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

Optimal. Leaf size=296 \[ \frac {4 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}+\frac {4 i b^{5/2} p^2 \text {Li}_2\left (1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{5 a^{5/2}}+\frac {4 i b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 a^{5/2}}-\frac {32 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 a^{5/2}}+\frac {8 b^{5/2} p^2 \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 a^{5/2}}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}-\frac {8 b^2 p^2}{15 a^2 x}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3} \]

[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)

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Rubi [A]  time = 0.32, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {2457, 2476, 2455, 325, 205, 2470, 12, 4920, 4854, 2402, 2315} \[ \frac {4 i b^{5/2} p^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}+\frac {4 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}-\frac {8 b^2 p^2}{15 a^2 x}+\frac {4 i b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 a^{5/2}}-\frac {32 b^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 a^{5/2}}+\frac {8 b^{5/2} p^2 \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 a^{5/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3} \]

Antiderivative was successfully verified.

[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 205

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

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 2315

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

Rule 2402

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 2455

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 2457

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 2470

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 2476

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 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[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 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[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^6} \, dx &=-\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 ) \operatorname {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*}

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Mathematica [C]  time = 0.29, size = 277, normalized size = 0.94 \[ -\frac {3 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {4 b p x^2 \left (a^{3/2} \log \left (c \left (a+b x^2\right )^p\right )-3 b^{3/2} x^3 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )-3 i b^{3/2} p x^3 \left (\text {Li}_2\left (\frac {\sqrt {b} x+i \sqrt {a}}{\sqrt {b} x-i \sqrt {a}}\right )+\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )-2 i \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )\right )\right )+6 b^{3/2} p x^3 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )-3 \sqrt {a} b x^2 \log \left (c \left (a+b x^2\right )^p\right )+2 \sqrt {a} b p x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {b x^2}{a}\right )\right )}{a^{5/2}}}{15 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(3*Log[c*(a + b*x^2)^p]^2 + (4*b*p*x^2*(6*b^(3/2)*p*x^3*ArcTan[(Sqrt[b]*x)/Sqrt[a]] + 2*Sqrt[a]*b*p*x^2*
Hypergeometric2F1[-1/2, 1, 1/2, -((b*x^2)/a)] + a^(3/2)*Log[c*(a + b*x^2)^p] - 3*Sqrt[a]*b*x^2*Log[c*(a + b*x^
2)^p] - 3*b^(3/2)*x^3*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p] - (3*I)*b^(3/2)*p*x^3*(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*Sqr
t[a] + Sqrt[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])))/a^(5/2))/x^5

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.97, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {p^{2} \log \left (b x^{2} + a\right )^{2}}{5 \, x^{5}} + \int \frac {5 \, b x^{2} \log \relax (c)^{2} + 5 \, a \log \relax (c)^{2} + 2 \, {\left ({\left (2 \, p^{2} + 5 \, p \log \relax (c)\right )} b x^{2} + 5 \, a p \log \relax (c)\right )} \log \left (b x^{2} + a\right )}{5 \, {\left (b x^{8} + a x^{6}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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