\(\int \frac {4 x-5 x^2-8 x^3+10 x^4+(-x+2 x^3) \log (5)+(4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)) \log (e^{-x^2} (10 e^{x^2}+x))}{10 e^{x^2}+x} \, dx\) [6838]

Optimal result

Integrand size = 91, antiderivative size = 23 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=x (4-5 x-\log (5)) \log \left (10+e^{-x^2} x\right ) \]

[Out]

(4-5*x-ln(5))*ln(10+x/exp(x^2))*x

Rubi [F]

\[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=\int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx \]

[In]

Int[(4*x - 5*x^2 - 8*x^3 + 10*x^4 + (-x + 2*x^3)*Log[5] + (4*x - 10*x^2 + E^x^2*(40 - 100*x - 10*Log[5]) - x*L
og[5])*Log[(10*E^x^2 + x)/E^x^2])/(10*E^x^2 + x),x]

[Out]

-1/20*((4 - 10*x - Log[5])^2*Log[10 + x/E^x^2]) + ((4 - Log[5])^2*Defer[Int][(10*E^x^2 + x)^(-1), x])/20 - 5*D
efer[Int][x^2/(10*E^x^2 + x), x] + ((34 + 8*Log[5] - Log[5]^2)*Defer[Int][x^2/(10*E^x^2 + x), x])/10 + 2*(4 -
Log[5])*Defer[Int][x^3/(10*E^x^2 + x), x] - (8 - Log[25])*Defer[Int][x^3/(10*E^x^2 + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x \left (4-5 x+10 x^3-\log (5)-x^2 (8-\log (25))\right )}{10 e^{x^2}+x}-(-4+10 x+\log (5)) \log \left (10+e^{-x^2} x\right )\right ) \, dx \\ & = \int \frac {x \left (4-5 x+10 x^3-\log (5)-x^2 (8-\log (25))\right )}{10 e^{x^2}+x} \, dx-\int (-4+10 x+\log (5)) \log \left (10+e^{-x^2} x\right ) \, dx \\ & = -\frac {1}{20} (4-10 x-\log (5))^2 \log \left (10+e^{-x^2} x\right )+\frac {1}{20} \int \frac {\left (1-2 x^2\right ) (4-10 x-\log (5))^2}{10 e^{x^2}+x} \, dx+\int \left (-\frac {5 x^2}{10 e^{x^2}+x}+\frac {10 x^4}{10 e^{x^2}+x}-\frac {x (-4+\log (5))}{10 e^{x^2}+x}+\frac {x^3 (-8+\log (25))}{10 e^{x^2}+x}\right ) \, dx \\ & = -\frac {1}{20} (4-10 x-\log (5))^2 \log \left (10+e^{-x^2} x\right )+\frac {1}{20} \int \left (-\frac {200 x^4}{10 e^{x^2}+x}+\frac {20 x (-4+\log (5))}{10 e^{x^2}+x}-\frac {40 x^3 (-4+\log (5))}{10 e^{x^2}+x}+\frac {(-4+\log (5))^2}{10 e^{x^2}+x}-\frac {2 x^2 \left (-34-8 \log (5)+\log ^2(5)\right )}{10 e^{x^2}+x}\right ) \, dx-5 \int \frac {x^2}{10 e^{x^2}+x} \, dx+10 \int \frac {x^4}{10 e^{x^2}+x} \, dx+(4-\log (5)) \int \frac {x}{10 e^{x^2}+x} \, dx+(-8+\log (25)) \int \frac {x^3}{10 e^{x^2}+x} \, dx \\ & = -\frac {1}{20} (4-10 x-\log (5))^2 \log \left (10+e^{-x^2} x\right )-5 \int \frac {x^2}{10 e^{x^2}+x} \, dx+(4-\log (5)) \int \frac {x}{10 e^{x^2}+x} \, dx+(2 (4-\log (5))) \int \frac {x^3}{10 e^{x^2}+x} \, dx+\frac {1}{20} (4-\log (5))^2 \int \frac {1}{10 e^{x^2}+x} \, dx+(-4+\log (5)) \int \frac {x}{10 e^{x^2}+x} \, dx+\frac {1}{10} \left (34+8 \log (5)-\log ^2(5)\right ) \int \frac {x^2}{10 e^{x^2}+x} \, dx+(-8+\log (25)) \int \frac {x^3}{10 e^{x^2}+x} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=\int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx \]

[In]

Integrate[(4*x - 5*x^2 - 8*x^3 + 10*x^4 + (-x + 2*x^3)*Log[5] + (4*x - 10*x^2 + E^x^2*(40 - 100*x - 10*Log[5])
- x*Log[5])*Log[(10*E^x^2 + x)/E^x^2])/(10*E^x^2 + x),x]

[Out]

Integrate[(4*x - 5*x^2 - 8*x^3 + 10*x^4 + (-x + 2*x^3)*Log[5] + (4*x - 10*x^2 + E^x^2*(40 - 100*x - 10*Log[5])
- x*Log[5])*Log[(10*E^x^2 + x)/E^x^2])/(10*E^x^2 + x), x]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(46\) vs. \(2(22)=44\).

