3.88.40 \(\int \frac {e^2 (-20 x^4-5 x^5) \log (4)+125 e x \log ^2(4)+(e^2 (-20 x^4-10 x^5) \log (4)-250 e x \log ^2(4)) \log (x)}{(e^2 (16 x^6+8 x^7+x^8)+e (-200 x^3-50 x^4) \log (4)+625 \log ^2(4)) \log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ \frac {x^2}{\left (-\frac {5}{e}+\frac {x^3 (4+x)}{5 \log (4)}\right ) \log (x)} \]

________________________________________________________________________________________

Rubi [F]  time = 0.68, 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^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^2*(-20*x^4 - 5*x^5)*Log[4] + 125*E*x*Log[4]^2 + (E^2*(-20*x^4 - 10*x^5)*Log[4] - 250*E*x*Log[4]^2)*Log[
x])/((E^2*(16*x^6 + 8*x^7 + x^8) + E*(-200*x^3 - 50*x^4)*Log[4] + 625*Log[4]^2)*Log[x]^2),x]

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.30, size = 27, normalized size = 0.90 \begin {gather*} \frac {5 e x^2 \log (4)}{\left (e x^3 (4+x)-25 \log (4)\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-20*x^4 - 5*x^5)*Log[4] + 125*E*x*Log[4]^2 + (E^2*(-20*x^4 - 10*x^5)*Log[4] - 250*E*x*Log[4]^2
)*Log[x])/((E^2*(16*x^6 + 8*x^7 + x^8) + E*(-200*x^3 - 50*x^4)*Log[4] + 625*Log[4]^2)*Log[x]^2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.74, size = 32, normalized size = 1.07 \begin {gather*} \frac {10 \, x^{2} e \log \relax (2)}{{\left ({\left (x^{4} + 4 \, x^{3}\right )} e - 50 \, \log \relax (2)\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1000*log(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*log(x)+500*log(2)^2*exp(1)*x+2*(-5*x^5
-20*x^4)*exp(1)^2*log(2))/(2500*log(2)^2+2*(-50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)
^2,x, algorithm="fricas")

[Out]

10*x^2*e*log(2)/(((x^4 + 4*x^3)*e - 50*log(2))*log(x))

________________________________________________________________________________________

giac [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(((-1000*log(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*log(x)+500*log(2)^2*exp(1)*x+2*(-5*x^5
-20*x^4)*exp(1)^2*log(2))/(2500*log(2)^2+2*(-50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)
^2,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.10, size = 35, normalized size = 1.17

method	result	size
risch	\(-\frac {10 x^{2} \ln \relax (2) {\mathrm e}}{\left (-x^{4} {\mathrm e}-4 x^{3} {\mathrm e}+50 \ln \relax (2)\right ) \ln \relax (x )}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-1000*ln(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*ln(2))*ln(x)+500*ln(2)^2*exp(1)*x+2*(-5*x^5-20*x^4)*e
xp(1)^2*ln(2))/(2500*ln(2)^2+2*(-50*x^4-200*x^3)*exp(1)*ln(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/ln(x)^2,x,method=_R
ETURNVERBOSE)

[Out]

-10*x^2*ln(2)*exp(1)/(-x^4*exp(1)-4*x^3*exp(1)+50*ln(2))/ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.48, size = 33, normalized size = 1.10 \begin {gather*} \frac {10 \, x^{2} e \log \relax (2)}{{\left (x^{4} e + 4 \, x^{3} e - 50 \, \log \relax (2)\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1000*log(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*log(x)+500*log(2)^2*exp(1)*x+2*(-5*x^5
-20*x^4)*exp(1)^2*log(2))/(2500*log(2)^2+2*(-50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)
^2,x, algorithm="maxima")

[Out]

10*x^2*e*log(2)/((x^4*e + 4*x^3*e - 50*log(2))*log(x))

________________________________________________________________________________________

mupad [B]  time = 6.46, size = 33, normalized size = 1.10 \begin {gather*} \frac {10\,x^2\,\mathrm {e}\,\ln \relax (2)}{\ln \relax (x)\,\left (\mathrm {e}\,x^4+4\,\mathrm {e}\,x^3-50\,\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(1000*x*exp(1)*log(2)^2 + 2*exp(2)*log(2)*(20*x^4 + 10*x^5)) - 500*x*exp(1)*log(2)^2 + 2*exp(2)*l
og(2)*(20*x^4 + 5*x^5))/(log(x)^2*(exp(2)*(16*x^6 + 8*x^7 + x^8) + 2500*log(2)^2 - 2*exp(1)*log(2)*(200*x^3 +
50*x^4))),x)

[Out]

(10*x^2*exp(1)*log(2))/(log(x)*(4*x^3*exp(1) - 50*log(2) + x^4*exp(1)))

________________________________________________________________________________________

sympy [A]  time = 0.17, size = 34, normalized size = 1.13 \begin {gather*} \frac {10 e x^{2} \log {\relax (2 )}}{\left (e x^{4} + 4 e x^{3} - 50 \log {\relax (2 )}\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1000*ln(2)**2*exp(1)*x+2*(-10*x**5-20*x**4)*exp(1)**2*ln(2))*ln(x)+500*ln(2)**2*exp(1)*x+2*(-5*x*
*5-20*x**4)*exp(1)**2*ln(2))/(2500*ln(2)**2+2*(-50*x**4-200*x**3)*exp(1)*ln(2)+(x**8+8*x**7+16*x**6)*exp(1)**2
)/ln(x)**2,x)

[Out]

10*E*x**2*log(2)/((E*x**4 + 4*E*x**3 - 50*log(2))*log(x))

________________________________________________________________________________________