3.95.47 \(\int \frac {-50 x^2+120 x^3+30 x^4+e^{.\frac {1}{5}/x} (-50 x+120 x^2+30 x^3)+(-10 x^2+20 x^3+e^{.\frac {1}{5}/x} (-10 x+20 x^2)) \log (-1+2 x)+(-50 x^2+25 x^3+150 x^4+e^{.\frac {1}{5}/x} (10-5 x-30 x^2)+(-10 x^2+5 x^3+30 x^4+e^{.\frac {1}{5}/x} (2-x-6 x^2)) \log (-1+2 x)) \log (\frac {2+3 x}{5 x+x \log (-1+2 x)})}{(-50 x^2+25 x^3+150 x^4+(-10 x^2+5 x^3+30 x^4) \log (-1+2 x)) \log ^2(\frac {2+3 x}{5 x+x \log (-1+2 x)})} \, dx\)

Optimal. Leaf size=33 \[ \frac {e^{\left .\frac {1}{5}\right /x}+x}{\log \left (\frac {3+\frac {2}{x}}{5+\log (-1+2 x)}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 21.00, 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 {-50 x^2+120 x^3+30 x^4+e^{\left .\frac {1}{5}\right /x} \left (-50 x+120 x^2+30 x^3\right )+\left (-10 x^2+20 x^3+e^{\left .\frac {1}{5}\right /x} \left (-10 x+20 x^2\right )\right ) \log (-1+2 x)+\left (-50 x^2+25 x^3+150 x^4+e^{\left .\frac {1}{5}\right /x} \left (10-5 x-30 x^2\right )+\left (-10 x^2+5 x^3+30 x^4+e^{\left .\frac {1}{5}\right /x} \left (2-x-6 x^2\right )\right ) \log (-1+2 x)\right ) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}{\left (-50 x^2+25 x^3+150 x^4+\left (-10 x^2+5 x^3+30 x^4\right ) \log (-1+2 x)\right ) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-50*x^2 + 120*x^3 + 30*x^4 + E^(1/(5*x))*(-50*x + 120*x^2 + 30*x^3) + (-10*x^2 + 20*x^3 + E^(1/(5*x))*(-1
0*x + 20*x^2))*Log[-1 + 2*x] + (-50*x^2 + 25*x^3 + 150*x^4 + E^(1/(5*x))*(10 - 5*x - 30*x^2) + (-10*x^2 + 5*x^
3 + 30*x^4 + E^(1/(5*x))*(2 - x - 6*x^2))*Log[-1 + 2*x])*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])])/((-50*x^2 + 2
5*x^3 + 150*x^4 + (-10*x^2 + 5*x^3 + 30*x^4)*Log[-1 + 2*x])*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])]^2),x]

[Out]

