3.38.59 \(\int \frac {e^3 (-100 x^3+24 x^4)-2 e^3 x^4 \log (4)+2 e^3 x^4 \log (4) \log (x)}{-15625+22500 x-10800 x^2+1728 x^3+(1875 x-1800 x^2+432 x^3) \log (4) \log (x)+(-75 x^2+36 x^3) \log ^2(4) \log ^2(x)+x^3 \log ^3(4) \log ^3(x)} \, dx\)

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

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Rubi [F]  time = 0.84, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^3 \left (-100 x^3+24 x^4\right )-2 e^3 x^4 \log (4)+2 e^3 x^4 \log (4) \log (x)}{-15625+22500 x-10800 x^2+1728 x^3+\left (1875 x-1800 x^2+432 x^3\right ) \log (4) \log (x)+\left (-75 x^2+36 x^3\right ) \log ^2(4) \log ^2(x)+x^3 \log ^3(4) \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^3*(-100*x^3 + 24*x^4) - 2*E^3*x^4*Log[4] + 2*E^3*x^4*Log[4]*Log[x])/(-15625 + 22500*x - 10800*x^2 + 172
8*x^3 + (1875*x - 1800*x^2 + 432*x^3)*Log[4]*Log[x] + (-75*x^2 + 36*x^3)*Log[4]^2*Log[x]^2 + x^3*Log[4]^3*Log[
x]^3),x]

[Out]

-50*E^3*Defer[Int][x^3/(-25 + 12*x + x*Log[4]*Log[x])^3, x] - 2*E^3*Log[4]*Defer[Int][x^4/(-25 + 12*x + x*Log[
4]*Log[x])^3, x] + 2*E^3*Defer[Int][x^3/(-25 + 12*x + x*Log[4]*Log[x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^3 x^3 (50+x (-12+\log (4))-x \log (4) \log (x))}{(25-12 x-x \log (4) \log (x))^3} \, dx\\ &=\left (2 e^3\right ) \int \frac {x^3 (50+x (-12+\log (4))-x \log (4) \log (x))}{(25-12 x-x \log (4) \log (x))^3} \, dx\\ &=\left (2 e^3\right ) \int \left (-\frac {x^3 (25+x \log (4))}{(-25+12 x+x \log (4) \log (x))^3}+\frac {x^3}{(-25+12 x+x \log (4) \log (x))^2}\right ) \, dx\\ &=-\left (\left (2 e^3\right ) \int \frac {x^3 (25+x \log (4))}{(-25+12 x+x \log (4) \log (x))^3} \, dx\right )+\left (2 e^3\right ) \int \frac {x^3}{(-25+12 x+x \log (4) \log (x))^2} \, dx\\ &=\left (2 e^3\right ) \int \frac {x^3}{(-25+12 x+x \log (4) \log (x))^2} \, dx-\left (2 e^3\right ) \int \left (\frac {25 x^3}{(-25+12 x+x \log (4) \log (x))^3}+\frac {x^4 \log (4)}{(-25+12 x+x \log (4) \log (x))^3}\right ) \, dx\\ &=\left (2 e^3\right ) \int \frac {x^3}{(-25+12 x+x \log (4) \log (x))^2} \, dx-\left (50 e^3\right ) \int \frac {x^3}{(-25+12 x+x \log (4) \log (x))^3} \, dx-\left (2 e^3 \log (4)\right ) \int \frac {x^4}{(-25+12 x+x \log (4) \log (x))^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.44, size = 20, normalized size = 0.69 \begin {gather*} \frac {e^3 x^4}{(-25+12 x+x \log (4) \log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(-100*x^3 + 24*x^4) - 2*E^3*x^4*Log[4] + 2*E^3*x^4*Log[4]*Log[x])/(-15625 + 22500*x - 10800*x^2
+ 1728*x^3 + (1875*x - 1800*x^2 + 432*x^3)*Log[4]*Log[x] + (-75*x^2 + 36*x^3)*Log[4]^2*Log[x]^2 + x^3*Log[4]^
3*Log[x]^3),x]

[Out]

(E^3*x^4)/(-25 + 12*x + x*Log[4]*Log[x])^2

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fricas [A]  time = 0.83, size = 46, normalized size = 1.59 \begin {gather*} \frac {x^{4} e^{3}}{4 \, x^{2} \log \relax (2)^{2} \log \relax (x)^{2} + 4 \, {\left (12 \, x^{2} - 25 \, x\right )} \log \relax (2) \log \relax (x) + 144 \, x^{2} - 600 \, x + 625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^4*exp(3)*log(2)*log(x)-4*x^4*exp(3)*log(2)+(24*x^4-100*x^3)*exp(3))/(8*x^3*log(2)^3*log(x)^3+4*
(36*x^3-75*x^2)*log(2)^2*log(x)^2+2*(432*x^3-1800*x^2+1875*x)*log(2)*log(x)+1728*x^3-10800*x^2+22500*x-15625),
x, algorithm="fricas")

[Out]

x^4*e^3/(4*x^2*log(2)^2*log(x)^2 + 4*(12*x^2 - 25*x)*log(2)*log(x) + 144*x^2 - 600*x + 625)

