3.62.41 \(\int \frac {3+e^{2 x} (3+6 x)+e^x (9+9 x)+(4 x-2 e^{2 x} x) \log (x)+(6 x+2 e^{2 x} x^2+e^x (6 x+6 x^2)) \log ^2(x)+(x^2+e^x (x^2+x^3)) \log ^4(x)}{9+6 x \log ^2(x)+x^2 \log ^4(x)} \, dx\)

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

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Rubi [F]  time = 0.96, 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 {3+e^{2 x} (3+6 x)+e^x (9+9 x)+\left (4 x-2 e^{2 x} x\right ) \log (x)+\left (6 x+2 e^{2 x} x^2+e^x \left (6 x+6 x^2\right )\right ) \log ^2(x)+\left (x^2+e^x \left (x^2+x^3\right )\right ) \log ^4(x)}{9+6 x \log ^2(x)+x^2 \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3 + E^(2*x)*(3 + 6*x) + E^x*(9 + 9*x) + (4*x - 2*E^(2*x)*x)*Log[x] + (6*x + 2*E^(2*x)*x^2 + E^x*(6*x + 6*
x^2))*Log[x]^2 + (x^2 + E^x*(x^2 + x^3))*Log[x]^4)/(9 + 6*x*Log[x]^2 + x^2*Log[x]^4),x]

[Out]

-E^x + x + E^x*(1 + x) + (E^(2*x)*(3*x + x^2*Log[x]^2))/(3 + x*Log[x]^2)^2 - 6*Defer[Int][(3 + x*Log[x]^2)^(-2
), x] + 4*Defer[Int][(x*Log[x])/(3 + x*Log[x]^2)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3+e^{2 x} (3+6 x)+e^x (9+9 x)+\left (4 x-2 e^{2 x} x\right ) \log (x)+\left (6 x+2 e^{2 x} x^2+e^x \left (6 x+6 x^2\right )\right ) \log ^2(x)+\left (x^2+e^x \left (x^2+x^3\right )\right ) \log ^4(x)}{\left (3+x \log ^2(x)\right )^2} \, dx\\ &=\int \left (e^x (1+x)+\frac {3}{\left (3+x \log ^2(x)\right )^2}+\frac {4 x \log (x)}{\left (3+x \log ^2(x)\right )^2}+\frac {6 x \log ^2(x)}{\left (3+x \log ^2(x)\right )^2}+\frac {x^2 \log ^4(x)}{\left (3+x \log ^2(x)\right )^2}+\frac {e^{2 x} \left (3+6 x-2 x \log (x)+2 x^2 \log ^2(x)\right )}{\left (3+x \log ^2(x)\right )^2}\right ) \, dx\\ &=3 \int \frac {1}{\left (3+x \log ^2(x)\right )^2} \, dx+4 \int \frac {x \log (x)}{\left (3+x \log ^2(x)\right )^2} \, dx+6 \int \frac {x \log ^2(x)}{\left (3+x \log ^2(x)\right )^2} \, dx+\int e^x (1+x) \, dx+\int \frac {x^2 \log ^4(x)}{\left (3+x \log ^2(x)\right )^2} \, dx+\int \frac {e^{2 x} \left (3+6 x-2 x \log (x)+2 x^2 \log ^2(x)\right )}{\left (3+x \log ^2(x)\right )^2} \, dx\\ &=e^x (1+x)+\frac {e^{2 x} \left (3 x+x^2 \log ^2(x)\right )}{\left (3+x \log ^2(x)\right )^2}+3 \int \frac {1}{\left (3+x \log ^2(x)\right )^2} \, dx+4 \int \frac {x \log (x)}{\left (3+x \log ^2(x)\right )^2} \, dx+6 \int \left (-\frac {3}{\left (3+x \log ^2(x)\right )^2}+\frac {1}{3+x \log ^2(x)}\right ) \, dx-\int e^x \, dx+\int \left (1+\frac {9}{\left (3+x \log ^2(x)\right )^2}-\frac {6}{3+x \log ^2(x)}\right ) \, dx\\ &=-e^x+x+e^x (1+x)+\frac {e^{2 x} \left (3 x+x^2 \log ^2(x)\right )}{\left (3+x \log ^2(x)\right )^2}+3 \int \frac {1}{\left (3+x \log ^2(x)\right )^2} \, dx+4 \int \frac {x \log (x)}{\left (3+x \log ^2(x)\right )^2} \, dx+9 \int \frac {1}{\left (3+x \log ^2(x)\right )^2} \, dx-18 \int \frac {1}{\left (3+x \log ^2(x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 26, normalized size = 0.96 \begin {gather*} x+e^x x+\frac {\left (-2+e^{2 x}\right ) x}{3+x \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + E^(2*x)*(3 + 6*x) + E^x*(9 + 9*x) + (4*x - 2*E^(2*x)*x)*Log[x] + (6*x + 2*E^(2*x)*x^2 + E^x*(6*
x + 6*x^2))*Log[x]^2 + (x^2 + E^x*(x^2 + x^3))*Log[x]^4)/(9 + 6*x*Log[x]^2 + x^2*Log[x]^4),x]

