3.92.57 \(\int \frac {150-30 e-225 x}{100 x^3+4 e^2 x^3-200 x^4+100 x^5+e (-40 x^3+40 x^4)} \, dx\)

Optimal. Leaf size=17 \[ 4+\frac {15}{4 x^2 (-5+e+5 x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 1998, 1594, 27, 12, 74} \begin {gather*} -\frac {15}{4 (-5 x-e+5) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(150 - 30*E - 225*x)/(100*x^3 + 4*E^2*x^3 - 200*x^4 + 100*x^5 + E*(-40*x^3 + 40*x^4)),x]

[Out]

-15/(4*(5 - E - 5*x)*x^2)

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 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 74

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

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 1998

Int[(u_)^(p_.)*(z_), x_Symbol] :> Int[ExpandToSum[z, x]*ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && BinomialQ[z,
x] && GeneralizedTrinomialQ[u, x] && EqQ[BinomialDegree[z, x] - GeneralizedTrinomialDegree[u, x], 0] &&  !(Bi
nomialMatchQ[z, x] && GeneralizedTrinomialMatchQ[u, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {150-30 e-225 x}{\left (100+4 e^2\right ) x^3-200 x^4+100 x^5+e \left (-40 x^3+40 x^4\right )} \, dx\\ &=\int \frac {30 (5-e)-225 x}{4 (5-e)^2 x^3-40 (5-e) x^4+100 x^5} \, dx\\ &=\int \frac {30 (5-e)-225 x}{x^3 \left (4 (5-e)^2-40 (5-e) x+100 x^2\right )} \, dx\\ &=\int \frac {30 (5-e)-225 x}{4 x^3 (-5+e+5 x)^2} \, dx\\ &=\frac {1}{4} \int \frac {30 (5-e)-225 x}{x^3 (-5+e+5 x)^2} \, dx\\ &=-\frac {15}{4 (5-e-5 x) x^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 15, normalized size = 0.88 \begin {gather*} \frac {15}{4 x^2 (-5+e+5 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(150 - 30*E - 225*x)/(100*x^3 + 4*E^2*x^3 - 200*x^4 + 100*x^5 + E*(-40*x^3 + 40*x^4)),x]

[Out]

15/(4*x^2*(-5 + E + 5*x))

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 21, normalized size = 1.24 \begin {gather*} \frac {15}{4 \, {\left (5 \, x^{3} + x^{2} e - 5 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*exp(1)-225*x+150)/(4*x^3*exp(1)^2+(40*x^4-40*x^3)*exp(1)+100*x^5-200*x^4+100*x^3),x, algorithm=
"fricas")

[Out]

15/4/(5*x^3 + x^2*e - 5*x^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*exp(1)-225*x+150)/(4*x^3*exp(1)^2+(40*x^4-40*x^3)*exp(1)+100*x^5-200*x^4+100*x^3),x, algorithm=
"giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -15/4*((100*exp(2)*exp(1)-500*exp(2)-100
*exp(1)^3+500*exp(1)^2)/(exp(2)^3-30*exp(2)^2*exp(1)+75*exp(2)^2+300*exp(2)*exp(1)^2-1500*exp(2)*exp(1)+1875*e
xp(2)-1000*exp(1)^3+7

________________________________________________________________________________________

maple [A]  time = 0.13, size = 15, normalized size = 0.88

method	result	size
gosper	\(\frac {15}{4 x^{2} \left (-5+{\mathrm e}+5 x \right )}\)	\(15\)
norman	\(\frac {15}{4 x^{2} \left (-5+{\mathrm e}+5 x \right )}\)	\(15\)
risch	\(\frac {15}{4 x^{2} \left (-5+{\mathrm e}+5 x \right )}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-30*exp(1)-225*x+150)/(4*x^3*exp(1)^2+(40*x^4-40*x^3)*exp(1)+100*x^5-200*x^4+100*x^3),x,method=_RETURNVER
BOSE)

[Out]

15/4/x^2/(-5+exp(1)+5*x)

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 18, normalized size = 1.06 \begin {gather*} \frac {15}{4 \, {\left (5 \, x^{3} + x^{2} {\left (e - 5\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*exp(1)-225*x+150)/(4*x^3*exp(1)^2+(40*x^4-40*x^3)*exp(1)+100*x^5-200*x^4+100*x^3),x, algorithm=
"maxima")

[Out]

15/4/(5*x^3 + x^2*(e - 5))

________________________________________________________________________________________

mupad [B]  time = 7.59, size = 20, normalized size = 1.18 \begin {gather*} \frac {15}{20\,x^3+\left (4\,\mathrm {e}-20\right )\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(225*x + 30*exp(1) - 150)/(4*x^3*exp(2) - exp(1)*(40*x^3 - 40*x^4) + 100*x^3 - 200*x^4 + 100*x^5),x)

[Out]

15/(x^2*(4*exp(1) - 20) + 20*x^3)

________________________________________________________________________________________

sympy [A]  time = 0.36, size = 15, normalized size = 0.88 \begin {gather*} \frac {15}{20 x^{3} + x^{2} \left (-20 + 4 e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*exp(1)-225*x+150)/(4*x**3*exp(1)**2+(40*x**4-40*x**3)*exp(1)+100*x**5-200*x**4+100*x**3),x)

[Out]

15/(20*x**3 + x**2*(-20 + 4*E))

________________________________________________________________________________________