3.52.1 \(\int \frac {e^x (84+6 e^5)+e^x (-70-70 x+14 x^2+e^5 (-5-5 x+x^2)) \log (x)}{1+(-10+2 x) \log (x)+(25-10 x+x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^x \left (2 x-\left (16+e^5\right ) x\right )}{-1+(5-x) \log (x)} \]

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Rubi [F]  time = 1.16, 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 {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(84 + 6*E^5) + E^x*(-70 - 70*x + 14*x^2 + E^5*(-5 - 5*x + x^2))*Log[x])/(1 + (-10 + 2*x)*Log[x] + (25
- 10*x + x^2)*Log[x]^2),x]

[Out]

6*(14 + E^5)*Defer[Int][E^x/(1 - 5*Log[x] + x*Log[x])^2, x] + 5*(14 + E^5)*Defer[Int][E^x/((-5 + x)*(1 - 5*Log
[x] + x*Log[x])^2), x] - (14 + E^5)*Defer[Int][(E^x*x)/(1 - 5*Log[x] + x*Log[x])^2, x] - 5*(14 + E^5)*Defer[In
t][E^x/((-5 + x)*(1 - 5*Log[x] + x*Log[x])), x] + (14 + E^5)*Defer[Int][(E^x*x)/(1 - 5*Log[x] + x*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (14+e^5\right ) \left (6+\left (-5-5 x+x^2\right ) \log (x)\right )}{(1+(-5+x) \log (x))^2} \, dx\\ &=\left (14+e^5\right ) \int \frac {e^x \left (6+\left (-5-5 x+x^2\right ) \log (x)\right )}{(1+(-5+x) \log (x))^2} \, dx\\ &=\left (14+e^5\right ) \int \left (\frac {e^x \left (-25+11 x-x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))^2}+\frac {e^x \left (-5-5 x+x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))}\right ) \, dx\\ &=\left (14+e^5\right ) \int \frac {e^x \left (-25+11 x-x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))^2} \, dx+\left (14+e^5\right ) \int \frac {e^x \left (-5-5 x+x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))} \, dx\\ &=\left (14+e^5\right ) \int \left (\frac {6 e^x}{(1-5 \log (x)+x \log (x))^2}+\frac {5 e^x}{(-5+x) (1-5 \log (x)+x \log (x))^2}-\frac {e^x x}{(1-5 \log (x)+x \log (x))^2}\right ) \, dx+\left (14+e^5\right ) \int \left (-\frac {5 e^x}{(-5+x) (1-5 \log (x)+x \log (x))}+\frac {e^x x}{1-5 \log (x)+x \log (x)}\right ) \, dx\\ &=\left (-14-e^5\right ) \int \frac {e^x x}{(1-5 \log (x)+x \log (x))^2} \, dx+\left (14+e^5\right ) \int \frac {e^x x}{1-5 \log (x)+x \log (x)} \, dx+\left (5 \left (14+e^5\right )\right ) \int \frac {e^x}{(-5+x) (1-5 \log (x)+x \log (x))^2} \, dx-\left (5 \left (14+e^5\right )\right ) \int \frac {e^x}{(-5+x) (1-5 \log (x)+x \log (x))} \, dx+\left (6 \left (14+e^5\right )\right ) \int \frac {e^x}{(1-5 \log (x)+x \log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 22, normalized size = 0.79 \begin {gather*} \frac {e^x \left (14+e^5\right ) x}{1-5 \log (x)+x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(84 + 6*E^5) + E^x*(-70 - 70*x + 14*x^2 + E^5*(-5 - 5*x + x^2))*Log[x])/(1 + (-10 + 2*x)*Log[x]
+ (25 - 10*x + x^2)*Log[x]^2),x]

[Out]

(E^x*(14 + E^5)*x)/(1 - 5*Log[x] + x*Log[x])

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fricas [A]  time = 0.58, size = 21, normalized size = 0.75 \begin {gather*} \frac {{\left (x e^{5} + 14 \, x\right )} e^{x}}{{\left (x - 5\right )} \log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84)*exp(x))/((x^2-10*x+25)*log(x)^2+(2*
x-10)*log(x)+1),x, algorithm="fricas")

[Out]

(x*e^5 + 14*x)*e^x/((x - 5)*log(x) + 1)

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giac [A]  time = 0.14, size = 25, normalized size = 0.89 \begin {gather*} \frac {x e^{\left (x + 5\right )} + 14 \, x e^{x}}{x \log \relax (x) - 5 \, \log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84)*exp(x))/((x^2-10*x+25)*log(x)^2+(2*
x-10)*log(x)+1),x, algorithm="giac")

[Out]

(x*e^(x + 5) + 14*x*e^x)/(x*log(x) - 5*log(x) + 1)

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maple [A]  time = 0.11, size = 21, normalized size = 0.75

method	result	size
norman	\(\frac {\left ({\mathrm e}^{5}+14\right ) x \,{\mathrm e}^{x}}{x \ln \relax (x )-5 \ln \relax (x )+1}\)	\(21\)
risch	\(\frac {\left ({\mathrm e}^{5}+14\right ) x \,{\mathrm e}^{x}}{x \ln \relax (x )-5 \ln \relax (x )+1}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*ln(x)+(6*exp(5)+84)*exp(x))/((x^2-10*x+25)*ln(x)^2+(2*x-10)*ln
(x)+1),x,method=_RETURNVERBOSE)

[Out]

(exp(5)+14)*x*exp(x)/(x*ln(x)-5*ln(x)+1)

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maxima [A]  time = 0.38, size = 18, normalized size = 0.64 \begin {gather*} \frac {x {\left (e^{5} + 14\right )} e^{x}}{{\left (x - 5\right )} \log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84)*exp(x))/((x^2-10*x+25)*log(x)^2+(2*
x-10)*log(x)+1),x, algorithm="maxima")

[Out]

x*(e^5 + 14)*e^x/((x - 5)*log(x) + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^x\,\left (6\,{\mathrm {e}}^5+84\right )-{\mathrm {e}}^x\,\ln \relax (x)\,\left (70\,x+{\mathrm {e}}^5\,\left (-x^2+5\,x+5\right )-14\,x^2+70\right )}{\left (x^2-10\,x+25\right )\,{\ln \relax (x)}^2+\left (2\,x-10\right )\,\ln \relax (x)+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(6*exp(5) + 84) - exp(x)*log(x)*(70*x + exp(5)*(5*x - x^2 + 5) - 14*x^2 + 70))/(log(x)*(2*x - 10)
+ log(x)^2*(x^2 - 10*x + 25) + 1),x)

[Out]

int((exp(x)*(6*exp(5) + 84) - exp(x)*log(x)*(70*x + exp(5)*(5*x - x^2 + 5) - 14*x^2 + 70))/(log(x)*(2*x - 10)
+ log(x)^2*(x^2 - 10*x + 25) + 1), x)

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sympy [A]  time = 0.34, size = 22, normalized size = 0.79 \begin {gather*} \frac {\left (14 x + x e^{5}\right ) e^{x}}{x \log {\relax (x )} - 5 \log {\relax (x )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-5*x-5)*exp(5)+14*x**2-70*x-70)*exp(x)*ln(x)+(6*exp(5)+84)*exp(x))/((x**2-10*x+25)*ln(x)**2+(
2*x-10)*ln(x)+1),x)

[Out]

(14*x + x*exp(5))*exp(x)/(x*log(x) - 5*log(x) + 1)

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