3.77.36 \(\int \frac {e^{-x} (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+(e^5 (-126 x-18 x^2)+e^5 (-63 x-9 x^2) \log (4)) \log (x^2))}{(125 x+75 x^2+15 x^3+x^4) \log ^2(x^2)} \, dx\)

Optimal. Leaf size=24 \[ \frac {9 e^{5-x} (2+\log (4))}{(5+x)^2 \log \left (x^2\right )} \]

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Rubi [F]  time = 1.86, 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 {e^{-x} \left (e^5 (-180-36 x)+e^5 (-90-18 x) \log (4)+\left (e^5 \left (-126 x-18 x^2\right )+e^5 \left (-63 x-9 x^2\right ) \log (4)\right ) \log \left (x^2\right )\right )}{\left (125 x+75 x^2+15 x^3+x^4\right ) \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^5*(-180 - 36*x) + E^5*(-90 - 18*x)*Log[4] + (E^5*(-126*x - 18*x^2) + E^5*(-63*x - 9*x^2)*Log[4])*Log[x^
2])/(E^x*(125*x + 75*x^2 + 15*x^3 + x^4)*Log[x^2]^2),x]

[Out]

(-18*(2 + Log[4])*Defer[Int][E^(5 - x)/(x*Log[x^2]^2), x])/25 + (18*(2 + Log[4])*Defer[Int][E^(5 - x)/((5 + x)
^2*Log[x^2]^2), x])/5 + (18*(2 + Log[4])*Defer[Int][E^(5 - x)/((5 + x)*Log[x^2]^2), x])/25 - 18*(2 + Log[4])*D
efer[Int][E^(5 - x)/((5 + x)^3*Log[x^2]), x] - 9*(2 + Log[4])*Defer[Int][E^(5 - x)/((5 + x)^2*Log[x^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 e^{5-x} (2+\log (4)) \left (-2 (5+x)-x (7+x) \log \left (x^2\right )\right )}{x (5+x)^3 \log ^2\left (x^2\right )} \, dx\\ &=(9 (2+\log (4))) \int \frac {e^{5-x} \left (-2 (5+x)-x (7+x) \log \left (x^2\right )\right )}{x (5+x)^3 \log ^2\left (x^2\right )} \, dx\\ &=(9 (2+\log (4))) \int \left (-\frac {2 e^{5-x}}{x (5+x)^2 \log ^2\left (x^2\right )}+\frac {e^{5-x} (-7-x)}{(5+x)^3 \log \left (x^2\right )}\right ) \, dx\\ &=(9 (2+\log (4))) \int \frac {e^{5-x} (-7-x)}{(5+x)^3 \log \left (x^2\right )} \, dx-(18 (2+\log (4))) \int \frac {e^{5-x}}{x (5+x)^2 \log ^2\left (x^2\right )} \, dx\\ &=(9 (2+\log (4))) \int \left (-\frac {2 e^{5-x}}{(5+x)^3 \log \left (x^2\right )}-\frac {e^{5-x}}{(5+x)^2 \log \left (x^2\right )}\right ) \, dx-(18 (2+\log (4))) \int \left (\frac {e^{5-x}}{25 x \log ^2\left (x^2\right )}-\frac {e^{5-x}}{5 (5+x)^2 \log ^2\left (x^2\right )}-\frac {e^{5-x}}{25 (5+x) \log ^2\left (x^2\right )}\right ) \, dx\\ &=-\left (\frac {1}{25} (18 (2+\log (4))) \int \frac {e^{5-x}}{x \log ^2\left (x^2\right )} \, dx\right )+\frac {1}{25} (18 (2+\log (4))) \int \frac {e^{5-x}}{(5+x) \log ^2\left (x^2\right )} \, dx+\frac {1}{5} (18 (2+\log (4))) \int \frac {e^{5-x}}{(5+x)^2 \log ^2\left (x^2\right )} \, dx-(9 (2+\log (4))) \int \frac {e^{5-x}}{(5+x)^2 \log \left (x^2\right )} \, dx-(18 (2+\log (4))) \int \frac {e^{5-x}}{(5+x)^3 \log \left (x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 24, normalized size = 1.00 \begin {gather*} \frac {9 e^{5-x} (2+\log (4))}{(5+x)^2 \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-180 - 36*x) + E^5*(-90 - 18*x)*Log[4] + (E^5*(-126*x - 18*x^2) + E^5*(-63*x - 9*x^2)*Log[4])*
Log[x^2])/(E^x*(125*x + 75*x^2 + 15*x^3 + x^4)*Log[x^2]^2),x]

[Out]

(9*E^(5 - x)*(2 + Log[4]))/((5 + x)^2*Log[x^2])

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fricas [A]  time = 0.87, size = 30, normalized size = 1.25 \begin {gather*} \frac {18 \, {\left (e^{5} \log \relax (2) + e^{5}\right )} e^{\left (-x\right )}}{{\left (x^{2} + 10 \, x + 25\right )} \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-9*x^2-63*x)*exp(5)*log(2)+(-18*x^2-126*x)*exp(5))*log(x^2)+2*(-18*x-90)*exp(5)*log(2)+(-36*x-1
80)*exp(5))/(x^4+15*x^3+75*x^2+125*x)/exp(x)/log(x^2)^2,x, algorithm="fricas")

