3.47.93 \(\int \frac {-98+68 x-2 x^2-62 x^3+16 x^4-2 x^6}{(49 x-14 x^2+x^3-14 x^4+2 x^5+x^7) \log ^2(e^{-\frac {15-5 x}{-7+x+x^3}} x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {2}{\log \left (e^{-\frac {5 (3-x)}{-7+x+x^3}} x\right )} \]

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Rubi [F]  time = 0.97, 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 {-98+68 x-2 x^2-62 x^3+16 x^4-2 x^6}{\left (49 x-14 x^2+x^3-14 x^4+2 x^5+x^7\right ) \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-98 + 68*x - 2*x^2 - 62*x^3 + 16*x^4 - 2*x^6)/((49*x - 14*x^2 + x^3 - 14*x^4 + 2*x^5 + x^7)*Log[x/E^((15
- 5*x)/(-7 + x + x^3))]^2),x]

[Out]

-2*Defer[Int][1/(x*Log[E^((5*(-3 + x))/(-7 + x + x^3))*x]^2), x] + 180*Defer[Int][1/((-7 + x + x^3)^2*Log[E^((
5*(-3 + x))/(-7 + x + x^3))*x]^2), x] - 20*Defer[Int][x/((-7 + x + x^3)^2*Log[E^((5*(-3 + x))/(-7 + x + x^3))*
x]^2), x] - 90*Defer[Int][x^2/((-7 + x + x^3)^2*Log[E^((5*(-3 + x))/(-7 + x + x^3))*x]^2), x] + 20*Defer[Int][
1/((-7 + x + x^3)*Log[E^((5*(-3 + x))/(-7 + x + x^3))*x]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-49+34 x-x^2-31 x^3+8 x^4-x^6\right )}{x \left (7-x-x^3\right )^2 \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx\\ &=2 \int \frac {-49+34 x-x^2-31 x^3+8 x^4-x^6}{x \left (7-x-x^3\right )^2 \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx\\ &=2 \int \left (-\frac {1}{x \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )}-\frac {5 \left (-18+2 x+9 x^2\right )}{\left (-7+x+x^3\right )^2 \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )}+\frac {10}{\left (-7+x+x^3\right ) \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )}\right ) \, dx\\ &=-\left (2 \int \frac {1}{x \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx\right )-10 \int \frac {-18+2 x+9 x^2}{\left (-7+x+x^3\right )^2 \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx+20 \int \frac {1}{\left (-7+x+x^3\right ) \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx\\ &=-\left (2 \int \frac {1}{x \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx\right )-10 \int \left (-\frac {18}{\left (-7+x+x^3\right )^2 \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )}+\frac {2 x}{\left (-7+x+x^3\right )^2 \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )}+\frac {9 x^2}{\left (-7+x+x^3\right )^2 \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )}\right ) \, dx+20 \int \frac {1}{\left (-7+x+x^3\right ) \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx\\ &=-\left (2 \int \frac {1}{x \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx\right )-20 \int \frac {x}{\left (-7+x+x^3\right )^2 \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx+20 \int \frac {1}{\left (-7+x+x^3\right ) \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx-90 \int \frac {x^2}{\left (-7+x+x^3\right )^2 \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx+180 \int \frac {1}{\left (-7+x+x^3\right )^2 \log ^2\left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 22, normalized size = 0.92 \begin {gather*} \frac {2}{\log \left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-98 + 68*x - 2*x^2 - 62*x^3 + 16*x^4 - 2*x^6)/((49*x - 14*x^2 + x^3 - 14*x^4 + 2*x^5 + x^7)*Log[x/E
^((15 - 5*x)/(-7 + x + x^3))]^2),x]

[Out]

2/Log[E^((5*(-3 + x))/(-7 + x + x^3))*x]

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fricas [A]  time = 0.58, size = 21, normalized size = 0.88 \begin {gather*} \frac {2}{\log \left (x e^{\left (\frac {5 \, {\left (x - 3\right )}}{x^{3} + x - 7}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^6+16*x^4-62*x^3-2*x^2+68*x-98)/(x^7+2*x^5-14*x^4+x^3-14*x^2+49*x)/log(x/exp((15-5*x)/(x^3+x-7)
))^2,x, algorithm="fricas")

