3.1.90 \(\int \frac {\log ^2(c (a+b x^2)^p)}{x^8} \, dx\) [90]

Optimal. Leaf size=338 \[ -\frac {8 b^2 p^2}{105 a^2 x^3}+\frac {64 b^3 p^2}{105 a^3 x}+\frac {184 b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{105 a^{7/2}}-\frac {4 i b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{7 a^{7/2}}-\frac {8 b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{7 a^{7/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{35 a x^5}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{21 a^2 x^3}-\frac {4 b^3 p \log \left (c \left (a+b x^2\right )^p\right )}{7 a^3 x}-\frac {4 b^{7/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{7 a^{7/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}-\frac {4 i b^{7/2} p^2 \text {Li}_2\left (1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{7 a^{7/2}} \]

[Out]

-8/105*b^2*p^2/a^2/x^3+64/105*b^3*p^2/a^3/x+184/105*b^(7/2)*p^2*arctan(x*b^(1/2)/a^(1/2))/a^(7/2)-4/7*I*b^(7/2
)*p^2*arctan(x*b^(1/2)/a^(1/2))^2/a^(7/2)-4/35*b*p*ln(c*(b*x^2+a)^p)/a/x^5+4/21*b^2*p*ln(c*(b*x^2+a)^p)/a^2/x^
3-4/7*b^3*p*ln(c*(b*x^2+a)^p)/a^3/x-4/7*b^(7/2)*p*arctan(x*b^(1/2)/a^(1/2))*ln(c*(b*x^2+a)^p)/a^(7/2)-1/7*ln(c
*(b*x^2+a)^p)^2/x^7-8/7*b^(7/2)*p^2*arctan(x*b^(1/2)/a^(1/2))*ln(2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/a^(7/2)-4/7*
I*b^(7/2)*p^2*polylog(2,1-2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/a^(7/2)

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Rubi [A]
time = 0.25, antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 11, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {2507, 2526, 2505, 331, 211, 2520, 12, 5040, 4964, 2449, 2352} \begin {gather*} -\frac {4 i b^{7/2} p^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{7 a^{7/2}}-\frac {4 b^{7/2} p \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{7 a^{7/2}}-\frac {4 i b^{7/2} p^2 \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{7 a^{7/2}}+\frac {184 b^{7/2} p^2 \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{105 a^{7/2}}-\frac {8 b^{7/2} p^2 \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{7 a^{7/2}}-\frac {4 b^3 p \log \left (c \left (a+b x^2\right )^p\right )}{7 a^3 x}+\frac {64 b^3 p^2}{105 a^3 x}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{21 a^2 x^3}-\frac {8 b^2 p^2}{105 a^2 x^3}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{35 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]^2/x^8,x]

[Out]

(-8*b^2*p^2)/(105*a^2*x^3) + (64*b^3*p^2)/(105*a^3*x) + (184*b^(7/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(105*a^(
7/2)) - (((4*I)/7)*b^(7/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2)/a^(7/2) - (8*b^(7/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt
[a]]*Log[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/(7*a^(7/2)) - (4*b*p*Log[c*(a + b*x^2)^p])/(35*a*x^5) + (4*b^2*
p*Log[c*(a + b*x^2)^p])/(21*a^2*x^3) - (4*b^3*p*Log[c*(a + b*x^2)^p])/(7*a^3*x) - (4*b^(7/2)*p*ArcTan[(Sqrt[b]
*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p])/(7*a^(7/2)) - Log[c*(a + b*x^2)^p]^2/(7*x^7) - (((4*I)/7)*b^(7/2)*p^2*PolyL
og[2, 1 - (2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/a^(7/2)