[Out]

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

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fricas [A]  time = 0.95, size = 39, normalized size = 1.44 \begin {gather*} \frac {{\left (x^{2} e^{x} + x^{2}\right )} \log \relax (x)^{2} + x e^{\left (2 \, x\right )} + 3 \, x e^{x} + x}{x \log \relax (x)^{2} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3+x^2)*exp(x)+x^2)*log(x)^4+(2*exp(x)^2*x^2+(6*x^2+6*x)*exp(x)+6*x)*log(x)^2+(-2*x*exp(x)^2+4*x
)*log(x)+(6*x+3)*exp(x)^2+(9*x+9)*exp(x)+3)/(x^2*log(x)^4+6*x*log(x)^2+9),x, algorithm="fricas")

[Out]

((x^2*e^x + x^2)*log(x)^2 + x*e^(2*x) + 3*x*e^x + x)/(x*log(x)^2 + 3)

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giac [A]  time = 0.16, size = 42, normalized size = 1.56 \begin {gather*} \frac {x^{2} e^{x} \log \relax (x)^{2} + x^{2} \log \relax (x)^{2} + x e^{\left (2 \, x\right )} + 3 \, x e^{x} + x}{x \log \relax (x)^{2} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3+x^2)*exp(x)+x^2)*log(x)^4+(2*exp(x)^2*x^2+(6*x^2+6*x)*exp(x)+6*x)*log(x)^2+(-2*x*exp(x)^2+4*x
)*log(x)+(6*x+3)*exp(x)^2+(9*x+9)*exp(x)+3)/(x^2*log(x)^4+6*x*log(x)^2+9),x, algorithm="giac")

[Out]

(x^2*e^x*log(x)^2 + x^2*log(x)^2 + x*e^(2*x) + 3*x*e^x + x)/(x*log(x)^2 + 3)

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maple [A]  time = 0.08, size = 25, normalized size = 0.93

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^3+x^2)*exp(x)+x^2)*ln(x)^4+(2*exp(x)^2*x^2+(6*x^2+6*x)*exp(x)+6*x)*ln(x)^2+(-2*x*exp(x)^2+4*x)*ln(x)+
(6*x+3)*exp(x)^2+(9*x+9)*exp(x)+3)/(x^2*ln(x)^4+6*x*ln(x)^2+9),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 42, normalized size = 1.56 \begin {gather*} \frac {x^{2} \log \relax (x)^{2} + x e^{\left (2 \, x\right )} + {\left (x^{2} \log \relax (x)^{2} + 3 \, x\right )} e^{x} + x}{x \log \relax (x)^{2} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3+x^2)*exp(x)+x^2)*log(x)^4+(2*exp(x)^2*x^2+(6*x^2+6*x)*exp(x)+6*x)*log(x)^2+(-2*x*exp(x)^2+4*x
)*log(x)+(6*x+3)*exp(x)^2+(9*x+9)*exp(x)+3)/(x^2*log(x)^4+6*x*log(x)^2+9),x, algorithm="maxima")

[Out]

(x^2*log(x)^2 + x*e^(2*x) + (x^2*log(x)^2 + 3*x)*e^x + x)/(x*log(x)^2 + 3)

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mupad [B]  time = 4.45, size = 36, normalized size = 1.33 \begin {gather*} \frac {x\,\left ({\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^x+x\,{\ln \relax (x)}^2+x\,{\mathrm {e}}^x\,{\ln \relax (x)}^2+1\right )}{x\,{\ln \relax (x)}^2+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^4*(exp(x)*(x^2 + x^3) + x^2) + exp(x)*(9*x + 9) + log(x)*(4*x - 2*x*exp(2*x)) + exp(2*x)*(6*x + 3)
+ log(x)^2*(6*x + 2*x^2*exp(2*x) + exp(x)*(6*x + 6*x^2)) + 3)/(6*x*log(x)^2 + x^2*log(x)^4 + 9),x)

[Out]

(x*(exp(2*x) + 3*exp(x) + x*log(x)^2 + x*exp(x)*log(x)^2 + 1))/(x*log(x)^2 + 3)

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sympy [A]  time = 0.37, size = 42, normalized size = 1.56 \begin {gather*} x - \frac {2 x}{x \log {\relax (x )}^{2} + 3} + \frac {x e^{2 x} + \left (x^{2} \log {\relax (x )}^{2} + 3 x\right ) e^{x}}{x \log {\relax (x )}^{2} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**3+x**2)*exp(x)+x**2)*ln(x)**4+(2*exp(x)**2*x**2+(6*x**2+6*x)*exp(x)+6*x)*ln(x)**2+(-2*x*exp(x)
**2+4*x)*ln(x)+(6*x+3)*exp(x)**2+(9*x+9)*exp(x)+3)/(x**2*ln(x)**4+6*x*ln(x)**2+9),x)

[Out]

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

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