[Out]

2/log(x*e^(5*(x - 3)/(x^3 + x - 7)))

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giac [A]  time = 0.16, size = 29, normalized size = 1.21 \begin {gather*} \frac {2 \, {\left (x^{3} + x - 7\right )}}{x^{3} \log \relax (x) + x \log \relax (x) + 5 \, x - 7 \, \log \relax (x) - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^6+16*x^4-62*x^3-2*x^2+68*x-98)/(x^7+2*x^5-14*x^4+x^3-14*x^2+49*x)/log(x/exp((15-5*x)/(x^3+x-7)
))^2,x, algorithm="giac")

[Out]

2*(x^3 + x - 7)/(x^3*log(x) + x*log(x) + 5*x - 7*log(x) - 15)

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maple [C]  time = 0.11, size = 168, normalized size = 7.00

method	result	size
risch	\(\frac {4 i}{\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right )^{2}+\pi \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right )^{3}+2 i \ln \relax (x )-2 i \ln \left ({\mathrm e}^{-\frac {5 \left (x -3\right )}{x^{3}+x -7}}\right )}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^6+16*x^4-62*x^3-2*x^2+68*x-98)/(x^7+2*x^5-14*x^4+x^3-14*x^2+49*x)/ln(x/exp((15-5*x)/(x^3+x-7)))^2,x,
method=_RETURNVERBOSE)

[Out]

4*I/(Pi*csgn(I*x)*csgn(I*exp(5*(x-3)/(x^3+x-7)))*csgn(I*x*exp(5*(x-3)/(x^3+x-7)))-Pi*csgn(I*x)*csgn(I*x*exp(5*
(x-3)/(x^3+x-7)))^2-Pi*csgn(I*exp(5*(x-3)/(x^3+x-7)))*csgn(I*x*exp(5*(x-3)/(x^3+x-7)))^2+Pi*csgn(I*x*exp(5*(x-
3)/(x^3+x-7)))^3+2*I*ln(x)-2*I*ln(exp(-5*(x-3)/(x^3+x-7))))

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maxima [A]  time = 0.44, size = 24, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (x^{3} + x - 7\right )}}{{\left (x^{3} + x - 7\right )} \log \relax (x) + 5 \, x - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^6+16*x^4-62*x^3-2*x^2+68*x-98)/(x^7+2*x^5-14*x^4+x^3-14*x^2+49*x)/log(x/exp((15-5*x)/(x^3+x-7)
))^2,x, algorithm="maxima")

[Out]

2*(x^3 + x - 7)/((x^3 + x - 7)*log(x) + 5*x - 15)

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mupad [B]  time = 3.26, size = 32, normalized size = 1.33 \begin {gather*} \frac {2\,x^3+2\,x-14}{5\,x-7\,\ln \relax (x)+x^3\,\ln \relax (x)+x\,\ln \relax (x)-15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^2 - 68*x + 62*x^3 - 16*x^4 + 2*x^6 + 98)/(log(x*exp((5*x - 15)/(x + x^3 - 7)))^2*(49*x - 14*x^2 + x^
3 - 14*x^4 + 2*x^5 + x^7)),x)

[Out]

(2*x + 2*x^3 - 14)/(5*x - 7*log(x) + x^3*log(x) + x*log(x) - 15)

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sympy [A]  time = 0.54, size = 17, normalized size = 0.71 \begin {gather*} \frac {2}{\log {\left (x e^{- \frac {15 - 5 x}{x^{3} + x - 7}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**6+16*x**4-62*x**3-2*x**2+68*x-98)/(x**7+2*x**5-14*x**4+x**3-14*x**2+49*x)/ln(x/exp((15-5*x)/(
x**3+x-7)))**2,x)

[Out]

2/log(x*exp(-(15 - 5*x)/(x**3 + x - 7)))

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