3.50.98 \(\int \frac {-225-30 x-4 x^2-6 x^3-x^4+e^{12} (-625-250 x-25 x^2)+e^9 (-2000-525 x+50 x^2+15 x^3)+e^3 (-1200-215 x+40 x^2+4 x^3-2 x^4)+e^6 (-2350-500 x+115 x^2+25 x^3-x^4)}{225 x+240 x^2+94 x^3+16 x^4+x^5+e^{12} (625 x+250 x^2+25 x^3)+e^9 (2000 x+1150 x^2+200 x^3+10 x^4)+e^6 (2350 x+1750 x^2+435 x^3+40 x^4+x^5)+e^3 (1200 x+1090 x^2+350 x^3+46 x^4+2 x^5)} \, dx\)

Optimal. Leaf size=35 \[ \frac {2+x+\frac {x}{3+x+e^3 (5+x)}}{1+e^3+\frac {x}{5}}-\log (x) \]

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Rubi [B]  time = 0.34, antiderivative size = 88, normalized size of antiderivative = 2.51, number of steps used = 2, number of rules used = 1, integrand size = 221, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.005, Rules used = {2074} \begin {gather*} -\frac {5 \left (1+20 e^3+40 e^6+25 e^9\right )}{\left (2+5 e^3+5 e^6\right ) \left (x+5 \left (1+e^3\right )\right )}-\frac {5 \left (3+5 e^3\right )}{\left (2+5 e^3+5 e^6\right ) \left (\left (1+e^3\right ) x+5 e^3+3\right )}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-225 - 30*x - 4*x^2 - 6*x^3 - x^4 + E^12*(-625 - 250*x - 25*x^2) + E^9*(-2000 - 525*x + 50*x^2 + 15*x^3)
+ E^3*(-1200 - 215*x + 40*x^2 + 4*x^3 - 2*x^4) + E^6*(-2350 - 500*x + 115*x^2 + 25*x^3 - x^4))/(225*x + 240*x^
2 + 94*x^3 + 16*x^4 + x^5 + E^12*(625*x + 250*x^2 + 25*x^3) + E^9*(2000*x + 1150*x^2 + 200*x^3 + 10*x^4) + E^6
*(2350*x + 1750*x^2 + 435*x^3 + 40*x^4 + x^5) + E^3*(1200*x + 1090*x^2 + 350*x^3 + 46*x^4 + 2*x^5)),x]

[Out]

(-5*(1 + 20*E^3 + 40*E^6 + 25*E^9))/((2 + 5*E^3 + 5*E^6)*(5*(1 + E^3) + x)) - (5*(3 + 5*E^3))/((2 + 5*E^3 + 5*
E^6)*(3 + 5*E^3 + (1 + E^3)*x)) - Log[x]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{x}+\frac {5 \left (1+20 e^3+40 e^6+25 e^9\right )}{\left (2+5 e^3+5 e^6\right ) \left (5+5 e^3+x\right )^2}+\frac {5 \left (3+8 e^3+5 e^6\right )}{\left (2+5 e^3+5 e^6\right ) \left (3+5 e^3+\left (1+e^3\right ) x\right )^2}\right ) \, dx\\ &=-\frac {5 \left (1+20 e^3+40 e^6+25 e^9\right )}{\left (2+5 e^3+5 e^6\right ) \left (5 \left (1+e^3\right )+x\right )}-\frac {5 \left (3+5 e^3\right )}{\left (2+5 e^3+5 e^6\right ) \left (3+5 e^3+\left (1+e^3\right ) x\right )}-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 59, normalized size = 1.69 \begin {gather*} -\frac {5 \left (9+2 x+5 e^6 (5+x)+e^3 (30+8 x)\right )}{15+8 x+x^2+5 e^6 (5+x)+e^3 \left (40+15 x+x^2\right )}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-225 - 30*x - 4*x^2 - 6*x^3 - x^4 + E^12*(-625 - 250*x - 25*x^2) + E^9*(-2000 - 525*x + 50*x^2 + 15
*x^3) + E^3*(-1200 - 215*x + 40*x^2 + 4*x^3 - 2*x^4) + E^6*(-2350 - 500*x + 115*x^2 + 25*x^3 - x^4))/(225*x +
240*x^2 + 94*x^3 + 16*x^4 + x^5 + E^12*(625*x + 250*x^2 + 25*x^3) + E^9*(2000*x + 1150*x^2 + 200*x^3 + 10*x^4)
+ E^6*(2350*x + 1750*x^2 + 435*x^3 + 40*x^4 + x^5) + E^3*(1200*x + 1090*x^2 + 350*x^3 + 46*x^4 + 2*x^5)),x]

