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

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

________________________________________________________________________________________

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} $$\begin{array}{c}\frac{x}{e^4{x}^2-2x+5}+\frac{\log (x)}{e}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{25+(-20+5e)x+4x^2+e^8{x}^4+e^4(10x^2-4x^3-e{x}^3)}{e^9{x}^5+e(25x-20x^2+4x^3)+e^5(10x^3-4x^4)}dx\\ & =\int (\frac{1}{ex}+\frac{1}{-5+2x-e^4{x}^2}-\frac{2(-5+x)}{{(5-2x+e^4{x}^2)}^2})dx\\ & =\frac{\log (x)}{e}-2\int \frac{-5+x}{{(5-2x+e^4{x}^2)}^2}dx+\int \frac{1}{-5+2x-e^4{x}^2}dx\\ & =\frac{x}{5-2x+e^4{x}^2}+\frac{\log (x)}{e}-2\mathrm{Subst}(\int \frac{1}{4(1-5e^4)-x^2}dx,x,2-2e^{4}x)+\int \frac{1}{5-2x+e^4{x}^2}dx\\ & =\frac{x}{5-2x+e^4{x}^2}+\frac{{\tan }^{-1}\left(\frac{1-e^{4}x}{\sqrt{-1+5e^4}}\right)}{\sqrt{-1+5e^4}}+\frac{\log (x)}{e}-2\mathrm{Subst}(\int \frac{1}{4(1-5e^4)-x^2}dx,x,-2+2e^{4}x)\\ & =\frac{x}{5-2x+e^4{x}^2}+\frac{\log (x)}{e}\end{array}\end{array}$$

________________________________________________________________________________________

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

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

________________________________________________________________________________________

fricas [A]  time = 0.84, size = 38, normalized size = 1.58 $$\begin{array}{c}\frac{xe+(x^2{e}^4-2x+5)\log (x)}{x^2{e}^5-(2x-5)e}\end{array}$$

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)

________________________________________________________________________________________

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

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

________________________________________________________________________________________

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

method	result	size
risch	$\frac{x}{x^2{\mathrm{e}}^4-2x+5}+\ln (x){\mathrm{e}}^{-1}$	$22$
norman	$\frac{\frac{x^2{\mathrm{e}}^4}{2}+\frac{5}{2}}{x^2{\mathrm{e}}^4-2x+5}+\ln (x){\mathrm{e}}^{-1}$	$36$
default	${\mathrm{e}}^{-1}(\ln (x)-\frac{(\sum _{\mathit{\text{\_R}}=\mathrm{RootOf}(25+{\mathit{\text{\_Z}}}^4{\mathrm{e}}^8-4{\mathit{\text{\_Z}}}^3{\mathrm{e}}^4-(-10{\mathrm{e}}^4-4){\mathit{\text{\_Z}}}^2-20\mathit{\text{\_Z}})}\frac{(-{\mathit{\text{\_R}}}^2{\mathrm{e}}^5+5\mathrm{e})\ln (x-\mathit{\text{\_R}})}{5-{\mathit{\text{\_R}}}^3{\mathrm{e}}^8+3{\mathit{\text{\_R}}}^2{\mathrm{e}}^4-5\mathit{\text{\_R}}{\mathrm{e}}^4-2\mathit{\text{\_R}}})}{4})$	$89$

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)

________________________________________________________________________________________

maxima [A]  time = 0.53, size = 21, normalized size = 0.88 $$\begin{array}{c}{e}^{(-1)}\log (x)+\frac{x}{x^2{e}^4-2x+5}\end{array}$$

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)

________________________________________________________________________________________

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

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))

________________________________________________________________________________________

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

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)

________________________________________________________________________________________