(E^(1/(5*x))*(10*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])] - 5*x*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])] - 30*x^2*
Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])] + 2*Log[-1 + 2*x]*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])] - x*Log[-1 + 2
*x]*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])] - 6*x^2*Log[-1 + 2*x]*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])]))/((1
- 2*x)*(2 + 3*x)*(5 + Log[-1 + 2*x])*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])]^2) + Defer[Int][1/((5 + Log[-1 + 2
*x])*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])]^2), x] + Defer[Int][1/((-1 + 2*x)*(5 + Log[-1 + 2*x])*Log[(2 + 3*x
)/(5*x + x*Log[-1 + 2*x])]^2), x] + 10*Defer[Int][1/((2 + 3*x)*(5 + Log[-1 + 2*x])*Log[(2 + 3*x)/(5*x + x*Log[
-1 + 2*x])]^2), x] + 2*Defer[Int][Log[-1 + 2*x]/((2 + 3*x)*(5 + Log[-1 + 2*x])*Log[(2 + 3*x)/(5*x + x*Log[-1 +
2*x])]^2), x] + Defer[Int][Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 x^2-120 x^3-30 x^4-10 e^{\left .\frac {1}{5}\right /x} x \left (-5+12 x+3 x^2\right )-10 x \left (e^{\left .\frac {1}{5}\right /x}+x\right ) (-1+2 x) \log (-1+2 x)+\left (e^{\left .\frac {1}{5}\right /x}-5 x^2\right ) \left (-2+x+6 x^2\right ) (5+\log (-1+2 x)) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}{5 x^2 \left (2-x-6 x^2\right ) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx\\ &=\frac {1}{5} \int \frac {50 x^2-120 x^3-30 x^4-10 e^{\left .\frac {1}{5}\right /x} x \left (-5+12 x+3 x^2\right )-10 x \left (e^{\left .\frac {1}{5}\right /x}+x\right ) (-1+2 x) \log (-1+2 x)+\left (e^{\left .\frac {1}{5}\right /x}-5 x^2\right ) \left (-2+x+6 x^2\right ) (5+\log (-1+2 x)) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}{x^2 \left (2-x-6 x^2\right ) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx\\ &=\frac {1}{5} \int \left (-\frac {50}{(-1+2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}+\frac {120 x}{(-1+2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}+\frac {30 x^2}{(-1+2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}+\frac {10 \log (-1+2 x)}{(2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}+\frac {5}{\log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}+\frac {e^{\left .\frac {1}{5}\right /x} \left (-50 x+120 x^2+30 x^3-10 x \log (-1+2 x)+20 x^2 \log (-1+2 x)+10 \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-5 x \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-30 x^2 \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )+2 \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-x \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-6 x^2 \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )\right )}{x^2 (-1+2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{\left .\frac {1}{5}\right /x} \left (-50 x+120 x^2+30 x^3-10 x \log (-1+2 x)+20 x^2 \log (-1+2 x)+10 \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-5 x \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-30 x^2 \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )+2 \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-x \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-6 x^2 \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )\right )}{x^2 (-1+2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+2 \int \frac {\log (-1+2 x)}{(2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+6 \int \frac {x^2}{(-1+2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx-10 \int \frac {1}{(-1+2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+24 \int \frac {x}{(-1+2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+\int \frac {1}{\log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx\\ &=\frac {e^{\left .\frac {1}{5}\right /x} \left (10 \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-5 x \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-30 x^2 \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )+2 \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-x \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-6 x^2 \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )\right )}{(1-2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}+2 \int \frac {\log (-1+2 x)}{(2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+6 \int \left (\frac {1}{6 (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}+\frac {1}{14 (-1+2 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}-\frac {4}{21 (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}\right ) \, dx-10 \int \left (\frac {2}{7 (-1+2 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}-\frac {3}{7 (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}\right ) \, dx+24 \int \left (\frac {1}{7 (-1+2 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}+\frac {2}{7 (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}\right ) \, dx+\int \frac {1}{\log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx\\ &=\frac {e^{\left .\frac {1}{5}\right /x} \left (10 \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-5 x \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-30 x^2 \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )+2 \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-x \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )-6 x^2 \log (-1+2 x) \log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )\right )}{(1-2 x) (2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )}+\frac {3}{7} \int \frac {1}{(-1+2 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx-\frac {8}{7} \int \frac {1}{(2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+2 \int \frac {\log (-1+2 x)}{(2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx-\frac {20}{7} \int \frac {1}{(-1+2 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+\frac {24}{7} \int \frac {1}{(-1+2 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+\frac {30}{7} \int \frac {1}{(2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+\frac {48}{7} \int \frac {1}{(2+3 x) (5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+\int \frac {1}{(5+\log (-1+2 x)) \log ^2\left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx+\int \frac {1}{\log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 35, normalized size = 1.06 \begin {gather*} \frac {e^{\left .\frac {1}{5}\right /x}+x}{\log \left (\frac {2+3 x}{5 x+x \log (-1+2 x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*x^2 + 120*x^3 + 30*x^4 + E^(1/(5*x))*(-50*x + 120*x^2 + 30*x^3) + (-10*x^2 + 20*x^3 + E^(1/(5*x
))*(-10*x + 20*x^2))*Log[-1 + 2*x] + (-50*x^2 + 25*x^3 + 150*x^4 + E^(1/(5*x))*(10 - 5*x - 30*x^2) + (-10*x^2
+ 5*x^3 + 30*x^4 + E^(1/(5*x))*(2 - x - 6*x^2))*Log[-1 + 2*x])*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])])/((-50*x
^2 + 25*x^3 + 150*x^4 + (-10*x^2 + 5*x^3 + 30*x^4)*Log[-1 + 2*x])*Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])]^2),x]

[Out]

(E^(1/(5*x)) + x)/Log[(2 + 3*x)/(5*x + x*Log[-1 + 2*x])]

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 32, normalized size = 0.97 \begin {gather*} \frac {x + e^{\left (\frac {1}{5 \, x}\right )}}{\log \left (\frac {3 \, x + 2}{x \log \left (2 \, x - 1\right ) + 5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-6*x^2-x+2)*exp(1/5/x)+30*x^4+5*x^3-10*x^2)*log(2*x-1)+(-30*x^2-5*x+10)*exp(1/5/x)+150*x^4+25*x^
3-50*x^2)*log((3*x+2)/(x*log(2*x-1)+5*x))+((20*x^2-10*x)*exp(1/5/x)+20*x^3-10*x^2)*log(2*x-1)+(30*x^3+120*x^2-
50*x)*exp(1/5/x)+30*x^4+120*x^3-50*x^2)/((30*x^4+5*x^3-10*x^2)*log(2*x-1)+150*x^4+25*x^3-50*x^2)/log((3*x+2)/(
x*log(2*x-1)+5*x))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.82, size = 34, normalized size = 1.03 \begin {gather*} -\frac {x + e^{\left (\frac {1}{5 \, x}\right )}}{\log \left (x \log \left (2 \, x - 1\right ) + 5 \, x\right ) - \log \left (3 \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-6*x^2-x+2)*exp(1/5/x)+30*x^4+5*x^3-10*x^2)*log(2*x-1)+(-30*x^2-5*x+10)*exp(1/5/x)+150*x^4+25*x^
3-50*x^2)*log((3*x+2)/(x*log(2*x-1)+5*x))+((20*x^2-10*x)*exp(1/5/x)+20*x^3-10*x^2)*log(2*x-1)+(30*x^3+120*x^2-
50*x)*exp(1/5/x)+30*x^4+120*x^3-50*x^2)/((30*x^4+5*x^3-10*x^2)*log(2*x-1)+150*x^4+25*x^3-50*x^2)/log((3*x+2)/(
x*log(2*x-1)+5*x))^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.67, size = 314, normalized size = 9.52

