\(\int \frac {2375+4 e^4+e^2 (-195-8 x)+200 x+4 x^2}{2500+2575 x+199 x^2+4 x^3+e^4 (4+4 x)+e^2 (-200-203 x-8 x^2)} \, dx\) [6717]

Optimal result

Integrand size = 65, antiderivative size = 22 \[ \int \frac {2375+4 e^4+e^2 (-195-8 x)+200 x+4 x^2}{2500+2575 x+199 x^2+4 x^3+e^4 (4+4 x)+e^2 \left (-200-203 x-8 x^2\right )} \, dx=5+\log \left (-4+x-5 \left (x+\frac {x}{-25+e^2-x}\right )\right ) \]

[Out]

5+ln(-4*x-4-5*x/(exp(2)-x-25))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {2099, 642} \[ \int \frac {2375+4 e^4+e^2 (-195-8 x)+200 x+4 x^2}{2500+2575 x+199 x^2+4 x^3+e^4 (4+4 x)+e^2 \left (-200-203 x-8 x^2\right )} \, dx=\log \left (4 x^2+\left (99-4 e^2\right ) x+4 \left (25-e^2\right )\right )-\log \left (x-e^2+25\right ) \]

[In]

Int[(2375 + 4*E^4 + E^2*(-195 - 8*x) + 200*x + 4*x^2)/(2500 + 2575*x + 199*x^2 + 4*x^3 + E^4*(4 + 4*x) + E^2*(
-200 - 203*x - 8*x^2)),x]

[Out]

-Log[25 - E^2 + x] + Log[4*(25 - E^2) + (99 - 4*E^2)*x + 4*x^2]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 2099

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{align*} \text {integral}& = \int \left (\frac {1}{-25+e^2-x}+\frac {99-4 e^2+8 x}{4 \left (25-e^2\right )+\left (99-4 e^2\right ) x+4 x^2}\right ) \, dx \\ & = -\log \left (25-e^2+x\right )+\int \frac {99-4 e^2+8 x}{4 \left (25-e^2\right )+\left (99-4 e^2\right ) x+4 x^2} \, dx \\ & = -\log \left (25-e^2+x\right )+\log \left (4 \left (25-e^2\right )+\left (99-4 e^2\right ) x+4 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {2375+4 e^4+e^2 (-195-8 x)+200 x+4 x^2}{2500+2575 x+199 x^2+4 x^3+e^4 (4+4 x)+e^2 \left (-200-203 x-8 x^2\right )} \, dx=-\log \left (25-e^2+x\right )+\log \left (100-4 e^2+99 x-4 e^2 x+4 x^2\right ) \]

[In]

Integrate[(2375 + 4*E^4 + E^2*(-195 - 8*x) + 200*x + 4*x^2)/(2500 + 2575*x + 199*x^2 + 4*x^3 + E^4*(4 + 4*x) +
E^2*(-200 - 203*x - 8*x^2)),x]

[Out]

-Log[25 - E^2 + x] + Log[100 - 4*E^2 + 99*x - 4*E^2*x + 4*x^2]

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\(-\ln \left (-{\mathrm e}^{2}+x +25\right )+\ln \left (-{\mathrm e}^{2} x +x^{2}-{\mathrm e}^{2}+\frac {99 x}{4}+25\right )\)	\(30\)
norman	\(-\ln \left ({\mathrm e}^{2}-x -25\right )+\ln \left (4 \,{\mathrm e}^{2} x -4 x^{2}+4 \,{\mathrm e}^{2}-99 x -100\right )\)	\(32\)
risch	\(-\ln \left (-{\mathrm e}^{2}+x +25\right )+\ln \left (4 x^{2}+\left (-4 \,{\mathrm e}^{2}+99\right ) x -4 \,{\mathrm e}^{2}+100\right )\)	\(32\)
default	\(-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4 \textit {\_Z}^{3}+\left (-8 \,{\mathrm e}^{2}+199\right ) \textit {\_Z}^{2}+\left (-203 \,{\mathrm e}^{2}+4 \,{\mathrm e}^{4}+2575\right ) \textit {\_Z} -200 \,{\mathrm e}^{2}+4 \,{\mathrm e}^{4}+2500\right )}{\sum }\frac {\left (-4 \,{\mathrm e}^{4}+8 \textit {\_R} \,{\mathrm e}^{2}-4 \textit {\_R}^{2}+195 \,{\mathrm e}^{2}-200 \textit {\_R} -2375\right ) \ln \left (x -\textit {\_R} \right )}{4 \,{\mathrm e}^{4}-16 \textit {\_R} \,{\mathrm e}^{2}+12 \textit {\_R}^{2}-203 \,{\mathrm e}^{2}+398 \textit {\_R} +2575}\right )\)	\(99\)

