3.74.41 \(\int \frac {125 x^3+50 x^4+5 x^5+e^{\frac {-20-12 x-125 x^2-25 x^3+e (25 x^2+5 x^3)}{25 x^2+5 x^3}} (200+120 x+24 x^2)}{125 x^4+50 x^5+5 x^6+e^{\frac {-20-12 x-125 x^2-25 x^3+e (25 x^2+5 x^3)}{25 x^2+5 x^3}} (125 x^3+50 x^4+5 x^5)} \, dx\)

Optimal. Leaf size=31 \[ \log \left (e^{-5+e+\frac {4 \left (-5+\frac {-5+2 x}{x}\right )}{5 x (5+x)}}+x\right ) \]

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Rubi [F]  time = 9.55, 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 {125 x^3+50 x^4+5 x^5+\exp \left (\frac {-20-12 x-125 x^2-25 x^3+e \left (25 x^2+5 x^3\right )}{25 x^2+5 x^3}\right ) \left (200+120 x+24 x^2\right )}{125 x^4+50 x^5+5 x^6+\exp \left (\frac {-20-12 x-125 x^2-25 x^3+e \left (25 x^2+5 x^3\right )}{25 x^2+5 x^3}\right ) \left (125 x^3+50 x^4+5 x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(125*x^3 + 50*x^4 + 5*x^5 + E^((-20 - 12*x - 125*x^2 - 25*x^3 + E*(25*x^2 + 5*x^3))/(25*x^2 + 5*x^3))*(200
+ 120*x + 24*x^2))/(125*x^4 + 50*x^5 + 5*x^6 + E^((-20 - 12*x - 125*x^2 - 25*x^3 + E*(25*x^2 + 5*x^3))/(25*x^
2 + 5*x^3))*(125*x^3 + 50*x^4 + 5*x^5)),x]

[Out]

Log[x] + (8*E^E*Defer[Int][1/(x^3*(E^E + E^(25/(5 + x) + 4/(x^2*(5 + x)) + 12/(5*x*(5 + x)) + (5*x)/(5 + x))*x
)), x])/5 + (8*E^E*Defer[Int][1/(x^2*(E^E + E^(25/(5 + x) + 4/(x^2*(5 + x)) + 12/(5*x*(5 + x)) + (5*x)/(5 + x)
)*x)), x])/25 - E^E*Defer[Int][1/(x*(E^E + E^(25/(5 + x) + 4/(x^2*(5 + x)) + 12/(5*x*(5 + x)) + (5*x)/(5 + x))
*x)), x] - (8*E^E*Defer[Int][1/((5 + x)^2*(E^E + E^(25/(5 + x) + 4/(x^2*(5 + x)) + 12/(5*x*(5 + x)) + (5*x)/(5
+ x))*x)), x])/25

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^{\frac {20+12 x+125 x^2+25 x^3}{5 x^2 (5+x)}} x^3 (5+x)^2+8 e^e \left (25+15 x+3 x^2\right )}{5 x^3 (5+x)^2 \left (e^e+e^{\frac {20+12 x+125 x^2+25 x^3}{5 x^2 (5+x)}} x\right )} \, dx\\ &=\frac {1}{5} \int \frac {5 e^{\frac {20+12 x+125 x^2+25 x^3}{5 x^2 (5+x)}} x^3 (5+x)^2+8 e^e \left (25+15 x+3 x^2\right )}{x^3 (5+x)^2 \left (e^e+e^{\frac {20+12 x+125 x^2+25 x^3}{5 x^2 (5+x)}} x\right )} \, dx\\ &=\frac {1}{5} \int \left (\frac {5}{x}-\frac {e^e \left (-200-120 x+101 x^2+50 x^3+5 x^4\right )}{x^3 (5+x)^2 \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )}\right ) \, dx\\ &=\log (x)-\frac {1}{5} e^e \int \frac {-200-120 x+101 x^2+50 x^3+5 x^4}{x^3 (5+x)^2 \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )} \, dx\\ &=\log (x)-\frac {1}{5} e^e \int \left (-\frac {8}{x^3 \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )}-\frac {8}{5 x^2 \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )}+\frac {5}{x \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )}+\frac {8}{5 (5+x)^2 \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )}\right ) \, dx\\ &=\log (x)+\frac {1}{25} \left (8 e^e\right ) \int \frac {1}{x^2 \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )} \, dx-\frac {1}{25} \left (8 e^e\right ) \int \frac {1}{(5+x)^2 \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )} \, dx-e^e \int \frac {1}{x \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )} \, dx+\frac {1}{5} \left (8 e^e\right ) \int \frac {1}{x^3 \left (e^e+\exp \left (\frac {25}{5+x}+\frac {4}{x^2 (5+x)}+\frac {12}{5 x (5+x)}+\frac {5 x}{5+x}\right ) x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 56, normalized size = 1.81 \begin {gather*} \frac {1}{5} \left (-\frac {4 (5+3 x)}{x^2 (5+x)}+5 \log \left (e^e+e^{5+\frac {4}{5 x^2}+\frac {8}{25 x}-\frac {8}{25 (5+x)}} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(125*x^3 + 50*x^4 + 5*x^5 + E^((-20 - 12*x - 125*x^2 - 25*x^3 + E*(25*x^2 + 5*x^3))/(25*x^2 + 5*x^3)
)*(200 + 120*x + 24*x^2))/(125*x^4 + 50*x^5 + 5*x^6 + E^((-20 - 12*x - 125*x^2 - 25*x^3 + E*(25*x^2 + 5*x^3))/
(25*x^2 + 5*x^3))*(125*x^3 + 50*x^4 + 5*x^5)),x]

