Integrand size = 18, antiderivative size = 190 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx=\frac {4 i \sqrt {b} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{\sqrt {a}}+\frac {8 \sqrt {b} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {a}}+\frac {4 \sqrt {b} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}+\frac {4 i \sqrt {b} p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {a}} \] Output:
4*I*b^(1/2)*p^2*arctan(b^(1/2)*x/a^(1/2))^2/a^(1/2)+8*b^(1/2)*p^2*arctan(b ^(1/2)*x/a^(1/2))*ln(2*a^(1/2)/(a^(1/2)+I*b^(1/2)*x))/a^(1/2)+4*b^(1/2)*p* arctan(b^(1/2)*x/a^(1/2))*ln(c*(b*x^2+a)^p)/a^(1/2)-ln(c*(b*x^2+a)^p)^2/x+ 4*I*b^(1/2)*p^2*polylog(2,1-2*a^(1/2)/(a^(1/2)+I*b^(1/2)*x))/a^(1/2)
Time = 0.07 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.91 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx=\frac {4 i \sqrt {b} p^2 x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2-\sqrt {a} \log ^2\left (c \left (a+b x^2\right )^p\right )+4 \sqrt {b} p x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (2 p \log \left (\frac {2 i}{i-\frac {\sqrt {b} x}{\sqrt {a}}}\right )+\log \left (c \left (a+b x^2\right )^p\right )\right )+4 i \sqrt {b} p^2 x \operatorname {PolyLog}\left (2,\frac {i \sqrt {a}+\sqrt {b} x}{-i \sqrt {a}+\sqrt {b} x}\right )}{\sqrt {a} x} \] Input:
Integrate[Log[c*(a + b*x^2)^p]^2/x^2,x]
Output:
((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[(Sqrt[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*PolyLo g[2, (I*Sqrt[a] + Sqrt[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])/(Sqrt[a]*x)
Time = 1.05 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2907, 2920, 27, 5455, 27, 5379, 27, 2849, 2752}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx\) |
| \(\Big \downarrow \) 2907 |
| \(\displaystyle 4 b p \int \frac {\log \left (c \left (b x^2+a\right )^p\right )}{b x^2+a}dx-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 2920 |
| \(\displaystyle 4 b p \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a} \sqrt {b}}-2 b p \int \frac {x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \left (b x^2+a\right )}dx\right )-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 b p \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a} \sqrt {b}}-\frac {2 \sqrt {b} p \int \frac {x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b x^2+a}dx}{\sqrt {a}}\right )-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 5455 |
| \(\displaystyle -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}+4 b p \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a} \sqrt {b}}-\frac {2 \sqrt {b} p \left (-\frac {\int \frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{i \sqrt {a}-\sqrt {b} x}dx}{\sqrt {a} \sqrt {b}}-\frac {i \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{2 b}\right )}{\sqrt {a}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}+4 b p \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a} \sqrt {b}}-\frac {2 \sqrt {b} p \left (-\frac {\int \frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{i \sqrt {a}-\sqrt {b} x}dx}{\sqrt {b}}-\frac {i \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{2 b}\right )}{\sqrt {a}}\right )\) |
| \(\Big \downarrow \) 5379 |
| \(\displaystyle -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}+4 b p \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a} \sqrt {b}}-\frac {2 \sqrt {b} p \left (-\frac {\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {b}}-\frac {\int \frac {a \log \left (\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{b x^2+a}dx}{\sqrt {a}}}{\sqrt {b}}-\frac {i \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{2 b}\right )}{\sqrt {a}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}+4 b p \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a} \sqrt {b}}-\frac {2 \sqrt {b} p \left (-\frac {\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {b}}-\sqrt {a} \int \frac {\log \left (\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{b x^2+a}dx}{\sqrt {b}}-\frac {i \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{2 b}\right )}{\sqrt {a}}\right )\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}+4 b p \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a} \sqrt {b}}-\frac {2 \sqrt {b} p \left (-\frac {\frac {i \sqrt {a} \int \frac {\log \left (\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}}d\frac {1}{i \sqrt {b} x+\sqrt {a}}}{\sqrt {b}}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {b}}}{\sqrt {b}}-\frac {i \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{2 b}\right )}{\sqrt {a}}\right )\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x}+4 b p \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{\sqrt {a} \sqrt {b}}-\frac {2 \sqrt {b} p \left (-\frac {\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {b}}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{2 \sqrt {b}}}{\sqrt {b}}-\frac {i \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{2 b}\right )}{\sqrt {a}}\right )\) |
