3.89.78 $\int \frac{100-20e^3-15x+8x^2+e^x(10+5x)+(50-10e^3+5e^x-5x+2x^2)\log (\frac{50-10e^3+5e^x-5x+2x^2}{2x})}{50-10e^3+5e^x-5x+2x^2}dx$

Optimal. Leaf size=33 $$x+x(2+\log (x+\frac{5(5-e^3-x+\frac{1}{2}(e^x+x))}{x}))$$

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Rubi [A]  time = 1.41, antiderivative size = 37, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 3, integrand size = 96, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.031, Rules used = {6741, 6742, 2548} $$\begin{array}{c}x\log \left(\frac{2x^2-5x+5e^x+10(5-e^3)}{2x}\right)+3x\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(100 - 20*E^3 - 15*x + 8*x^2 + E^x*(10 + 5*x) + (50 - 10*E^3 + 5*E^x - 5*x + 2*x^2)*Log[(50 - 10*E^3 + 5*E
^x - 5*x + 2*x^2)/(2*x)])/(50 - 10*E^3 + 5*E^x - 5*x + 2*x^2),x]

[Out]

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

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{100(1-\frac{e^3}{5})-15x+8x^2+e^x(10+5x)+(50-10e^3+5e^x-5x+2x^2)\log \left(\frac{50-10e^3+5e^x-5x+2x^2}{2x}\right)}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx\\ & =\int (2+x+\frac{x(-5(11-2e^3)+9x-2x^2)}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}+\log \left(\frac{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}{2x}\right))dx\\ & =2x+\frac{x^2}{2}+\int \frac{x(-5(11-2e^3)+9x-2x^2)}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx+\int \log \left(\frac{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}{2x}\right)dx\\ & =2x+\frac{x^2}{2}+x\log \left(\frac{5e^x+10(5-e^3)-5x+2x^2}{2x}\right)-\int \frac{10e^3+5e^x(-1+x)+2(-25+x^2)}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx+\int (\frac{2x^3}{-5e^x-50(1-\frac{e^3}{5})+5x-2x^2}+\frac{5(-11+2e^3)x}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}+\frac{9x^2}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2})dx\\ & =2x+\frac{x^2}{2}+x\log \left(\frac{5e^x+10(5-e^3)-5x+2x^2}{2x}\right)+2\int \frac{x^3}{-5e^x-50(1-\frac{e^3}{5})+5x-2x^2}dx+9\int \frac{x^2}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx-\left(5(11-2e^3)\right)\int \frac{x}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx-\int (-1+x+\frac{x(-5(11-2e^3)+9x-2x^2)}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2})dx\\ & =3x+x\log \left(\frac{5e^x+10(5-e^3)-5x+2x^2}{2x}\right)+2\int \frac{x^3}{-5e^x-50(1-\frac{e^3}{5})+5x-2x^2}dx+9\int \frac{x^2}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx-\left(5(11-2e^3)\right)\int \frac{x}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx-\int \frac{x(-5(11-2e^3)+9x-2x^2)}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx\\ & =3x+x\log \left(\frac{5e^x+10(5-e^3)-5x+2x^2}{2x}\right)+2\int \frac{x^3}{-5e^x-50(1-\frac{e^3}{5})+5x-2x^2}dx+9\int \frac{x^2}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx-\left(5(11-2e^3)\right)\int \frac{x}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}dx-\int (\frac{2x^3}{-5e^x-50(1-\frac{e^3}{5})+5x-2x^2}+\frac{5(-11+2e^3)x}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2}+\frac{9x^2}{5e^x+50(1-\frac{e^3}{5})-5x+2x^2})dx\\ & =3x+x\log \left(\frac{5e^x+10(5-e^3)-5x+2x^2}{2x}\right)\end{array}\end{array}$$

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Mathematica [A]  time = 0.04, size = 34, normalized size = 1.03 $$\begin{array}{c}3x+x\log \left(\frac{50-10e^3+5e^x-5x+2x^2}{2x}\right)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(100 - 20*E^3 - 15*x + 8*x^2 + E^x*(10 + 5*x) + (50 - 10*E^3 + 5*E^x - 5*x + 2*x^2)*Log[(50 - 10*E^3
+ 5*E^x - 5*x + 2*x^2)/(2*x)])/(50 - 10*E^3 + 5*E^x - 5*x + 2*x^2),x]

[Out]

3*x + x*Log[(50 - 10*E^3 + 5*E^x - 5*x + 2*x^2)/(2*x)]

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fricas [A]  time = 0.60, size = 30, normalized size = 0.91 $$\begin{array}{c}x\log \left(\frac{2x^2-5x-10e^3+5e^x+50}{2x}\right)+3x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)-10*exp(3)+2*x^2-5*x+50)*log(1/2*(5*exp(x)-10*exp(3)+2*x^2-5*x+50)/x)+(5*x+10)*exp(x)-20*e
xp(3)+8*x^2-15*x+100)/(5*exp(x)-10*exp(3)+2*x^2-5*x+50),x, algorithm="fricas")

