3.36.97 \(\int \frac {-115 x^2+18 x^4-8 x^5-3 x^6+e^4 (-90-60 x-10 x^2)}{90 x^2+60 x^3+10 x^4} \, dx\)

Optimal. Leaf size=33 \[ \frac {1}{2} x \left (-3+x-\frac {x^2}{5}\right )+\frac {e^4+x-\frac {2 x}{3+x}}{x} \]

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Rubi [A]  time = 0.09, antiderivative size = 34, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1594, 27, 12, 1620} \begin {gather*} -\frac {x^3}{10}+\frac {x^2}{2}-\frac {3 x}{2}-\frac {2}{x+3}+\frac {e^4}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-115*x^2 + 18*x^4 - 8*x^5 - 3*x^6 + E^4*(-90 - 60*x - 10*x^2))/(90*x^2 + 60*x^3 + 10*x^4),x]

[Out]

E^4/x - (3*x)/2 + x^2/2 - x^3/10 - 2/(3 + 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 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} &=\int \frac {-115 x^2+18 x^4-8 x^5-3 x^6+e^4 \left (-90-60 x-10 x^2\right )}{x^2 \left (90+60 x+10 x^2\right )} \, dx\\ &=\int \frac {-115 x^2+18 x^4-8 x^5-3 x^6+e^4 \left (-90-60 x-10 x^2\right )}{10 x^2 (3+x)^2} \, dx\\ &=\frac {1}{10} \int \frac {-115 x^2+18 x^4-8 x^5-3 x^6+e^4 \left (-90-60 x-10 x^2\right )}{x^2 (3+x)^2} \, dx\\ &=\frac {1}{10} \int \left (-15-\frac {10 e^4}{x^2}+10 x-3 x^2+\frac {20}{(3+x)^2}\right ) \, dx\\ &=\frac {e^4}{x}-\frac {3 x}{2}+\frac {x^2}{2}-\frac {x^3}{10}-\frac {2}{3+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 40, normalized size = 1.21 \begin {gather*} \frac {e^4}{x}-\frac {2}{3+x}-\frac {36 (3+x)}{5}+\frac {7}{5} (3+x)^2-\frac {1}{10} (3+x)^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-115*x^2 + 18*x^4 - 8*x^5 - 3*x^6 + E^4*(-90 - 60*x - 10*x^2))/(90*x^2 + 60*x^3 + 10*x^4),x]

[Out]

E^4/x - 2/(3 + x) - (36*(3 + x))/5 + (7*(3 + x)^2)/5 - (3 + x)^3/10

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fricas [A]  time = 0.94, size = 35, normalized size = 1.06 \begin {gather*} -\frac {x^{5} - 2 \, x^{4} + 45 \, x^{2} - 10 \, {\left (x + 3\right )} e^{4} + 20 \, x}{10 \, {\left (x^{2} + 3 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2-60*x-90)*exp(4)-3*x^6-8*x^5+18*x^4-115*x^2)/(10*x^4+60*x^3+90*x^2),x, algorithm="fricas")

[Out]

-1/10*(x^5 - 2*x^4 + 45*x^2 - 10*(x + 3)*e^4 + 20*x)/(x^2 + 3*x)

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giac [A]  time = 0.24, size = 36, normalized size = 1.09 \begin {gather*} -\frac {1}{10} \, x^{3} + \frac {1}{2} \, x^{2} - \frac {3}{2} \, x + \frac {x e^{4} - 2 \, x + 3 \, e^{4}}{x^{2} + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2-60*x-90)*exp(4)-3*x^6-8*x^5+18*x^4-115*x^2)/(10*x^4+60*x^3+90*x^2),x, algorithm="giac")

[Out]

-1/10*x^3 + 1/2*x^2 - 3/2*x + (x*e^4 - 2*x + 3*e^4)/(x^2 + 3*x)

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maple [A]  time = 0.05, size = 28, normalized size = 0.85

method	result	size
default	\(-\frac {x^{3}}{10}+\frac {x^{2}}{2}-\frac {3 x}{2}-\frac {2}{3+x}+\frac {{\mathrm e}^{4}}{x}\)	\(28\)
norman	\(\frac {\left (\frac {23}{2}+{\mathrm e}^{4}\right ) x +\frac {x^{4}}{5}-\frac {x^{5}}{10}+3 \,{\mathrm e}^{4}}{\left (3+x \right ) x}\)	\(31\)
gosper	\(\frac {-x^{5}+2 x^{4}+10 x \,{\mathrm e}^{4}+30 \,{\mathrm e}^{4}+115 x}{10 x \left (3+x \right )}\)	\(34\)
risch	\(-\frac {x^{3}}{10}+\frac {x^{2}}{2}-\frac {3 x}{2}+\frac {\frac {\left (10 \,{\mathrm e}^{4}-20\right ) x}{10}+3 \,{\mathrm e}^{4}}{\left (3+x \right ) x}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*x^2-60*x-90)*exp(4)-3*x^6-8*x^5+18*x^4-115*x^2)/(10*x^4+60*x^3+90*x^2),x,method=_RETURNVERBOSE)

[Out]

-1/10*x^3+1/2*x^2-3/2*x-2/(3+x)+exp(4)/x

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maxima [A]  time = 0.35, size = 35, normalized size = 1.06 \begin {gather*} -\frac {1}{10} \, x^{3} + \frac {1}{2} \, x^{2} - \frac {3}{2} \, x + \frac {x {\left (e^{4} - 2\right )} + 3 \, e^{4}}{x^{2} + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2-60*x-90)*exp(4)-3*x^6-8*x^5+18*x^4-115*x^2)/(10*x^4+60*x^3+90*x^2),x, algorithm="maxima")

[Out]

-1/10*x^3 + 1/2*x^2 - 3/2*x + (x*(e^4 - 2) + 3*e^4)/(x^2 + 3*x)

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mupad [B]  time = 2.12, size = 35, normalized size = 1.06 \begin {gather*} \frac {3\,{\mathrm {e}}^4+x\,\left ({\mathrm {e}}^4-2\right )}{x^2+3\,x}-\frac {3\,x}{2}+\frac {x^2}{2}-\frac {x^3}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4)*(60*x + 10*x^2 + 90) + 115*x^2 - 18*x^4 + 8*x^5 + 3*x^6)/(90*x^2 + 60*x^3 + 10*x^4),x)

[Out]

(3*exp(4) + x*(exp(4) - 2))/(3*x + x^2) - (3*x)/2 + x^2/2 - x^3/10

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sympy [A]  time = 0.26, size = 32, normalized size = 0.97 \begin {gather*} - \frac {x^{3}}{10} + \frac {x^{2}}{2} - \frac {3 x}{2} - \frac {x \left (2 - e^{4}\right ) - 3 e^{4}}{x^{2} + 3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x**2-60*x-90)*exp(4)-3*x**6-8*x**5+18*x**4-115*x**2)/(10*x**4+60*x**3+90*x**2),x)

[Out]

-x**3/10 + x**2/2 - 3*x/2 - (x*(2 - exp(4)) - 3*exp(4))/(x**2 + 3*x)

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