(4) +
x*log(log(x^2) - 15*x) - x^2)*(log(x^2)*(x*exp(4) - x^2) - 15*x^2*exp(4) + 15*x^3 + log(log(x^2) - 15*x)*(x*l
og(x^2) - 15*x^2)) - log(x^2)*log(5)^2*(x*exp(4) - x^2)),x)

[Out]

log(log(x*(exp(4) - x + log(log(x^2) - 15*x))) - log(5)^2)

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sympy [A]  time = 41.04, size = 27, normalized size = 1.00 \begin {gather*} \log {\left (\log {\left (- x^{2} + x \log {\left (- 15 x + \log {\left (x^{2} \right )} \right )} + x e^{4} \right )} - \log {\relax (5 )}^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x**2)-15*x)*ln(ln(x**2)-15*x)+(exp(4)-2*x)*ln(x**2)-15*x*exp(4)+30*x**2-15*x+2)/(((x*ln(x**2)-1
5*x**2)*ln(ln(x**2)-15*x)+(x*exp(4)-x**2)*ln(x**2)-15*x**2*exp(4)+15*x**3)*ln(x*ln(ln(x**2)-15*x)+x*exp(4)-x**
2)+(-x*ln(5)**2*ln(x**2)+15*x**2*ln(5)**2)*ln(ln(x**2)-15*x)+(-x*exp(4)+x**2)*ln(5)**2*ln(x**2)+(15*x**2*exp(4
)-15*x**3)*ln(5)**2),x)

[Out]

log(log(-x**2 + x*log(-15*x + log(x**2)) + x*exp(4)) - log(5)**2)

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