3.53.60 \(\int \frac {-15 e^x+6 e^{2 x} x}{50+e^x (45-20 e^2+20 x^2)+e^{2 x} (9+2 e^4+9 x^2+2 x^4+e^2 (-9-4 x^2))+(-20 e^x+e^{2 x} (-9+4 e^2-4 x^2)) \log (\log (4))+2 e^{2 x} \log ^2(\log (4))} \, dx\)

Optimal. Leaf size=29 \[ 4+\log \left (2+\frac {3}{-3+e^2-5 e^{-x}-x^2+\log (\log (4))}\right ) \]

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Rubi [A]  time = 3.35, antiderivative size = 55, normalized size of antiderivative = 1.90, number of steps used = 6, number of rules used = 4, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6741, 12, 6742, 6684} \begin {gather*} \log \left (2 e^x x^2+e^x \left (3-2 \left (e^2+\log (\log (4))\right )\right )+10\right )-\log \left (e^x x^2+e^x \left (3-e^2-\log (\log (4))\right )+5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*E^x + 6*E^(2*x)*x)/(50 + E^x*(45 - 20*E^2 + 20*x^2) + E^(2*x)*(9 + 2*E^4 + 9*x^2 + 2*x^4 + E^2*(-9 -
4*x^2)) + (-20*E^x + E^(2*x)*(-9 + 4*E^2 - 4*x^2))*Log[Log[4]] + 2*E^(2*x)*Log[Log[4]]^2),x]

[Out]

-Log[5 + E^x*x^2 + E^x*(3 - E^2 - Log[Log[4]])] + Log[10 + 2*E^x*x^2 + E^x*(3 - 2*(E^2 + Log[Log[4]]))]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^x \left (-5+2 e^x x\right )}{50+e^x \left (45-20 e^2+20 x^2\right )+e^{2 x} \left (9+2 e^4+9 x^2+2 x^4+e^2 \left (-9-4 x^2\right )\right )+\left (-20 e^x+e^{2 x} \left (-9+4 e^2-4 x^2\right )\right ) \log (\log (4))+2 e^{2 x} \log ^2(\log (4))} \, dx\\ &=3 \int \frac {e^x \left (-5+2 e^x x\right )}{50+e^x \left (45-20 e^2+20 x^2\right )+e^{2 x} \left (9+2 e^4+9 x^2+2 x^4+e^2 \left (-9-4 x^2\right )\right )+\left (-20 e^x+e^{2 x} \left (-9+4 e^2-4 x^2\right )\right ) \log (\log (4))+2 e^{2 x} \log ^2(\log (4))} \, dx\\ &=3 \int \left (\frac {e^x \left (-3+e^2-2 x-x^2+\log (\log (4))\right )}{3 \left (5+e^x x^2+3 e^x \left (1+\frac {1}{3} \left (-e^2-\log (\log (4))\right )\right )\right )}+\frac {e^x \left (3-2 e^2+4 x+2 x^2-2 \log (\log (4))\right )}{3 \left (10+2 e^x x^2+3 e^x \left (1-\frac {2}{3} \left (e^2+\log (\log (4))\right )\right )\right )}\right ) \, dx\\ &=\int \frac {e^x \left (-3+e^2-2 x-x^2+\log (\log (4))\right )}{5+e^x x^2+3 e^x \left (1+\frac {1}{3} \left (-e^2-\log (\log (4))\right )\right )} \, dx+\int \frac {e^x \left (3-2 e^2+4 x+2 x^2-2 \log (\log (4))\right )}{10+2 e^x x^2+3 e^x \left (1-\frac {2}{3} \left (e^2+\log (\log (4))\right )\right )} \, dx\\ &=-\log \left (5+e^x x^2+e^x \left (3-e^2-\log (\log (4))\right )\right )+\log \left (10+2 e^x x^2+e^x \left (3-2 \left (e^2+\log (\log (4))\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 3.63, size = 72, normalized size = 2.48 \begin {gather*} 3 \left (\frac {1}{3} \log \left (10+3 e^x-2 e^{2+x}+2 e^x x^2-2 e^x \log (\log (4))\right )-\frac {1}{3} \log \left (5+3 e^x-e^{2+x}+e^x x^2-e^x \log (\log (4))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*E^x + 6*E^(2*x)*x)/(50 + E^x*(45 - 20*E^2 + 20*x^2) + E^(2*x)*(9 + 2*E^4 + 9*x^2 + 2*x^4 + E^2*
(-9 - 4*x^2)) + (-20*E^x + E^(2*x)*(-9 + 4*E^2 - 4*x^2))*Log[Log[4]] + 2*E^(2*x)*Log[Log[4]]^2),x]