[Out]

(-5*(9 + 2*x + 5*E^6*(5 + x) + E^3*(30 + 8*x)))/(15 + 8*x + x^2 + 5*E^6*(5 + x) + E^3*(40 + 15*x + x^2)) - Log
[x]

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fricas [B]  time = 0.76, size = 80, normalized size = 2.29 \begin {gather*} -\frac {25 \, {\left (x + 5\right )} e^{6} + 10 \, {\left (4 \, x + 15\right )} e^{3} + {\left (x^{2} + 5 \, {\left (x + 5\right )} e^{6} + {\left (x^{2} + 15 \, x + 40\right )} e^{3} + 8 \, x + 15\right )} \log \relax (x) + 10 \, x + 45}{x^{2} + 5 \, {\left (x + 5\right )} e^{6} + {\left (x^{2} + 15 \, x + 40\right )} e^{3} + 8 \, x + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^2-250*x-625)*exp(3)^4+(15*x^3+50*x^2-525*x-2000)*exp(3)^3+(-x^4+25*x^3+115*x^2-500*x-2350)*e
xp(3)^2+(-2*x^4+4*x^3+40*x^2-215*x-1200)*exp(3)-x^4-6*x^3-4*x^2-30*x-225)/((25*x^3+250*x^2+625*x)*exp(3)^4+(10
*x^4+200*x^3+1150*x^2+2000*x)*exp(3)^3+(x^5+40*x^4+435*x^3+1750*x^2+2350*x)*exp(3)^2+(2*x^5+46*x^4+350*x^3+109
0*x^2+1200*x)*exp(3)+x^5+16*x^4+94*x^3+240*x^2+225*x),x, algorithm="fricas")

[Out]

-(25*(x + 5)*e^6 + 10*(4*x + 15)*e^3 + (x^2 + 5*(x + 5)*e^6 + (x^2 + 15*x + 40)*e^3 + 8*x + 15)*log(x) + 10*x
+ 45)/(x^2 + 5*(x + 5)*e^6 + (x^2 + 15*x + 40)*e^3 + 8*x + 15)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^2-250*x-625)*exp(3)^4+(15*x^3+50*x^2-525*x-2000)*exp(3)^3+(-x^4+25*x^3+115*x^2-500*x-2350)*e
xp(3)^2+(-2*x^4+4*x^3+40*x^2-215*x-1200)*exp(3)-x^4-6*x^3-4*x^2-30*x-225)/((25*x^3+250*x^2+625*x)*exp(3)^4+(10
*x^4+200*x^3+1150*x^2+2000*x)*exp(3)^3+(x^5+40*x^4+435*x^3+1750*x^2+2350*x)*exp(3)^2+(2*x^5+46*x^4+350*x^3+109
0*x^2+1200*x)*exp(3)+x^5+16*x^4+94*x^3+240*x^2+225*x),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.43, size = 65, normalized size = 1.86