[Out]

((-4*(5 + 3*x))/(x^2*(5 + x)) + 5*Log[E^E + E^(5 + 4/(5*x^2) + 8/(25*x) - 8/(25*(5 + x)))*x])/5

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fricas [A]  time = 0.59, size = 45, normalized size = 1.45 \begin {gather*} \log \left (x + e^{\left (-\frac {25 \, x^{3} + 125 \, x^{2} - 5 \, {\left (x^{3} + 5 \, x^{2}\right )} e + 12 \, x + 20}{5 \, {\left (x^{3} + 5 \, x^{2}\right )}}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x^2+120*x+200)*exp(((5*x^3+25*x^2)*exp(1)-25*x^3-125*x^2-12*x-20)/(5*x^3+25*x^2))+5*x^5+50*x^4+
125*x^3)/((5*x^5+50*x^4+125*x^3)*exp(((5*x^3+25*x^2)*exp(1)-25*x^3-125*x^2-12*x-20)/(5*x^3+25*x^2))+5*x^6+50*x
^5+125*x^4),x, algorithm="fricas")

[Out]

log(x + e^(-1/5*(25*x^3 + 125*x^2 - 5*(x^3 + 5*x^2)*e + 12*x + 20)/(x^3 + 5*x^2)))

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giac [A]  time = 0.56, size = 46, normalized size = 1.48 \begin {gather*} \log \left (x + e^{\left (\frac {5 \, x^{3} e - 25 \, x^{3} + 25 \, x^{2} e - 125 \, x^{2} - 12 \, x - 20}{5 \, {\left (x^{3} + 5 \, x^{2}\right )}}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x^2+120*x+200)*exp(((5*x^3+25*x^2)*exp(1)-25*x^3-125*x^2-12*x-20)/(5*x^3+25*x^2))+5*x^5+50*x^4+
125*x^3)/((5*x^5+50*x^4+125*x^3)*exp(((5*x^3+25*x^2)*exp(1)-25*x^3-125*x^2-12*x-20)/(5*x^3+25*x^2))+5*x^6+50*x
^5+125*x^4),x, algorithm="giac")

[Out]

log(x + e^(1/5*(5*x^3*e - 25*x^3 + 25*x^2*e - 125*x^2 - 12*x - 20)/(x^3 + 5*x^2)))

