3.100.62 \(\int \frac {10 x+2 x^2+2 \sqrt [3]{e} x^2+4 e^{2/3} x^2+(-10 x-x^2+\sqrt [3]{e} (-20 x-4 x^2)) \log (x^2)+(25+10 x+x^2) \log ^2(x^2)}{4 e^{2/3} x^2+\sqrt [3]{e} (-20 x-4 x^2) \log (x^2)+(25+10 x+x^2) \log ^2(x^2)} \, dx\)

Optimal. Leaf size=27 \[ -4+x+\frac {x}{2 \sqrt [3]{e}-\frac {(5+x) \log \left (x^2\right )}{x}} \]

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Rubi [F]  time = 1.39, 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 {10 x+2 x^2+2 \sqrt [3]{e} x^2+4 e^{2/3} x^2+\left (-10 x-x^2+\sqrt [3]{e} \left (-20 x-4 x^2\right )\right ) \log \left (x^2\right )+\left (25+10 x+x^2\right ) \log ^2\left (x^2\right )}{4 e^{2/3} x^2+\sqrt [3]{e} \left (-20 x-4 x^2\right ) \log \left (x^2\right )+\left (25+10 x+x^2\right ) \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10*x + 2*x^2 + 2*E^(1/3)*x^2 + 4*E^(2/3)*x^2 + (-10*x - x^2 + E^(1/3)*(-20*x - 4*x^2))*Log[x^2] + (25 + 1
0*x + x^2)*Log[x^2]^2)/(4*E^(2/3)*x^2 + E^(1/3)*(-20*x - 4*x^2)*Log[x^2] + (25 + 10*x + x^2)*Log[x^2]^2),x]

[Out]

x + 50*E^(1/3)*Defer[Int][(2*E^(1/3)*x - 5*Log[x^2] - x*Log[x^2])^(-2), x] + 10*(1 - E^(1/3))*Defer[Int][x/(2*
E^(1/3)*x - 5*Log[x^2] - x*Log[x^2])^2, x] + 2*Defer[Int][x^2/(2*E^(1/3)*x - 5*Log[x^2] - x*Log[x^2])^2, x] -
250*E^(1/3)*Defer[Int][1/((5 + x)*(2*E^(1/3)*x - 5*Log[x^2] - x*Log[x^2])^2), x] + 5*Defer[Int][(2*E^(1/3)*x -
 5*Log[x^2] - x*Log[x^2])^(-1), x] + Defer[Int][x/(2*E^(1/3)*x - 5*Log[x^2] - x*Log[x^2]), x] - 25*Defer[Int][
1/((5 + x)*(2*E^(1/3)*x - 5*Log[x^2] - x*Log[x^2])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 x+\left (2+2 \sqrt [3]{e}\right ) x^2+4 e^{2/3} x^2+\left (-10 x-x^2+\sqrt [3]{e} \left (-20 x-4 x^2\right )\right ) \log \left (x^2\right )+\left (25+10 x+x^2\right ) \log ^2\left (x^2\right )}{4 e^{2/3} x^2+\sqrt [3]{e} \left (-20 x-4 x^2\right ) \log \left (x^2\right )+\left (25+10 x+x^2\right ) \log ^2\left (x^2\right )} \, dx\\ &=\int \frac {10 x+\left (2+2 \sqrt [3]{e}+4 e^{2/3}\right ) x^2+\left (-10 x-x^2+\sqrt [3]{e} \left (-20 x-4 x^2\right )\right ) \log \left (x^2\right )+\left (25+10 x+x^2\right ) \log ^2\left (x^2\right )}{4 e^{2/3} x^2+\sqrt [3]{e} \left (-20 x-4 x^2\right ) \log \left (x^2\right )+\left (25+10 x+x^2\right ) \log ^2\left (x^2\right )} \, dx\\ &=\int \frac {2 x \left (5+\left (1+\sqrt [3]{e}+2 e^{2/3}\right ) x\right )-x \left (10+x+4 \sqrt [3]{e} (5+x)\right ) \log \left (x^2\right )+(5+x)^2 \log ^2\left (x^2\right )}{\left (2 \sqrt [3]{e} x-(5+x) \log \left (x^2\right )\right )^2} \, dx\\ &=\int \left (1+\frac {2 x \left (25+5 \left (2-\sqrt [3]{e}\right ) x+x^2\right )}{(5+x) \left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2}+\frac {x (10+x)}{(5+x) \left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )}\right ) \, dx\\ &=x+2 \int \frac {x \left (25+5 \left (2-\sqrt [3]{e}\right ) x+x^2\right )}{(5+x) \left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2} \, dx+\int \frac {x (10+x)}{(5+x) \left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )} \, dx\\ &=x+2 \int \left (\frac {25 \sqrt [3]{e}}{\left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2}-\frac {5 \left (-1+\sqrt [3]{e}\right ) x}{\left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2}+\frac {x^2}{\left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2}-\frac {125 \sqrt [3]{e}}{(5+x) \left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2}\right ) \, dx+\int \left (\frac {5}{2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )}+\frac {x}{2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )}-\frac {25}{(5+x) \left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )}\right ) \, dx\\ &=x+2 \int \frac {x^2}{\left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2} \, dx+5 \int \frac {1}{2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )} \, dx-25 \int \frac {1}{(5+x) \left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )} \, dx+\left (10 \left (1-\sqrt [3]{e}\right )\right ) \int \frac {x}{\left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2} \, dx+\left (50 \sqrt [3]{e}\right ) \int \frac {1}{\left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2} \, dx-\left (250 \sqrt [3]{e}\right ) \int \frac {1}{(5+x) \left (2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )\right )^2} \, dx+\int \frac {x}{2 \sqrt [3]{e} x-5 \log \left (x^2\right )-x \log \left (x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.61, size = 26, normalized size = 0.96 \begin {gather*} x+\frac {x^2}{2 \sqrt [3]{e} x-(5+x) \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x + 2*x^2 + 2*E^(1/3)*x^2 + 4*E^(2/3)*x^2 + (-10*x - x^2 + E^(1/3)*(-20*x - 4*x^2))*Log[x^2] + (
25 + 10*x + x^2)*Log[x^2]^2)/(4*E^(2/3)*x^2 + E^(1/3)*(-20*x - 4*x^2)*Log[x^2] + (25 + 10*x + x^2)*Log[x^2]^2)
,x]

