3.17.97 \(\int \frac {25-20 x+5 e x+4 x^2+e^8 x^4+e^4 (10 x^2-4 x^3-e x^3)}{e^9 x^5+e (25 x-20 x^2+4 x^3)+e^5 (10 x^3-4 x^4)} \, dx\)

Optimal. Leaf size=24 \[ 5+\frac {x}{5-2 x+e^4 x^2}+\frac {\log (x)}{e} \]

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Rubi [A]  time = 0.17, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 5, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 2074, 618, 204, 638} \begin {gather*} \frac {x}{e^4 x^2-2 x+5}+\frac {\log (x)}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 - 20*x + 5*E*x + 4*x^2 + E^8*x^4 + E^4*(10*x^2 - 4*x^3 - E*x^3))/(E^9*x^5 + E*(25*x - 20*x^2 + 4*x^3)
+ E^5*(10*x^3 - 4*x^4)),x]

[Out]

x/(5 - 2*x + E^4*x^2) + Log[x]/E

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

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 \frac {25+(-20+5 e) x+4 x^2+e^8 x^4+e^4 \left (10 x^2-4 x^3-e x^3\right )}{e^9 x^5+e \left (25 x-20 x^2+4 x^3\right )+e^5 \left (10 x^3-4 x^4\right )} \, dx\\ &=\int \left (\frac {1}{e x}+\frac {1}{-5+2 x-e^4 x^2}-\frac {2 (-5+x)}{\left (5-2 x+e^4 x^2\right )^2}\right ) \, dx\\ &=\frac {\log (x)}{e}-2 \int \frac {-5+x}{\left (5-2 x+e^4 x^2\right )^2} \, dx+\int \frac {1}{-5+2 x-e^4 x^2} \, dx\\ &=\frac {x}{5-2 x+e^4 x^2}+\frac {\log (x)}{e}-2 \operatorname {Subst}\left (\int \frac {1}{4 \left (1-5 e^4\right )-x^2} \, dx,x,2-2 e^4 x\right )+\int \frac {1}{5-2 x+e^4 x^2} \, dx\\ &=\frac {x}{5-2 x+e^4 x^2}+\frac {\tan ^{-1}\left (\frac {1-e^4 x}{\sqrt {-1+5 e^4}}\right )}{\sqrt {-1+5 e^4}}+\frac {\log (x)}{e}-2 \operatorname {Subst}\left (\int \frac {1}{4 \left (1-5 e^4\right )-x^2} \, dx,x,-2+2 e^4 x\right )\\ &=\frac {x}{5-2 x+e^4 x^2}+\frac {\log (x)}{e}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.00 \begin {gather*} \frac {\frac {e x}{5-2 x+e^4 x^2}+\log (x)}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 - 20*x + 5*E*x + 4*x^2 + E^8*x^4 + E^4*(10*x^2 - 4*x^3 - E*x^3))/(E^9*x^5 + E*(25*x - 20*x^2 + 4
*x^3) + E^5*(10*x^3 - 4*x^4)),x]

[Out]

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

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fricas [A]  time = 0.84, size = 38, normalized size = 1.58 \begin {gather*} \frac {x e + {\left (x^{2} e^{4} - 2 \, x + 5\right )} \log \relax (x)}{x^{2} e^{5} - {\left (2 \, x - 5\right )} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4*exp(2)^4+(-x^3*exp(1)-4*x^3+10*x^2)*exp(2)^2+5*x*exp(1)+4*x^2-20*x+25)/(x^5*exp(1)*exp(2)^4+(-4
*x^4+10*x^3)*exp(1)*exp(2)^2+(4*x^3-20*x^2+25*x)*exp(1)),x, algorithm="fricas")

[Out]

(x*e + (x^2*e^4 - 2*x + 5)*log(x))/(x^2*e^5 - (2*x - 5)*e)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4*exp(2)^4+(-x^3*exp(1)-4*x^3+10*x^2)*exp(2)^2+5*x*exp(1)+4*x^2-20*x+25)/(x^5*exp(1)*exp(2)^4+(-4
*x^4+10*x^3)*exp(1)*exp(2)^2+(4*x^3-20*x^2+25*x)*exp(1)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 0.72Not invertible Error: Bad Argument Value