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maple [A]  time = 0.17, size = 48, normalized size = 1.55

method	result	size
norman	\(\ln \left (x +{\mathrm e}^{\frac {\left (5 x^{3}+25 x^{2}\right ) {\mathrm e}-25 x^{3}-125 x^{2}-12 x -20}{5 x^{3}+25 x^{2}}}\right )\)	\(48\)
risch	\(\frac {-\frac {12 x}{5}-4}{x^{2} \left (5+x \right )}-\frac {\left (5 x^{3}+25 x^{2}\right ) {\mathrm e}-25 x^{3}-125 x^{2}-12 x -20}{5 x^{3}+25 x^{2}}+\ln \left ({\mathrm e}^{\frac {5 x^{3} {\mathrm e}+25 x^{2} {\mathrm e}-25 x^{3}-125 x^{2}-12 x -20}{5 x^{2} \left (5+x \right )}}+x \right )\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((24*x^2+120*x+200)*exp(((5*x^3+25*x^2)*exp(1)-25*x^3-125*x^2-12*x-20)/(5*x^3+25*x^2))+5*x^5+50*x^4+125*x^
3)/((5*x^5+50*x^4+125*x^3)*exp(((5*x^3+25*x^2)*exp(1)-25*x^3-125*x^2-12*x-20)/(5*x^3+25*x^2))+5*x^6+50*x^5+125
*x^4),x,method=_RETURNVERBOSE)

[Out]

ln(x+exp(((5*x^3+25*x^2)*exp(1)-25*x^3-125*x^2-12*x-20)/(5*x^3+25*x^2)))

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maxima [B]  time = 0.54, size = 65, normalized size = 2.10 \begin {gather*} -\frac {633 \, x^{2} + 60 \, x + 100}{25 \, {\left (x^{3} + 5 \, x^{2}\right )}} + \log \left ({\left (x e^{\left (\frac {25}{x + 5} + \frac {8}{25 \, x} + \frac {4}{5 \, x^{2}} + 5\right )} + e^{\left (\frac {633}{25 \, {\left (x + 5\right )}} + e\right )}\right )} e^{\left (-e\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x^2+120*x+200)*exp(((5*x^3+25*x^2)*exp(1)-25*x^3-125*x^2-12*x-20)/(5*x^3+25*x^2))+5*x^5+50*x^4+
125*x^3)/((5*x^5+50*x^4+125*x^3)*exp(((5*x^3+25*x^2)*exp(1)-25*x^3-125*x^2-12*x-20)/(5*x^3+25*x^2))+5*x^6+50*x
^5+125*x^4),x, algorithm="maxima")

[Out]

-1/25*(633*x^2 + 60*x + 100)/(x^3 + 5*x^2) + log((x*e^(25/(x + 5) + 8/25/x + 4/5/x^2 + 5) + e^(633/25/(x + 5)
+ e))*e^(-e))

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mupad [B]  time = 5.06, size = 43, normalized size = 1.39 \begin {gather*} \ln \left (x+{\mathrm {e}}^{-\frac {12\,x-25\,x^2\,\mathrm {e}-5\,x^3\,\mathrm {e}+125\,x^2+25\,x^3+20}{5\,x^2\,\left (x+5\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((125*x^3 + 50*x^4 + 5*x^5 + exp(-(12*x - exp(1)*(25*x^2 + 5*x^3) + 125*x^2 + 25*x^3 + 20)/(25*x^2 + 5*x^3)
)*(120*x + 24*x^2 + 200))/(exp(-(12*x - exp(1)*(25*x^2 + 5*x^3) + 125*x^2 + 25*x^3 + 20)/(25*x^2 + 5*x^3))*(12
5*x^3 + 50*x^4 + 5*x^5) + 125*x^4 + 50*x^5 + 5*x^6),x)

[Out]

log(x + exp(-(12*x - 25*x^2*exp(1) - 5*x^3*exp(1) + 125*x^2 + 25*x^3 + 20)/(5*x^2*(x + 5))))

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sympy [A]  time = 0.58, size = 42, normalized size = 1.35 \begin {gather*} \log {\left (x + e^{\frac {- 25 x^{3} - 125 x^{2} - 12 x + e \left (5 x^{3} + 25 x^{2}\right ) - 20}{5 x^{3} + 25 x^{2}}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x**2+120*x+200)*exp(((5*x**3+25*x**2)*exp(1)-25*x**3-125*x**2-12*x-20)/(5*x**3+25*x**2))+5*x**5
+50*x**4+125*x**3)/((5*x**5+50*x**4+125*x**3)*exp(((5*x**3+25*x**2)*exp(1)-25*x**3-125*x**2-12*x-20)/(5*x**3+2
5*x**2))+5*x**6+50*x**5+125*x**4),x)

[Out]

log(x + exp((-25*x**3 - 125*x**2 - 12*x + E*(5*x**3 + 25*x**2) - 20)/(5*x**3 + 25*x**2)))

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