3.95.98 \(\int \frac {(16 x+16 x^2+e^{2 x} (-x-x^2)) \log (-\frac {1}{x+x^2})+(9+18 x) \log ^{18}(-\frac {1}{x+x^2})+\log ^9(-\frac {1}{x+x^2}) (e^x (-9-18 x)+e^x (x+x^2) \log (-\frac {1}{x+x^2}))}{(8 x+8 x^2) \log (-\frac {1}{x+x^2})} \, dx\)

Optimal. Leaf size=28 \[ 2 x-\frac {1}{16} \left (-e^x+\log ^9\left (-\frac {1}{x+x^2}\right )\right )^2 \]

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Rubi [F]  time = 2.98, 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 {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{\left (8 x+8 x^2\right ) \log \left (-\frac {1}{x+x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((16*x + 16*x^2 + E^(2*x)*(-x - x^2))*Log[-(x + x^2)^(-1)] + (9 + 18*x)*Log[-(x + x^2)^(-1)]^18 + Log[-(x
+ x^2)^(-1)]^9*(E^x*(-9 - 18*x) + E^x*(x + x^2)*Log[-(x + x^2)^(-1)]))/((8*x + 8*x^2)*Log[-(x + x^2)^(-1)]),x]

[Out]

-1/16*E^(2*x) + 2*x + (9*Defer[Int][(E^x*Log[-(1/(x*(1 + x)))]^8)/(-1 - x), x])/8 - (9*Defer[Int][(E^x*Log[-(1
/(x*(1 + x)))]^8)/x, x])/8 + Defer[Int][E^x*Log[-(1/(x*(1 + x)))]^9, x]/8 + (9*Defer[Int][Log[-(1/(x*(1 + x)))
]^17/x, x])/8 + (9*Defer[Int][Log[-(1/(x*(1 + x)))]^17/(1 + x), x])/8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (16 x+16 x^2+e^{2 x} \left (-x-x^2\right )\right ) \log \left (-\frac {1}{x+x^2}\right )+(9+18 x) \log ^{18}\left (-\frac {1}{x+x^2}\right )+\log ^9\left (-\frac {1}{x+x^2}\right ) \left (e^x (-9-18 x)+e^x \left (x+x^2\right ) \log \left (-\frac {1}{x+x^2}\right )\right )}{x (8+8 x) \log \left (-\frac {1}{x+x^2}\right )} \, dx\\ &=\int \frac {-\left (\left (-16+e^{2 x}\right ) x (1+x)\right )-9 e^x (1+2 x) \log ^8\left (-\frac {1}{x+x^2}\right )+e^x x (1+x) \log ^9\left (-\frac {1}{x+x^2}\right )+9 (1+2 x) \log ^{17}\left (-\frac {1}{x+x^2}\right )}{8 x (1+x)} \, dx\\ &=\frac {1}{8} \int \frac {-\left (\left (-16+e^{2 x}\right ) x (1+x)\right )-9 e^x (1+2 x) \log ^8\left (-\frac {1}{x+x^2}\right )+e^x x (1+x) \log ^9\left (-\frac {1}{x+x^2}\right )+9 (1+2 x) \log ^{17}\left (-\frac {1}{x+x^2}\right )}{x (1+x)} \, dx\\ &=\frac {1}{8} \int \left (-e^{2 x}+\frac {e^x \log ^8\left (-\frac {1}{x (1+x)}\right ) \left (-9-18 x+x \log \left (-\frac {1}{x+x^2}\right )+x^2 \log \left (-\frac {1}{x+x^2}\right )\right )}{x (1+x)}+\frac {16 x+16 x^2+9 \log ^{17}\left (-\frac {1}{x+x^2}\right )+18 x \log ^{17}\left (-\frac {1}{x+x^2}\right )}{x (1+x)}\right ) \, dx\\ &=-\left (\frac {1}{8} \int e^{2 x} \, dx\right )+\frac {1}{8} \int \frac {e^x \log ^8\left (-\frac {1}{x (1+x)}\right ) \left (-9-18 x+x \log \left (-\frac {1}{x+x^2}\right )+x^2 \log \left (-\frac {1}{x+x^2}\right )\right )}{x (1+x)} \, dx+\frac {1}{8} \int \frac {16 x+16 x^2+9 \log ^{17}\left (-\frac {1}{x+x^2}\right )+18 x \log ^{17}\left (-\frac {1}{x+x^2}\right )}{x (1+x)} \, dx\\ &=-\frac {e^{2 x}}{16}+\frac {1}{8} \int \left (16+\frac {9 (1+2 