3.56.38 \(\int \frac {-40+5 x-16 x^2-4 x^3+x^5+(5 x+3 x^3) \log (\frac {\log (18)}{x})}{25 x^3+10 x^5+x^7} \, dx\)

Optimal. Leaf size=31 \[ \frac {\frac {4}{x}-x^2-\log \left (\frac {\log (18)}{x}\right )}{x \left (5+x^2\right )} \]

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Rubi [B]  time = 0.53, antiderivative size = 85, normalized size of antiderivative = 2.74, number of steps used = 32, number of rules used = 17, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.327, Rules used = {1594, 28, 6742, 199, 203, 266, 44, 290, 325, 288, 2357, 2304, 2323, 2324, 12, 4848, 2391} \begin {gather*} -\frac {9 x}{10 \left (x^2+5\right )}-\frac {4}{5 \left (x^2+5\right )}+\frac {1}{2 \left (x^2+5\right ) x}+\frac {4}{5 x^2}+\frac {x \log \left (\frac {\log (18)}{x}\right )}{5 \left (x^2+5\right )}-\frac {1}{10 x}-\frac {\log \left (\frac {\log (18)}{x}\right )}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40 + 5*x - 16*x^2 - 4*x^3 + x^5 + (5*x + 3*x^3)*Log[Log[18]/x])/(25*x^3 + 10*x^5 + x^7),x]

[Out]

4/(5*x^2) - 1/(10*x) - 4/(5*(5 + x^2)) + 1/(2*x*(5 + x^2)) - (9*x)/(10*(5 + x^2)) - Log[Log[18]/x]/(5*x) + (x*
Log[Log[18]/x])/(5*(5 + x^2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, 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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2323

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(q + 1
)*(a + b*Log[c*x^n]))/(2*d*(q + 1)), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*Log[c*
x^n]), x], x] + Dist[(b*n)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1), x], x]) /; FreeQ[{a, b, c, d, e, n}, x] &&
LtQ[q, -1]