[Out]

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

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fricas [B]  time = 0.75, size = 133, normalized size = 4.59 \begin {gather*} \log \left (2 \, x^{2} - 2 \, e^{2} - 2 \, \log \left (2 \, \log \relax (2)\right ) + 3\right ) - \log \left (x^{2} - e^{2} - \log \left (2 \, \log \relax (2)\right ) + 3\right ) + \log \left (-\frac {{\left (2 \, x^{2} - 2 \, e^{2} + 3\right )} e^{x} - 2 \, e^{x} \log \left (2 \, \log \relax (2)\right ) + 10}{2 \, x^{2} - 2 \, e^{2} - 2 \, \log \left (2 \, \log \relax (2)\right ) + 3}\right ) - \log \left (-\frac {{\left (x^{2} - e^{2} + 3\right )} e^{x} - e^{x} \log \left (2 \, \log \relax (2)\right ) + 5}{x^{2} - e^{2} - \log \left (2 \, \log \relax (2)\right ) + 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(x)^2-15*exp(x))/(2*exp(x)^2*log(2*log(2))^2+((4*exp(2)-4*x^2-9)*exp(x)^2-20*exp(x))*log(2*l
og(2))+(2*exp(2)^2+(-4*x^2-9)*exp(2)+2*x^4+9*x^2+9)*exp(x)^2+(-20*exp(2)+20*x^2+45)*exp(x)+50),x, algorithm="f
ricas")

[Out]

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

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giac [B]  time = 0.66, size = 68, normalized size = 2.34 \begin {gather*} \log \left (2 \, x^{2} e^{x} - 2 \, e^{x} \log \relax (2) - 2 \, e^{x} \log \left (\log \relax (2)\right ) - 2 \, e^{\left (x + 2\right )} + 3 \, e^{x} + 10\right ) - \log \left (x^{2} e^{x} - e^{x} \log \relax (2) - e^{x} \log \left (\log \relax (2)\right ) - e^{\left (x + 2\right )} + 3 \, e^{x} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(x)^2-15*exp(x))/(2*exp(x)^2*log(2*log(2))^2+((4*exp(2)-4*x^2-9)*exp(x)^2-20*exp(x))*log(2*l
og(2))+(2*exp(2)^2+(-4*x^2-9)*exp(2)+2*x^4+9*x^2+9)*exp(x)^2+(-20*exp(2)+20*x^2+45)*exp(x)+50),x, algorithm="g
iac")

[Out]

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

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maple [B]  time = 0.77, size = 60, normalized size = 2.07