[Out]

x*log(1/2*(2*x^2 - 5*x - 10*e^3 + 5*e^x + 50)/x) + 3*x

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giac [A]  time = 0.23, size = 30, normalized size = 0.91 $$\begin{array}{c}x\log \left(\frac{2x^2-5x-10e^3+5e^x+50}{2x}\right)+3x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)-10*exp(3)+2*x^2-5*x+50)*log(1/2*(5*exp(x)-10*exp(3)+2*x^2-5*x+50)/x)+(5*x+10)*exp(x)-20*e
xp(3)+8*x^2-15*x+100)/(5*exp(x)-10*exp(3)+2*x^2-5*x+50),x, algorithm="giac")

[Out]

x*log(1/2*(2*x^2 - 5*x - 10*e^3 + 5*e^x + 50)/x) + 3*x

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maple [A]  time = 0.28, size = 31, normalized size = 0.94

method	result	size
norman	$x\ln \left(\frac{5{\mathrm{e}}^x-10{\mathrm{e}}^3+2x^2-5x+50}{2x}\right)+3x$	$31$
risch	$x\ln (-\frac{x^2}{5}+{\mathrm{e}}^3+\frac{x}{2}-\frac{{\mathrm{e}}^x}{2}-5)-x\ln (x)-\frac{i\pi x\mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}\left(i(-\frac{x^2}{5}+{\mathrm{e}}^3+\frac{x}{2}-\frac{{\mathrm{e}}^x}{2}-5)\right)\mathrm{csgn}\left(\frac{i(-\frac{x^2}{5}+{\mathrm{e}}^3+\frac{x}{2}-\frac{{\mathrm{e}}^x}{2}-5)}{x}\right)}{2}+\frac{i\pi x\mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}{\left(\frac{i(-\frac{x^2}{5}+{\mathrm{e}}^3+\frac{x}{2}-\frac{{\mathrm{e}}^x}{2}-5)}{x}\right)}^2}{2}+\frac{i\pi x\mathrm{csgn}\left(i(-\frac{x^2}{5}+{\mathrm{e}}^3+\frac{x}{2}-\frac{{\mathrm{e}}^x}{2}-5)\right)\mathrm{csgn}{\left(\frac{i(-\frac{x^2}{5}+{\mathrm{e}}^3+\frac{x}{2}-\frac{{\mathrm{e}}^x}{2}-5)}{x}\right)}^2}{2}+\frac{i\pi x\mathrm{csgn}{\left(\frac{i(-\frac{x^2}{5}+{\mathrm{e}}^3+\frac{x}{2}-\frac{{\mathrm{e}}^x}{2}-5)}{x}\right)}^3}{2}-i\pi x\mathrm{csgn}{\left(\frac{i(-\frac{x^2}{5}+{\mathrm{e}}^3+\frac{x}{2}-\frac{{\mathrm{e}}^x}{2}-5)}{x}\right)}^2+i\pi x+x\ln (5)+3x$	$240$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(x)-10*exp(3)+2*x^2-5*x+50)*ln(1/2*(5*exp(x)-10*exp(3)+2*x^2-5*x+50)/x)+(5*x+10)*exp(x)-20*exp(3)+8
*x^2-15*x+100)/(5*exp(x)-10*exp(3)+2*x^2-5*x+50),x,method=_RETURNVERBOSE)

[Out]

x*ln(1/2*(5*exp(x)-10*exp(3)+2*x^2-5*x+50)/x)+3*x

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maxima [A]  time = 0.49, size = 34, normalized size = 1.03 $$\begin{array}{c}-x(\log (2)-3)+x\log (2x^2-5x-10e^3+5e^x+50)-x\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)-10*exp(3)+2*x^2-5*x+50)*log(1/2*(5*exp(x)-10*exp(3)+2*x^2-5*x+50)/x)+(5*x+10)*exp(x)-20*e
xp(3)+8*x^2-15*x+100)/(5*exp(x)-10*exp(3)+2*x^2-5*x+50),x, algorithm="maxima")

[Out]

-x*(log(2) - 3) + x*log(2*x^2 - 5*x - 10*e^3 + 5*e^x + 50) - x*log(x)

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mupad [B]  time = 5.67, size = 25, normalized size = 0.76 $$\begin{array}{c}x(\ln \left(\frac{\frac{5{\mathrm{e}}^x}{2}-5{\mathrm{e}}^3-\frac{5x}{2}+x^2+25}{x}\right)+3)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(5*x + 10) - 20*exp(3) - 15*x + log(((5*exp(x))/2 - 5*exp(3) - (5*x)/2 + x^2 + 25)/x)*(5*exp(x) -
10*exp(3) - 5*x + 2*x^2 + 50) + 8*x^2 + 100)/(5*exp(x) - 10*exp(3) - 5*x + 2*x^2 + 50),x)

[Out]

x*(log(((5*exp(x))/2 - 5*exp(3) - (5*x)/2 + x^2 + 25)/x) + 3)

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sympy [A]  time = 0.52, size = 29, normalized size = 0.88 $$\begin{array}{c}x\log \left(\frac{x^2-\frac{5x}{2}+\frac{5e^x}{2}-5e^3+25}{x}\right)+3x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)-10*exp(3)+2*x**2-5*x+50)*ln(1/2*(5*exp(x)-10*exp(3)+2*x**2-5*x+50)/x)+(5*x+10)*exp(x)-20*
exp(3)+8*x**2-15*x+100)/(5*exp(x)-10*exp(3)+2*x**2-5*x+50),x)

[Out]

x*log((x**2 - 5*x/2 + 5*exp(x)/2 - 5*exp(3) + 25)/x) + 3*x

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