3.7.45 \(\int \frac {(-24-12 x) \log ^2(5)+e^x (32+4 x-4 x^2) \log ^2(5)}{9 x^3+6 x^4+x^5+e^{2 x} (16 x^3-8 x^4+x^5)+e^x (-24 x^3-2 x^4+2 x^5)} \, dx\)

Optimal. Leaf size=24 \[ \frac {4 \log ^2(5)}{x^2 \left (3-e^x (4-x)+x\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 1.89, 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 {(-24-12 x) \log ^2(5)+e^x \left (32+4 x-4 x^2\right ) \log ^2(5)}{9 x^3+6 x^4+x^5+e^{2 x} \left (16 x^3-8 x^4+x^5\right )+e^x \left (-24 x^3-2 x^4+2 x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-24 - 12*x)*Log[5]^2 + E^x*(32 + 4*x - 4*x^2)*Log[5]^2)/(9*x^3 + 6*x^4 + x^5 + E^(2*x)*(16*x^3 - 8*x^4 +
x^5) + E^x*(-24*x^3 - 2*x^4 + 2*x^5)),x]

[Out]

(7*Log[5]^2*Defer[Int][1/((-4 + x)*(3 - 4*E^x + x + E^x*x)^2), x])/4 + 5*Log[5]^2*Defer[Int][1/(x^2*(3 - 4*E^x
+ x + E^x*x)^2), x] + (9*Log[5]^2*Defer[Int][1/(x*(3 - 4*E^x + x + E^x*x)^2), x])/4 - (Log[5]^2*Defer[Int][1/
((-4 + x)*(3 - 4*E^x + x + E^x*x)), x])/4 - 8*Log[5]^2*Defer[Int][1/(x^3*(3 - 4*E^x + x + E^x*x)), x] - 3*Log[
5]^2*Defer[Int][1/(x^2*(3 - 4*E^x + x + E^x*x)), x] + (Log[5]^2*Defer[Int][1/(x*(3 - 4*E^x + x + E^x*x)), x])/
4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (-3 (2+x)-e^x \left (-8-x+x^2\right )\right ) \log ^2(5)}{x^3 \left (3+e^x (-4+x)+x\right )^2} \, dx\\ &=\left (4 \log ^2(5)\right ) \int \frac {-3 (2+x)-e^x \left (-8-x+x^2\right )}{x^3 \left (3+e^x (-4+x)+x\right )^2} \, dx\\ &=\left (4 \log ^2(5)\right ) \int \left (-\frac {-8-x+x^2}{(-4+x) x^3 \left (3-4 e^x+x+e^x x\right )}+\frac {-5-x+x^2}{(-4+x) x^2 \left (3-4 e^x+x+e^x x\right )^2}\right ) \, dx\\ &=-\left (\left (4 \log ^2(5)\right ) \int \frac {-8-x+x^2}{(-4+x) x^3 \left (3-4 e^x+x+e^x x\right )} \, dx\right )+\left (4 \log ^2(5)\right ) \int \frac {-5-x+x^2}{(-4+x) x^2 \left (3-4 e^x+x+e^x x\right )^2} \, dx\\ &=\left (4 \log ^2(5)\right ) \int \left (\frac {7}{16 (-4+x) \left (3-4 e^x+x+e^x x\right )^2}+\frac {5}{4 x^2 \left (3-4 e^x+x+e^x x\right )^2}+\frac {9}{16 x \left (3-4 e^x+x+e^x x\right )^2}\right ) \, dx-\left (4 \log ^2(5)\right ) \int \left (\frac {1}{16 (-4+x) \left (3-4 e^x+x+e^x x\right )}+\frac {2}{x^3 \left (3-4 e^x+x+e^x x\right )}+\frac {3}{4 x^2 \left (3-4 e^x+x+e^x x\right )}-\frac {1}{16 x \left (3-4 e^x+x+e^x x\right )}\right ) \, dx\\ &=-\left (\frac {1}{4} \log ^2(5) \int \frac {1}{(-4+x) \left (3-4 e^x+x+e^x x\right )} \, dx\right )+\frac {1}{4} \log ^2(5) \int \frac {1}{x \left (3-4 e^x+x+e^x x\right )} \, dx+\frac {1}{4} \left (7 \log ^2(5)\right ) \int \frac {1}{(-4+x) \left (3-4 e^x+x+e^x x\right )^2} \, dx+\frac {1}{4} \left (9 \log ^2(5)\right ) \int \frac {1}{x \left (3-4 e^x+x+e^x x\right )^2} \, dx-\left (3 \log ^2(5)\right ) \int \frac {1}{x^2 \left (3-4 e^x+x+e^x x\right )} \, dx+\left (5 \log ^2(5)\right ) \int \frac {1}{x^2 \left (3-4 e^x+x+e^x x\right )^2} \, dx-\left (8 \log ^2(5)\right ) \int \frac {1}{x^3 \left (3-4 e^x+x+e^x x\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.36, size = 21, normalized size = 0.88 \begin {gather*} \frac {4 \log ^2(5)}{x^2 \left (3+e^x (-4+x)+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-24 - 12*x)*Log[5]^2 + E^x*(32 + 4*x - 4*x^2)*Log[5]^2)/(9*x^3 + 6*x^4 + x^5 + E^(2*x)*(16*x^3 - 8
*x^4 + x^5) + E^x*(-24*x^3 - 2*x^4 + 2*x^5)),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 29, normalized size = 1.21 \begin {gather*} \frac {4 \, \log \relax (5)^{2}}{x^{3} + 3 \, x^{2} + {\left (x^{3} - 4 \, x^{2}\right )} e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+4*x+32)*log(5)^2*exp(x)+(-12*x-24)*log(5)^2)/((x^5-8*x^4+16*x^3)*exp(x)^2+(2*x^5-2*x^4-24*x
^3)*exp(x)+x^5+6*x^4+9*x^3),x, algorithm="fricas")

