3.35.100 \(\int \frac {2 e^{-e^3+i \pi +x} (125-50 x+5 x^2)+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+(2 e^{-e^3+i \pi +x} (145+90 x-50 x^2+5 x^3)+2 e^{-e^3+i \pi +x} (-25-20 x+5 x^2) \log (-5+x)) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+(50+30 x-18 x^2+2 x^3) \log (-5+x)+(-5-4 x+x^2) \log ^2(-5+x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {10 e^{-e^3+i \pi +x} \log (1+x)}{-5+x+\log (-5+x)} \]

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Rubi [F]  time = 11.71, 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 {2 e^{-e^3+i \pi +x} \left (125-50 x+5 x^2\right )+2 e^{-e^3+i \pi +x} (-25+5 x) \log (-5+x)+\left (2 e^{-e^3+i \pi +x} \left (145+90 x-50 x^2+5 x^3\right )+2 e^{-e^3+i \pi +x} \left (-25-20 x+5 x^2\right ) \log (-5+x)\right ) \log (1+x)}{-125-50 x+60 x^2-14 x^3+x^4+\left (50+30 x-18 x^2+2 x^3\right ) \log (-5+x)+\left (-5-4 x+x^2\right ) \log ^2(-5+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*E^(-E^3 + I*Pi + x)*(125 - 50*x + 5*x^2) + 2*E^(-E^3 + I*Pi + x)*(-25 + 5*x)*Log[-5 + x] + (2*E^(-E^3 +
I*Pi + x)*(145 + 90*x - 50*x^2 + 5*x^3) + 2*E^(-E^3 + I*Pi + x)*(-25 - 20*x + 5*x^2)*Log[-5 + x])*Log[1 + x])
/(-125 - 50*x + 60*x^2 - 14*x^3 + x^4 + (50 + 30*x - 18*x^2 + 2*x^3)*Log[-5 + x] + (-5 - 4*x + x^2)*Log[-5 + x
]^2),x]

[Out]

-10*Defer[Int][E^(-E^3 + x)/((1 + x)*(-5 + x + Log[-5 + x])), x] + 60*Defer[Int][(E^(-E^3 + x)*Log[1 + x])/(-5
+ x + Log[-5 + x])^2, x] + 10*Defer[Int][(E^(-E^3 + x)*Log[1 + x])/((-5 + x)*(-5 + x + Log[-5 + x])^2), x] -
10*Defer[Int][(E^(-E^3 + x)*x*Log[1 + x])/(-5 + x + Log[-5 + x])^2, x] - 10*Defer[Int][(E^(-E^3 + x)*Log[-5 +
x]*Log[1 + x])/(-5 + x + Log[-5 + x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 e^{-e^3+x} \left ((-5+x)^2+\left (29+18 x-10 x^2+x^3\right ) \log (1+x)+(-5+x) \log (-5+x) (1+(1+x) \log (1+x))\right )}{\left (5+4 x-x^2\right ) (5-x-\log (-5+x))^2} \, dx\\ &=10 \int \frac {e^{-e^3+x} \left ((-5+x)^2+\left (29+18 x-10 x^2+x^3\right ) \log (1+x)+(-5+x) \log (-5+x) (1+(1+x) \log (1+x))\right )}{\left (5+4 x-x^2\right ) (5-x-\log (-5+x))^2} \, dx\\ &=10 \int \left (-\frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))}-\frac {e^{-e^3+x} \left (29-11 x+x^2-5 \log (-5+x)+x \log (-5+x)\right ) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}\right ) \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \frac {e^{-e^3+x} \left (29-11 x+x^2-5 \log (-5+x)+x \log (-5+x)\right ) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \left (\frac {29 e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}-\frac {11 e^{-e^3+x} x \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}+\frac {e^{-e^3+x} x^2 \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}-\frac {5 e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}+\frac {e^{-e^3+x} x \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}\right ) \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \frac {e^{-e^3+x} x^2 \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx-10 \int \frac {e^{-e^3+x} x \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx+50 \int \frac {e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx+110 \int \frac {e^{-e^3+x} x \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx-290 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \left (\frac {5 e^{-e^3+x} \log (1+x)}{(-5+x+\log (-5+x))^2}+\frac {25 e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}+\frac {e^{-e^3+x} x \log (1+x)}{(-5+x+\log (-5+x))^2}\right ) \, dx-10 \int \left (\frac {e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x+\log (-5+x))^2}+\frac {5 e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}\right ) \, dx+50 \int \frac {e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx+110 \int \left (\frac {e^{-e^3+x} \log (1+x)}{(-5+x+\log (-5+x))^2}+\frac {5 e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2}\right ) \, dx-290 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx\\ &=-\left (10 \int \frac {e^{-e^3+x}}{(1+x) (-5+x+\log (-5+x))} \, dx\right )-10 \int \frac {e^{-e^3+x} x \log (1+x)}{(-5+x+\log (-5+x))^2} \, dx-10 \int \frac {e^{-e^3+x} \log (-5+x) \log (1+x)}{(-5+x+\log (-5+x))^2} \, dx-50 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x+\log (-5+x))^2} \, dx+110 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x+\log (-5+x))^2} \, dx-250 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx-290 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx+550 \int \frac {e^{-e^3+x} \log (1+x)}{(-5+x) (-5+x+\log (-5+x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.67, size = 24, normalized size = 0.83 \begin {gather*} -\frac {10 e^{-e^3+x} \log (1+x)}{-5+x+\log (-5+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^(-E^3 + I*Pi + x)*(125 - 50*x + 5*x^2) + 2*E^(-E^3 + I*Pi + x)*(-25 + 5*x)*Log[-5 + x] + (2*E^(
-E^3 + I*Pi + x)*(145 + 90*x - 50*x^2 + 5*x^3) + 2*E^(-E^3 + I*Pi + x)*(-25 - 20*x + 5*x^2)*Log[-5 + x])*Log[1
+ x])/(-125 - 50*x + 60*x^2 - 14*x^3 + x^4 + (50 + 30*x - 18*x^2 + 2*x^3)*Log[-5 + x] + (-5 - 4*x + x^2)*Log[
-5 + x]^2),x]

