3.42.50 \(\int \frac {30 x^3-30 x^4+e^{e^x} (-9-5 x^2+e^x (9 x-9 x^2+5 x^3-5 x^4))+(-27 x^2-15 x^4) \log (\frac {1}{9} (45+25 x^2))}{9 x^2+5 x^4} \, dx\)

Optimal. Leaf size=28 \[ \frac {(1-x) \left (e^{e^x}+3 x \log \left (5+\frac {25 x^2}{9}\right )\right )}{x} \]

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Rubi [F]  time = 0.80, 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 {30 x^3-30 x^4+e^{e^x} \left (-9-5 x^2+e^x \left (9 x-9 x^2+5 x^3-5 x^4\right )\right )+\left (-27 x^2-15 x^4\right ) \log \left (\frac {1}{9} \left (45+25 x^2\right )\right )}{9 x^2+5 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(30*x^3 - 30*x^4 + E^E^x*(-9 - 5*x^2 + E^x*(9*x - 9*x^2 + 5*x^3 - 5*x^4)) + (-27*x^2 - 15*x^4)*Log[(45 + 2
5*x^2)/9])/(9*x^2 + 5*x^4),x]

[Out]

-E^E^x - 3*x*Log[5 + (25*x^2)/9] + 3*Log[9 + 5*x^2] - Defer[Int][E^E^x/x^2, x] + Defer[Int][E^(E^x + x)/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {30 x^3-30 x^4+e^{e^x} \left (-9-5 x^2+e^x \left (9 x-9 x^2+5 x^3-5 x^4\right )\right )+\left (-27 x^2-15 x^4\right ) \log \left (\frac {1}{9} \left (45+25 x^2\right )\right )}{x^2 \left (9+5 x^2\right )} \, dx\\ &=\int \left (e^{e^x+x} \left (-1+\frac {1}{x}\right )-\frac {e^{e^x}}{x^2}-\frac {30 (-1+x) x}{9+5 x^2}-3 \log \left (5+\frac {25 x^2}{9}\right )\right ) \, dx\\ &=-\left (3 \int \log \left (5+\frac {25 x^2}{9}\right ) \, dx\right )-30 \int \frac {(-1+x) x}{9+5 x^2} \, dx+\int e^{e^x+x} \left (-1+\frac {1}{x}\right ) \, dx-\int \frac {e^{e^x}}{x^2} \, dx\\ &=-6 x-3 x \log \left (5+\frac {25 x^2}{9}\right )-6 \int \frac {-9-5 x}{9+5 x^2} \, dx+\frac {50}{3} \int \frac {x^2}{5+\frac {25 x^2}{9}} \, dx+\int \left (-e^{e^x+x}+\frac {e^{e^x+x}}{x}\right ) \, dx-\int \frac {e^{e^x}}{x^2} \, dx\\ &=-3 x \log \left (5+\frac {25 x^2}{9}\right )-30 \int \frac {1}{5+\frac {25 x^2}{9}} \, dx+30 \int \frac {x}{9+5 x^2} \, dx+54 \int \frac {1}{9+5 x^2} \, dx-\int e^{e^x+x} \, dx-\int \frac {e^{e^x}}{x^2} \, dx+\int \frac {e^{e^x+x}}{x} \, dx\\ &=-3 x \log \left (5+\frac {25 x^2}{9}\right )+3 \log \left (9+5 x^2\right )-\int \frac {e^{e^x}}{x^2} \, dx+\int \frac {e^{e^x+x}}{x} \, dx-\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=-e^{e^x}-3 x \log \left (5+\frac {25 x^2}{9}\right )+3 \log \left (9+5 x^2\right )-\int \frac {e^{e^x}}{x^2} \, dx+\int \frac {e^{e^x+x}}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 38, normalized size = 1.36 \begin {gather*} -e^{e^x} \left (1-\frac {1}{x}\right )-3 x \log \left (5+\frac {25 x^2}{9}\right )+3 \log \left (9+5 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30*x^3 - 30*x^4 + E^E^x*(-9 - 5*x^2 + E^x*(9*x - 9*x^2 + 5*x^3 - 5*x^4)) + (-27*x^2 - 15*x^4)*Log[(
45 + 25*x^2)/9])/(9*x^2 + 5*x^4),x]

[Out]

-(E^E^x*(1 - x^(-1))) - 3*x*Log[5 + (25*x^2)/9] + 3*Log[9 + 5*x^2]

