3.54.17 \(\int \frac {-4 x^4-2 x^5+(4 x^2+4 x^3+2 x^4+2 x^5) \log (2+x^2)+(4 x^2+4 x^3+2 x^4+2 x^5+e^x (-4+4 x+2 x^3+x^4)) \log ^2(2+x^2)}{(2 x^2+x^4) \log ^2(2+x^2)} \, dx\)

Optimal. Leaf size=26 \[ (2+x) \left (x+\frac {e^x+\frac {x^2}{\log \left (2+x^2\right )}}{x}\right ) \]

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Rubi [F]  time = 1.24, 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 {-4 x^4-2 x^5+\left (4 x^2+4 x^3+2 x^4+2 x^5\right ) \log \left (2+x^2\right )+\left (4 x^2+4 x^3+2 x^4+2 x^5+e^x \left (-4+4 x+2 x^3+x^4\right )\right ) \log ^2\left (2+x^2\right )}{\left (2 x^2+x^4\right ) \log ^2\left (2+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

E^x + (2*E^x)/x + (1 + x)^2 - 2/Log[2 + x^2] + (2 + x^2)/Log[2 + x^2] - LogIntegral[2 + x^2] - 4*Defer[Int][Lo
g[2 + x^2]^(-2), x] + 8*Defer[Int][1/((2 + x^2)*Log[2 + x^2]^2), x] + 2*Defer[Int][(1 + x)/Log[2 + x^2], x]

Rubi steps

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

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Mathematica [A]  time = 1.09, size = 26, normalized size = 1.00 \begin {gather*} \frac {(2+x) \left (e^x+x^2+\frac {x^2}{\log \left (2+x^2\right )}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.08, size = 43, normalized size = 1.65 \begin {gather*} \frac {x^{3} + 2 \, x^{2} + {\left (x^{3} + 2 \, x^{2} + {\left (x + 2\right )} e^{x}\right )} \log \left (x^{2} + 2\right )}{x \log \left (x^{2} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^4+2*x^3+4*x-4)*exp(x)+2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)^2+(2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2
)-2*x^5-4*x^4)/(x^4+2*x^2)/log(x^2+2)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.51, size = 62, normalized size = 2.38 \begin {gather*} \frac {x^{3} \log \left (x^{2} + 2\right ) + x^{3} + 2 \, x^{2} \log \left (x^{2} + 2\right ) + x e^{x} \log \left (x^{2} + 2\right ) + 2 \, x^{2} + 2 \, e^{x} \log \left (x^{2} + 2\right )}{x \log \left (x^{2} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^4+2*x^3+4*x-4)*exp(x)+2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)^2+(2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2
)-2*x^5-4*x^4)/(x^4+2*x^2)/log(x^2+2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.19, size = 36, normalized size = 1.38

method	result	size
risch	\(\frac {x^{3}+2 x^{2}+{\mathrm e}^{x} x +2 \,{\mathrm e}^{x}}{x}+\frac {x \left (2+x \right )}{\ln \left (x^{2}+2\right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^4+2*x^3+4*x-4)*exp(x)+2*x^5+2*x^4+4*x^3+4*x^2)*ln(x^2+2)^2+(2*x^5+2*x^4+4*x^3+4*x^2)*ln(x^2+2)-2*x^5-
4*x^4)/(x^4+2*x^2)/ln(x^2+2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 43, normalized size = 1.65 \begin {gather*} \frac {x^{3} + 2 \, x^{2} + {\left (x^{3} + 2 \, x^{2} + {\left (x + 2\right )} e^{x}\right )} \log \left (x^{2} + 2\right )}{x \log \left (x^{2} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^4+2*x^3+4*x-4)*exp(x)+2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)^2+(2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2
)-2*x^5-4*x^4)/(x^4+2*x^2)/log(x^2+2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.15, size = 39, normalized size = 1.50 \begin {gather*} 2\,x+{\mathrm {e}}^x+\frac {2\,{\mathrm {e}}^x}{x}+\frac {x^2}{\ln \left (x^2+2\right )}+x^2+\frac {2\,x}{\ln \left (x^2+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.34, size = 27, normalized size = 1.04 \begin {gather*} x^{2} + 2 x + \frac {x^{2} + 2 x}{\log {\left (x^{2} + 2 \right )}} + \frac {\left (x + 2\right ) e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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