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

Optimal. Leaf size=190 \[ -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}+\frac {4 \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a}}+\frac {4 i \sqrt {b} p^2 \text {Li}_2\left (1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{\sqrt {a}}+\frac {4 i \sqrt {b} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{\sqrt {a}}+\frac {8 \sqrt {b} p^2 \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a}} \]

[Out]

-ln(c*(b*x^2+a)^p)^2/x+4*I*p^2*arctan(x*b^(1/2)/a^(1/2))^2*b^(1/2)/a^(1/2)+4*p*arctan(x*b^(1/2)/a^(1/2))*ln(c*
(b*x^2+a)^p)*b^(1/2)/a^(1/2)+8*p^2*arctan(x*b^(1/2)/a^(1/2))*ln(2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))*b^(1/2)/a^(1/
2)+4*I*p^2*polylog(2,1-2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))*b^(1/2)/a^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2457, 205, 2470, 12, 4920, 4854, 2402, 2315} \[ \frac {4 i \sqrt {b} p^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {a}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}+\frac {4 \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a}}+\frac {4 i \sqrt {b} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{\sqrt {a}}+\frac {8 \sqrt {b} p^2 \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a}} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]^2/x^2,x]

[Out]

((4*I)*Sqrt[b]*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2)/Sqrt[a] + (8*Sqrt[b]*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[(2*
Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/Sqrt[a] + (4*Sqrt[b]*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p])/Sq
rt[a] - Log[c*(a + b*x^2)^p]^2/x + ((4*I)*Sqrt[b]*p^2*PolyLog[2, 1 - (2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/Sqr
t[a]

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

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Mathematica [A]  time = 0.05, size = 173, normalized size = 0.91 \[ \frac {-\sqrt {a} \log ^2\left (c \left (a+b x^2\right )^p\right )+4 \sqrt {b} p x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (\log \left (c \left (a+b x^2\right )^p\right )+2 p \log \left (\frac {2 i}{-\frac {\sqrt {b} x}{\sqrt {a}}+i}\right )\right )+4 i \sqrt {b} p^2 x \text {Li}_2\left (\frac {\sqrt {b} x+i \sqrt {a}}{\sqrt {b} x-i \sqrt {a}}\right )+4 i \sqrt {b} p^2 x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{\sqrt {a} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*I)*Sqrt[b]*p^2*x*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2 - Sqrt[a]*Log[c*(a + b*x^2)^p]^2 + 4*Sqrt[b]*p*x*ArcTan[(Sq
rt[b]*x)/Sqrt[a]]*(2*p*Log[(2*I)/(I - (Sqrt[b]*x)/Sqrt[a])] + Log[c*(a + b*x^2)^p]) + (4*I)*Sqrt[b]*p^2*x*Poly
Log[2, (I*Sqrt[a] + Sqrt[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])/(Sqrt[a]*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.86, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {p^{2} \log \left (b x^{2} + a\right )^{2}}{x} + \int \frac {b x^{2} \log \relax (c)^{2} + a \log \relax (c)^{2} + 2 \, {\left ({\left (2 \, p^{2} + p \log \relax (c)\right )} b x^{2} + a p \log \relax (c)\right )} \log \left (b x^{2} + a\right )}{b x^{4} + a x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-p^2*log(b*x^2 + a)^2/x + integrate((b*x^2*log(c)^2 + a*log(c)^2 + 2*((2*p^2 + p*log(c))*b*x^2 + a*p*log(c))*l
og(b*x^2 + a))/(b*x^4 + a*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^2)^p)^2/x^2,x)

[Out]

int(log(c*(a + b*x^2)^p)^2/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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