3.18.40 \(\int \frac {e^x (-128+128 x) \log ^2(5)}{(x^2+e^x x \log ^2(5)) \log ^3(\frac {x \log ^2(5)}{9 x+9 e^x \log ^2(5)})} \, dx\)

Optimal. Leaf size=22 \[ \frac {64}{\log ^2\left (\frac {x}{9 \left (e^x+\frac {x}{\log ^2(5)}\right )}\right )} \]

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Rubi [A]  time = 0.27, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12, 6686} \begin {gather*} \frac {64}{\log ^2\left (\frac {x \log ^2(5)}{9 \left (x+e^x \log ^2(5)\right )}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-128 + 128*x)*Log[5]^2)/((x^2 + E^x*x*Log[5]^2)*Log[(x*Log[5]^2)/(9*x + 9*E^x*Log[5]^2)]^3),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log ^2(5) \int \frac {e^x (-128+128 x)}{\left (x^2+e^x x \log ^2(5)\right ) \log ^3\left (\frac {x \log ^2(5)}{9 x+9 e^x \log ^2(5)}\right )} \, dx\\ &=\frac {64}{\log ^2\left (\frac {x \log ^2(5)}{9 \left (x+e^x \log ^2(5)\right )}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 26, normalized size = 1.18 \begin {gather*} \frac {64}{\log ^2\left (\frac {x \log ^2(5)}{9 x+9 e^x \log ^2(5)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-128 + 128*x)*Log[5]^2)/((x^2 + E^x*x*Log[5]^2)*Log[(x*Log[5]^2)/(9*x + 9*E^x*Log[5]^2)]^3),x]

[Out]

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

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fricas [A]  time = 0.51, size = 23, normalized size = 1.05 \begin {gather*} \frac {64}{\log \left (\frac {x \log \relax (5)^{2}}{9 \, {\left (e^{x} \log \relax (5)^{2} + x\right )}}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((128*x-128)*log(5)^2*exp(x)/(x*log(5)^2*exp(x)+x^2)/log(x*log(5)^2/(9*log(5)^2*exp(x)+9*x))^3,x, alg
orithm="fricas")

[Out]

64/log(1/9*x*log(5)^2/(e^x*log(5)^2 + x))^2

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giac [B]  time = 0.35, size = 69, normalized size = 3.14 \begin {gather*} \frac {64 \, \log \relax (5)^{2}}{\log \relax (5)^{2} \log \left (x \log \relax (5)^{2}\right )^{2} - 2 \, \log \relax (5)^{2} \log \left (x \log \relax (5)^{2}\right ) \log \left (9 \, e^{x} \log \relax (5)^{2} + 9 \, x\right ) + \log \relax (5)^{2} \log \left (9 \, e^{x} \log \relax (5)^{2} + 9 \, x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((128*x-128)*log(5)^2*exp(x)/(x*log(5)^2*exp(x)+x^2)/log(x*log(5)^2/(9*log(5)^2*exp(x)+9*x))^3,x, alg
orithm="giac")

[Out]

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

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maple [C]  time = 0.11, size = 155, normalized size = 7.05

method	result	size
risch	\(-\frac {256}{\left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (5)^{2} {\mathrm e}^{x}+x}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (5)^{2} {\mathrm e}^{x}+x}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (5)^{2} {\mathrm e}^{x}+x}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (5)^{2} {\mathrm e}^{x}+x}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (5)^{2} {\mathrm e}^{x}+x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (5)^{2} {\mathrm e}^{x}+x}\right )^{3}-4 i \ln \relax (3)+4 i \ln \left (\ln \relax (5)\right )+2 i \ln \relax (x )-2 i \ln \left (\ln \relax (5)^{2} {\mathrm e}^{x}+x \right )\right )^{2}}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((128*x-128)*ln(5)^2*exp(x)/(x*ln(5)^2*exp(x)+x^2)/ln(x*ln(5)^2/(9*ln(5)^2*exp(x)+9*x))^3,x,method=_RETURNV
ERBOSE)

[Out]

-256/(Pi*csgn(I*x)*csgn(I/(ln(5)^2*exp(x)+x))*csgn(I*x/(ln(5)^2*exp(x)+x))-Pi*csgn(I*x)*csgn(I*x/(ln(5)^2*exp(
x)+x))^2-Pi*csgn(I/(ln(5)^2*exp(x)+x))*csgn(I*x/(ln(5)^2*exp(x)+x))^2+Pi*csgn(I*x/(ln(5)^2*exp(x)+x))^3-4*I*ln
(3)+4*I*ln(ln(5))+2*I*ln(x)-2*I*ln(ln(5)^2*exp(x)+x))^2

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maxima [B]  time = 0.60, size = 126, normalized size = 5.73 \begin {gather*} \frac {64 \, \log \relax (5)^{2}}{4 \, \log \relax (5)^{2} \log \relax (3)^{2} + \log \relax (5)^{2} \log \left (e^{x} \log \relax (5)^{2} + x\right )^{2} + \log \relax (5)^{2} \log \relax (x)^{2} - 8 \, \log \relax (5)^{2} \log \relax (3) \log \left (\log \relax (5)\right ) + 4 \, \log \relax (5)^{2} \log \left (\log \relax (5)\right )^{2} + 2 \, {\left (2 \, \log \relax (5)^{2} \log \relax (3) - \log \relax (5)^{2} \log \relax (x) - 2 \, \log \relax (5)^{2} \log \left (\log \relax (5)\right )\right )} \log \left (e^{x} \log \relax (5)^{2} + x\right ) - 4 \, {\left (\log \relax (5)^{2} \log \relax (3) - \log \relax (5)^{2} \log \left (\log \relax (5)\right )\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((128*x-128)*log(5)^2*exp(x)/(x*log(5)^2*exp(x)+x^2)/log(x*log(5)^2/(9*log(5)^2*exp(x)+9*x))^3,x, alg
orithm="maxima")

[Out]

64*log(5)^2/(4*log(5)^2*log(3)^2 + log(5)^2*log(e^x*log(5)^2 + x)^2 + log(5)^2*log(x)^2 - 8*log(5)^2*log(3)*lo
g(log(5)) + 4*log(5)^2*log(log(5))^2 + 2*(2*log(5)^2*log(3) - log(5)^2*log(x) - 2*log(5)^2*log(log(5)))*log(e^
x*log(5)^2 + x) - 4*(log(5)^2*log(3) - log(5)^2*log(log(5)))*log(x))

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mupad [B]  time = 1.47, size = 25, normalized size = 1.14 \begin {gather*} \frac {64}{{\ln \left (\frac {x\,{\ln \relax (5)}^2}{9\,x+9\,{\mathrm {e}}^x\,{\ln \relax (5)}^2}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*log(5)^2*(128*x - 128))/(log((x*log(5)^2)/(9*x + 9*exp(x)*log(5)^2))^3*(x^2 + x*exp(x)*log(5)^2)),
x)

[Out]

64/log((x*log(5)^2)/(9*x + 9*exp(x)*log(5)^2))^2

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sympy [A]  time = 0.23, size = 24, normalized size = 1.09 \begin {gather*} \frac {64}{\log {\left (\frac {x \log {\relax (5 )}^{2}}{9 x + 9 e^{x} \log {\relax (5 )}^{2}} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((128*x-128)*ln(5)**2*exp(x)/(x*ln(5)**2*exp(x)+x**2)/ln(x*ln(5)**2/(9*ln(5)**2*exp(x)+9*x))**3,x)

[Out]

64/log(x*log(5)**2/(9*x + 9*exp(x)*log(5)**2))**2

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