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giac [A]  time = 0.15, size = 47, normalized size = 1.62 \begin {gather*} \frac {x^{4} e^{3}}{4 \, x^{2} \log \relax (2)^{2} \log \relax (x)^{2} + 48 \, x^{2} \log \relax (2) \log \relax (x) - 100 \, x \log \relax (2) \log \relax (x) + 144 \, x^{2} - 600 \, x + 625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^4*exp(3)*log(2)*log(x)-4*x^4*exp(3)*log(2)+(24*x^4-100*x^3)*exp(3))/(8*x^3*log(2)^3*log(x)^3+4*
(36*x^3-75*x^2)*log(2)^2*log(x)^2+2*(432*x^3-1800*x^2+1875*x)*log(2)*log(x)+1728*x^3-10800*x^2+22500*x-15625),
x, algorithm="giac")

[Out]

x^4*e^3/(4*x^2*log(2)^2*log(x)^2 + 48*x^2*log(2)*log(x) - 100*x*log(2)*log(x) + 144*x^2 - 600*x + 625)

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maple [A]  time = 0.23, size = 21, normalized size = 0.72

method	result	size
norman	\(\frac {x^{4} {\mathrm e}^{3}}{\left (2 x \ln \relax (2) \ln \relax (x )+12 x -25\right )^{2}}\)	\(21\)
risch	\(\frac {x^{4} {\mathrm e}^{3}}{\left (2 x \ln \relax (2) \ln \relax (x )+12 x -25\right )^{2}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^4*exp(3)*ln(2)*ln(x)-4*x^4*exp(3)*ln(2)+(24*x^4-100*x^3)*exp(3))/(8*x^3*ln(2)^3*ln(x)^3+4*(36*x^3-75*
x^2)*ln(2)^2*ln(x)^2+2*(432*x^3-1800*x^2+1875*x)*ln(2)*ln(x)+1728*x^3-10800*x^2+22500*x-15625),x,method=_RETUR
NVERBOSE)

[Out]

x^4*exp(3)/(2*x*ln(2)*ln(x)+12*x-25)^2

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maxima [A]  time = 0.48, size = 48, normalized size = 1.66 \begin {gather*} \frac {x^{4} e^{3}}{4 \, x^{2} \log \relax (2)^{2} \log \relax (x)^{2} + 144 \, x^{2} + 4 \, {\left (12 \, x^{2} \log \relax (2) - 25 \, x \log \relax (2)\right )} \log \relax (x) - 600 \, x + 625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^4*exp(3)*log(2)*log(x)-4*x^4*exp(3)*log(2)+(24*x^4-100*x^3)*exp(3))/(8*x^3*log(2)^3*log(x)^3+4*
(36*x^3-75*x^2)*log(2)^2*log(x)^2+2*(432*x^3-1800*x^2+1875*x)*log(2)*log(x)+1728*x^3-10800*x^2+22500*x-15625),
x, algorithm="maxima")

[Out]

x^4*e^3/(4*x^2*log(2)^2*log(x)^2 + 144*x^2 + 4*(12*x^2*log(2) - 25*x*log(2))*log(x) - 600*x + 625)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^3\,\left (100\,x^3-24\,x^4\right )+4\,x^4\,{\mathrm {e}}^3\,\ln \relax (2)-4\,x^4\,{\mathrm {e}}^3\,\ln \relax (2)\,\ln \relax (x)}{22500\,x-10800\,x^2+1728\,x^3-4\,{\ln \relax (2)}^2\,{\ln \relax (x)}^2\,\left (75\,x^2-36\,x^3\right )+8\,x^3\,{\ln \relax (2)}^3\,{\ln \relax (x)}^3+2\,\ln \relax (2)\,\ln \relax (x)\,\left (432\,x^3-1800\,x^2+1875\,x\right )-15625} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3)*(100*x^3 - 24*x^4) + 4*x^4*exp(3)*log(2) - 4*x^4*exp(3)*log(2)*log(x))/(22500*x - 10800*x^2 + 172
8*x^3 - 4*log(2)^2*log(x)^2*(75*x^2 - 36*x^3) + 8*x^3*log(2)^3*log(x)^3 + 2*log(2)*log(x)*(1875*x - 1800*x^2 +
432*x^3) - 15625),x)

[Out]

int(-(exp(3)*(100*x^3 - 24*x^4) + 4*x^4*exp(3)*log(2) - 4*x^4*exp(3)*log(2)*log(x))/(22500*x - 10800*x^2 + 172
8*x^3 - 4*log(2)^2*log(x)^2*(75*x^2 - 36*x^3) + 8*x^3*log(2)^3*log(x)^3 + 2*log(2)*log(x)*(1875*x - 1800*x^2 +
432*x^3) - 15625), x)

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sympy [A]  time = 0.26, size = 49, normalized size = 1.69 \begin {gather*} \frac {x^{4} e^{3}}{4 x^{2} \log {\relax (2 )}^{2} \log {\relax (x )}^{2} + 144 x^{2} - 600 x + \left (48 x^{2} \log {\relax (2 )} - 100 x \log {\relax (2 )}\right ) \log {\relax (x )} + 625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**4*exp(3)*ln(2)*ln(x)-4*x**4*exp(3)*ln(2)+(24*x**4-100*x**3)*exp(3))/(8*x**3*ln(2)**3*ln(x)**3+
4*(36*x**3-75*x**2)*ln(2)**2*ln(x)**2+2*(432*x**3-1800*x**2+1875*x)*ln(2)*ln(x)+1728*x**3-10800*x**2+22500*x-1
5625),x)

[Out]

x**4*exp(3)/(4*x**2*log(2)**2*log(x)**2 + 144*x**2 - 600*x + (48*x**2*log(2) - 100*x*log(2))*log(x) + 625)

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