[Out]

18*(e^5*log(2) + e^5)*e^(-x)/((x^2 + 10*x + 25)*log(x^2))

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giac [A]  time = 0.20, size = 42, normalized size = 1.75 \begin {gather*} \frac {18 \, {\left (e^{\left (-x + 5\right )} \log \relax (2) + e^{\left (-x + 5\right )}\right )}}{x^{2} \log \left (x^{2}\right ) + 10 \, x \log \left (x^{2}\right ) + 25 \, \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-9*x^2-63*x)*exp(5)*log(2)+(-18*x^2-126*x)*exp(5))*log(x^2)+2*(-18*x-90)*exp(5)*log(2)+(-36*x-1
80)*exp(5))/(x^4+15*x^3+75*x^2+125*x)/exp(x)/log(x^2)^2,x, algorithm="giac")

[Out]

18*(e^(-x + 5)*log(2) + e^(-x + 5))/(x^2*log(x^2) + 10*x*log(x^2) + 25*log(x^2))

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maple [C]  time = 0.09, size = 76, normalized size = 3.17

method	result	size
risch	\(\frac {36 i \left (1+\ln \relax (2)\right ) {\mathrm e}^{5-x}}{\left (4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right ) \left (x^{2}+10 x +25\right )}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(-9*x^2-63*x)*exp(5)*ln(2)+(-18*x^2-126*x)*exp(5))*ln(x^2)+2*(-18*x-90)*exp(5)*ln(2)+(-36*x-180)*exp(5
))/(x^4+15*x^3+75*x^2+125*x)/exp(x)/ln(x^2)^2,x,method=_RETURNVERBOSE)

[Out]

36*I/(4*I*ln(x)+Pi*csgn(I*x^2)^3+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2)/(x^2+10*x+25)*(1+ln(
2))*exp(5-x)

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maxima [A]  time = 0.51, size = 26, normalized size = 1.08 \begin {gather*} \frac {9 \, {\left (\log \relax (2) + 1\right )} e^{\left (-x + 5\right )}}{{\left (x^{2} + 10 \, x + 25\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-9*x^2-63*x)*exp(5)*log(2)+(-18*x^2-126*x)*exp(5))*log(x^2)+2*(-18*x-90)*exp(5)*log(2)+(-36*x-1
80)*exp(5))/(x^4+15*x^3+75*x^2+125*x)/exp(x)/log(x^2)^2,x, algorithm="maxima")

[Out]

9*(log(2) + 1)*e^(-x + 5)/((x^2 + 10*x + 25)*log(x))

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mupad [B]  time = 5.49, size = 87, normalized size = 3.62 \begin {gather*} \frac {9\,\left (\ln \relax (2)+1\right )\,\left (7\,x\,{\mathrm {e}}^{5-x}+x^2\,{\mathrm {e}}^{5-x}-7\,x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5-x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\right )}{{\left (x+5\right )}^3}+\frac {9\,\left (\ln \relax (2)+1\right )\,\left (10\,{\mathrm {e}}^{5-x}+2\,x\,{\mathrm {e}}^{5-x}\right )}{\ln \left (x^2\right )\,{\left (x+5\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(log(x^2)*(exp(5)*(126*x + 18*x^2) + 2*exp(5)*log(2)*(63*x + 9*x^2)) + exp(5)*(36*x + 180) + 2*e
xp(5)*log(2)*(18*x + 90)))/(log(x^2)^2*(125*x + 75*x^2 + 15*x^3 + x^4)),x)

[Out]

(9*(log(2) + 1)*(7*x*exp(5 - x) + x^2*exp(5 - x) - 7*x*exp(-x)*exp(5) - x^2*exp(-x)*exp(5)))/(x + 5)^3 + (9*(l
og(2) + 1)*(10*exp(5 - x) + 2*x*exp(5 - x)))/(log(x^2)*(x + 5)^3)

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sympy [A]  time = 0.39, size = 39, normalized size = 1.62 \begin {gather*} \frac {\left (18 e^{5} \log {\relax (2 )} + 18 e^{5}\right ) e^{- x}}{x^{2} \log {\left (x^{2} \right )} + 10 x \log {\left (x^{2} \right )} + 25 \log {\left (x^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-9*x**2-63*x)*exp(5)*ln(2)+(-18*x**2-126*x)*exp(5))*ln(x**2)+2*(-18*x-90)*exp(5)*ln(2)+(-36*x-1
80)*exp(5))/(x**4+15*x**3+75*x**2+125*x)/exp(x)/ln(x**2)**2,x)

[Out]

(18*exp(5)*log(2) + 18*exp(5))*exp(-x)/(x**2*log(x**2) + 10*x*log(x**2) + 25*log(x**2))

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