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

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

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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 {gather*} x \log \left (\frac {2 x^2-5 x+5 e^x+10 \left (5-e^3\right )}{2 x}\right )+3 x \end {gather*}

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

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

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 {gather*} x \log \left (\frac {2 \, x^{2} - 5 \, x - 10 \, e^{3} + 5 \, e^{x} + 50}{2 \, x}\right ) + 3 \, x \end {gather*}

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 {gather*} x \log \left (\frac {2 \, x^{2} - 5 \, x - 10 \, e^{3} + 5 \, e^{x} + 50}{2 \, x}\right ) + 3 \, x \end {gather*}

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	\(x \ln \left (\frac {5 \,{\mathrm e}^{x}-10 \,{\mathrm e}^{3}+2 x^{2}-5 x +50}{2 x}\right )+3 x\)	\(31\)
risch	\(x \ln \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )-x \ln \relax (x )-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right )^{3}}{2}-i \pi x \mathrm {csgn}\left (\frac {i \left (-\frac {x^{2}}{5}+{\mathrm e}^{3}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}-5\right )}{x}\right )^{2}+i \pi x +x \ln \relax (5)+3 x\)	\(240\)

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 {gather*} -x {\left (\log \relax (2) - 3\right )} + x \log \left (2 \, x^{2} - 5 \, x - 10 \, e^{3} + 5 \, e^{x} + 50\right ) - x \log \relax (x) \end {gather*}

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 {gather*} x\,\left (\ln \left (\frac {\frac {5\,{\mathrm {e}}^x}{2}-5\,{\mathrm {e}}^3-\frac {5\,x}{2}+x^2+25}{x}\right )+3\right ) \end {gather*}

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 {gather*} x \log {\left (\frac {x^{2} - \frac {5 x}{2} + \frac {5 e^{x}}{2} - 5 e^{3} + 25}{x} \right )} + 3 x \end {gather*}

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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