Rule 2324

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2),
x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-40+5 x-16 x^2-4 x^3+x^5+\left (5 x+3 x^3\right ) \log \left (\frac {\log (18)}{x}\right )}{x^3 \left (25+10 x^2+x^4\right )} \, dx\\ &=\int \frac {-40+5 x-16 x^2-4 x^3+x^5+\left (5 x+3 x^3\right ) \log \left (\frac {\log (18)}{x}\right )}{x^3 \left (5+x^2\right )^2} \, dx\\ &=\int \left (-\frac {4}{\left (5+x^2\right )^2}-\frac {40}{x^3 \left (5+x^2\right )^2}+\frac {5}{x^2 \left (5+x^2\right )^2}-\frac {16}{x \left (5+x^2\right )^2}+\frac {x^2}{\left (5+x^2\right )^2}+\frac {\left (5+3 x^2\right ) \log \left (\frac {\log (18)}{x}\right )}{x^2 \left (5+x^2\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {1}{\left (5+x^2\right )^2} \, dx\right )+5 \int \frac {1}{x^2 \left (5+x^2\right )^2} \, dx-16 \int \frac {1}{x \left (5+x^2\right )^2} \, dx-40 \int \frac {1}{x^3 \left (5+x^2\right )^2} \, dx+\int \frac {x^2}{\left (5+x^2\right )^2} \, dx+\int \frac {\left (5+3 x^2\right ) \log \left (\frac {\log (18)}{x}\right )}{x^2 \left (5+x^2\right )^2} \, dx\\ &=\frac {1}{2 x \left (5+x^2\right )}-\frac {9 x}{10 \left (5+x^2\right )}-\frac {2}{5} \int \frac {1}{5+x^2} \, dx+\frac {1}{2} \int \frac {1}{5+x^2} \, dx+\frac {3}{2} \int \frac {1}{x^2 \left (5+x^2\right )} \, dx-8 \operatorname {Subst}\left (\int \frac {1}{x (5+x)^2} \, dx,x,x^2\right )-20 \operatorname {Subst}\left (\int \frac {1}{x^2 (5+x)^2} \, dx,x,x^2\right )+\int \left (\frac {\log \left (\frac {\log (18)}{x}\right )}{5 x^2}+\frac {2 \log \left (\frac {\log (18)}{x}\right )}{\left (5+x^2\right )^2}-\frac {\log \left (\frac {\log (18)}{x}\right )}{5 \left (5+x^2\right )}\right ) \, dx\\ &=-\frac {3}{10 x}+\frac {1}{2 x \left (5+x^2\right )}-\frac {9 x}{10 \left (5+x^2\right )}+\frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{10 \sqrt {5}}+\frac {1}{5} \int \frac {\log \left (\frac {\log (18)}{x}\right )}{x^2} \, dx-\frac {1}{5} \int \frac {\log \left (\frac {\log (18)}{x}\right )}{5+x^2} \, dx-\frac {3}{10} \int \frac {1}{5+x^2} \, dx+2 \int \frac {\log \left (\frac {\log (18)}{x}\right )}{\left (5+x^2\right )^2} \, dx-8 \operatorname {Subst}\left (\int \left (\frac {1}{25 x}-\frac {1}{5 (5+x)^2}-\frac {1}{25 (5+x)}\right ) \, dx,x,x^2\right )-20 \operatorname {Subst}\left (\int \left (\frac {1}{25 x^2}-\frac {2}{125 x}+\frac {1}{25 (5+x)^2}+\frac {2}{125 (5+x)}\right ) \, dx,x,x^2\right )\\ &=\frac {4}{5 x^2}-\frac {1}{10 x}-\frac {4}{5 \left (5+x^2\right )}+\frac {1}{2 x \left (5+x^2\right )}-\frac {9 x}{10 \left (5+x^2\right )}-\frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{5 \sqrt {5}}-\frac {\log \left (\frac {\log (18)}{x}\right )}{5 x}+\frac {x \log \left (\frac {\log (18)}{x}\right )}{5 \left (5+x^2\right )}-\frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right ) \log \left (\frac {\log (18)}{x}\right )}{5 \sqrt {5}}+\frac {1}{5} \int \frac {1}{5+x^2} \, dx-\frac {1}{5} \int \frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5} x} \, dx+\frac {1}{5} \int \frac {\log \left (\frac {\log (18)}{x}\right )}{5+x^2} \, dx\\ &=\frac {4}{5 x^2}-\frac {1}{10 x}-\frac {4}{5 \left (5+x^2\right )}+\frac {1}{2 x \left (5+x^2\right )}-\frac {9 x}{10 \left (5+x^2\right )}-\frac {\log \left (\frac {\log (18)}{x}\right )}{5 x}+\frac {x \log \left (\frac {\log (18)}{x}\right )}{5 \left (5+x^2\right )}+\frac {1}{5} \int \frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{\sqrt {5} x} \, dx-\frac {\int \frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{x} \, dx}{5 \sqrt {5}}\\ &=\frac {4}{5 x^2}-\frac {1}{10 x}-\frac {4}{5 \left (5+x^2\right )}+\frac {1}{2 x \left (5+x^2\right )}-\frac {9 x}{10 \left (5+x^2\right )}-\frac {\log \left (\frac {\log (18)}{x}\right )}{5 x}+\frac {x \log \left (\frac {\log (18)}{x}\right )}{5 \left (5+x^2\right )}-\frac {i \int \frac {\log \left (1-\frac {i x}{\sqrt {5}}\right )}{x} \, dx}{10 \sqrt {5}}+\frac {i \int \frac {\log \left (1+\frac {i x}{\sqrt {5}}\right )}{x} \, dx}{10 \sqrt {5}}+\frac {\int \frac {\tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )}{x} \, dx}{5 \sqrt {5}}\\ &=\frac {4}{5 x^2}-\frac {1}{10 x}-\frac {4}{5 \left (5+x^2\right )}+\frac {1}{2 x \left (5+x^2\right )}-\frac {9 x}{10 \left (5+x^2\right )}-\frac {\log \left (\frac {\log (18)}{x}\right )}{5 x}+\frac {x \log \left (\frac {\log (18)}{x}\right )}{5 \left (5+x^2\right )}-\frac {i \text {Li}_2\left (-\frac {i x}{\sqrt {5}}\right )}{10 \sqrt {5}}+\frac {i \text {Li}_2\left (\frac {i x}{\sqrt {5}}\right )}{10 \sqrt {5}}+\frac {i \int \frac {\log \left (1-\frac {i x}{\sqrt {5}}\right )}{x} \, dx}{10 \sqrt {5}}-\frac {i \int \frac {\log \left (1+\frac {i x}{\sqrt {5}}\right )}{x} \, dx}{10 \sqrt {5}}\\ &=\frac {4}{5 x^2}-\frac {1}{10 x}-\frac {4}{5 \left (5+x^2\right )}+\frac {1}{2 x \left (5+x^2\right )}-\frac {9 x}{10 \left (5+x^2\right )}-\frac {\log \left (\frac {\log (18)}{x}\right )}{5 x}+\frac {x \log \left (\frac {\log (18)}{x}\right )}{5 \left (5+x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.39, size = 188, normalized size = 6.06 \begin {gather*} \frac {1}{50} \left (\frac {40}{x^2}-\frac {40}{5+x^2}-\frac {50 x}{5+x^2}-10 \sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )-i \sqrt {5} \log \left (\sqrt {5}+i x\right )+i \sqrt {5} \log \left (i \sqrt {5}+x\right )+4 i \sqrt {5} \log \left (1-\frac {i x}{\sqrt {5}}\right )-4 i \sqrt {5} \log \left (1+\frac {i x}{\sqrt {5}}\right )+\frac {5 i \log \left (\frac {\log (18)}{x}\right )}{\sqrt {5}+i x}-\frac {10 \log \left (\frac {\log (18)}{x}\right )}{x}+\frac {5 \log \left (\frac {\log (18)}{x}\right )}{i \sqrt {5}+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40 + 5*x - 16*x^2 - 4*x^3 + x^5 + (5*x + 3*x^3)*Log[Log[18]/x])/(25*x^3 + 10*x^5 + x^7),x]