x) \log ^{17}\left (-\frac {1}{x (1+x)}\right )}{x (1+x)}\right ) \, dx+\frac {1}{8} \int \frac {e^x \log ^8\left (-\frac {1}{x (1+x)}\right ) \left (-9 (1+2 x)+x (1+x) \log \left (-\frac {1}{x+x^2}\right )\right )}{x (1+x)} \, dx\\ &=-\frac {e^{2 x}}{16}+2 x+\frac {1}{8} \int \left (\frac {9 e^x (-1-2 x) \log ^8\left (-\frac {1}{x (1+x)}\right )}{x (1+x)}+e^x \log ^9\left (-\frac {1}{x (1+x)}\right )\right ) \, dx+\frac {9}{8} \int \frac {(1+2 x) \log ^{17}\left (-\frac {1}{x (1+x)}\right )}{x (1+x)} \, dx\\ &=-\frac {e^{2 x}}{16}+2 x+\frac {1}{8} \int e^x \log ^9\left (-\frac {1}{x (1+x)}\right ) \, dx+\frac {9}{8} \int \frac {e^x (-1-2 x) \log ^8\left (-\frac {1}{x (1+x)}\right )}{x (1+x)} \, dx+\frac {9}{8} \int \left (\frac {\log ^{17}\left (-\frac {1}{x (1+x)}\right )}{x}+\frac {\log ^{17}\left (-\frac {1}{x (1+x)}\right )}{1+x}\right ) \, dx\\ &=-\frac {e^{2 x}}{16}+2 x+\frac {1}{8} \int e^x \log ^9\left (-\frac {1}{x (1+x)}\right ) \, dx+\frac {9}{8} \int \frac {\log ^{17}\left (-\frac {1}{x (1+x)}\right )}{x} \, dx+\frac {9}{8} \int \frac {\log ^{17}\left (-\frac {1}{x (1+x)}\right )}{1+x} \, dx+\frac {9}{8} \int \left (\frac {e^x \log ^8\left (-\frac {1}{x (1+x)}\right )}{-1-x}-\frac {e^x \log ^8\left (-\frac {1}{x (1+x)}\right )}{x}\right ) \, dx\\ &=-\frac {e^{2 x}}{16}+2 x+\frac {1}{8} \int e^x \log ^9\left (-\frac {1}{x (1+x)}\right ) \, dx+\frac {9}{8} \int \frac {e^x \log ^8\left (-\frac {1}{x (1+x)}\right )}{-1-x} \, dx-\frac {9}{8} \int \frac {e^x \log ^8\left (-\frac {1}{x (1+x)}\right )}{x} \, dx+\frac {9}{8} \int \frac {\log ^{17}\left (-\frac {1}{x (1+x)}\right )}{x} \, dx+\frac {9}{8} \int \frac {\log ^{17}\left (-\frac {1}{x (1+x)}\right )}{1+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 53, normalized size = 1.89 \begin {gather*} \frac {1}{8} e^x \log ^9\left (-\frac {1}{x+x^2}\right )+\frac {1}{8} \left (-\frac {e^{2 x}}{2}+16 x-\frac {1}{2} \log ^{18}\left (-\frac {1}{x+x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((16*x + 16*x^2 + E^(2*x)*(-x - x^2))*Log[-(x + x^2)^(-1)] + (9 + 18*x)*Log[-(x + x^2)^(-1)]^18 + Lo
g[-(x + x^2)^(-1)]^9*(E^x*(-9 - 18*x) + E^x*(x + x^2)*Log[-(x + x^2)^(-1)]))/((8*x + 8*x^2)*Log[-(x + x^2)^(-1
)]),x]

[Out]

(E^x*Log[-(x + x^2)^(-1)]^9)/8 + (-1/2*E^(2*x) + 16*x - Log[-(x + x^2)^(-1)]^18/2)/8

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fricas [A]  time = 0.66, size = 40, normalized size = 1.43 \begin {gather*} -\frac {1}{16} \, \log \left (-\frac {1}{x^{2} + x}\right )^{18} + \frac {1}{8} \, e^{x} \log \left (-\frac {1}{x^{2} + x}\right )^{9} + 2 \, x - \frac {1}{16} \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x+9)*log(-1/(x^2+x))^18+((x^2+x)*exp(x)*log(-1/(x^2+x))+(-18*x-9)*exp(x))*log(-1/(x^2+x))^9+((-
x^2-x)*exp(x)^2+16*x^2+16*x)*log(-1/(x^2+x)))/(8*x^2+8*x)/log(-1/(x^2+x)),x, algorithm="fricas")

