3.66.84 \(\int \frac {x-e^{5 x} x-5 x^2+(1-x^2+e^{5 x} (-1+x^2)) \log ^2(e^{-5 x} (x-e^{5 x} x))}{(-x^2+e^{5 x} x^2) \log ^2(e^{-5 x} (x-e^{5 x} x))} \, dx\)

Optimal. Leaf size=20 \[ 1+\frac {1}{x}+x+\frac {1}{\log \left (-x+e^{-5 x} x\right )} \]

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Rubi [F]  time = 1.90, 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 {x-e^{5 x} x-5 x^2+\left (1-x^2+e^{5 x} \left (-1+x^2\right )\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )}{\left (-x^2+e^{5 x} x^2\right ) \log ^2\left (e^{-5 x} \left (x-e^{5 x} x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x - E^(5*x)*x - 5*x^2 + (1 - x^2 + E^(5*x)*(-1 + x^2))*Log[(x - E^(5*x)*x)/E^(5*x)]^2)/((-x^2 + E^(5*x)*x
^2)*Log[(x - E^(5*x)*x)/E^(5*x)]^2),x]

[Out]

x^(-1) + x + Defer[Int][1/((1 - E^x)*Log[-x + x/E^(5*x)]^2), x] + 4*Defer[Int][1/((1 + E^x + E^(2*x) + E^(3*x)
+ E^(4*x))*Log[-x + x/E^(5*x)]^2), x] + 3*Defer[Int][E^x/((1 + E^x + E^(2*x) + E^(3*x) + E^(4*x))*Log[-x + x/
E^(5*x)]^2), x] + 2*Defer[Int][E^(2*x)/((1 + E^x + E^(2*x) + E^(3*x) + E^(4*x))*Log[-x + x/E^(5*x)]^2), x] + D
efer[Int][E^(3*x)/((1 + E^x + E^(2*x) + E^(3*x) + E^(4*x))*Log[-x + x/E^(5*x)]^2), x] - Defer[Int][1/(x*Log[-x
+ x/E^(5*x)]^2), x]

Rubi steps

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

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Mathematica [A]  time = 0.53, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{x}+x+\frac {1}{\log \left (\left (-1+e^{-5 x}\right ) x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - E^(5*x)*x - 5*x^2 + (1 - x^2 + E^(5*x)*(-1 + x^2))*Log[(x - E^(5*x)*x)/E^(5*x)]^2)/((-x^2 + E^(
5*x)*x^2)*Log[(x - E^(5*x)*x)/E^(5*x)]^2),x]

[Out]

x^(-1) + x + Log[(-1 + E^(-5*x))*x]^(-1)

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fricas [B]  time = 0.52, size = 48, normalized size = 2.40 \begin {gather*} \frac {{\left (x^{2} + 1\right )} \log \left (-{\left (x e^{\left (5 \, x\right )} - x\right )} e^{\left (-5 \, x\right )}\right ) + x}{x \log \left (-{\left (x e^{\left (5 \, x\right )} - x\right )} e^{\left (-5 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-1)*exp(5*x)-x^2+1)*log((-x*exp(5*x)+x)/exp(5*x))^2-x*exp(5*x)-5*x^2+x)/(x^2*exp(5*x)-x^2)/log
((-x*exp(5*x)+x)/exp(5*x))^2,x, algorithm="fricas")

[Out]

((x^2 + 1)*log(-(x*e^(5*x) - x)*e^(-5*x)) + x)/(x*log(-(x*e^(5*x) - x)*e^(-5*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((((x^2-1)*exp(5*x)-x^2+1)*log((-x*exp(5*x)+x)/exp(5*x))^2-x*exp(5*x)-5*x^2+x)/(x^2*exp(5*x)-x^2)/log
((-x*exp(5*x)+x)/exp(5*x))^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.25, size = 64, normalized size = 3.20

