\(\int \frac {\log ^2(c (a+b x^2)^p)}{x^6} \, dx\) [89]

Optimal result

Integrand size = 18, antiderivative size = 296 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=-\frac {8 b^2 p^2}{15 a^2 x}-\frac {32 b^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 a^{5/2}}+\frac {4 i b^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 a^{5/2}}+\frac {8 b^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}+\frac {4 b^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {4 i b^{5/2} p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}} \]

[Out]

-8/15*b^2*p^2/a^2/x-32/15*b^(5/2)*p^2*arctan(x*b^(1/2)/a^(1/2))/a^(5/2)+4/5*I*b^(5/2)*p^2*arctan(x*b^(1/2)/a^(
1/2))^2/a^(5/2)-4/15*b*p*ln(c*(b*x^2+a)^p)/a/x^3+4/5*b^2*p*ln(c*(b*x^2+a)^p)/a^2/x+4/5*b^(5/2)*p*arctan(x*b^(1
/2)/a^(1/2))*ln(c*(b*x^2+a)^p)/a^(5/2)-1/5*ln(c*(b*x^2+a)^p)^2/x^5+8/5*b^(5/2)*p^2*arctan(x*b^(1/2)/a^(1/2))*l
n(2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/a^(5/2)+4/5*I*b^(5/2)*p^2*polylog(2,1-2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/a^(5
/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {2507, 2526, 2505, 331, 211, 2520, 12, 5040, 4964, 2449, 2352} \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\frac {4 b^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}+\frac {4 i b^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 a^{5/2}}-\frac {32 b^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 a^{5/2}}+\frac {8 b^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}}+\frac {4 i b^{5/2} p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{5 a^{5/2}}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}-\frac {8 b^2 p^2}{15 a^2 x}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3} \]

[In]

Int[Log[c*(a + b*x^2)^p]^2/x^6,x]

[Out]

(-8*b^2*p^2)/(15*a^2*x) - (32*b^(5/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(15*a^(5/2)) + (((4*I)/5)*b^(5/2)*p^2*A
rcTan[(Sqrt[b]*x)/Sqrt[a]]^2)/a^(5/2) + (8*b^(5/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[(2*Sqrt[a])/(Sqrt[a] +
I*Sqrt[b]*x)])/(5*a^(5/2)) - (4*b*p*Log[c*(a + b*x^2)^p])/(15*a*x^3) + (4*b^2*p*Log[c*(a + b*x^2)^p])/(5*a^2*x
) + (4*b^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p])/(5*a^(5/2)) - Log[c*(a + b*x^2)^p]^2/(5*x^5
) + (((4*I)/5)*b^(5/2)*p^2*PolyLog[2, 1 - (2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/a^(5/2)

Rule 12

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

Rule 211

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

Rule 331

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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2505

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

Rule 2507

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_)*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[(f*x)
^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])^q/(f*(m + 1))), x] - Dist[b*e*n*p*(q/(f^n*(m + 1))), Int[(f*x)^(m + n)*
((a + b*Log[c*(d + e*x^n)^p])^(q - 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && IGtQ[q, 1]
&& IntegerQ[n] && NeQ[m, -1]

Rule 2520

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[u*(x^(n - 1)/(d + e*x^n)
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

Rule 2526

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] && IntegerQ[s]