[In]

int((4*exp(2)^2+(-8*x-195)*exp(2)+4*x^2+200*x+2375)/((4+4*x)*exp(2)^2+(-8*x^2-203*x-200)*exp(2)+4*x^3+199*x^2+
2575*x+2500),x,method=_RETURNVERBOSE)

[Out]

-ln(-exp(2)+x+25)+ln(-exp(2)*x+x^2-exp(2)+99/4*x+25)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {2375+4 e^4+e^2 (-195-8 x)+200 x+4 x^2}{2500+2575 x+199 x^2+4 x^3+e^4 (4+4 x)+e^2 \left (-200-203 x-8 x^2\right )} \, dx=\log \left (4 \, x^{2} - 4 \, {\left (x + 1\right )} e^{2} + 99 \, x + 100\right ) - \log \left (x - e^{2} + 25\right ) \]

[In]

integrate((4*exp(2)^2+(-8*x-195)*exp(2)+4*x^2+200*x+2375)/((4+4*x)*exp(2)^2+(-8*x^2-203*x-200)*exp(2)+4*x^3+19
9*x^2+2575*x+2500),x, algorithm="fricas")

[Out]

log(4*x^2 - 4*(x + 1)*e^2 + 99*x + 100) - log(x - e^2 + 25)

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {2375+4 e^4+e^2 (-195-8 x)+200 x+4 x^2}{2500+2575 x+199 x^2+4 x^3+e^4 (4+4 x)+e^2 \left (-200-203 x-8 x^2\right )} \, dx=- \log {\left (x - e^{2} + 25 \right )} + \log {\left (x^{2} + x \left (\frac {99}{4} - e^{2}\right ) - e^{2} + 25 \right )} \]

[In]

integrate((4*exp(2)**2+(-8*x-195)*exp(2)+4*x**2+200*x+2375)/((4+4*x)*exp(2)**2+(-8*x**2-203*x-200)*exp(2)+4*x*
*3+199*x**2+2575*x+2500),x)

[Out]

-log(x - exp(2) + 25) + log(x**2 + x*(99/4 - exp(2)) - exp(2) + 25)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {2375+4 e^4+e^2 (-195-8 x)+200 x+4 x^2}{2500+2575 x+199 x^2+4 x^3+e^4 (4+4 x)+e^2 \left (-200-203 x-8 x^2\right )} \, dx=\log \left (4 \, x^{2} - x {\left (4 \, e^{2} - 99\right )} - 4 \, e^{2} + 100\right ) - \log \left (x - e^{2} + 25\right ) \]

[In]

integrate((4*exp(2)^2+(-8*x-195)*exp(2)+4*x^2+200*x+2375)/((4+4*x)*exp(2)^2+(-8*x^2-203*x-200)*exp(2)+4*x^3+19
9*x^2+2575*x+2500),x, algorithm="maxima")

[Out]

log(4*x^2 - x*(4*e^2 - 99) - 4*e^2 + 100) - log(x - e^2 + 25)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {2375+4 e^4+e^2 (-195-8 x)+200 x+4 x^2}{2500+2575 x+199 x^2+4 x^3+e^4 (4+4 x)+e^2 \left (-200-203 x-8 x^2\right )} \, dx=\log \left ({\left | 4 \, x^{2} - 4 \, x e^{2} + 99 \, x - 4 \, e^{2} + 100 \right |}\right ) - \log \left ({\left | x - e^{2} + 25 \right |}\right ) \]

[In]

integrate((4*exp(2)^2+(-8*x-195)*exp(2)+4*x^2+200*x+2375)/((4+4*x)*exp(2)^2+(-8*x^2-203*x-200)*exp(2)+4*x^3+19
9*x^2+2575*x+2500),x, algorithm="giac")

[Out]

log(abs(4*x^2 - 4*x*e^2 + 99*x - 4*e^2 + 100)) - log(abs(x - e^2 + 25))

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {2375+4 e^4+e^2 (-195-8 x)+200 x+4 x^2}{2500+2575 x+199 x^2+4 x^3+e^4 (4+4 x)+e^2 \left (-200-203 x-8 x^2\right )} \, dx=\ln \left (\frac {99\,x}{4}-{\mathrm {e}}^2-x\,{\mathrm {e}}^2+x^2+25\right )-\ln \left (x-{\mathrm {e}}^2+25\right ) \]

[In]

int((200*x + 4*exp(4) + 4*x^2 - exp(2)*(8*x + 195) + 2375)/(2575*x - exp(2)*(203*x + 8*x^2 + 200) + 199*x^2 +
4*x^3 + exp(4)*(4*x + 4) + 2500),x)

[Out]

log((99*x)/4 - exp(2) - x*exp(2) + x^2 + 25) - log(x - exp(2) + 25)