[Out]

x + x^2/(2*E^(1/3)*x - (5 + x)*Log[x^2])

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fricas [A]  time = 1.06, size = 42, normalized size = 1.56 \begin {gather*} \frac {2 \, x^{2} e^{\frac {1}{3}} + x^{2} - {\left (x^{2} + 5 \, x\right )} \log \left (x^{2}\right )}{2 \, x e^{\frac {1}{3}} - {\left (x + 5\right )} \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+10*x+25)*log(x^2)^2+((-4*x^2-20*x)*exp(1/3)-x^2-10*x)*log(x^2)+4*x^2*exp(1/3)^2+2*x^2*exp(1/3)
+2*x^2+10*x)/((x^2+10*x+25)*log(x^2)^2+(-4*x^2-20*x)*exp(1/3)*log(x^2)+4*x^2*exp(1/3)^2),x, algorithm="fricas"
)

[Out]

(2*x^2*e^(1/3) + x^2 - (x^2 + 5*x)*log(x^2))/(2*x*e^(1/3) - (x + 5)*log(x^2))

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giac [B]  time = 0.39, size = 49, normalized size = 1.81 \begin {gather*} \frac {2 \, x^{2} e^{\frac {1}{3}} - x^{2} \log \left (x^{2}\right ) + x^{2} - 5 \, x \log \left (x^{2}\right )}{2 \, x e^{\frac {1}{3}} - x \log \left (x^{2}\right ) - 5 \, \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+10*x+25)*log(x^2)^2+((-4*x^2-20*x)*exp(1/3)-x^2-10*x)*log(x^2)+4*x^2*exp(1/3)^2+2*x^2*exp(1/3)
+2*x^2+10*x)/((x^2+10*x+25)*log(x^2)^2+(-4*x^2-20*x)*exp(1/3)*log(x^2)+4*x^2*exp(1/3)^2),x, algorithm="giac")

