3.63.55 \(\int \frac {40+2 x^5+(5-x^5) \log (\frac {25-10 x^5+x^{10}}{x^8})}{-30 x^2+6 x^7} \, dx\)

Optimal. Leaf size=17 \[ \frac {\log \left (\left (-\frac {5}{x^4}+x\right )^2\right )}{6 x} \]

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Rubi [A]  time = 1.30, antiderivative size = 21, normalized size of antiderivative = 1.24, number of steps used = 19, number of rules used = 11, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1593, 6725, 453, 294, 634, 618, 204, 628, 31, 2525, 12} \begin {gather*} \frac {\log \left (\frac {\left (5-x^5\right )^2}{x^8}\right )}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 + 2*x^5 + (5 - x^5)*Log[(25 - 10*x^5 + x^10)/x^8])/(-30*x^2 + 6*x^7),x]

[Out]

Log[(5 - x^5)^2/x^8]/(6*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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 294

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt
[-(a/b), n]], k, u}, Simp[u = Int[(r*Cos[((2*k - 1)*m*Pi)/n] + s*Cos[((2*k - 1)*(m + 1)*Pi)/n]*x)/(r^2 + 2*r*s
*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x]; (r^(m + 1)*Int[1/(r - s*x), x])/(a*n*s^m) - Dist[(2*(-r)^(m + 1))/(a*
n*s^m), Sum[u, {k, 1, (n - 1)/2}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 1)/2, 0] && IGtQ[m, 0] && LtQ[m, n
- 1] && NegQ[a/b]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1593