method	result	size
risch	\(\frac {\left (-5 \,{\mathrm e}^{6}-8 \,{\mathrm e}^{3}-2\right ) x -25 \,{\mathrm e}^{6}-30 \,{\mathrm e}^{3}-9}{x \,{\mathrm e}^{6}+\frac {x^{2} {\mathrm e}^{3}}{5}+5 \,{\mathrm e}^{6}+3 x \,{\mathrm e}^{3}+\frac {x^{2}}{5}+8 \,{\mathrm e}^{3}+\frac {8 x}{5}+3}-\ln \relax (x )\)	\(65\)
norman	\(\frac {-\frac {\left (50 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{9}+65 \,{\mathrm e}^{6}+10\right ) x}{{\mathrm e}^{3}+1}-125 \,{\mathrm e}^{6}-45-150 \,{\mathrm e}^{3}}{5 x \,{\mathrm e}^{6}+x^{2} {\mathrm e}^{3}+25 \,{\mathrm e}^{6}+15 x \,{\mathrm e}^{3}+x^{2}+40 \,{\mathrm e}^{3}+8 x +15}-\ln \relax (x )\)	\(84\)
default	\(\frac {5 \left (\munderset {\textit {\_R} =\RootOf \left (\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) \textit {\_Z}^{4}+\left (46 \,{\mathrm e}^{3}+10 \,{\mathrm e}^{9}+40 \,{\mathrm e}^{6}+16\right ) \textit {\_Z}^{3}+\left (350 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{12}+200 \,{\mathrm e}^{9}+435 \,{\mathrm e}^{6}+94\right ) \textit {\_Z}^{2}+\left (1090 \,{\mathrm e}^{3}+250 \,{\mathrm e}^{12}+1150 \,{\mathrm e}^{9}+1750 \,{\mathrm e}^{6}+240\right ) \textit {\_Z} +625 \,{\mathrm e}^{12}+2000 \,{\mathrm e}^{9}+2350 \,{\mathrm e}^{6}+1200 \,{\mathrm e}^{3}+225\right )}{\sum }\frac {\left (42+\left (2+10 \,{\mathrm e}^{3}+5 \,{\mathrm e}^{9}+13 \,{\mathrm e}^{6}\right ) \textit {\_R}^{2}+2 \left (9+39 \,{\mathrm e}^{3}+25 \,{\mathrm e}^{9}+55 \,{\mathrm e}^{6}\right ) \textit {\_R} +175 \,{\mathrm e}^{3}+125 \,{\mathrm e}^{9}+250 \,{\mathrm e}^{6}\right ) \ln \left (x -\textit {\_R} \right )}{120+25 \textit {\_R} \,{\mathrm e}^{12}+15 \textit {\_R}^{2} {\mathrm e}^{9}+2 \textit {\_R}^{3} {\mathrm e}^{6}+125 \,{\mathrm e}^{12}+200 \textit {\_R} \,{\mathrm e}^{9}+60 \textit {\_R}^{2} {\mathrm e}^{6}+4 \textit {\_R}^{3} {\mathrm e}^{3}+575 \,{\mathrm e}^{9}+435 \textit {\_R} \,{\mathrm e}^{6}+69 \textit {\_R}^{2} {\mathrm e}^{3}+2 \textit {\_R}^{3}+875 \,{\mathrm e}^{6}+350 \textit {\_R} \,{\mathrm e}^{3}+24 \textit {\_R}^{2}+545 \,{\mathrm e}^{3}+94 \textit {\_R}}\right )}{2}-\ln \relax (x )\)	\(246\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-25*x^2-250*x-625)*exp(3)^4+(15*x^3+50*x^2-525*x-2000)*exp(3)^3+(-x^4+25*x^3+115*x^2-500*x-2350)*exp(3)^
2+(-2*x^4+4*x^3+40*x^2-215*x-1200)*exp(3)-x^4-6*x^3-4*x^2-30*x-225)/((25*x^3+250*x^2+625*x)*exp(3)^4+(10*x^4+2
00*x^3+1150*x^2+2000*x)*exp(3)^3+(x^5+40*x^4+435*x^3+1750*x^2+2350*x)*exp(3)^2+(2*x^5+46*x^4+350*x^3+1090*x^2+
1200*x)*exp(3)+x^5+16*x^4+94*x^3+240*x^2+225*x),x,method=_RETURNVERBOSE)

[Out]

((-5*exp(6)-8*exp(3)-2)*x-25*exp(6)-30*exp(3)-9)/(x*exp(6)+1/5*x^2*exp(3)+5*exp(6)+3*x*exp(3)+1/5*x^2+8*exp(3)
+8/5*x+3)-ln(x)

