3.8.35 \(\int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log (2+e^{15+x})+(-18-9 e^{15+x}) \log ^2(2+e^{15+x})}{16 x^2+8 e^{15+x} x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac {18 \left (2+\frac {1}{16} \left (-3+x+\log ^2\left (2+e^{15+x}\right )\right )\right )}{x} \]

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Rubi [F]  time = 0.45, 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 {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-522 - 261*E^(15 + x) + 18*E^(15 + x)*x*Log[2 + E^(15 + x)] + (-18 - 9*E^(15 + x))*Log[2 + E^(15 + x)]^2)
/(16*x^2 + 8*E^(15 + x)*x^2),x]

[Out]

261/(8*x) - (9*Log[2 + E^(15 + x)]*Defer[Int][1/((2 + E^(15 + x))*x), x])/2 + (9*Defer[Int][Log[2 + E^(15 + x)
]/x, x])/4 - (9*Defer[Int][Log[2 + E^(15 + x)]^2/x^2, x])/8 + (9*Defer[Int][(E^(15 + x)*Defer[Int][(2*x + E^(1
5 + x)*x)^(-1), x])/(2 + E^(15 + x)), x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 \left (-29+\frac {2 e^{15+x} x \log \left (2+e^{15+x}\right )}{2+e^{15+x}}-\log ^2\left (2+e^{15+x}\right )\right )}{8 x^2} \, dx\\ &=\frac {9}{8} \int \frac {-29+\frac {2 e^{15+x} x \log \left (2+e^{15+x}\right )}{2+e^{15+x}}-\log ^2\left (2+e^{15+x}\right )}{x^2} \, dx\\ &=\frac {9}{8} \int \left (-\frac {4 \log \left (2+e^{15+x}\right )}{\left (2+e^{15+x}\right ) x}+\frac {-29+2 x \log \left (2+e^{15+x}\right )-\log ^2\left (2+e^{15+x}\right )}{x^2}\right ) \, dx\\ &=\frac {9}{8} \int \frac {-29+2 x \log \left (2+e^{15+x}\right )-\log ^2\left (2+e^{15+x}\right )}{x^2} \, dx-\frac {9}{2} \int \frac {\log \left (2+e^{15+x}\right )}{\left (2+e^{15+x}\right ) x} \, dx\\ &=\frac {9}{8} \int \left (-\frac {29}{x^2}+\frac {2 \log \left (2+e^{15+x}\right )}{x}-\frac {\log ^2\left (2+e^{15+x}\right )}{x^2}\right ) \, dx+\frac {9}{2} \int \frac {e^{15+x} \int \frac {1}{2 x+e^{15+x} x} \, dx}{2+e^{15+x}} \, dx-\frac {1}{2} \left (9 \log \left (2+e^{15+x}\right )\right ) \int \frac {1}{\left (2+e^{15+x}\right ) x} \, dx\\ &=\frac {261}{8 x}-\frac {9}{8} \int \frac {\log ^2\left (2+e^{15+x}\right )}{x^2} \, dx+\frac {9}{4} \int \frac {\log \left (2+e^{15+x}\right )}{x} \, dx+\frac {9}{2} \int \frac {e^{15+x} \int \frac {1}{2 x+e^{15+x} x} \, dx}{2+e^{15+x}} \, dx-\frac {1}{2} \left (9 \log \left (2+e^{15+x}\right )\right ) \int \frac {1}{\left (2+e^{15+x}\right ) x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 24, normalized size = 1.00 \begin {gather*} \frac {9}{8} \left (\frac {29}{x}+\frac {\log ^2\left (2+e^{15+x}\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-522 - 261*E^(15 + x) + 18*E^(15 + x)*x*Log[2 + E^(15 + x)] + (-18 - 9*E^(15 + x))*Log[2 + E^(15 +
x)]^2)/(16*x^2 + 8*E^(15 + x)*x^2),x]

[Out]

(9*(29/x + Log[2 + E^(15 + x)]^2/x))/8

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fricas [A]  time = 0.60, size = 16, normalized size = 0.67 \begin {gather*} \frac {9 \, {\left (\log \left (e^{\left (x + 15\right )} + 2\right )^{2} + 29\right )}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+15)+2)-261*exp(x+15)-522)/(8*x^2*exp(
x+15)+16*x^2),x, algorithm="fricas")

[Out]

9/8*(log(e^(x + 15) + 2)^2 + 29)/x

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giac [A]  time = 0.38, size = 16, normalized size = 0.67 \begin {gather*} \frac {9 \, {\left (\log \left (e^{\left (x + 15\right )} + 2\right )^{2} + 29\right )}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+15)+2)-261*exp(x+15)-522)/(8*x^2*exp(
x+15)+16*x^2),x, algorithm="giac")

[Out]

9/8*(log(e^(x + 15) + 2)^2 + 29)/x

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maple [A]  time = 0.03, size = 18, normalized size = 0.75

method	result	size
norman	\(\frac {\frac {261}{8}+\frac {9 \ln \left ({\mathrm e}^{x +15}+2\right )^{2}}{8}}{x}\)	\(18\)
risch	\(\frac {9 \ln \left ({\mathrm e}^{x +15}+2\right )^{2}}{8 x}+\frac {261}{8 x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-9*exp(x+15)-18)*ln(exp(x+15)+2)^2+18*x*exp(x+15)*ln(exp(x+15)+2)-261*exp(x+15)-522)/(8*x^2*exp(x+15)+16
*x^2),x,method=_RETURNVERBOSE)

[Out]

(261/8+9/8*ln(exp(x+15)+2)^2)/x

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maxima [A]  time = 0.57, size = 16, normalized size = 0.67 \begin {gather*} \frac {9 \, {\left (\log \left (e^{\left (x + 15\right )} + 2\right )^{2} + 29\right )}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+15)+2)-261*exp(x+15)-522)/(8*x^2*exp(
x+15)+16*x^2),x, algorithm="maxima")

[Out]

9/8*(log(e^(x + 15) + 2)^2 + 29)/x

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mupad [B]  time = 0.64, size = 17, normalized size = 0.71 \begin {gather*} \frac {9\,\left ({\ln \left ({\mathrm {e}}^{15}\,{\mathrm {e}}^x+2\right )}^2+29\right )}{8\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(261*exp(x + 15) + log(exp(x + 15) + 2)^2*(9*exp(x + 15) + 18) - 18*x*log(exp(x + 15) + 2)*exp(x + 15) +
522)/(8*x^2*exp(x + 15) + 16*x^2),x)

[Out]

(9*(log(exp(15)*exp(x) + 2)^2 + 29))/(8*x)

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sympy [A]  time = 0.18, size = 19, normalized size = 0.79 \begin {gather*} \frac {9 \log {\left (e^{x + 15} + 2 \right )}^{2}}{8 x} + \frac {261}{8 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*exp(x+15)-18)*ln(exp(x+15)+2)**2+18*x*exp(x+15)*ln(exp(x+15)+2)-261*exp(x+15)-522)/(8*x**2*exp(
x+15)+16*x**2),x)

[Out]

9*log(exp(x + 15) + 2)**2/(8*x) + 261/(8*x)

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