3.48.4 \(\int \frac {2 e^{3+8 e^5} x}{e^9-15 e^6 x+75 e^3 x^2-125 x^3} \, dx\)

Optimal. Leaf size=22 \[ \frac {e^{8 e^5} x^2}{\left (-e^3+5 x\right )^2} \]

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Rubi [B]  time = 0.04, antiderivative size = 45, normalized size of antiderivative = 2.05, number of steps used = 3, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 2074} \begin {gather*} \frac {e^{6+8 e^5}}{25 \left (e^3-5 x\right )^2}-\frac {2 e^{3+8 e^5}}{25 \left (e^3-5 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^(3 + 8*E^5)*x)/(E^9 - 15*E^6*x + 75*E^3*x^2 - 125*x^3),x]

[Out]

E^(6 + 8*E^5)/(25*(E^3 - 5*x)^2) - (2*E^(3 + 8*E^5))/(25*(E^3 - 5*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\left (2 e^{3+8 e^5}\right ) \int \frac {x}{e^9-15 e^6 x+75 e^3 x^2-125 x^3} \, dx\\ &=\left (2 e^{3+8 e^5}\right ) \int \left (\frac {e^3}{5 \left (e^3-5 x\right )^3}-\frac {1}{5 \left (e^3-5 x\right )^2}\right ) \, dx\\ &=\frac {e^{6+8 e^5}}{25 \left (e^3-5 x\right )^2}-\frac {2 e^{3+8 e^5}}{25 \left (e^3-5 x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.32 \begin {gather*} -\frac {e^{3+8 e^5} \left (e^3-10 x\right )}{25 \left (e^3-5 x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^(3 + 8*E^5)*x)/(E^9 - 15*E^6*x + 75*E^3*x^2 - 125*x^3),x]

[Out]

-1/25*(E^(3 + 8*E^5)*(E^3 - 10*x))/(E^3 - 5*x)^2

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fricas [A]  time = 1.03, size = 32, normalized size = 1.45 \begin {gather*} \frac {{\left (10 \, x - e^{3}\right )} e^{\left (8 \, e^{5} + 3\right )}}{25 \, {\left (25 \, x^{2} - 10 \, x e^{3} + e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(3)*exp(4*exp(5))^2/(exp(3)^3-15*x*exp(3)^2+75*x^2*exp(3)-125*x^3),x, algorithm="fricas")

[Out]

1/25*(10*x - e^3)*e^(8*e^5 + 3)/(25*x^2 - 10*x*e^3 + e^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, x e^{\left (8 \, e^{5} + 3\right )}}{125 \, x^{3} - 75 \, x^{2} e^{3} + 15 \, x e^{6} - e^{9}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(3)*exp(4*exp(5))^2/(exp(3)^3-15*x*exp(3)^2+75*x^2*exp(3)-125*x^3),x, algorithm="giac")

[Out]

integrate(-2*x*e^(8*e^5 + 3)/(125*x^3 - 75*x^2*e^3 + 15*x*e^6 - e^9), x)

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maple [A]  time = 0.07, size = 32, normalized size = 1.45

method	result	size
norman	\(\frac {\frac {2 \,{\mathrm e}^{3} {\mathrm e}^{8 \,{\mathrm e}^{5}} x}{5}-\frac {{\mathrm e}^{6} {\mathrm e}^{8 \,{\mathrm e}^{5}}}{25}}{\left ({\mathrm e}^{3}-5 x \right )^{2}}\)	\(32\)
risch	\(\frac {2 \left (\frac {x}{5}-\frac {{\mathrm e}^{3}}{50}\right ) {\mathrm e}^{3+8 \,{\mathrm e}^{5}}}{{\mathrm e}^{6}-10 x \,{\mathrm e}^{3}+25 x^{2}}\)	\(33\)
gosper	\(-\frac {{\mathrm e}^{8 \,{\mathrm e}^{5}} {\mathrm e}^{3} \left ({\mathrm e}^{3}-10 x \right )}{25 \left ({\mathrm e}^{6}-10 x \,{\mathrm e}^{3}+25 x^{2}\right )}\)	\(35\)
default	\(-\frac {2 \,{\mathrm e}^{3} {\mathrm e}^{8 \,{\mathrm e}^{5}} \left (\munderset {\textit {\_R} =\RootOf \left (-{\mathrm e}^{9}+15 \textit {\_Z} \,{\mathrm e}^{6}-75 \textit {\_Z}^{2} {\mathrm e}^{3}+125 \textit {\_Z}^{3}\right )}{\sum }\frac {\textit {\_R} \ln \left (x -\textit {\_R} \right )}{{\mathrm e}^{6}-10 \textit {\_R} \,{\mathrm e}^{3}+25 \textit {\_R}^{2}}\right )}{15}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*exp(3)*exp(4*exp(5))^2/(exp(3)^3-15*x*exp(3)^2+75*x^2*exp(3)-125*x^3),x,method=_RETURNVERBOSE)

[Out]

(2/5*exp(3)*exp(exp(5))^8*x-1/25*exp(3)^2*exp(exp(5))^8)/(exp(3)-5*x)^2

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maxima [A]  time = 0.38, size = 32, normalized size = 1.45 \begin {gather*} \frac {{\left (10 \, x - e^{3}\right )} e^{\left (8 \, e^{5} + 3\right )}}{25 \, {\left (25 \, x^{2} - 10 \, x e^{3} + e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(3)*exp(4*exp(5))^2/(exp(3)^3-15*x*exp(3)^2+75*x^2*exp(3)-125*x^3),x, algorithm="maxima")

[Out]

1/25*(10*x - e^3)*e^(8*e^5 + 3)/(25*x^2 - 10*x*e^3 + e^6)

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mupad [B]  time = 0.12, size = 37, normalized size = 1.68 \begin {gather*} -\frac {\frac {{\mathrm {e}}^{8\,{\mathrm {e}}^5+6}}{25}-\frac {2\,x\,{\mathrm {e}}^{8\,{\mathrm {e}}^5+3}}{5}}{25\,x^2-10\,{\mathrm {e}}^3\,x+{\mathrm {e}}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(8*exp(5))*exp(3))/(exp(9) - 15*x*exp(6) + 75*x^2*exp(3) - 125*x^3),x)

[Out]

-(exp(8*exp(5) + 6)/25 - (2*x*exp(8*exp(5) + 3))/5)/(exp(6) - 10*x*exp(3) + 25*x^2)

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sympy [B]  time = 0.15, size = 36, normalized size = 1.64 \begin {gather*} - \frac {2 \left (- 10 x + e^{3}\right ) e^{3} e^{8 e^{5}}}{1250 x^{2} - 500 x e^{3} + 50 e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(3)*exp(4*exp(5))**2/(exp(3)**3-15*x*exp(3)**2+75*x**2*exp(3)-125*x**3),x)

[Out]

-2*(-10*x + exp(3))*exp(3)*exp(8*exp(5))/(1250*x**2 - 500*x*exp(3) + 50*exp(6))

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