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

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {40+2 x^5+\left (5-x^5\right ) \log \left (\frac {25-10 x^5+x^{10}}{x^8}\right )}{x^2 \left (-30+6 x^5\right )} \, dx\\ &=\int \left (\frac {20+x^5}{3 x^2 \left (-5+x^5\right )}-\frac {\log \left (\frac {\left (-5+x^5\right )^2}{x^8}\right )}{6 x^2}\right ) \, dx\\ &=-\left (\frac {1}{6} \int \frac {\log \left (\frac {\left (-5+x^5\right )^2}{x^8}\right )}{x^2} \, dx\right )+\frac {1}{3} \int \frac {20+x^5}{x^2 \left (-5+x^5\right )} \, dx\\ &=\frac {4}{3 x}+\frac {\log \left (\frac {\left (5-x^5\right )^2}{x^8}\right )}{6 x}-\frac {1}{6} \int \frac {2 \left (-20-x^5\right )}{x^2 \left (5-x^5\right )} \, dx+\frac {5}{3} \int \frac {x^3}{-5+x^5} \, dx\\ &=\frac {4}{3 x}+\frac {\log \left (\frac {\left (5-x^5\right )^2}{x^8}\right )}{6 x}-\frac {1}{3} \int \frac {-20-x^5}{x^2 \left (5-x^5\right )} \, dx-\frac {\int \frac {1}{\sqrt [5]{5}-x} \, dx}{3 \sqrt [5]{5}}+\frac {2 \int \frac {\frac {1}{4} \sqrt [5]{5} \left (1+\sqrt {5}\right )+\frac {1}{4} \left (-1+\sqrt {5}\right ) x}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right ) x+x^2} \, dx}{3 \sqrt [5]{5}}+\frac {2 \int \frac {\frac {1}{4} \sqrt [5]{5} \left (1-\sqrt {5}\right )+\frac {1}{4} \left (-1-\sqrt {5}\right ) x}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right ) x+x^2} \, dx}{3 \sqrt [5]{5}}\\ &=\frac {\log \left (\sqrt [5]{5}-x\right )}{3 \sqrt [5]{5}}+\frac {\log \left (\frac {\left (5-x^5\right )^2}{x^8}\right )}{6 x}+\frac {5}{3} \int \frac {x^3}{5-x^5} \, dx-\frac {\left (1-\sqrt {5}\right ) \int \frac {\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right )+2 x}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right ) x+x^2} \, dx}{12 \sqrt [5]{5}}-\frac {1}{12} \left (-5+\sqrt {5}\right ) \int \frac {1}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right ) x+x^2} \, dx-\frac {\left (1+\sqrt {5}\right ) \int \frac {\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right )+2 x}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right ) x+x^2} \, dx}{12 \sqrt [5]{5}}+\frac {1}{12} \left (5+\sqrt {5}\right ) \int \frac {1}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right ) x+x^2} \, dx\\ &=\frac {\log \left (\sqrt [5]{5}-x\right )}{3 \sqrt [5]{5}}-\frac {\left (1-\sqrt {5}\right ) \log \left (2\ 5^{2/5}+\sqrt [5]{5} x-5^{7/10} x+2 x^2\right )}{12 \sqrt [5]{5}}-\frac {\left (1+\sqrt {5}\right ) \log \left (2\ 5^{2/5}+\sqrt [5]{5} x+5^{7/10} x+2 x^2\right )}{12 \sqrt [5]{5}}+\frac {\log \left (\frac {\left (5-x^5\right )^2}{x^8}\right )}{6 x}+\frac {\int \frac {1}{\sqrt [5]{5}-x} \, dx}{3 \sqrt [5]{5}}-\frac {2 \int \frac {\frac {1}{4} \sqrt [5]{5} \left (1+\sqrt {5}\right )+\frac {1}{4} \left (-1+\sqrt {5}\right ) x}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right ) x+x^2} \, dx}{3 \sqrt [5]{5}}-\frac {2 \int \frac {\frac {1}{4} \sqrt [5]{5} \left (1-\sqrt {5}\right )+\frac {1}{4} \left (-1-\sqrt {5}\right ) x}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right ) x+x^2} \, dx}{3 \sqrt [5]{5}}-\frac {1}{6} \left (5-\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{2} 5^{9/10} \left (1-\sqrt {5}\right )-x^2} \, dx,x,\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right )+2 x\right )-\frac {1}{6} \left (5+\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{2} 5^{2/5} \left (5+\sqrt {5}\right )-x^2} \, dx,x,\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right )+2 x\right )\\ &=\frac {\sqrt {5+\sqrt {5}} \tan ^{-1}\left (\frac {\sqrt [5]{5}-5^{7/10}+4 x}{\sqrt [5]{5} \sqrt {2 \left (5+\sqrt {5}\right )}}\right )}{3 \sqrt {2} \sqrt [5]{5}}+\frac {\sqrt [20]{5} \sqrt {-1+\sqrt {5}} \tan ^{-1}\left (\frac {\sqrt [5]{5}+5^{7/10}+4 x}{5^{9/20} \sqrt {2 \left (-1+\sqrt {5}\right )}}\right )}{3 \sqrt {2}}-\frac {\left (1-\sqrt {5}\right ) \log \left (2\ 5^{2/5}+\sqrt [5]{5} x-5^{7/10} x+2 x^2\right )}{12 \sqrt [5]{5}}-\frac {\left (1+\sqrt {5}\right ) \log \left (2\ 5^{2/5}+\sqrt [5]{5} x+5^{7/10} x+2 x^2\right )}{12 \sqrt [5]{5}}+\frac {\log \left (\frac {\left (5-x^5\right )^2}{x^8}\right )}{6 x}+\frac {\left (1-\sqrt {5}\right ) \int \frac {\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right )+2 x}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right ) x+x^2} \, dx}{12 \sqrt [5]{5}}+\frac {1}{12} \left (-5+\sqrt {5}\right ) \int \frac {1}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right ) x+x^2} \, dx+\frac {\left (1+\sqrt {5}\right ) \int \frac {\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right )+2 x}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right ) x+x^2} \, dx}{12 \sqrt [5]{5}}-\frac {1}{12} \left (5+\sqrt {5}\right ) \int \frac {1}{5^{2/5}+\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right ) x+x^2} \, dx\\ &=\frac {\sqrt {5+\sqrt {5}} \tan ^{-1}\left (\frac {\sqrt [5]{5}-5^{7/10}+4 x}{\sqrt [5]{5} \sqrt {2 \left (5+\sqrt {5}\right )}}\right )}{3 \sqrt {2} \sqrt [5]{5}}+\frac {\sqrt [20]{5} \sqrt {-1+\sqrt {5}} \tan ^{-1}\left (\frac {\sqrt [5]{5}+5^{7/10}+4 x}{5^{9/20} \sqrt {2 \left (-1+\sqrt {5}\right )}}\right )}{3 \sqrt {2}}+\frac {\log \left (\frac {\left (5-x^5\right )^2}{x^8}\right )}{6 x}+\frac {1}{6} \left (5-\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{2} 5^{9/10} \left (1-\sqrt {5}\right )-x^2} \, dx,x,\frac {1}{2} \sqrt [5]{5} \left (1+\sqrt {5}\right )+2 x\right )+\frac {1}{6} \left (5+\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{2} 5^{2/5} \left (5+\sqrt {5}\right )-x^2} \, dx,x,\frac {1}{2} \sqrt [5]{5} \left (1-\sqrt {5}\right )+2 x\right )\\ &=\frac {\log \left (\frac {\left (5-x^5\right )^2}{x^8}\right )}{6 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.12 \begin {gather*} \frac {\log \left (\frac {\left (-5+x^5\right )^2}{x^8}\right )}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 + 2*x^5 + (5 - x^5)*Log[(25 - 10*x^5 + x^10)/x^8])/(-30*x^2 + 6*x^7),x]