Time = 0.53 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04

method	result	size
norman	\(\left (-\ln \left (5\right )+4\right ) x \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right )-5 \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right ) x^{2}\)	\(47\)
parallelrisch	\(-\ln \left (5\right ) \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right ) x -5 \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right ) x^{2}+4 \ln \left (\left (10 \,{\mathrm e}^{x^{2}}+x \right ) {\mathrm e}^{-x^{2}}\right ) x\)	\(63\)
risch	\(\left (x \ln \left (5\right )+5 x^{2}-4 x \right ) \ln \left ({\mathrm e}^{x^{2}}\right )-\ln \left (5\right ) x \ln \left (10 \,{\mathrm e}^{x^{2}}+x \right )-5 x^{2} \ln \left (10 \,{\mathrm e}^{x^{2}}+x \right )+4 x \ln \left (10 \,{\mathrm e}^{x^{2}}+x \right )+\frac {i \ln \left (5\right ) \pi x \,\operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )}{2}+\frac {5 i \pi \,x^{2} \operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )}{2}+2 i \pi x \,\operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2}-\frac {i \ln \left (5\right ) \pi x \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )}{2}-2 i \pi x \,\operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )+2 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )+\frac {5 i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{3}}{2}-\frac {i \ln \left (5\right ) \pi x \,\operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2}}{2}-\frac {5 i \pi \,x^{2} \operatorname {csgn}\left (i \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2}}{2}-2 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{3}+\frac {i \ln \left (5\right ) \pi x \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{3}}{2}-\frac {5 i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}} \left (10 \,{\mathrm e}^{x^{2}}+x \right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )}{2}\)	\(503\)

[In]

int((((-10*ln(5)-100*x+40)*exp(x^2)-x*ln(5)-10*x^2+4*x)*ln((10*exp(x^2)+x)/exp(x^2))+(2*x^3-x)*ln(5)+10*x^4-8*
x^3-5*x^2+4*x)/(10*exp(x^2)+x),x,method=_RETURNVERBOSE)

[Out]

(-ln(5)+4)*x*ln((10*exp(x^2)+x)/exp(x^2))-5*ln((10*exp(x^2)+x)/exp(x^2))*x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=-{\left (5 \, x^{2} + x \log \left (5\right ) - 4 \, x\right )} \log \left ({\left (x + 10 \, e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}\right ) \]

[In]

integrate((((-10*log(5)-100*x+40)*exp(x^2)-x*log(5)-10*x^2+4*x)*log((10*exp(x^2)+x)/exp(x^2))+(2*x^3-x)*log(5)
+10*x^4-8*x^3-5*x^2+4*x)/(10*exp(x^2)+x),x, algorithm="fricas")

[Out]

-(5*x^2 + x*log(5) - 4*x)*log((x + 10*e^(x^2))*e^(-x^2))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=\left (- 5 x^{2} - x \log {\left (5 \right )} + 4 x\right ) \log {\left (\left (x + 10 e^{x^{2}}\right ) e^{- x^{2}} \right )} \]

[In]

integrate((((-10*ln(5)-100*x+40)*exp(x**2)-x*ln(5)-10*x**2+4*x)*ln((10*exp(x**2)+x)/exp(x**2))+(2*x**3-x)*ln(5
)+10*x**4-8*x**3-5*x**2+4*x)/(10*exp(x**2)+x),x)

[Out]

(-5*x**2 - x*log(5) + 4*x)*log((x + 10*exp(x**2))*exp(-x**2))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=5 \, x^{4} + x^{3} {\left (\log \left (5\right ) - 4\right )} - {\left (5 \, x^{2} + x {\left (\log \left (5\right ) - 4\right )}\right )} \log \left (x + 10 \, e^{\left (x^{2}\right )}\right ) \]

[In]

integrate((((-10*log(5)-100*x+40)*exp(x^2)-x*log(5)-10*x^2+4*x)*log((10*exp(x^2)+x)/exp(x^2))+(2*x^3-x)*log(5)
+10*x^4-8*x^3-5*x^2+4*x)/(10*exp(x^2)+x),x, algorithm="maxima")

[Out]

5*x^4 + x^3*(log(5) - 4) - (5*x^2 + x*(log(5) - 4))*log(x + 10*e^(x^2))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (21) = 42\).

Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.70 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=-5 \, x^{2} \log \left ({\left (x + 10 \, e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}\right ) - x \log \left (5\right ) \log \left ({\left (x + 10 \, e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}\right ) + 4 \, x \log \left ({\left (x + 10 \, e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}\right ) \]

[In]

integrate((((-10*log(5)-100*x+40)*exp(x^2)-x*log(5)-10*x^2+4*x)*log((10*exp(x^2)+x)/exp(x^2))+(2*x^3-x)*log(5)
+10*x^4-8*x^3-5*x^2+4*x)/(10*exp(x^2)+x),x, algorithm="giac")

[Out]

-5*x^2*log((x + 10*e^(x^2))*e^(-x^2)) - x*log(5)*log((x + 10*e^(x^2))*e^(-x^2)) + 4*x*log((x + 10*e^(x^2))*e^(
-x^2))

Mupad [B] (verification not implemented)

Time = 13.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {4 x-5 x^2-8 x^3+10 x^4+\left (-x+2 x^3\right ) \log (5)+\left (4 x-10 x^2+e^{x^2} (40-100 x-10 \log (5))-x \log (5)\right ) \log \left (e^{-x^2} \left (10 e^{x^2}+x\right )\right )}{10 e^{x^2}+x} \, dx=-\left (5\,x^2+\left (\ln \left (5\right )-4\right )\,x\right )\,\left (\ln \left (x+10\,{\mathrm {e}}^{x^2}\right )-x^2\right ) \]

[In]

int(-(log(exp(-x^2)*(x + 10*exp(x^2)))*(x*log(5) - 4*x + exp(x^2)*(100*x + 10*log(5) - 40) + 10*x^2) - 4*x + 5
*x^2 + 8*x^3 - 10*x^4 + log(5)*(x - 2*x^3))/(x + 10*exp(x^2)),x)

[Out]

-(x*(log(5) - 4) + 5*x^2)*(log(x + 10*exp(x^2)) - x^2)