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*} \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^8} \, dx &=-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}+\frac {1}{7} (4 b p) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6 \left (a+b x^2\right )} \, dx\\ &=-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}+\frac {1}{7} (4 b p) \int \left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{a x^6}-\frac {b \log \left (c \left (a+b x^2\right )^p\right )}{a^2 x^4}+\frac {b^2 \log \left (c \left (a+b x^2\right )^p\right )}{a^3 x^2}-\frac {b^3 \log \left (c \left (a+b x^2\right )^p\right )}{a^3 \left (a+b x^2\right )}\right ) \, dx\\ &=-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}+\frac {(4 b p) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx}{7 a}-\frac {\left (4 b^2 p\right ) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx}{7 a^2}+\frac {\left (4 b^3 p\right ) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx}{7 a^3}-\frac {\left (4 b^4 p\right ) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{a+b x^2} \, dx}{7 a^3}\\ &=-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{35 a x^5}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{21 a^2 x^3}-\frac {4 b^3 p \log \left (c \left (a+b x^2\right )^p\right )}{7 a^3 x}-\frac {4 b^{7/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{7 a^{7/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}+\frac {\left (8 b^2 p^2\right ) \int \frac {1}{x^4 \left (a+b x^2\right )} \, dx}{35 a}-\frac {\left (8 b^3 p^2\right ) \int \frac {1}{x^2 \left (a+b x^2\right )} \, dx}{21 a^2}+\frac {\left (8 b^4 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{7 a^3}+\frac {\left (8 b^5 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}{7 a^3}\\ &=-\frac {8 b^2 p^2}{105 a^2 x^3}+\frac {8 b^3 p^2}{21 a^3 x}+\frac {8 b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{7 a^{7/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{35 a x^5}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{21 a^2 x^3}-\frac {4 b^3 p \log \left (c \left (a+b x^2\right )^p\right )}{7 a^3 x}-\frac {4 b^{7/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{7 a^{7/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}-\frac {\left (8 b^3 p^2\right ) \int \frac {1}{x^2 \left (a+b x^2\right )} \, dx}{35 a^2}+\frac {\left (8 b^4 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{21 a^3}+\frac {\left (8 b^{9/2} p^2\right ) \int \frac {x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a+b x^2} \, dx}{7 a^{7/2}}\\ &=-\frac {8 b^2 p^2}{105 a^2 x^3}+\frac {64 b^3 p^2}{105 a^3 x}+\frac {32 b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{21 a^{7/2}}-\frac {4 i b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{7 a^{7/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{35 a x^5}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{21 a^2 x^3}-\frac {4 b^3 p \log \left (c \left (a+b x^2\right )^p\right )}{7 a^3 x}-\frac {4 b^{7/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{7 a^{7/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}-\frac {\left (8 b^4 p^2\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{i-\frac {\sqrt {b} x}{\sqrt {a}}} \, dx}{7 a^4}+\frac {\left (8 b^4 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{35 a^3}\\ &=-\frac {8 b^2 p^2}{105 a^2 x^3}+\frac {64 b^3 p^2}{105 a^3 x}+\frac {184 b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{105 a^{7/2}}-\frac {4 i b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{7 a^{7/2}}-\frac {8 b^{7/2} 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 )}{7 a^{7/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{35 a x^5}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{21 a^2 x^3}-\frac {4 b^3 p \log \left (c \left (a+b x^2\right )^p\right )}{7 a^3 x}-\frac {4 b^{7/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{7 a^{7/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}+\frac {\left (8 b^4 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}{7 a^4}\\ &=-\frac {8 b^2 p^2}{105 a^2 x^3}+\frac {64 b^3 p^2}{105 a^3 x}+\frac {184 b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{105 a^{7/2}}-\frac {4 i b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{7 a^{7/2}}-\frac {8 b^{7/2} 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 )}{7 a^{7/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{35 a x^5}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{21 a^2 x^3}-\frac {4 b^3 p \log \left (c \left (a+b x^2\right )^p\right )}{7 a^3 x}-\frac {4 b^{7/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{7 a^{7/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}-\frac {\left (8 i b^{7/2} 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 )}{7 a^{7/2}}\\ &=-\frac {8 b^2 p^2}{105 a^2 x^3}+\frac {64 b^3 p^2}{105 a^3 x}+\frac {184 b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{105 a^{7/2}}-\frac {4 i b^{7/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{7 a^{7/2}}-\frac {8 b^{7/2} 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 )}{7 a^{7/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{35 a x^5}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{21 a^2 x^3}-\frac {4 b^3 p \log \left (c \left (a+b x^2\right )^p\right )}{7 a^3 x}-\frac {4 b^{7/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{7 a^{7/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}-\frac {4 i b^{7/2} p^2 \text {Li}_2\left (1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{7 a^{7/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.16, size = 334, normalized size = 0.99 \begin {gather*} -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{7 x^7}+\frac {4 b p \left (30 b^{5/2} p x^5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )-2 a^{3/2} b p x^2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {b x^2}{a}\right )+10 \sqrt {a} b^2 p x^4 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {b x^2}{a}\right )-3 a^{5/2} \log \left (c \left (a+b x^2\right )^p\right )+5 a^{3/2} b x^2 \log \left (c \left (a+b x^2\right )^p\right )-15 \sqrt {a} b^2 x^4 \log \left (c \left (a+b x^2\right )^p\right )-15 b^{5/2} x^5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )-15 i b^{5/2} p x^5 \left (\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )-2 i \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )\right )+\text {Li}_2\left (\frac {i \sqrt {a}+\sqrt {b} x}{-i \sqrt {a}+\sqrt {b} x}\right )\right )\right )}{105 a^{7/2} x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]^2/x^8,x]

[Out]

-1/7*Log[c*(a + b*x^2)^p]^2/x^7 + (4*b*p*(30*b^(5/2)*p*x^5*ArcTan[(Sqrt[b]*x)/Sqrt[a]] - 2*a^(3/2)*b*p*x^2*Hyp
ergeometric2F1[-3/2, 1, -1/2, -((b*x^2)/a)] + 10*Sqrt[a]*b^2*p*x^4*Hypergeometric2F1[-1/2, 1, 1/2, -((b*x^2)/a
)] - 3*a^(5/2)*Log[c*(a + b*x^2)^p] + 5*a^(3/2)*b*x^2*Log[c*(a + b*x^2)^p] - 15*Sqrt[a]*b^2*x^4*Log[c*(a + b*x
^2)^p] - 15*b^(5/2)*x^5*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p] - (15*I)*b^(5/2)*p*x^5*(ArcTan[(Sqrt[
b]*x)/Sqrt[a]]*(ArcTan[(Sqrt[b]*x)/Sqrt[a]] - (2*I)*Log[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)]) + PolyLog[2, (I*
Sqrt[a] + Sqrt[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])))/(105*a^(7/2)*x^5)

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{x^{8}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^2+a)^p)^2/x^8,x)

[Out]

int(ln(c*(b*x^2+a)^p)^2/x^8,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)^2/x^8,x, algorithm="maxima")

[Out]

-1/7*p^2*log(b*x^2 + a)^2/x^7 + integrate(1/7*(7*b*x^2*log(c)^2 + 7*a*log(c)^2 + 2*((2*p^2 + 7*p*log(c))*b*x^2
+ 7*a*p*log(c))*log(b*x^2 + a))/(b*x^10 + a*x^8), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)^2/x^8,x, algorithm="fricas")

[Out]

integral(log((b*x^2 + a)^p*c)^2/x^8, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{x^{8}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**2+a)**p)**2/x**8,x)

[Out]

Integral(log(c*(a + b*x**2)**p)**2/x**8, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)^2/x^8,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2}{x^8} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^2)^p)^2/x^8,x)

[Out]

int(log(c*(a + b*x^2)^p)^2/x^8, x)

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