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fricas [A]  time = 0.65, size = 30, normalized size = 1.07 \begin {gather*} -\frac {{\left (x - 1\right )} e^{\left (e^{x}\right )} + 3 \, {\left (x^{2} - x\right )} \log \left (\frac {25}{9} \, x^{2} + 5\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^4+5*x^3-9*x^2+9*x)*exp(x)-5*x^2-9)*exp(exp(x))+(-15*x^4-27*x^2)*log(25/9*x^2+5)-30*x^4+30*x^
3)/(5*x^4+9*x^2),x, algorithm="fricas")

[Out]

-((x - 1)*e^(e^x) + 3*(x^2 - x)*log(25/9*x^2 + 5))/x

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giac [B]  time = 0.15, size = 52, normalized size = 1.86 \begin {gather*} -\frac {{\left (3 \, x^{2} e^{x} \log \left (\frac {25}{9} \, x^{2} + 5\right ) - 3 \, x e^{x} \log \left (5 \, x^{2} + 9\right ) + x e^{\left (x + e^{x}\right )} - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^4+5*x^3-9*x^2+9*x)*exp(x)-5*x^2-9)*exp(exp(x))+(-15*x^4-27*x^2)*log(25/9*x^2+5)-30*x^4+30*x^
3)/(5*x^4+9*x^2),x, algorithm="giac")

[Out]

-(3*x^2*e^x*log(25/9*x^2 + 5) - 3*x*e^x*log(5*x^2 + 9) + x*e^(x + e^x) - e^(x + e^x))*e^(-x)/x

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maple [A]  time = 0.08, size = 34, normalized size = 1.21

method	result	size
risch	\(-3 x \ln \left (\frac {25 x^{2}}{9}+5\right )+3 \ln \left (5 x^{2}+9\right )-\frac {\left (x -1\right ) {\mathrm e}^{{\mathrm e}^{x}}}{x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x^4+5*x^3-9*x^2+9*x)*exp(x)-5*x^2-9)*exp(exp(x))+(-15*x^4-27*x^2)*ln(25/9*x^2+5)-30*x^4+30*x^3)/(5*x
^4+9*x^2),x,method=_RETURNVERBOSE)

[Out]

-3*x*ln(25/9*x^2+5)+3*ln(5*x^2+9)-(x-1)/x*exp(exp(x))

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maxima [B]  time = 0.53, size = 53, normalized size = 1.89 \begin {gather*} -6 \, x - \frac {3 \, x^{2} {\left (\log \relax (5) - 2 \, \log \relax (3) - 2\right )} + 3 \, x^{2} \log \left (5 \, x^{2} + 9\right ) + {\left (x - 1\right )} e^{\left (e^{x}\right )}}{x} + 3 \, \log \left (5 \, x^{2} + 9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^4+5*x^3-9*x^2+9*x)*exp(x)-5*x^2-9)*exp(exp(x))+(-15*x^4-27*x^2)*log(25/9*x^2+5)-30*x^4+30*x^
3)/(5*x^4+9*x^2),x, algorithm="maxima")

[Out]

-6*x - (3*x^2*(log(5) - 2*log(3) - 2) + 3*x^2*log(5*x^2 + 9) + (x - 1)*e^(e^x))/x + 3*log(5*x^2 + 9)

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mupad [B]  time = 3.30, size = 31, normalized size = 1.11 \begin {gather*} 3\,\ln \left (x^2+\frac {9}{5}\right )-3\,x\,\ln \left (\frac {25\,x^2}{9}+5\right )-\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x-1\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((25*x^2)/9 + 5)*(27*x^2 + 15*x^4) + exp(exp(x))*(5*x^2 - exp(x)*(9*x - 9*x^2 + 5*x^3 - 5*x^4) + 9) -
30*x^3 + 30*x^4)/(9*x^2 + 5*x^4),x)

[Out]

3*log(x^2 + 9/5) - 3*x*log((25*x^2)/9 + 5) - (exp(exp(x))*(x - 1))/x

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sympy [A]  time = 0.44, size = 32, normalized size = 1.14 \begin {gather*} - 3 x \log {\left (\frac {25 x^{2}}{9} + 5 \right )} + 3 \log {\left (5 x^{2} + 9 \right )} + \frac {\left (1 - x\right ) e^{e^{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x**4+5*x**3-9*x**2+9*x)*exp(x)-5*x**2-9)*exp(exp(x))+(-15*x**4-27*x**2)*ln(25/9*x**2+5)-30*x**
4+30*x**3)/(5*x**4+9*x**2),x)

[Out]

-3*x*log(25*x**2/9 + 5) + 3*log(5*x**2 + 9) + (1 - x)*exp(exp(x))/x

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