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 + c^2*x
^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {1}{5} (4 b p) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^4 \left (a+b x^2\right )} \, dx \\ & = -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {1}{5} (4 b p) \int \left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{a x^4}-\frac {b \log \left (c \left (a+b x^2\right )^p\right )}{a^2 x^2}+\frac {b^2 \log \left (c \left (a+b x^2\right )^p\right )}{a^2 \left (a+b x^2\right )}\right ) \, dx \\ & = -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {(4 b p) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx}{5 a}-\frac {\left (4 b^2 p\right ) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx}{5 a^2}+\frac {\left (4 b^3 p\right ) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{a+b x^2} \, dx}{5 a^2} \\ & = -\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}+\frac {4 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {\left (8 b^2 p^2\right ) \int \frac {1}{x^2 \left (a+b x^2\right )} \, dx}{15 a}-\frac {\left (8 b^3 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{5 a^2}-\frac {\left (8 b^4 p^2\right ) \int \frac {x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \left (a+b x^2\right )} \, dx}{5 a^2} \\ & = -\frac {8 b^2 p^2}{15 a^2 x}-\frac {8 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 a^{5/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}+\frac {4 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}-\frac {\left (8 b^3 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{15 a^2}-\frac {\left (8 b^{7/2} p^2\right ) \int \frac {x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a+b x^2} \, dx}{5 a^{5/2}} \\ & = -\frac {8 b^2 p^2}{15 a^2 x}-\frac {32 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 a^{5/2}}+\frac {4 i b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 a^{5/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}+\frac {4 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {\left (8 b^3 p^2\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{i-\frac {\sqrt {b} x}{\sqrt {a}}} \, dx}{5 a^3} \\ & = -\frac {8 b^2 p^2}{15 a^2 x}-\frac {32 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 a^{5/2}}+\frac {4 i b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 a^{5/2}}+\frac {8 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}+\frac {4 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}-\frac {\left (8 b^3 p^2\right ) \int \frac {\log \left (\frac {2}{1+\frac {i \sqrt {b} x}{\sqrt {a}}}\right )}{1+\frac {b x^2}{a}} \, dx}{5 a^3} \\ & = -\frac {8 b^2 p^2}{15 a^2 x}-\frac {32 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 a^{5/2}}+\frac {4 i b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 a^{5/2}}+\frac {8 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}+\frac {4 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {\left (8 i b^{5/2} p^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+\frac {i \sqrt {b} x}{\sqrt {a}}}\right )}{5 a^{5/2}} \\ & = -\frac {8 b^2 p^2}{15 a^2 x}-\frac {32 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 a^{5/2}}+\frac {4 i b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 a^{5/2}}+\frac {8 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}+\frac {4 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {4 i b^{5/2} p^2 \text {Li}_2\left (1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.12 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.99 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {4}{5} b p \left (-\frac {2 b^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2}}-\frac {2 b p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {b x^2}{a}\right )}{3 a^2 x}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 a x^3}+\frac {b \log \left (c \left (a+b x^2\right )^p\right )}{a^2 x}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{a^{5/2}}+\frac {p \left (i b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2+2 b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 i \sqrt {a}}{i \sqrt {a}-\sqrt {b} x}\right )+i b^{3/2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a}+\sqrt {b} x}{i \sqrt {a}-\sqrt {b} x}\right )\right )}{a^{5/2}}\right ) \]

[In]

Integrate[Log[c*(a + b*x^2)^p]^2/x^6,x]

[Out]

-1/5*Log[c*(a + b*x^2)^p]^2/x^5 + (4*b*p*((-2*b^(3/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(5/2) - (2*b*p*Hypergeo
metric2F1[-1/2, 1, 1/2, -((b*x^2)/a)])/(3*a^2*x) - Log[c*(a + b*x^2)^p]/(3*a*x^3) + (b*Log[c*(a + b*x^2)^p])/(
a^2*x) + (b^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p])/a^(5/2) + (p*(I*b^(3/2)*ArcTan[(Sqrt[b]*x)
/Sqrt[a]]^2 + 2*b^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[((2*I)*Sqrt[a])/(I*Sqrt[a] - Sqrt[b]*x)] + I*b^(3/2)*P
olyLog[2, -((I*Sqrt[a] + Sqrt[b]*x)/(I*Sqrt[a] - Sqrt[b]*x))]))/a^(5/2)))/5