method	result	size
risch	\(\frac {2 i \left ({\mathrm e}^{\frac {1}{5 x}}+x \right )}{\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{\ln \left (2 x -1\right )+5}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{x \left (\ln \left (2 x -1\right )+5\right )}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{x \left (\ln \left (2 x -1\right )+5\right )}\right )^{2}+\pi \,\mathrm {csgn}\left (i \left (\frac {2}{3}+x \right )\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (2 x -1\right )+5}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{\ln \left (2 x -1\right )+5}\right )-\pi \,\mathrm {csgn}\left (i \left (\frac {2}{3}+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{\ln \left (2 x -1\right )+5}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{\ln \left (2 x -1\right )+5}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{\ln \left (2 x -1\right )+5}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{\ln \left (2 x -1\right )+5}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{\ln \left (2 x -1\right )+5}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{x \left (\ln \left (2 x -1\right )+5\right )}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (\frac {2}{3}+x \right )}{x \left (\ln \left (2 x -1\right )+5\right )}\right )^{3}-2 i \ln \relax (x )+2 i \ln \left (\frac {2}{3}+x \right )+2 i \ln \relax (3)-2 i \ln \left (\ln \left (2 x -1\right )+5\right )}\)	\(314\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-6*x^2-x+2)*exp(1/5/x)+30*x^4+5*x^3-10*x^2)*ln(2*x-1)+(-30*x^2-5*x+10)*exp(1/5/x)+150*x^4+25*x^3-50*x^
2)*ln((3*x+2)/(x*ln(2*x-1)+5*x))+((20*x^2-10*x)*exp(1/5/x)+20*x^3-10*x^2)*ln(2*x-1)+(30*x^3+120*x^2-50*x)*exp(
1/5/x)+30*x^4+120*x^3-50*x^2)/((30*x^4+5*x^3-10*x^2)*ln(2*x-1)+150*x^4+25*x^3-50*x^2)/ln((3*x+2)/(x*ln(2*x-1)+
5*x))^2,x,method=_RETURNVERBOSE)

[Out]

2*I*(exp(1/5/x)+x)/(Pi*csgn(I/x)*csgn(I*(2/3+x)/(ln(2*x-1)+5))*csgn(I/x/(ln(2*x-1)+5)*(2/3+x))-Pi*csgn(I/x)*cs
gn(I/x/(ln(2*x-1)+5)*(2/3+x))^2+Pi*csgn(I*(2/3+x))*csgn(I/(ln(2*x-1)+5))*csgn(I*(2/3+x)/(ln(2*x-1)+5))-Pi*csgn
(I*(2/3+x))*csgn(I*(2/3+x)/(ln(2*x-1)+5))^2-Pi*csgn(I/(ln(2*x-1)+5))*csgn(I*(2/3+x)/(ln(2*x-1)+5))^2+Pi*csgn(I
*(2/3+x)/(ln(2*x-1)+5))^3-Pi*csgn(I*(2/3+x)/(ln(2*x-1)+5))*csgn(I/x/(ln(2*x-1)+5)*(2/3+x))^2+Pi*csgn(I/x/(ln(2
*x-1)+5)*(2/3+x))^3-2*I*ln(x)+2*I*ln(2/3+x)+2*I*ln(3)-2*I*ln(ln(2*x-1)+5))

