3.75.1 \(\int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{32 x-32 x^4+8 x^7} \, dx\)

Optimal. Leaf size=30 \[ 4+\frac {2+x}{16 \left (\frac {2}{x^2}-x\right )}+\log (75)-\log \left (\frac {3}{x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 24, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1594, 28, 1829, 21, 29} \begin {gather*} \frac {x \left (x^2+2 x\right )}{16 \left (2-x^3\right )}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 + 4*x^2 - 29*x^3 + x^5 + 8*x^6)/(32*x - 32*x^4 + 8*x^7),x]

[Out]

(x*(2*x + x^2))/(16*(2 - x^3)) + 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 1829

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = Polynomi
alQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1
)*x^m*Pq, a + b*x^n, x], i}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[x^m*(a + b*x^n)^(p + 1)*Expand
ToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)*Coeff[R, x, i]*x^(i - m))/a, {i, 0, n - 1}], x], x], x] - S
imp[(x*R*(a + b*x^n)^(p + 1))/(a^2*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]]] /; FreeQ[{a, b}, x] && PolyQ[Pq,
x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{x \left (32-32 x^3+8 x^6\right )} \, dx\\ &=8 \int \frac {32+4 x^2-29 x^3+x^5+8 x^6}{x \left (-16+8 x^3\right )^2} \, dx\\ &=\frac {x \left (2 x+x^2\right )}{16 \left (2-x^3\right )}+\frac {1}{384} \int \frac {-6144+3072 x^3}{x \left (-16+8 x^3\right )} \, dx\\ &=\frac {x \left (2 x+x^2\right )}{16 \left (2-x^3\right )}+\int \frac {1}{x} \, dx\\ &=\frac {x \left (2 x+x^2\right )}{16 \left (2-x^3\right )}+\log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{8} \left (\frac {-1-x^2}{-2+x^3}+8 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 + 4*x^2 - 29*x^3 + x^5 + 8*x^6)/(32*x - 32*x^4 + 8*x^7),x]

[Out]

((-1 - x^2)/(-2 + x^3) + 8*Log[x])/8

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 23, normalized size = 0.77 \begin {gather*} -\frac {x^{2} - 8 \, {\left (x^{3} - 2\right )} \log \relax (x) + 1}{8 \, {\left (x^{3} - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^6+x^5-29*x^3+4*x^2+32)/(8*x^7-32*x^4+32*x),x, algorithm="fricas")

[Out]

-1/8*(x^2 - 8*(x^3 - 2)*log(x) + 1)/(x^3 - 2)

________________________________________________________________________________________

giac [A]  time = 0.14, size = 18, normalized size = 0.60 \begin {gather*} -\frac {x^{2} + 1}{8 \, {\left (x^{3} - 2\right )}} + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^6+x^5-29*x^3+4*x^2+32)/(8*x^7-32*x^4+32*x),x, algorithm="giac")

[Out]

-1/8*(x^2 + 1)/(x^3 - 2) + log(abs(x))

________________________________________________________________________________________

maple [A]  time = 0.03, size = 19, normalized size = 0.63

method	result	size
norman	\(\frac {-\frac {x^{2}}{8}-\frac {1}{8}}{x^{3}-2}+\ln \relax (x )\)	\(19\)
risch	\(\frac {-\frac {x^{2}}{8}-\frac {1}{8}}{x^{3}-2}+\ln \relax (x )\)	\(19\)
default	\(\ln \relax (x )+\frac {-x^{2}-1}{8 x^{3}-16}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^6+x^5-29*x^3+4*x^2+32)/(8*x^7-32*x^4+32*x),x,method=_RETURNVERBOSE)

[Out]

(-1/8*x^2-1/8)/(x^3-2)+ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 17, normalized size = 0.57 \begin {gather*} -\frac {x^{2} + 1}{8 \, {\left (x^{3} - 2\right )}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^6+x^5-29*x^3+4*x^2+32)/(8*x^7-32*x^4+32*x),x, algorithm="maxima")

[Out]

-1/8*(x^2 + 1)/(x^3 - 2) + log(x)

________________________________________________________________________________________

mupad [B]  time = 0.05, size = 19, normalized size = 0.63 \begin {gather*} \ln \relax (x)-\frac {\frac {x^2}{8}+\frac {1}{8}}{x^3-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2 - 29*x^3 + x^5 + 8*x^6 + 32)/(32*x - 32*x^4 + 8*x^7),x)

[Out]

log(x) - (x^2/8 + 1/8)/(x^3 - 2)

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 15, normalized size = 0.50 \begin {gather*} \frac {- x^{2} - 1}{8 x^{3} - 16} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**6+x**5-29*x**3+4*x**2+32)/(8*x**7-32*x**4+32*x),x)

[Out]

(-x**2 - 1)/(8*x**3 - 16) + log(x)

________________________________________________________________________________________