[Out]

Log[(-5 + x^5)^2/x^8]/(6*x)

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fricas [A]  time = 1.16, size = 20, normalized size = 1.18 \begin {gather*} \frac {\log \left (\frac {x^{10} - 10 \, x^{5} + 25}{x^{8}}\right )}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^5+5)*log((x^10-10*x^5+25)/x^8)+2*x^5+40)/(6*x^7-30*x^2),x, algorithm="fricas")

[Out]

1/6*log((x^10 - 10*x^5 + 25)/x^8)/x

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giac [A]  time = 0.50, size = 20, normalized size = 1.18 \begin {gather*} \frac {\log \left (\frac {x^{10} - 10 \, x^{5} + 25}{x^{8}}\right )}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^5+5)*log((x^10-10*x^5+25)/x^8)+2*x^5+40)/(6*x^7-30*x^2),x, algorithm="giac")

[Out]

1/6*log((x^10 - 10*x^5 + 25)/x^8)/x

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maple [A]  time = 0.34, size = 21, normalized size = 1.24

method	result	size
default	\(\frac {\ln \left (\frac {x^{10}-10 x^{5}+25}{x^{8}}\right )}{6 x}\)	\(21\)
norman	\(\frac {\ln \left (\frac {x^{10}-10 x^{5}+25}{x^{8}}\right )}{6 x}\)	\(21\)
risch	\(\frac {\ln \left (\frac {x^{10}-10 x^{5}+25}{x^{8}}\right )}{6 x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^5+5)*ln((x^10-10*x^5+25)/x^8)+2*x^5+40)/(6*x^7-30*x^2),x,method=_RETURNVERBOSE)

[Out]

1/6/x*ln((x^10-10*x^5+25)/x^8)

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maxima [A]  time = 0.74, size = 23, normalized size = 1.35 \begin {gather*} \frac {\log \left (x^{5} - 5\right ) - 4 \, \log \relax (x) - 4}{3 \, x} + \frac {4}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^5+5)*log((x^10-10*x^5+25)/x^8)+2*x^5+40)/(6*x^7-30*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.09, size = 20, normalized size = 1.18 \begin {gather*} \frac {\ln \left (\frac {x^{10}-10\,x^5+25}{x^8}\right )}{6\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^5 - log((x^10 - 10*x^5 + 25)/x^8)*(x^5 - 5) + 40)/(30*x^2 - 6*x^7),x)

[Out]

log((x^10 - 10*x^5 + 25)/x^8)/(6*x)

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sympy [A]  time = 0.16, size = 17, normalized size = 1.00 \begin {gather*} \frac {\log {\left (\frac {x^{10} - 10 x^{5} + 25}{x^{8}} \right )}}{6 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**5+5)*ln((x**10-10*x**5+25)/x**8)+2*x**5+40)/(6*x**7-30*x**2),x)

[Out]

log((x**10 - 10*x**5 + 25)/x**8)/(6*x)

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