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 33, normalized size = 1.00 \begin {gather*} \frac {x + e^{\left (\frac {1}{5 \, x}\right )}}{\log \left (3 \, x + 2\right ) - \log \relax (x) - \log \left (\log \left (2 \, x - 1\right ) + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-6*x^2-x+2)*exp(1/5/x)+30*x^4+5*x^3-10*x^2)*log(2*x-1)+(-30*x^2-5*x+10)*exp(1/5/x)+150*x^4+25*x^
3-50*x^2)*log((3*x+2)/(x*log(2*x-1)+5*x))+((20*x^2-10*x)*exp(1/5/x)+20*x^3-10*x^2)*log(2*x-1)+(30*x^3+120*x^2-
50*x)*exp(1/5/x)+30*x^4+120*x^3-50*x^2)/((30*x^4+5*x^3-10*x^2)*log(2*x-1)+150*x^4+25*x^3-50*x^2)/log((3*x+2)/(
x*log(2*x-1)+5*x))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\ln \left (2\,x-1\right )\,\left ({\mathrm {e}}^{\frac {1}{5\,x}}\,\left (10\,x-20\,x^2\right )+10\,x^2-20\,x^3\right )-{\mathrm {e}}^{\frac {1}{5\,x}}\,\left (30\,x^3+120\,x^2-50\,x\right )+50\,x^2-120\,x^3-30\,x^4+\ln \left (\frac {3\,x+2}{5\,x+x\,\ln \left (2\,x-1\right )}\right )\,\left (\ln \left (2\,x-1\right )\,\left ({\mathrm {e}}^{\frac {1}{5\,x}}\,\left (6\,x^2+x-2\right )+10\,x^2-5\,x^3-30\,x^4\right )+{\mathrm {e}}^{\frac {1}{5\,x}}\,\left (30\,x^2+5\,x-10\right )+50\,x^2-25\,x^3-150\,x^4\right )}{{\ln \left (\frac {3\,x+2}{5\,x+x\,\ln \left (2\,x-1\right )}\right )}^2\,\left (\ln \left (2\,x-1\right )\,\left (30\,x^4+5\,x^3-10\,x^2\right )-50\,x^2+25\,x^3+150\,x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*x - 1)*(exp(1/(5*x))*(10*x - 20*x^2) + 10*x^2 - 20*x^3) - exp(1/(5*x))*(120*x^2 - 50*x + 30*x^3) +
50*x^2 - 120*x^3 - 30*x^4 + log((3*x + 2)/(5*x + x*log(2*x - 1)))*(log(2*x - 1)*(exp(1/(5*x))*(x + 6*x^2 - 2)
+ 10*x^2 - 5*x^3 - 30*x^4) + exp(1/(5*x))*(5*x + 30*x^2 - 10) + 50*x^2 - 25*x^3 - 150*x^4))/(log((3*x + 2)/(5
*x + x*log(2*x - 1)))^2*(log(2*x - 1)*(5*x^3 - 10*x^2 + 30*x^4) - 50*x^2 + 25*x^3 + 150*x^4)),x)

[Out]

int(-(log(2*x - 1)*(exp(1/(5*x))*(10*x - 20*x^2) + 10*x^2 - 20*x^3) - exp(1/(5*x))*(120*x^2 - 50*x + 30*x^3) +
50*x^2 - 120*x^3 - 30*x^4 + log((3*x + 2)/(5*x + x*log(2*x - 1)))*(log(2*x - 1)*(exp(1/(5*x))*(x + 6*x^2 - 2)
+ 10*x^2 - 5*x^3 - 30*x^4) + exp(1/(5*x))*(5*x + 30*x^2 - 10) + 50*x^2 - 25*x^3 - 150*x^4))/(log((3*x + 2)/(5
*x + x*log(2*x - 1)))^2*(log(2*x - 1)*(5*x^3 - 10*x^2 + 30*x^4) - 50*x^2 + 25*x^3 + 150*x^4)), x)

________________________________________________________________________________________

sympy [A]  time = 1.68, size = 44, normalized size = 1.33 \begin {gather*} \frac {x}{\log {\left (\frac {3 x + 2}{x \log {\left (2 x - 1 \right )} + 5 x} \right )}} + \frac {e^{\frac {1}{5 x}}}{\log {\left (\frac {3 x + 2}{x \log {\left (2 x - 1 \right )} + 5 x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-6*x**2-x+2)*exp(1/5/x)+30*x**4+5*x**3-10*x**2)*ln(2*x-1)+(-30*x**2-5*x+10)*exp(1/5/x)+150*x**4+
25*x**3-50*x**2)*ln((3*x+2)/(x*ln(2*x-1)+5*x))+((20*x**2-10*x)*exp(1/5/x)+20*x**3-10*x**2)*ln(2*x-1)+(30*x**3+
120*x**2-50*x)*exp(1/5/x)+30*x**4+120*x**3-50*x**2)/((30*x**4+5*x**3-10*x**2)*ln(2*x-1)+150*x**4+25*x**3-50*x*
*2)/ln((3*x+2)/(x*ln(2*x-1)+5*x))**2,x)

[Out]

x/log((3*x + 2)/(x*log(2*x - 1) + 5*x)) + exp(1/(5*x))/log((3*x + 2)/(x*log(2*x - 1) + 5*x))

________________________________________________________________________________________