Input:
Int[Log[c*(a + b*x^2)^p]^2/x^2,x]
Output:
-(Log[c*(a + b*x^2)^p]^2/x) + 4*b*p*((ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p])/(Sqrt[a]*Sqrt[b]) - (2*Sqrt[b]*p*(((-1/2*I)*ArcTan[(Sqrt[b]* x)/Sqrt[a]]^2)/b - ((ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/Sqrt[b] + ((I/2)*PolyLog[2, 1 - (2*Sqrt[a])/(Sqrt[a] + I* Sqrt[b]*x)])/Sqrt[b])/Sqrt[b]))/Sqrt[a])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-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]
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] - Simp[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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.) *(x_)^2), x_Symbol] :> With[{u = IntHide[1/(f + g*x^2), x]}, Simp[u*(a + b* Log[c*(d + e*x^n)^p]), x] - Simp[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]
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] + Simp[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 ]
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] - Si mp[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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.79 (sec) , antiderivative size = 446, normalized size of antiderivative = 2.35
| method | result | size |
| risch | \(-\frac {{\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}^{2}}{x}-\frac {4 p^{2} b \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (b \,x^{2}+a \right )}{\sqrt {a b}}+\frac {4 p b \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{\sqrt {a b}}+p^{2} \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (b \,x^{2}+a \right )-b \left (\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{\underline {\hspace {1.25 ex}}\alpha b}+\frac {2 \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 )}{a}+\frac {2 \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 )}{a}\right )}{\underline {\hspace {1.25 ex}}\alpha }\right )+\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 )}{x}+\frac {2 p b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}\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}}{4 x}\) | \(446\) |
Input:
int(ln(c*(b*x^2+a)^p)^2/x^2,x,method=_RETURNVERBOSE)
Output:
-1/x*ln((b*x^2+a)^p)^2-4*p^2*b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*ln(b*x^ 2+a)+4*p*b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*ln((b*x^2+a)^p)+p^2*sum(1/_ alpha*(2*ln(x-_alpha)*ln(b*x^2+a)-b*(1/_alpha/b*ln(x-_alpha)^2+2*_alpha/a* ln(x-_alpha)*ln(1/2*(x+_alpha)/_alpha)+2*_alpha/a*dilog(1/2*(x+_alpha)/_al pha))),_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/x* ln((b*x^2+a)^p)+2*p*b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))-1/4*(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
\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(log(c*(b*x^2+a)^p)^2/x^2,x, algorithm="fricas")
Output:
integral(log((b*x^2 + a)^p*c)^2/x^2, x)
\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx=\int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{x^{2}}\, dx \] Input:
integrate(ln(c*(b*x**2+a)**p)**2/x**2,x)
Output:
Integral(log(c*(a + b*x**2)**p)**2/x**2, x)
\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(log(c*(b*x^2+a)^p)^2/x^2,x, algorithm="maxima")
Output:
-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))*log(b*x^2 + a))/(b*x^4 + a*x^2), x)
\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(log(c*(b*x^2+a)^p)^2/x^2,x, algorithm="giac")
Output:
integrate(log((b*x^2 + a)^p*c)^2/x^2, x)
Timed out. \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx=\int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2}{x^2} \,d x \] Input:
int(log(c*(a + b*x^2)^p)^2/x^2,x)
Output:
int(log(c*(a + b*x^2)^p)^2/x^2, x)
\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx=\frac {8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) p^{2} x -4 \left (\int \frac {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}{b \,x^{4}+a \,x^{2}}d x \right ) a^{2} p x -{\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} a -4 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a p}{a x} \] Input:
int(log(c*(b*x^2+a)^p)^2/x^2,x)
Output:
(8*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*p**2*x - 4*int(log((a + b *x**2)**p*c)/(a*x**2 + b*x**4),x)*a**2*p*x - log((a + b*x**2)**p*c)**2*a - 4*log((a + b*x**2)**p*c)*a*p)/(a*x)