3.90.46 \(\int \frac {100-40 x^2+4 x^4+e^9 (5 x+x^3)}{100 x-40 x^3+4 x^5} \, dx\)

Optimal. Leaf size=25 \[ -1-\frac {e^9 x}{4 \left (-5+x^2\right )}-\log (x)+\log \left (x^2\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 20, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1594, 28, 1805, 21, 29} \begin {gather*} \frac {e^9 x}{4 \left (5-x^2\right )}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(100 - 40*x^2 + 4*x^4 + E^9*(5*x + x^3))/(100*x - 40*x^3 + 4*x^5),x]

[Out]

(E^9*x)/(4*(5 - x^2)) + Log[x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {100-40 x^2+4 x^4+e^9 \left (5 x+x^3\right )}{x \left (100-40 x^2+4 x^4\right )} \, dx\\ &=4 \int \frac {100-40 x^2+4 x^4+e^9 \left (5 x+x^3\right )}{x \left (-20+4 x^2\right )^2} \, dx\\ &=\frac {e^9 x}{4 \left (5-x^2\right )}+\frac {1}{10} \int \frac {-200+40 x^2}{x \left (-20+4 x^2\right )} \, dx\\ &=\frac {e^9 x}{4 \left (5-x^2\right )}+\int \frac {1}{x} \, dx\\ &=\frac {e^9 x}{4 \left (5-x^2\right )}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.72 \begin {gather*} -\frac {e^9 x}{4 \left (-5+x^2\right )}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 - 40*x^2 + 4*x^4 + E^9*(5*x + x^3))/(100*x - 40*x^3 + 4*x^5),x]

[Out]

-1/4*(E^9*x)/(-5 + x^2) + Log[x]

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fricas [A]  time = 0.52, size = 23, normalized size = 0.92 \begin {gather*} -\frac {x e^{9} - 4 \, {\left (x^{2} - 5\right )} \log \relax (x)}{4 \, {\left (x^{2} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+5*x)*exp(9)+4*x^4-40*x^2+100)/(4*x^5-40*x^3+100*x),x, algorithm="fricas")

[Out]

-1/4*(x*e^9 - 4*(x^2 - 5)*log(x))/(x^2 - 5)

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giac [A]  time = 0.16, size = 16, normalized size = 0.64 \begin {gather*} -\frac {x e^{9}}{4 \, {\left (x^{2} - 5\right )}} + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+5*x)*exp(9)+4*x^4-40*x^2+100)/(4*x^5-40*x^3+100*x),x, algorithm="giac")

[Out]

-1/4*x*e^9/(x^2 - 5) + log(abs(x))

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maple [A]  time = 0.06, size = 16, normalized size = 0.64

method	result	size
default	\(\ln \relax (x )-\frac {x \,{\mathrm e}^{9}}{4 \left (x^{2}-5\right )}\)	\(16\)
norman	\(\ln \relax (x )-\frac {x \,{\mathrm e}^{9}}{4 \left (x^{2}-5\right )}\)	\(16\)
risch	\(\ln \relax (x )-\frac {x \,{\mathrm e}^{9}}{4 \left (x^{2}-5\right )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+5*x)*exp(9)+4*x^4-40*x^2+100)/(4*x^5-40*x^3+100*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)-1/4*x*exp(9)/(x^2-5)

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maxima [A]  time = 0.35, size = 15, normalized size = 0.60 \begin {gather*} -\frac {x e^{9}}{4 \, {\left (x^{2} - 5\right )}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+5*x)*exp(9)+4*x^4-40*x^2+100)/(4*x^5-40*x^3+100*x),x, algorithm="maxima")

[Out]

-1/4*x*e^9/(x^2 - 5) + log(x)

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mupad [B]  time = 0.08, size = 17, normalized size = 0.68 \begin {gather*} \ln \relax (x)-\frac {x\,{\mathrm {e}}^9}{4\,\left (x^2-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(9)*(5*x + x^3) - 40*x^2 + 4*x^4 + 100)/(100*x - 40*x^3 + 4*x^5),x)

[Out]

log(x) - (x*exp(9))/(4*(x^2 - 5))

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sympy [A]  time = 0.22, size = 14, normalized size = 0.56 \begin {gather*} - \frac {x e^{9}}{4 x^{2} - 20} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+5*x)*exp(9)+4*x**4-40*x**2+100)/(4*x**5-40*x**3+100*x),x)

[Out]

-x*exp(9)/(4*x**2 - 20) + log(x)

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