method	result	size
norman	\(-\ln \left (-{\mathrm e}^{x} x^{2}+{\mathrm e}^{2} {\mathrm e}^{x}+{\mathrm e}^{x} \ln \left (2 \ln \relax (2)\right )-3 \,{\mathrm e}^{x}-5\right )+\ln \left (-2 \,{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{2} {\mathrm e}^{x}+2 \,{\mathrm e}^{x} \ln \left (2 \ln \relax (2)\right )-3 \,{\mathrm e}^{x}-10\right )\)	\(60\)
risch	\(-\ln \left (x^{2}-{\mathrm e}^{2}-\ln \left (\ln \relax (2)\right )-\ln \relax (2)+3\right )+\ln \left (-2 x^{2}+2 \,{\mathrm e}^{2}+2 \ln \left (\ln \relax (2)\right )+2 \ln \relax (2)-3\right )+\ln \left ({\mathrm e}^{x}-\frac {10}{-2 x^{2}+2 \,{\mathrm e}^{2}+2 \ln \left (\ln \relax (2)\right )+2 \ln \relax (2)-3}\right )-\ln \left ({\mathrm e}^{x}-\frac {5}{-x^{2}+{\mathrm e}^{2}+\ln \left (\ln \relax (2)\right )+\ln \relax (2)-3}\right )\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x*exp(x)^2-15*exp(x))/(2*exp(x)^2*ln(2*ln(2))^2+((4*exp(2)-4*x^2-9)*exp(x)^2-20*exp(x))*ln(2*ln(2))+(2*
exp(2)^2+(-4*x^2-9)*exp(2)+2*x^4+9*x^2+9)*exp(x)^2+(-20*exp(2)+20*x^2+45)*exp(x)+50),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.49, size = 139, normalized size = 4.79 \begin {gather*} \log \left (2 \, x^{2} - 2 \, e^{2} - 2 \, \log \relax (2) - 2 \, \log \left (\log \relax (2)\right ) + 3\right ) - \log \left (x^{2} - e^{2} - \log \relax (2) - \log \left (\log \relax (2)\right ) + 3\right ) + \log \left (\frac {{\left (2 \, x^{2} - 2 \, e^{2} - 2 \, \log \relax (2) - 2 \, \log \left (\log \relax (2)\right ) + 3\right )} e^{x} + 10}{2 \, x^{2} - 2 \, e^{2} - 2 \, \log \relax (2) - 2 \, \log \left (\log \relax (2)\right ) + 3}\right ) - \log \left (\frac {{\left (x^{2} - e^{2} - \log \relax (2) - \log \left (\log \relax (2)\right ) + 3\right )} e^{x} + 5}{x^{2} - e^{2} - \log \relax (2) - \log \left (\log \relax (2)\right ) + 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(x)^2-15*exp(x))/(2*exp(x)^2*log(2*log(2))^2+((4*exp(2)-4*x^2-9)*exp(x)^2-20*exp(x))*log(2*l
og(2))+(2*exp(2)^2+(-4*x^2-9)*exp(2)+2*x^4+9*x^2+9)*exp(x)^2+(-20*exp(2)+20*x^2+45)*exp(x)+50),x, algorithm="m
axima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {15\,{\mathrm {e}}^x-6\,x\,{\mathrm {e}}^{2\,x}}{2\,{\ln \left (2\,\ln \relax (2)\right )}^2\,{\mathrm {e}}^{2\,x}-\ln \left (2\,\ln \relax (2)\right )\,\left (20\,{\mathrm {e}}^x+{\mathrm {e}}^{2\,x}\,\left (4\,x^2-4\,{\mathrm {e}}^2+9\right )\right )+{\mathrm {e}}^{2\,x}\,\left (2\,{\mathrm {e}}^4-{\mathrm {e}}^2\,\left (4\,x^2+9\right )+9\,x^2+2\,x^4+9\right )+{\mathrm {e}}^x\,\left (20\,x^2-20\,{\mathrm {e}}^2+45\right )+50} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*exp(x) - 6*x*exp(2*x))/(2*log(2*log(2))^2*exp(2*x) - log(2*log(2))*(20*exp(x) + exp(2*x)*(4*x^2 - 4*e
xp(2) + 9)) + exp(2*x)*(2*exp(4) - exp(2)*(4*x^2 + 9) + 9*x^2 + 2*x^4 + 9) + exp(x)*(20*x^2 - 20*exp(2) + 45)
+ 50),x)

[Out]

int(-(15*exp(x) - 6*x*exp(2*x))/(2*log(2*log(2))^2*exp(2*x) - log(2*log(2))*(20*exp(x) + exp(2*x)*(4*x^2 - 4*e
xp(2) + 9)) + exp(2*x)*(2*exp(4) - exp(2)*(4*x^2 + 9) + 9*x^2 + 2*x^4 + 9) + exp(x)*(20*x^2 - 20*exp(2) + 45)
+ 50), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(x)**2-15*exp(x))/(2*exp(x)**2*ln(2*ln(2))**2+((4*exp(2)-4*x**2-9)*exp(x)**2-20*exp(x))*ln(2
*ln(2))+(2*exp(2)**2+(-4*x**2-9)*exp(2)+2*x**4+9*x**2+9)*exp(x)**2+(-20*exp(2)+20*x**2+45)*exp(x)+50),x)

[Out]

Timed out

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