[Out]

(40/x^2 - 40/(5 + x^2) - (50*x)/(5 + x^2) - 10*Sqrt[5]*ArcTan[x/Sqrt[5]] - I*Sqrt[5]*Log[Sqrt[5] + I*x] + I*Sq
rt[5]*Log[I*Sqrt[5] + x] + (4*I)*Sqrt[5]*Log[1 - (I*x)/Sqrt[5]] - (4*I)*Sqrt[5]*Log[1 + (I*x)/Sqrt[5]] + ((5*I
)*Log[Log[18]/x])/(Sqrt[5] + I*x) - (10*Log[Log[18]/x])/x + (5*Log[Log[18]/x])/(I*Sqrt[5] + x))/50

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fricas [A]  time = 0.95, size = 27, normalized size = 0.87 \begin {gather*} -\frac {x^{3} + x \log \left (\frac {\log \left (18\right )}{x}\right ) - 4}{x^{4} + 5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+5*x)*log(log(18)/x)+x^5-4*x^3-16*x^2+5*x-40)/(x^7+10*x^5+25*x^3),x, algorithm="fricas")

[Out]

-(x^3 + x*log(log(18)/x) - 4)/(x^4 + 5*x^2)

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giac [B]  time = 0.12, size = 88, normalized size = 2.84 \begin {gather*} \frac {5 \, {\left (\frac {\log \left (18\right )^{3}}{{\left (\log \left (18\right )^{2} + \frac {5 \, \log \left (18\right )^{2}}{x^{2}}\right )} x} - \frac {\log \left (18\right )}{x}\right )} \log \left (\frac {\log \left (18\right )}{x}\right ) + \frac {4 \, \log \left (18\right )^{3} - \frac {25 \, \log \left (18\right )^{3}}{x}}{\log \left (18\right )^{2} + \frac {5 \, \log \left (18\right )^{2}}{x^{2}}} + \frac {20 \, \log \left (18\right )}{x^{2}}}{25 \, \log \left (18\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+5*x)*log(log(18)/x)+x^5-4*x^3-16*x^2+5*x-40)/(x^7+10*x^5+25*x^3),x, algorithm="giac")

[Out]

1/25*(5*(log(18)^3/((log(18)^2 + 5*log(18)^2/x^2)*x) - log(18)/x)*log(log(18)/x) + (4*log(18)^3 - 25*log(18)^3
/x)/(log(18)^2 + 5*log(18)^2/x^2) + 20*log(18)/x^2)/log(18)

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maple [A]  time = 0.22, size = 29, normalized size = 0.94

method	result	size
norman	\(\frac {4-x^{3}-\ln \left (\frac {\ln \left (18\right )}{x}\right ) x}{x^{2} \left (x^{2}+5\right )}\)	\(29\)
risch	\(-\frac {\ln \left (\frac {\ln \relax (2)+2 \ln \relax (3)}{x}\right )}{x \left (x^{2}+5\right )}-\frac {x^{3}-4}{x^{2} \left (x^{2}+5\right )}\)	\(43\)
derivativedivides	\(-\ln \left (18\right ) \left (-\frac {4}{5 \ln \left (18\right ) x^{2}}+\frac {\ln \left (18\right )}{\left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right ) x}-\frac {4 \ln \left (18\right )}{25 \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right )}+\frac {\ln \left (\frac {\ln \left (18\right )}{x}\right )}{5 \ln \left (18\right ) x}+\frac {i \ln \left (18\right ) \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}+\ln \left (18\right )}{\ln \left (18\right )}\right ) \sqrt {5}}{50 \ln \left (18\right )^{2}+\frac {250 \ln \left (18\right )^{2}}{x^{2}}}+\frac {i \ln \left (18\right ) \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}+\ln \left (18\right )}{\ln \left (18\right )}\right ) \sqrt {5}}{10 \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right ) x^{2}}-\frac {i \ln \left (18\right ) \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (-\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}-\ln \left (18\right )}{\ln \left (18\right )}\right ) \sqrt {5}}{50 \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right )}-\frac {i \ln \left (18\right ) \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (-\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}-\ln \left (18\right )}{\ln \left (18\right )}\right ) \sqrt {5}}{10 \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right ) x^{2}}-\frac {\ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (18\right )}{5 x \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right )}-\frac {i \sqrt {5}\, \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}+\ln \left (18\right )}{\ln \left (18\right )}\right )}{50 \ln \left (18\right )}+\frac {i \sqrt {5}\, \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (-\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}-\ln \left (18\right )}{\ln \left (18\right )}\right )}{50 \ln \left (18\right )}\right )\)	\(396\)
default	\(-\ln \left (18\right ) \left (-\frac {4}{5 \ln \left (18\right ) x^{2}}+\frac {\ln \left (18\right )}{\left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right ) x}-\frac {4 \ln \left (18\right )}{25 \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right )}+\frac {\ln \left (\frac {\ln \left (18\right )}{x}\right )}{5 \ln \left (18\right ) x}+\frac {i \ln \left (18\right ) \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}+\ln \left (18\right )}{\ln \left (18\right )}\right ) \sqrt {5}}{50 \ln \left (18\right )^{2}+\frac {250 \ln \left (18\right )^{2}}{x^{2}}}+\frac {i \ln \left (18\right ) \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}+\ln \left (18\right )}{\ln \left (18\right )}\right ) \sqrt {5}}{10 \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right ) x^{2}}-\frac {i \ln \left (18\right ) \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (-\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}-\ln \left (18\right )}{\ln \left (18\right )}\right ) \sqrt {5}}{50 \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right )}-\frac {i \ln \left (18\right ) \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (-\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}-\ln \left (18\right )}{\ln \left (18\right )}\right ) \sqrt {5}}{10 \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right ) x^{2}}-\frac {\ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (18\right )}{5 x \left (\ln \left (18\right )^{2}+\frac {5 \ln \left (18\right )^{2}}{x^{2}}\right )}-\frac {i \sqrt {5}\, \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}+\ln \left (18\right )}{\ln \left (18\right )}\right )}{50 \ln \left (18\right )}+\frac {i \sqrt {5}\, \ln \left (\frac {\ln \left (18\right )}{x}\right ) \ln \left (-\frac {\frac {i \ln \left (18\right ) \sqrt {5}}{x}-\ln \left (18\right )}{\ln \left (18\right )}\right )}{50 \ln \left (18\right )}\right )\)	\(396\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^3+5*x)*ln(ln(18)/x)+x^5-4*x^3-16*x^2+5*x-40)/(x^7+10*x^5+25*x^3),x,method=_RETURNVERBOSE)