method	result	size
norman	\(\frac {x +x^{2} \ln \left (\left (-x \,{\mathrm e}^{5 x}+x \right ) {\mathrm e}^{-5 x}\right )+\ln \left (\left (-x \,{\mathrm e}^{5 x}+x \right ) {\mathrm e}^{-5 x}\right )}{x \ln \left (\left (-x \,{\mathrm e}^{5 x}+x \right ) {\mathrm e}^{-5 x}\right )}\)	\(64\)
risch	\(\frac {x^{2}+1}{x}+\frac {2 i}{\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-5 x} \left ({\mathrm e}^{5 x}-1\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )^{2}+\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-5 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-5 x} \left ({\mathrm e}^{5 x}-1\right )\right )-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{5 x}-1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-5 x} \left ({\mathrm e}^{5 x}-1\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-5 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-5 x} \left ({\mathrm e}^{5 x}-1\right )\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{-5 x} \left ({\mathrm e}^{5 x}-1\right )\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-5 x} \left ({\mathrm e}^{5 x}-1\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )^{2}-\pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )^{3}+2 \pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{5 x}-1\right ) {\mathrm e}^{-5 x}\right )^{2}-2 \pi -2 i \ln \left ({\mathrm e}^{5 x}\right )+2 i \ln \relax (x )+2 i \ln \left ({\mathrm e}^{5 x}-1\right )}\)	\(286\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2-1)*exp(5*x)-x^2+1)*ln((-x*exp(5*x)+x)/exp(5*x))^2-x*exp(5*x)-5*x^2+x)/(x^2*exp(5*x)-x^2)/ln((-x*exp
(5*x)+x)/exp(5*x))^2,x,method=_RETURNVERBOSE)

[Out]

(x+x^2*ln((-x*exp(5*x)+x)/exp(5*x))+ln((-x*exp(5*x)+x)/exp(5*x)))/x/ln((-x*exp(5*x)+x)/exp(5*x))

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maxima [B]  time = 0.52, size = 100, normalized size = 5.00 \begin {gather*} \frac {5 \, x^{3} - {\left (x^{2} + 1\right )} \log \relax (x) - {\left (x^{2} + 1\right )} \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) - {\left (x^{2} + 1\right )} \log \left (-e^{x} + 1\right ) + 4 \, x}{5 \, x^{2} - x \log \relax (x) - x \log \left (e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1\right ) - x \log \left (-e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-1)*exp(5*x)-x^2+1)*log((-x*exp(5*x)+x)/exp(5*x))^2-x*exp(5*x)-5*x^2+x)/(x^2*exp(5*x)-x^2)/log
((-x*exp(5*x)+x)/exp(5*x))^2,x, algorithm="maxima")

[Out]

(5*x^3 - (x^2 + 1)*log(x) - (x^2 + 1)*log(e^(4*x) + e^(3*x) + e^(2*x) + e^x + 1) - (x^2 + 1)*log(-e^x + 1) + 4
*x)/(5*x^2 - x*log(x) - x*log(e^(4*x) + e^(3*x) + e^(2*x) + e^x + 1) - x*log(-e^x + 1))

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mupad [B]  time = 4.22, size = 18, normalized size = 0.90 \begin {gather*} x+\frac {1}{\ln \left (x\,{\mathrm {e}}^{-5\,x}-x\right )}+\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - x*exp(5*x) + log(exp(-5*x)*(x - x*exp(5*x)))^2*(exp(5*x)*(x^2 - 1) - x^2 + 1) - 5*x^2)/(log(exp(-5*x)
*(x - x*exp(5*x)))^2*(x^2*exp(5*x) - x^2)),x)

[Out]

x + 1/log(x*exp(-5*x) - x) + 1/x

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sympy [A]  time = 0.20, size = 20, normalized size = 1.00 \begin {gather*} x + \frac {1}{\log {\left (\left (- x e^{5 x} + x\right ) e^{- 5 x} \right )}} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-1)*exp(5*x)-x**2+1)*ln((-x*exp(5*x)+x)/exp(5*x))**2-x*exp(5*x)-5*x**2+x)/(x**2*exp(5*x)-x**2
)/ln((-x*exp(5*x)+x)/exp(5*x))**2,x)

[Out]

x + 1/log((-x*exp(5*x) + x)*exp(-5*x)) + 1/x

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