3.69.38 $\int \frac{4x-5x^2-8x^3+10x^4+(-x+2x^3)\log (5)+(4x-10x^2+e^{x^2}(40-100x-10\log (5))-x\log (5))\log (e^{-x^2}(10e^{x^2}+x))}{10e^{x^2}+x}dx$

Optimal. Leaf size=23 $$x(4-5x-\log (5))\log (10+e^{-x^2}x)$$

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Rubi [F]  time = 1.30, 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 = {} $$\begin{array}{c}\int \frac{4x-5x^2-8x^3+10x^4+(-x+2x^3)\log (5)+(4x-10x^2+e^{x^2}(40-100x-10\log (5))-x\log (5))\log \left(e^{-x^2}(10e^{x^2}+x)\right)}{10e^{x^2}+x}dx\end{array}$$

Verification is not applicable to the result.

[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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{x(4-5x+10x^3-\log (5)-x^2(8-\log (25)))}{10e^{x^2}+x}-(-4+10x+\log (5))\log (10+e^{-x^2}x))dx\\ & =\int \frac{x(4-5x+10x^3-\log (5)-x^2(8-\log (25)))}{10e^{x^2}+x}dx-\int (-4+10x+\log (5))\log (10+e^{-x^2}x)dx\\ & =-\frac{1}{20}(4-10x-\log (5){)}^2\log (10+e^{-x^2}x)+\frac{1}{20}\int \frac{(1-2x^2)(4-10x-\log (5){)}^2}{10e^{x^2}+x}dx+\int (-\frac{5x^2}{10e^{x^2}+x}+\frac{10x^4}{10e^{x^2}+x}-\frac{x(-4+\log (5))}{10e^{x^2}+x}+\frac{x^3(-8+\log (25))}{10e^{x^2}+x})dx\\ & =-\frac{1}{20}(4-10x-\log (5){)}^2\log (10+e^{-x^2}x)+\frac{1}{20}\int (-\frac{200x^4}{10e^{x^2}+x}+\frac{20x(-4+\log (5))}{10e^{x^2}+x}-\frac{40x^3(-4+\log (5))}{10e^{x^2}+x}+\frac{(-4+\log (5){)}^2}{10e^{x^2}+x}-\frac{2x^2(-34-8\log (5)+{\log }^2(5))}{10e^{x^2}+x})dx-5\int \frac{x^2}{10e^{x^2}+x}dx+10\int \frac{x^4}{10e^{x^2}+x}dx+(4-\log (5))\int \frac{x}{10e^{x^2}+x}dx+(-8+\log (25))\int \frac{x^3}{10e^{x^2}+x}dx\\ & =-\frac{1}{20}(4-10x-\log (5){)}^2\log (10+e^{-x^2}x)-5\int \frac{x^2}{10e^{x^2}+x}dx+(4-\log (5))\int \frac{x}{10e^{x^2}+x}dx+(2(4-\log (5)))\int \frac{x^3}{10e^{x^2}+x}dx+\frac{1}{20}(4-\log (5){)}^2\int \frac{1}{10e^{x^2}+x}dx+(-4+\log (5))\int \frac{x}{10e^{x^2}+x}dx+\frac{1}{10}(34+8\log (5)-{\log }^2(5))\int \frac{x^2}{10e^{x^2}+x}dx+(-8+\log (25))\int \frac{x^3}{10e^{x^2}+x}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.32, size = 22, normalized size = 0.96 $$\begin{array}{c}-x(-4+5x+\log (5))\log (10+e^{-x^2}x)\end{array}$$

Antiderivative was successfully verified.

[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]

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

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fricas [A]  time = 1.11, size = 31, normalized size = 1.35 $$\begin{array}{c}-(5x^2+x\log (5)-4x)\log \left((x+10e^{\left(x^2\right)})e^{(-x^2)}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [B]  time = 0.19, size = 57, normalized size = 2.48 $$\begin{array}{c}5x^4+x^3\log (5)-4x^3-5x^2\log (x+10e^{\left(x^2\right)})-x\log (5)\log (x+10e^{\left(x^2\right)})+4x\log (x+10e^{\left(x^2\right)})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [B]  time = 0.18, size = 47, normalized size = 2.04

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.49, size = 37, normalized size = 1.61 $$\begin{array}{c}5x^4+x^3(\log (5)-4)-(5x^2+x(\log (5)-4))\log (x+10e^{\left(x^2\right)})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 4.38, size = 29, normalized size = 1.26 $$\begin{array}{c}-(5x^2+(\ln (5)-4)x)(\ln (x+10{\mathrm{e}}^{x^2})-x^2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.74, size = 27, normalized size = 1.17 $$\begin{array}{c}(-5x^2-x\log (5)+4x)\log \left((x+10e^{x^2})e^{-x^2}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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