[Out]

(2*x^2*e^(1/3) - x^2*log(x^2) + x^2 - 5*x*log(x^2))/(2*x*e^(1/3) - x*log(x^2) - 5*log(x^2))

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maple [A]  time = 0.10, size = 28, normalized size = 1.04




method result size



risch \(x +\frac {x^{2}}{2 x \,{\mathrm e}^{\frac {1}{3}}-x \ln \left (x^{2}\right )-5 \ln \left (x^{2}\right )}\) \(28\)
norman \(\frac {25 \ln \left (x^{2}\right )-10 x \,{\mathrm e}^{\frac {1}{3}}+\left (2 \,{\mathrm e}^{\frac {1}{3}}+1\right ) x^{2}-x^{2} \ln \left (x^{2}\right )}{2 x \,{\mathrm e}^{\frac {1}{3}}-x \ln \left (x^{2}\right )-5 \ln \left (x^{2}\right )}\) \(54\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+10*x+25)*ln(x^2)^2+((-4*x^2-20*x)*exp(1/3)-x^2-10*x)*ln(x^2)+4*x^2*exp(1/3)^2+2*x^2*exp(1/3)+2*x^2+1
0*x)/((x^2+10*x+25)*ln(x^2)^2+(-4*x^2-20*x)*exp(1/3)*ln(x^2)+4*x^2*exp(1/3)^2),x,method=_RETURNVERBOSE)

[Out]

x+x^2/(2*x*exp(1/3)-x*ln(x^2)-5*ln(x^2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+10*x+25)*log(x^2)^2+((-4*x^2-20*x)*exp(1/3)-x^2-10*x)*log(x^2)+4*x^2*exp(1/3)^2+2*x^2*exp(1/3)
+2*x^2+10*x)/((x^2+10*x+25)*log(x^2)^2+(-4*x^2-20*x)*exp(1/3)*log(x^2)+4*x^2*exp(1/3)^2),x, algorithm="maxima"
)

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [B]  time = 7.54, size = 73, normalized size = 2.70 \begin {gather*} \frac {25\,\ln \left (x^2\right )\,{\mathrm {e}}^{1/3}+10\,x\,\ln \left (x^2\right )-10\,x\,{\mathrm {e}}^{2/3}+2\,x^2\,\ln \left (x^2\right )-4\,x^2\,{\mathrm {e}}^{1/3}-2\,x^2+5\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{1/3}}{2\,\left (5\,\ln \left (x^2\right )+x\,\ln \left (x^2\right )-2\,x\,{\mathrm {e}}^{1/3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + log(x^2)^2*(10*x + x^2 + 25) + 2*x^2*exp(1/3) + 4*x^2*exp(2/3) + 2*x^2 - log(x^2)*(10*x + exp(1/3)
*(20*x + 4*x^2) + x^2))/(log(x^2)^2*(10*x + x^2 + 25) + 4*x^2*exp(2/3) - log(x^2)*exp(1/3)*(20*x + 4*x^2)),x)

[Out]

(25*log(x^2)*exp(1/3) + 10*x*log(x^2) - 10*x*exp(2/3) + 2*x^2*log(x^2) - 4*x^2*exp(1/3) - 2*x^2 + 5*x*log(x^2)
*exp(1/3))/(2*(5*log(x^2) + x*log(x^2) - 2*x*exp(1/3)))

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sympy [A]  time = 0.20, size = 20, normalized size = 0.74 \begin {gather*} - \frac {x^{2}}{- 2 x e^{\frac {1}{3}} + \left (x + 5\right ) \log {\left (x^{2} \right )}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+10*x+25)*ln(x**2)**2+((-4*x**2-20*x)*exp(1/3)-x**2-10*x)*ln(x**2)+4*x**2*exp(1/3)**2+2*x**2*e
xp(1/3)+2*x**2+10*x)/((x**2+10*x+25)*ln(x**2)**2+(-4*x**2-20*x)*exp(1/3)*ln(x**2)+4*x**2*exp(1/3)**2),x)

[Out]

-x**2/(-2*x*exp(1/3) + (x + 5)*log(x**2)) + x

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