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.13 (sec) , antiderivative size = 568, normalized size of antiderivative = 1.92

method	result	size
risch	\(-\frac {{\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}^{2}}{5 x^{5}}-\frac {4 p b \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{15 a \,x^{3}}+\frac {4 p \,b^{2} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5 a^{2} x}-\frac {4 p^{2} b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (b \,x^{2}+a \right )}{5 a^{2} \sqrt {a b}}+\frac {4 p \,b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5 a^{2} \sqrt {a b}}-\frac {8 b^{2} p^{2}}{15 a^{2} x}-\frac {32 p^{2} b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{15 a^{2} \sqrt {a b}}+\frac {4 p^{2} b \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (b \,x^{2}+a \right )-2 b \left (\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{4 \underline {\hspace {1.25 ex}}\alpha b}+\frac {\underline {\hspace {1.25 ex}}\alpha \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )}{2 a}+\frac {\underline {\hspace {1.25 ex}}\alpha \operatorname {dilog}\left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )}{2 a}\right )\right ) b}{2 a^{2} \underline {\hspace {1.25 ex}}\alpha }\right )}{5}+\left (i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )\right ) \left (-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5 x^{5}}+\frac {2 p b \left (-\frac {1}{3 a \,x^{3}}+\frac {b}{a^{2} x}+\frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}\right )}{5}\right )-\frac {{\left (i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )\right )}^{2}}{20 x^{5}}\)	\(568\)

[In]

int(ln(c*(b*x^2+a)^p)^2/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/5*ln((b*x^2+a)^p)^2/x^5-4/15*p*b*ln((b*x^2+a)^p)/a/x^3+4/5*p*b^2*ln((b*x^2+a)^p)/a^2/x-4/5*p^2*b^3/a^2/(a*b
)^(1/2)*arctan(b*x/(a*b)^(1/2))*ln(b*x^2+a)+4/5*p*b^3/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*ln((b*x^2+a)^p)-
8/15*b^2*p^2/a^2/x-32/15*p^2*b^3/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))+4/5*p^2*b*Sum(1/2*(ln(x-_alpha)*ln(b*
x^2+a)-2*b*(1/4/_alpha/b*ln(x-_alpha)^2+1/2*_alpha/a*ln(x-_alpha)*ln(1/2*(x+_alpha)/_alpha)+1/2*_alpha/a*dilog
(1/2*(x+_alpha)/_alpha)))/a^2*b/_alpha,_alpha=RootOf(_Z^2*b+a))+(I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p
)^2-I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*csgn(I*c*(b*x^2
+a)^p)^2*csgn(I*c)+2*ln(c))*(-1/5/x^5*ln((b*x^2+a)^p)+2/5*p*b*(-1/3/a/x^3+b/a^2/x+b^2/a^2/(a*b)^(1/2)*arctan(b
*x/(a*b)^(1/2))))-1/20*(I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*
x^2+a)^p)*csgn(I*c)-I*Pi*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+2*ln(c))^2/x^5

Fricas [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{6}} \,d x } \]

[In]

integrate(log(c*(b*x^2+a)^p)^2/x^6,x, algorithm="fricas")

[Out]

integral(log((b*x^2 + a)^p*c)^2/x^6, x)

Sympy [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{x^{6}}\, dx \]

[In]

integrate(ln(c*(b*x**2+a)**p)**2/x**6,x)

[Out]

Integral(log(c*(a + b*x**2)**p)**2/x**6, x)

Maxima [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{6}} \,d x } \]

[In]

integrate(log(c*(b*x^2+a)^p)^2/x^6,x, algorithm="maxima")

[Out]

-1/5*p^2*log(b*x^2 + a)^2/x^5 + integrate(1/5*(5*b*x^2*log(c)^2 + 5*a*log(c)^2 + 2*((2*p^2 + 5*p*log(c))*b*x^2
+ 5*a*p*log(c))*log(b*x^2 + a))/(b*x^8 + a*x^6), x)

Giac [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{6}} \,d x } \]

[In]

integrate(log(c*(b*x^2+a)^p)^2/x^6,x, algorithm="giac")

[Out]

integrate(log((b*x^2 + a)^p*c)^2/x^6, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2}{x^6} \,d x \]

[In]

int(log(c*(a + b*x^2)^p)^2/x^6,x)

[Out]

int(log(c*(a + b*x^2)^p)^2/x^6, x)