[Out]

(-10*E^(-E^3 + x)*Log[1 + x])/(-5 + x + Log[-5 + x])

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fricas [A]  time = 0.75, size = 24, normalized size = 0.83 \begin {gather*} -\frac {5 \, e^{\left (x - e^{3} + \log \relax (2)\right )} \log \left (x + 1\right )}{x + \log \left (x - 5\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-(5*x^2-20*x-25)*exp(log(2)+x-exp(3))*log(x-5)-(5*x^3-50*x^2+90*x+145)*exp(log(2)+x-exp(3)))*log(x
+1)-(5*x-25)*exp(log(2)+x-exp(3))*log(x-5)-(5*x^2-50*x+125)*exp(log(2)+x-exp(3)))/((x^2-4*x-5)*log(x-5)^2+(2*x
^3-18*x^2+30*x+50)*log(x-5)+x^4-14*x^3+60*x^2-50*x-125),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 22, normalized size = 0.76 \begin {gather*} -\frac {10 \, e^{\left (x - e^{3}\right )} \log \left (x + 1\right )}{x + \log \left (x - 5\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-(5*x^2-20*x-25)*exp(log(2)+x-exp(3))*log(x-5)-(5*x^3-50*x^2+90*x+145)*exp(log(2)+x-exp(3)))*log(x
+1)-(5*x-25)*exp(log(2)+x-exp(3))*log(x-5)-(5*x^2-50*x+125)*exp(log(2)+x-exp(3)))/((x^2-4*x-5)*log(x-5)^2+(2*x
^3-18*x^2+30*x+50)*log(x-5)+x^4-14*x^3+60*x^2-50*x-125),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 23, normalized size = 0.79

method	result	size
risch	\(-\frac {10 \,{\mathrm e}^{x -{\mathrm e}^{3}} \ln \left (x +1\right )}{-5+x +\ln \left (x -5\right )}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-(5*x^2-20*x-25)*exp(ln(2)+x-exp(3))*ln(x-5)-(5*x^3-50*x^2+90*x+145)*exp(ln(2)+x-exp(3)))*ln(x+1)-(5*x-2
5)*exp(ln(2)+x-exp(3))*ln(x-5)-(5*x^2-50*x+125)*exp(ln(2)+x-exp(3)))/((x^2-4*x-5)*ln(x-5)^2+(2*x^3-18*x^2+30*x
+50)*ln(x-5)+x^4-14*x^3+60*x^2-50*x-125),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.58, size = 29, normalized size = 1.00 \begin {gather*} -\frac {10 \, e^{x} \log \left (x + 1\right )}{x e^{\left (e^{3}\right )} + e^{\left (e^{3}\right )} \log \left (x - 5\right ) - 5 \, e^{\left (e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-(5*x^2-20*x-25)*exp(log(2)+x-exp(3))*log(x-5)-(5*x^3-50*x^2+90*x+145)*exp(log(2)+x-exp(3)))*log(x
+1)-(5*x-25)*exp(log(2)+x-exp(3))*log(x-5)-(5*x^2-50*x+125)*exp(log(2)+x-exp(3)))/((x^2-4*x-5)*log(x-5)^2+(2*x
^3-18*x^2+30*x+50)*log(x-5)+x^4-14*x^3+60*x^2-50*x-125),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.44, size = 22, normalized size = 0.76 \begin {gather*} -\frac {10\,\ln \left (x+1\right )\,{\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^x}{x+\ln \left (x-5\right )-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x - exp(3) + log(2))*(5*x^2 - 50*x + 125) + log(x + 1)*(exp(x - exp(3) + log(2))*(90*x - 50*x^2 + 5*x
^3 + 145) - log(x - 5)*exp(x - exp(3) + log(2))*(20*x - 5*x^2 + 25)) + log(x - 5)*exp(x - exp(3) + log(2))*(5*
x - 25))/(50*x - log(x - 5)*(30*x - 18*x^2 + 2*x^3 + 50) + log(x - 5)^2*(4*x - x^2 + 5) - 60*x^2 + 14*x^3 - x^
4 + 125),x)

[Out]

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

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sympy [A]  time = 0.55, size = 22, normalized size = 0.76 \begin {gather*} - \frac {10 e^{x - e^{3}} \log {\left (x + 1 \right )}}{x + \log {\left (x - 5 \right )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-(5*x**2-20*x-25)*exp(ln(2)+x-exp(3))*ln(x-5)-(5*x**3-50*x**2+90*x+145)*exp(ln(2)+x-exp(3)))*ln(x+
1)-(5*x-25)*exp(ln(2)+x-exp(3))*ln(x-5)-(5*x**2-50*x+125)*exp(ln(2)+x-exp(3)))/((x**2-4*x-5)*ln(x-5)**2+(2*x**
3-18*x**2+30*x+50)*ln(x-5)+x**4-14*x**3+60*x**2-50*x-125),x)

[Out]

-10*exp(x - exp(3))*log(x + 1)/(x + log(x - 5) - 5)

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