[Out]

-1/16*log(-1/(x^2 + x))^18 + 1/8*e^x*log(-1/(x^2 + x))^9 + 2*x - 1/16*e^(2*x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x+9)*log(-1/(x^2+x))^18+((x^2+x)*exp(x)*log(-1/(x^2+x))+(-18*x-9)*exp(x))*log(-1/(x^2+x))^9+((-
x^2-x)*exp(x)^2+16*x^2+16*x)*log(-1/(x^2+x)))/(8*x^2+8*x)/log(-1/(x^2+x)),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (18 x +9\right ) \ln \left (-\frac {1}{x^{2}+x}\right )^{18}+\left (\left (x^{2}+x \right ) {\mathrm e}^{x} \ln \left (-\frac {1}{x^{2}+x}\right )+\left (-18 x -9\right ) {\mathrm e}^{x}\right ) \ln \left (-\frac {1}{x^{2}+x}\right )^{9}+\left (\left (-x^{2}-x \right ) {\mathrm e}^{2 x}+16 x^{2}+16 x \right ) \ln \left (-\frac {1}{x^{2}+x}\right )}{\left (8 x^{2}+8 x \right ) \ln \left (-\frac {1}{x^{2}+x}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((18*x+9)*ln(-1/(x^2+x))^18+((x^2+x)*exp(x)*ln(-1/(x^2+x))+(-18*x-9)*exp(x))*ln(-1/(x^2+x))^9+((-x^2-x)*ex
p(x)^2+16*x^2+16*x)*ln(-1/(x^2+x)))/(8*x^2+8*x)/ln(-1/(x^2+x)),x)

[Out]

int(((18*x+9)*ln(-1/(x^2+x))^18+((x^2+x)*exp(x)*ln(-1/(x^2+x))+(-18*x-9)*exp(x))*ln(-1/(x^2+x))^9+((-x^2-x)*ex
p(x)^2+16*x^2+16*x)*ln(-1/(x^2+x)))/(8*x^2+8*x)/ln(-1/(x^2+x)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{16} \, \log \relax (x)^{18} - \frac {21879}{8} \, \log \relax (x)^{8} \log \left (-x - 1\right )^{10} - 1989 \, \log \relax (x)^{7} \log \left (-x - 1\right )^{11} - \frac {4641}{4} \, \log \relax (x)^{6} \log \left (-x - 1\right )^{12} - \frac {1071}{2} \, \log \relax (x)^{5} \log \left (-x - 1\right )^{13} - \frac {765}{4} \, \log \relax (x)^{4} \log \left (-x - 1\right )^{14} - 51 \, \log \relax (x)^{3} \log \left (-x - 1\right )^{15} - \frac {153}{16} \, \log \relax (x)^{2} \log \left (-x - 1\right )^{16} - \frac {9}{8} \, \log \relax (x) \log \left (-x - 1\right )^{17} - \frac {1}{16} \, \log \left (-x - 1\right )^{18} - \frac {1}{8} \, e^{x} \log \relax (x)^{9} - \frac {1}{8} \, {\left (24310 \, \log \relax (x)^{9} + e^{x}\right )} \log \left (-x - 1\right )^{9} - \frac {9}{8} \, {\left (2431 \, \log \relax (x)^{10} + e^{x} \log \relax (x)\right )} \log \left (-x - 1\right )^{8} - \frac {9}{2} \, {\left (442 \, \log \relax (x)^{11} + e^{x} \log \relax (x)^{2}\right )} \log \left (-x - 1\right )^{7} - \frac {21}{4} \, {\left (221 \, \log \relax (x)^{12} + 2 \, e^{x} \log \relax (x)^{3}\right )} \log \left (-x - 1\right )^{6} - \frac {63}{4} \, {\left (34 \, \log \relax (x)^{13} + e^{x} \log \relax (x)^{4}\right )} \log \left (-x - 1\right )^{5} - \frac {9}{4} \, {\left (85 \, \log \relax (x)^{14} + 7 \, e^{x} \log \relax (x)^{5}\right )} \log \left (-x - 1\right )^{4} - \frac {3}{2} \, {\left (34 \, \log \relax (x)^{15} + 7 \, e^{x} \log \relax (x)^{6}\right )} \log \left (-x - 1\right )^{3} - \frac {9}{16} \, {\left (17 \, \log \relax (x)^{16} + 8 \, e^{x} \log \relax (x)^{7}\right )} \log \left (-x - 1\right )^{2} + \frac {1}{8} \, e^{\left (-2\right )} E_{1}\left (-2 \, x - 2\right ) - \frac {9}{8} \, {\left (\log \relax (x)^{17} + e^{x} \log \relax (x)^{8}\right )} \log \left (-x - 1\right ) + 2 \, x - \frac {1}{8} \, \int \frac {x e^{\left (2 \, x\right )}}{x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x+9)*log(-1/(x^2+x))^18+((x^2+x)*exp(x)*log(-1/(x^2+x))+(-18*x-9)*exp(x))*log(-1/(x^2+x))^9+((-
x^2-x)*exp(x)^2+16*x^2+16*x)*log(-1/(x^2+x)))/(8*x^2+8*x)/log(-1/(x^2+x)),x, algorithm="maxima")