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maxima [A]  time = 0.35, size = 61, normalized size = 1.74 \begin {gather*} -\frac {5 \, {\left (x {\left (5 \, e^{6} + 8 \, e^{3} + 2\right )} + 25 \, e^{6} + 30 \, e^{3} + 9\right )}}{x^{2} {\left (e^{3} + 1\right )} + x {\left (5 \, e^{6} + 15 \, e^{3} + 8\right )} + 25 \, e^{6} + 40 \, e^{3} + 15} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^2-250*x-625)*exp(3)^4+(15*x^3+50*x^2-525*x-2000)*exp(3)^3+(-x^4+25*x^3+115*x^2-500*x-2350)*e
xp(3)^2+(-2*x^4+4*x^3+40*x^2-215*x-1200)*exp(3)-x^4-6*x^3-4*x^2-30*x-225)/((25*x^3+250*x^2+625*x)*exp(3)^4+(10
*x^4+200*x^3+1150*x^2+2000*x)*exp(3)^3+(x^5+40*x^4+435*x^3+1750*x^2+2350*x)*exp(3)^2+(2*x^5+46*x^4+350*x^3+109
0*x^2+1200*x)*exp(3)+x^5+16*x^4+94*x^3+240*x^2+225*x),x, algorithm="maxima")

[Out]

-5*(x*(5*e^6 + 8*e^3 + 2) + 25*e^6 + 30*e^3 + 9)/(x^2*(e^3 + 1) + x*(5*e^6 + 15*e^3 + 8) + 25*e^6 + 40*e^3 + 1
5) - log(x)

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mupad [B]  time = 3.61, size = 61, normalized size = 1.74 \begin {gather*} -\ln \relax (x)-\frac {150\,{\mathrm {e}}^3+125\,{\mathrm {e}}^6+x\,\left (40\,{\mathrm {e}}^3+25\,{\mathrm {e}}^6+10\right )+45}{\left ({\mathrm {e}}^3+1\right )\,x^2+\left (15\,{\mathrm {e}}^3+5\,{\mathrm {e}}^6+8\right )\,x+40\,{\mathrm {e}}^3+25\,{\mathrm {e}}^6+15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(30*x + exp(12)*(250*x + 25*x^2 + 625) + exp(6)*(500*x - 115*x^2 - 25*x^3 + x^4 + 2350) + exp(9)*(525*x -
50*x^2 - 15*x^3 + 2000) + exp(3)*(215*x - 40*x^2 - 4*x^3 + 2*x^4 + 1200) + 4*x^2 + 6*x^3 + x^4 + 225)/(225*x
+ exp(12)*(625*x + 250*x^2 + 25*x^3) + exp(6)*(2350*x + 1750*x^2 + 435*x^3 + 40*x^4 + x^5) + exp(9)*(2000*x +
1150*x^2 + 200*x^3 + 10*x^4) + exp(3)*(1200*x + 1090*x^2 + 350*x^3 + 46*x^4 + 2*x^5) + 240*x^2 + 94*x^3 + 16*x
^4 + x^5),x)

[Out]

- log(x) - (150*exp(3) + 125*exp(6) + x*(40*exp(3) + 25*exp(6) + 10) + 45)/(40*exp(3) + 25*exp(6) + x*(15*exp(
3) + 5*exp(6) + 8) + x^2*(exp(3) + 1) + 15)

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sympy [B]  time = 6.36, size = 63, normalized size = 1.80 \begin {gather*} - \frac {x \left (10 + 40 e^{3} + 25 e^{6}\right ) + 45 + 150 e^{3} + 125 e^{6}}{x^{2} \left (1 + e^{3}\right ) + x \left (8 + 15 e^{3} + 5 e^{6}\right ) + 15 + 40 e^{3} + 25 e^{6}} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x**2-250*x-625)*exp(3)**4+(15*x**3+50*x**2-525*x-2000)*exp(3)**3+(-x**4+25*x**3+115*x**2-500*x
-2350)*exp(3)**2+(-2*x**4+4*x**3+40*x**2-215*x-1200)*exp(3)-x**4-6*x**3-4*x**2-30*x-225)/((25*x**3+250*x**2+62
5*x)*exp(3)**4+(10*x**4+200*x**3+1150*x**2+2000*x)*exp(3)**3+(x**5+40*x**4+435*x**3+1750*x**2+2350*x)*exp(3)**
2+(2*x**5+46*x**4+350*x**3+1090*x**2+1200*x)*exp(3)+x**5+16*x**4+94*x**3+240*x**2+225*x),x)

[Out]

-(x*(10 + 40*exp(3) + 25*exp(6)) + 45 + 150*exp(3) + 125*exp(6))/(x**2*(1 + exp(3)) + x*(8 + 15*exp(3) + 5*exp
(6)) + 15 + 40*exp(3) + 25*exp(6)) - log(x)

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