3.58.78 \(\int \frac {-250+75 x+e (-25 x^2+30 x^3-x^4)}{e (400 x^3-160 x^4+16 x^5)} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{16} \left (x+\frac {-5+\frac {25}{e x^2}+x^2}{5-x}-\log (x)\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 6, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 1594, 27, 1620} \begin {gather*} \frac {5}{16 e x^2}+\frac {1+20 e}{16 e (5-x)}+\frac {1}{16 e x}-\frac {\log (x)}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-250 + 75*x + E*(-25*x^2 + 30*x^3 - x^4))/(E*(400*x^3 - 160*x^4 + 16*x^5)),x]

[Out]

(1 + 20*E)/(16*E*(5 - x)) + 5/(16*E*x^2) + 1/(16*E*x) - Log[x]/16

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{400 x^3-160 x^4+16 x^5} \, dx}{e}\\ &=\frac {\int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{x^3 \left (400-160 x+16 x^2\right )} \, dx}{e}\\ &=\frac {\int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{16 (-5+x)^2 x^3} \, dx}{e}\\ &=\frac {\int \frac {-250+75 x+e \left (-25 x^2+30 x^3-x^4\right )}{(-5+x)^2 x^3} \, dx}{16 e}\\ &=\frac {\int \left (\frac {1+20 e}{(-5+x)^2}-\frac {10}{x^3}-\frac {1}{x^2}-\frac {e}{x}\right ) \, dx}{16 e}\\ &=\frac {1+20 e}{16 e (5-x)}+\frac {5}{16 e x^2}+\frac {1}{16 e x}-\frac {\log (x)}{16}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 0.97 \begin {gather*} -\frac {\frac {5 \left (5+4 e x^2\right )}{(-5+x) x^2}+e \log (x)}{16 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-250 + 75*x + E*(-25*x^2 + 30*x^3 - x^4))/(E*(400*x^3 - 160*x^4 + 16*x^5)),x]

[Out]

-1/16*((5*(5 + 4*E*x^2))/((-5 + x)*x^2) + E*Log[x])/E

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fricas [A]  time = 1.64, size = 38, normalized size = 1.23 \begin {gather*} -\frac {{\left (20 \, x^{2} e + {\left (x^{3} - 5 \, x^{2}\right )} e \log \relax (x) + 25\right )} e^{\left (-1\right )}}{16 \, {\left (x^{3} - 5 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+30*x^3-25*x^2)*exp(1)+75*x-250)/(16*x^5-160*x^4+400*x^3)/exp(1),x, algorithm="fricas")

[Out]

-1/16*(20*x^2*e + (x^3 - 5*x^2)*e*log(x) + 25)*e^(-1)/(x^3 - 5*x^2)

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giac [A]  time = 0.16, size = 30, normalized size = 0.97 \begin {gather*} -\frac {1}{16} \, {\left (e \log \left ({\left | x \right |}\right ) + \frac {5 \, {\left (4 \, x^{2} e + 5\right )}}{{\left (x - 5\right )} x^{2}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+30*x^3-25*x^2)*exp(1)+75*x-250)/(16*x^5-160*x^4+400*x^3)/exp(1),x, algorithm="giac")

[Out]

-1/16*(e*log(abs(x)) + 5*(4*x^2*e + 5)/((x - 5)*x^2))*e^(-1)

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maple [A]  time = 0.09, size = 26, normalized size = 0.84

method	result	size
risch	\(\frac {{\mathrm e}^{-1} \left (-\frac {5 x^{2} {\mathrm e}}{4}-\frac {25}{16}\right )}{x^{2} \left (x -5\right )}-\frac {\ln \relax (x )}{16}\)	\(26\)
norman	\(\frac {-\frac {5 x^{2}}{4}-\frac {25 \,{\mathrm e}^{-1}}{16}}{x^{2} \left (x -5\right )}-\frac {\ln \relax (x )}{16}\)	\(27\)
default	\(\frac {{\mathrm e}^{-1} \left (\frac {5}{x^{2}}+\frac {1}{x}-{\mathrm e} \ln \relax (x )-\frac {20 \,{\mathrm e}+1}{x -5}\right )}{16}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^4+30*x^3-25*x^2)*exp(1)+75*x-250)/(16*x^5-160*x^4+400*x^3)/exp(1),x,method=_RETURNVERBOSE)

[Out]

exp(-1)*(-5/4*x^2*exp(1)-25/16)/x^2/(x-5)-1/16*ln(x)

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maxima [A]  time = 0.34, size = 32, normalized size = 1.03 \begin {gather*} -\frac {1}{16} \, {\left (e \log \relax (x) + \frac {5 \, {\left (4 \, x^{2} e + 5\right )}}{x^{3} - 5 \, x^{2}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+30*x^3-25*x^2)*exp(1)+75*x-250)/(16*x^5-160*x^4+400*x^3)/exp(1),x, algorithm="maxima")

[Out]

-1/16*(e*log(x) + 5*(4*x^2*e + 5)/(x^3 - 5*x^2))*e^(-1)

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mupad [B]  time = 4.17, size = 32, normalized size = 1.03 \begin {gather*} \frac {20\,\mathrm {e}\,x^2+25}{80\,x^2\,\mathrm {e}-16\,x^3\,\mathrm {e}}-\frac {\ln \relax (x)}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-1)*(exp(1)*(25*x^2 - 30*x^3 + x^4) - 75*x + 250))/(400*x^3 - 160*x^4 + 16*x^5),x)

[Out]

(20*x^2*exp(1) + 25)/(80*x^2*exp(1) - 16*x^3*exp(1)) - log(x)/16

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sympy [A]  time = 0.52, size = 32, normalized size = 1.03 \begin {gather*} - \frac {20 e x^{2} + 25}{16 e x^{3} - 80 e x^{2}} - \frac {\log {\relax (x )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**4+30*x**3-25*x**2)*exp(1)+75*x-250)/(16*x**5-160*x**4+400*x**3)/exp(1),x)

[Out]

-(20*E*x**2 + 25)/(16*E*x**3 - 80*E*x**2) - log(x)/16

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