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maple [A]  time = 0.15, size = 22, normalized size = 0.92

method	result	size
risch	\(\frac {x}{x^{2} {\mathrm e}^{4}-2 x +5}+\ln \relax (x ) {\mathrm e}^{-1}\)	\(22\)
norman	\(\frac {\frac {x^{2} {\mathrm e}^{4}}{2}+\frac {5}{2}}{x^{2} {\mathrm e}^{4}-2 x +5}+\ln \relax (x ) {\mathrm e}^{-1}\)	\(36\)
default	\({\mathrm e}^{-1} \left (\ln \relax (x )-\frac {\left (\munderset {\textit {\_R} =\RootOf \left (25+\textit {\_Z}^{4} {\mathrm e}^{8}-4 \textit {\_Z}^{3} {\mathrm e}^{4}-\left (-10 \,{\mathrm e}^{4}-4\right ) \textit {\_Z}^{2}-20 \textit {\_Z} \right )}{\sum }\frac {\left (-\textit {\_R}^{2} {\mathrm e}^{5}+5 \,{\mathrm e}\right ) \ln \left (x -\textit {\_R} \right )}{5-\textit {\_R}^{3} {\mathrm e}^{8}+3 \textit {\_R}^{2} {\mathrm e}^{4}-5 \textit {\_R} \,{\mathrm e}^{4}-2 \textit {\_R}}\right )}{4}\right )\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*exp(2)^4+(-x^3*exp(1)-4*x^3+10*x^2)*exp(2)^2+5*x*exp(1)+4*x^2-20*x+25)/(x^5*exp(1)*exp(2)^4+(-4*x^4+1
0*x^3)*exp(1)*exp(2)^2+(4*x^3-20*x^2+25*x)*exp(1)),x,method=_RETURNVERBOSE)

[Out]

x/(x^2*exp(4)-2*x+5)+ln(x)*exp(-1)

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maxima [A]  time = 0.53, size = 21, normalized size = 0.88 \begin {gather*} e^{\left (-1\right )} \log \relax (x) + \frac {x}{x^{2} e^{4} - 2 \, x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4*exp(2)^4+(-x^3*exp(1)-4*x^3+10*x^2)*exp(2)^2+5*x*exp(1)+4*x^2-20*x+25)/(x^5*exp(1)*exp(2)^4+(-4
*x^4+10*x^3)*exp(1)*exp(2)^2+(4*x^3-20*x^2+25*x)*exp(1)),x, algorithm="maxima")

[Out]

e^(-1)*log(x) + x/(x^2*e^4 - 2*x + 5)

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mupad [B]  time = 0.18, size = 28, normalized size = 1.17 \begin {gather*} {\mathrm {e}}^{-1}\,\ln \relax (x)+\frac {x\,\mathrm {e}}{{\mathrm {e}}^5\,x^2-2\,\mathrm {e}\,x+5\,\mathrm {e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*exp(1) - exp(4)*(x^3*exp(1) - 10*x^2 + 4*x^3) - 20*x + x^4*exp(8) + 4*x^2 + 25)/(exp(1)*(25*x - 20*x^
2 + 4*x^3) + exp(5)*(10*x^3 - 4*x^4) + x^5*exp(9)),x)

[Out]

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

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sympy [A]  time = 1.36, size = 19, normalized size = 0.79 \begin {gather*} \frac {x}{x^{2} e^{4} - 2 x + 5} + \frac {\log {\relax (x )}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4*exp(2)**4+(-x**3*exp(1)-4*x**3+10*x**2)*exp(2)**2+5*x*exp(1)+4*x**2-20*x+25)/(x**5*exp(1)*exp(
2)**4+(-4*x**4+10*x**3)*exp(1)*exp(2)**2+(4*x**3-20*x**2+25*x)*exp(1)),x)

[Out]

x/(x**2*exp(4) - 2*x + 5) + exp(-1)*log(x)

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