3.61.7 \(\int \frac {-20 x-4 x^3+3 x^2 \log (e^{2 x+\frac {2 (-5+2 x)}{x}})}{\log ^3(e^{2 x+\frac {2 (-5+2 x)}{x}})} \, dx\)

Optimal. Leaf size=19 \[ \frac {x^3}{\log ^2\left (e^{4-\frac {10}{x}+2 x}\right )} \]

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Rubi [F]  time = 0.34, 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 {-20 x-4 x^3+3 x^2 \log \left (e^{2 x+\frac {2 (-5+2 x)}{x}}\right )}{\log ^3\left (e^{2 x+\frac {2 (-5+2 x)}{x}}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-20*x - 4*x^3 + 3*x^2*Log[E^(2*x + (2*(-5 + 2*x))/x)])/Log[E^(2*x + (2*(-5 + 2*x))/x)]^3,x]

[Out]

-20*Defer[Int][x/Log[E^(2*(2 - 5/x + x))]^3, x] - 4*Defer[Int][x^3/Log[E^(2*(2 - 5/x + x))]^3, x] + 3*Defer[In
t][x^2/Log[E^(2*(2 - 5/x + x))]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20 x-4 x^3+3 x^2 \log \left (e^{2 x+\frac {2 (-5+2 x)}{x}}\right )}{\log ^3\left (e^{\frac {2 \left (-5+2 x+x^2\right )}{x}}\right )} \, dx\\ &=\int \left (-\frac {4 x \left (5+x^2\right )}{\log ^3\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )}+\frac {3 x^2}{\log ^2\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )}\right ) \, dx\\ &=3 \int \frac {x^2}{\log ^2\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )} \, dx-4 \int \frac {x \left (5+x^2\right )}{\log ^3\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )} \, dx\\ &=3 \int \frac {x^2}{\log ^2\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )} \, dx-4 \int \left (\frac {5 x}{\log ^3\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )}+\frac {x^3}{\log ^3\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )}\right ) \, dx\\ &=3 \int \frac {x^2}{\log ^2\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )} \, dx-4 \int \frac {x^3}{\log ^3\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )} \, dx-20 \int \frac {x}{\log ^3\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.32, size = 46, normalized size = 2.42 \begin {gather*} \frac {1}{4} \left (\frac {10}{x}-2 x+\frac {4 x^3}{\log ^2\left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )}+\log \left (e^{2 \left (2-\frac {5}{x}+x\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x - 4*x^3 + 3*x^2*Log[E^(2*x + (2*(-5 + 2*x))/x)])/Log[E^(2*x + (2*(-5 + 2*x))/x)]^3,x]

[Out]

(10/x - 2*x + (4*x^3)/Log[E^(2*(2 - 5/x + x))]^2 + Log[E^(2*(2 - 5/x + x))])/4

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fricas [B]  time = 0.56, size = 45, normalized size = 2.37 \begin {gather*} \frac {x^{5} + 4 \, x^{4} + 16 \, x^{3} - 24 \, x^{2} - 80 \, x + 100}{4 \, {\left (x^{4} + 4 \, x^{3} - 6 \, x^{2} - 20 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(exp((2*x-5)/x)^2*exp(x)^2)-4*x^3-20*x)/log(exp((2*x-5)/x)^2*exp(x)^2)^3,x, algorithm="fri
cas")

[Out]

1/4*(x^5 + 4*x^4 + 16*x^3 - 24*x^2 - 80*x + 100)/(x^4 + 4*x^3 - 6*x^2 - 20*x + 25)

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giac [A]  time = 0.16, size = 31, normalized size = 1.63 \begin {gather*} \frac {1}{4} \, x + \frac {22 \, x^{3} - 4 \, x^{2} - 105 \, x + 100}{4 \, {\left (x^{2} + 2 \, x - 5\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(exp((2*x-5)/x)^2*exp(x)^2)-4*x^3-20*x)/log(exp((2*x-5)/x)^2*exp(x)^2)^3,x, algorithm="gia
c")

[Out]

1/4*x + 1/4*(22*x^3 - 4*x^2 - 105*x + 100)/(x^2 + 2*x - 5)^2

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maple [C]  time = 0.23, size = 298, normalized size = 15.68

method	result	size
risch	\(-\frac {4 x^{3}}{\left (\pi \mathrm {csgn}\left (i {\mathrm e}^{\frac {2 x -5}{x}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {4 x -10}{x}}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {2 x -5}{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {4 x -10}{x}}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{\frac {4 x -10}{x}}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {4 x -10}{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {2 x^{2}+4 x -10}{x}}\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {4 x -10}{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {2 x^{2}+4 x -10}{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+\pi \mathrm {csgn}\left (i {\mathrm e}^{\frac {2 x^{2}+4 x -10}{x}}\right )^{3}-\pi \mathrm {csgn}\left (i {\mathrm e}^{\frac {2 x^{2}+4 x -10}{x}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+\pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+4 i \ln \left ({\mathrm e}^{x}\right )+4 i \ln \left ({\mathrm e}^{\frac {2 x -5}{x}}\right )\right )^{2}}\)	\(298\)
default	\(\text {Expression too large to display}\)	\(9401\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2*ln(exp((2*x-5)/x)^2*exp(x)^2)-4*x^3-20*x)/ln(exp((2*x-5)/x)^2*exp(x)^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.35, size = 41, normalized size = 2.16 \begin {gather*} \frac {1}{4} \, x + \frac {22 \, x^{3} - 4 \, x^{2} - 105 \, x + 100}{4 \, {\left (x^{4} + 4 \, x^{3} - 6 \, x^{2} - 20 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(exp((2*x-5)/x)^2*exp(x)^2)-4*x^3-20*x)/log(exp((2*x-5)/x)^2*exp(x)^2)^3,x, algorithm="max
ima")

[Out]

1/4*x + 1/4*(22*x^3 - 4*x^2 - 105*x + 100)/(x^4 + 4*x^3 - 6*x^2 - 20*x + 25)

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mupad [B]  time = 4.43, size = 29, normalized size = 1.53 \begin {gather*} \frac {x}{4}-\frac {-\frac {11\,x^3}{2}+x^2+\frac {105\,x}{4}-25}{{\left (x^2+2\,x-5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x - 3*x^2*log(exp(2*x)*exp((2*(2*x - 5))/x)) + 4*x^3)/log(exp(2*x)*exp((2*(2*x - 5))/x))^3,x)

[Out]

x/4 - ((105*x)/4 + x^2 - (11*x^3)/2 - 25)/(2*x + x^2 - 5)^2

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sympy [B]  time = 0.21, size = 37, normalized size = 1.95 \begin {gather*} \frac {x}{4} + \frac {22 x^{3} - 4 x^{2} - 105 x + 100}{4 x^{4} + 16 x^{3} - 24 x^{2} - 80 x + 100} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2*ln(exp((2*x-5)/x)**2*exp(x)**2)-4*x**3-20*x)/ln(exp((2*x-5)/x)**2*exp(x)**2)**3,x)

[Out]

x/4 + (22*x**3 - 4*x**2 - 105*x + 100)/(4*x**4 + 16*x**3 - 24*x**2 - 80*x + 100)

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