3.25.88 \(\int \frac {450+180 x+30 x^2+e^3 (-270 x^2-60 x^3)}{25 x^2+10 x^3+x^4+e^6 x^6+e^3 (-10 x^4-2 x^5)} \, dx\)

Optimal. Leaf size=27 \[ \frac {2}{e^6}-\frac {30 (3+x)}{x \left (5+x-e^3 x^2\right )} \]

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Rubi [A]  time = 0.18, antiderivative size = 29, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 4, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2074, 638, 618, 206} \begin {gather*} -\frac {6 \left (3 e^3 x+2\right )}{-e^3 x^2+x+5}-\frac {18}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(450 + 180*x + 30*x^2 + E^3*(-270*x^2 - 60*x^3))/(25*x^2 + 10*x^3 + x^4 + E^6*x^6 + E^3*(-10*x^4 - 2*x^5))
,x]

[Out]

-18/x - (6*(2 + 3*E^3*x))/(5 + x - E^3*x^2)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 \left (\frac {18}{x^2}-\frac {6 \left (-2+30 e^3+7 e^3 x\right )}{\left (-5-x+e^3 x^2\right )^2}-\frac {18 e^3}{-5-x+e^3 x^2}\right ) \, dx\\ &=-\frac {18}{x}-6 \int \frac {-2+30 e^3+7 e^3 x}{\left (-5-x+e^3 x^2\right )^2} \, dx-\left (18 e^3\right ) \int \frac {1}{-5-x+e^3 x^2} \, dx\\ &=-\frac {18}{x}-\frac {6 \left (2+3 e^3 x\right )}{5+x-e^3 x^2}+\left (18 e^3\right ) \int \frac {1}{-5-x+e^3 x^2} \, dx+\left (36 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+20 e^3-x^2} \, dx,x,-1+2 e^3 x\right )\\ &=-\frac {18}{x}-\frac {6 \left (2+3 e^3 x\right )}{5+x-e^3 x^2}-\frac {36 e^3 \tanh ^{-1}\left (\frac {1-2 e^3 x}{\sqrt {1+20 e^3}}\right )}{\sqrt {1+20 e^3}}-\left (36 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+20 e^3-x^2} \, dx,x,-1+2 e^3 x\right )\\ &=-\frac {18}{x}-\frac {6 \left (2+3 e^3 x\right )}{5+x-e^3 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.81 \begin {gather*} -\frac {30 (3+x)}{5 x+x^2-e^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(450 + 180*x + 30*x^2 + E^3*(-270*x^2 - 60*x^3))/(25*x^2 + 10*x^3 + x^4 + E^6*x^6 + E^3*(-10*x^4 - 2
*x^5)),x]

[Out]

(-30*(3 + x))/(5*x + x^2 - E^3*x^3)

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fricas [A]  time = 0.73, size = 22, normalized size = 0.81 \begin {gather*} \frac {30 \, {\left (x + 3\right )}}{x^{3} e^{3} - x^{2} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-60*x^3-270*x^2)*exp(3)+30*x^2+180*x+450)/(x^6*exp(3)^2+(-2*x^5-10*x^4)*exp(3)+x^4+10*x^3+25*x^2),
x, algorithm="fricas")

[Out]

30*(x + 3)/(x^3*e^3 - x^2 - 5*x)

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giac [A]  time = 0.40, size = 30, normalized size = 1.11 \begin {gather*} -\frac {18}{x} + 2.79551404258800 \times 10^{9} \, \log \left (x + 0.474661298947000\right ) - 2.79550903378200 \times 10^{9} \, \log \left (x + 0.474661296102000\right ) + 2.89251442174200 \times 10^{9} \, \log \left (x - 0.524448364156000\right ) - 2.89252337572800 \times 10^{9} \, \log \left (x - 0.524448367629000\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-60*x^3-270*x^2)*exp(3)+30*x^2+180*x+450)/(x^6*exp(3)^2+(-2*x^5-10*x^4)*exp(3)+x^4+10*x^3+25*x^2),
x, algorithm="giac")

[Out]

-18/x + 2.79551404258800e9*log(x + 0.474661298947000) - 2.79550903378200e9*log(x + 0.474661296102000) + 2.8925
1442174200e9*log(x - 0.524448364156000) - 2.89252337572800e9*log(x - 0.524448367629000)

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maple [A]  time = 0.12, size = 22, normalized size = 0.81

method	result	size
gosper	\(\frac {30 x +90}{x \left (x^{2} {\mathrm e}^{3}-x -5\right )}\)	\(22\)
norman	\(\frac {30 x +90}{x \left (x^{2} {\mathrm e}^{3}-x -5\right )}\)	\(23\)
risch	\(\frac {30 x +90}{x \left (x^{2} {\mathrm e}^{3}-x -5\right )}\)	\(23\)
default	\(-\frac {18}{x}+3 \left (\munderset {\textit {\_R} =\RootOf \left (25+\textit {\_Z}^{4} {\mathrm e}^{6}-2 \textit {\_Z}^{3} {\mathrm e}^{3}+\left (-10 \,{\mathrm e}^{3}+1\right ) \textit {\_Z}^{2}+10 \textit {\_Z} \right )}{\sum }\frac {\left (-3 \textit {\_R}^{2} {\mathrm e}^{6}-4 \textit {\_R} \,{\mathrm e}^{3}-15 \,{\mathrm e}^{3}+2\right ) \ln \left (x -\textit {\_R} \right )}{5+2 \textit {\_R}^{3} {\mathrm e}^{6}-3 \textit {\_R}^{2} {\mathrm e}^{3}-10 \textit {\_R} \,{\mathrm e}^{3}+\textit {\_R}}\right )\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-60*x^3-270*x^2)*exp(3)+30*x^2+180*x+450)/(x^6*exp(3)^2+(-2*x^5-10*x^4)*exp(3)+x^4+10*x^3+25*x^2),x,meth
od=_RETURNVERBOSE)

[Out]

30*(3+x)/x/(x^2*exp(3)-x-5)

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maxima [A]  time = 0.38, size = 22, normalized size = 0.81 \begin {gather*} \frac {30 \, {\left (x + 3\right )}}{x^{3} e^{3} - x^{2} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-60*x^3-270*x^2)*exp(3)+30*x^2+180*x+450)/(x^6*exp(3)^2+(-2*x^5-10*x^4)*exp(3)+x^4+10*x^3+25*x^2),
x, algorithm="maxima")

[Out]

30*(x + 3)/(x^3*e^3 - x^2 - 5*x)

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mupad [B]  time = 1.49, size = 20, normalized size = 0.74 \begin {gather*} -\frac {30\,\left (x+3\right )}{x\,\left (-{\mathrm {e}}^3\,x^2+x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((180*x - exp(3)*(270*x^2 + 60*x^3) + 30*x^2 + 450)/(x^6*exp(6) - exp(3)*(10*x^4 + 2*x^5) + 25*x^2 + 10*x^3
+ x^4),x)

[Out]

-(30*(x + 3))/(x*(x - x^2*exp(3) + 5))

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sympy [A]  time = 0.94, size = 20, normalized size = 0.74 \begin {gather*} - \frac {- 30 x - 90}{x^{3} e^{3} - x^{2} - 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-60*x**3-270*x**2)*exp(3)+30*x**2+180*x+450)/(x**6*exp(3)**2+(-2*x**5-10*x**4)*exp(3)+x**4+10*x**3
+25*x**2),x)

[Out]

-(-30*x - 90)/(x**3*exp(3) - x**2 - 5*x)

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