[Out]

-1/16*log(x)^18 - 21879/8*log(x)^8*log(-x - 1)^10 - 1989*log(x)^7*log(-x - 1)^11 - 4641/4*log(x)^6*log(-x - 1)
^12 - 1071/2*log(x)^5*log(-x - 1)^13 - 765/4*log(x)^4*log(-x - 1)^14 - 51*log(x)^3*log(-x - 1)^15 - 153/16*log
(x)^2*log(-x - 1)^16 - 9/8*log(x)*log(-x - 1)^17 - 1/16*log(-x - 1)^18 - 1/8*e^x*log(x)^9 - 1/8*(24310*log(x)^
9 + e^x)*log(-x - 1)^9 - 9/8*(2431*log(x)^10 + e^x*log(x))*log(-x - 1)^8 - 9/2*(442*log(x)^11 + e^x*log(x)^2)*
log(-x - 1)^7 - 21/4*(221*log(x)^12 + 2*e^x*log(x)^3)*log(-x - 1)^6 - 63/4*(34*log(x)^13 + e^x*log(x)^4)*log(-
x - 1)^5 - 9/4*(85*log(x)^14 + 7*e^x*log(x)^5)*log(-x - 1)^4 - 3/2*(34*log(x)^15 + 7*e^x*log(x)^6)*log(-x - 1)
^3 - 9/16*(17*log(x)^16 + 8*e^x*log(x)^7)*log(-x - 1)^2 + 1/8*e^(-2)*exp_integral_e(1, -2*x - 2) - 9/8*(log(x)
^17 + e^x*log(x)^8)*log(-x - 1) + 2*x - 1/8*integrate(x*e^(2*x)/(x + 1), x)

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mupad [B]  time = 7.48, size = 40, normalized size = 1.43 \begin {gather*} -\frac {{\ln \left (-\frac {1}{x^2+x}\right )}^{18}}{16}+\frac {{\mathrm {e}}^x\,{\ln \left (-\frac {1}{x^2+x}\right )}^9}{8}+2\,x-\frac {{\mathrm {e}}^{2\,x}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-1/(x + x^2))^18*(18*x + 9) - log(-1/(x + x^2))^9*(exp(x)*(18*x + 9) - exp(x)*log(-1/(x + x^2))*(x +
x^2)) + log(-1/(x + x^2))*(16*x - exp(2*x)*(x + x^2) + 16*x^2))/(log(-1/(x + x^2))*(8*x + 8*x^2)),x)

[Out]

2*x - exp(2*x)/16 - log(-1/(x + x^2))^18/16 + (exp(x)*log(-1/(x + x^2))^9)/8

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sympy [A]  time = 10.43, size = 39, normalized size = 1.39 \begin {gather*} 2 x - \frac {e^{2 x}}{16} + \frac {e^{x} \log {\left (- \frac {1}{x^{2} + x} \right )}^{9}}{8} - \frac {\log {\left (- \frac {1}{x^{2} + x} \right )}^{18}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x+9)*ln(-1/(x**2+x))**18+((x**2+x)*exp(x)*ln(-1/(x**2+x))+(-18*x-9)*exp(x))*ln(-1/(x**2+x))**9+
((-x**2-x)*exp(x)**2+16*x**2+16*x)*ln(-1/(x**2+x)))/(8*x**2+8*x)/ln(-1/(x**2+x)),x)

[Out]

2*x - exp(2*x)/16 + exp(x)*log(-1/(x**2 + x))**9/8 - log(-1/(x**2 + x))**18/16

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