[Out]

4*log(5)^2/(x^3 + 3*x^2 + (x^3 - 4*x^2)*e^x)

________________________________________________________________________________________

giac [A]  time = 0.33, size = 30, normalized size = 1.25 \begin {gather*} \frac {4 \, \log \relax (5)^{2}}{x^{3} e^{x} + x^{3} - 4 \, x^{2} e^{x} + 3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+4*x+32)*log(5)^2*exp(x)+(-12*x-24)*log(5)^2)/((x^5-8*x^4+16*x^3)*exp(x)^2+(2*x^5-2*x^4-24*x
^3)*exp(x)+x^5+6*x^4+9*x^3),x, algorithm="giac")

[Out]

4*log(5)^2/(x^3*e^x + x^3 - 4*x^2*e^x + 3*x^2)

________________________________________________________________________________________

maple [A]  time = 0.13, size = 23, normalized size = 0.96

method	result	size
norman	\(\frac {4 \ln \relax (5)^{2}}{x^{2} \left ({\mathrm e}^{x} x -4 \,{\mathrm e}^{x}+x +3\right )}\)	\(23\)
risch	\(\frac {4 \ln \relax (5)^{2}}{x^{2} \left ({\mathrm e}^{x} x -4 \,{\mathrm e}^{x}+x +3\right )}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2+4*x+32)*ln(5)^2*exp(x)+(-12*x-24)*ln(5)^2)/((x^5-8*x^4+16*x^3)*exp(x)^2+(2*x^5-2*x^4-24*x^3)*exp(
x)+x^5+6*x^4+9*x^3),x,method=_RETURNVERBOSE)

[Out]

4*ln(5)^2/x^2/(exp(x)*x-4*exp(x)+x+3)

________________________________________________________________________________________

maxima [A]  time = 0.68, size = 29, normalized size = 1.21 \begin {gather*} \frac {4 \, \log \relax (5)^{2}}{x^{3} + 3 \, x^{2} + {\left (x^{3} - 4 \, x^{2}\right )} e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+4*x+32)*log(5)^2*exp(x)+(-12*x-24)*log(5)^2)/((x^5-8*x^4+16*x^3)*exp(x)^2+(2*x^5-2*x^4-24*x
^3)*exp(x)+x^5+6*x^4+9*x^3),x, algorithm="maxima")

[Out]

4*log(5)^2/(x^3 + 3*x^2 + (x^3 - 4*x^2)*e^x)

________________________________________________________________________________________

mupad [B]  time = 0.21, size = 20, normalized size = 0.83 \begin {gather*} \frac {4\,{\ln \relax (5)}^2}{x^2\,\left (x+{\mathrm {e}}^x\,\left (x-4\right )+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)^2*(12*x + 24) - exp(x)*log(5)^2*(4*x - 4*x^2 + 32))/(exp(2*x)*(16*x^3 - 8*x^4 + x^5) - exp(x)*(24
*x^3 + 2*x^4 - 2*x^5) + 9*x^3 + 6*x^4 + x^5),x)

[Out]

(4*log(5)^2)/(x^2*(x + exp(x)*(x - 4) + 3))

________________________________________________________________________________________

sympy [A]  time = 0.21, size = 26, normalized size = 1.08 \begin {gather*} \frac {4 \log {\relax (5 )}^{2}}{x^{3} + 3 x^{2} + \left (x^{3} - 4 x^{2}\right ) e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2+4*x+32)*ln(5)**2*exp(x)+(-12*x-24)*ln(5)**2)/((x**5-8*x**4+16*x**3)*exp(x)**2+(2*x**5-2*x*
*4-24*x**3)*exp(x)+x**5+6*x**4+9*x**3),x)

[Out]

4*log(5)**2/(x**3 + 3*x**2 + (x**3 - 4*x**2)*exp(x))

________________________________________________________________________________________