[Out]

(4-x^3-ln(ln(18)/x)*x)/x^2/(x^2+5)

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maxima [B]  time = 0.62, size = 88, normalized size = 2.84 \begin {gather*} -\frac {35 \, x^{2} + 2 \, {\left (16 \, x^{3} + 80 \, x - 25\right )} \log \relax (x) + 80 \, x + 50 \, \log \left (2 \, \log \relax (3) + \log \relax (2)\right ) - 50}{50 \, {\left (x^{3} + 5 \, x\right )}} - \frac {3 \, x^{2} + 10}{10 \, {\left (x^{3} + 5 \, x\right )}} + \frac {4 \, {\left (2 \, x^{2} + 5\right )}}{5 \, {\left (x^{4} + 5 \, x^{2}\right )}} + \frac {16}{25} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+5*x)*log(log(18)/x)+x^5-4*x^3-16*x^2+5*x-40)/(x^7+10*x^5+25*x^3),x, algorithm="maxima")

[Out]

-1/50*(35*x^2 + 2*(16*x^3 + 80*x - 25)*log(x) + 80*x + 50*log(2*log(3) + log(2)) - 50)/(x^3 + 5*x) - 1/10*(3*x
^2 + 10)/(x^3 + 5*x) + 4/5*(2*x^2 + 5)/(x^4 + 5*x^2) + 16/25*log(x)

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mupad [B]  time = 3.65, size = 27, normalized size = 0.87 \begin {gather*} -\frac {x^3+x\,\ln \left (\frac {\ln \left (18\right )}{x}\right )-4}{x^4+5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x - 16*x^2 - 4*x^3 + x^5 + log(log(18)/x)*(5*x + 3*x^3) - 40)/(25*x^3 + 10*x^5 + x^7),x)

[Out]

-(x^3 + x*log(log(18)/x) - 4)/(5*x^2 + x^4)

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sympy [A]  time = 0.17, size = 26, normalized size = 0.84 \begin {gather*} \frac {4 - x^{3}}{x^{4} + 5 x^{2}} - \frac {\log {\left (\frac {\log {\left (18 \right )}}{x} \right )}}{x^{3} + 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**3+5*x)*ln(ln(18)/x)+x**5-4*x**3-16*x**2+5*x-40)/(x**7+10*x**5+25*x**3),x)

[Out]

(4 - x**3)/(x**4 + 5*x**2) - log(log(18)/x)/(x**3 + 5*x)

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