\(\int \frac {\log ^2(c (a+b x^2)^p)}{x^4} \, dx\) [88]

Optimal result

Integrand size = 18, antiderivative size = 254 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx=\frac {8 b^{3/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 a^{3/2}}-\frac {4 i b^{3/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{3 a^{3/2}}-\frac {8 b^{3/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{3 a^{3/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{3 a x}-\frac {4 b^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 a^{3/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{3 x^3}-\frac {4 i b^{3/2} p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{3 a^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.81 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx=\frac {-4 i b^{3/2} p^2 x^3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2-4 b^{3/2} p x^3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-2 p+2 p \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )+\log \left (c \left (a+b x^2\right )^p\right )\right )-\sqrt {a} \log \left (c \left (a+b x^2\right )^p\right ) \left (4 b p x^2+a \log \left (c \left (a+b x^2\right )^p\right )\right )-4 i b^{3/2} p^2 x^3 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a}+\sqrt {b} x}{-i \sqrt {a}+\sqrt {b} x}\right )}{3 a^{3/2} x^3} \] Input:

Integrate[Log[c*(a + b*x^2)^p]^2/x^4,x]
 

Output:

((-4*I)*b^(3/2)*p^2*x^3*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2 - 4*b^(3/2)*p*x^3*Ar 
cTan[(Sqrt[b]*x)/Sqrt[a]]*(-2*p + 2*p*Log[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b] 
*x)] + Log[c*(a + b*x^2)^p]) - Sqrt[a]*Log[c*(a + b*x^2)^p]*(4*b*p*x^2 + a 
*Log[c*(a + b*x^2)^p]) - (4*I)*b^(3/2)*p^2*x^3*PolyLog[2, (I*Sqrt[a] + Sqr 
t[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])/(3*a^(3/2)*x^3)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2907, 2926, 2009}

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.

\(\displaystyle \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx\)

\(\Big \downarrow \) 2907

\(\displaystyle \frac {4}{3} b p \int \frac {\log \left (c \left (b x^2+a\right )^p\right )}{x^2 \left (b x^2+a\right )}dx-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{3 x^3}\)

\(\Big \downarrow \) 2926

\(\displaystyle \frac {4}{3} b p \int \left (\frac {\log \left (c \left (b x^2+a\right )^p\right )}{a x^2}-\frac {b \log \left (c \left (b x^2+a\right )^p\right )}{a \left (b x^2+a\right )}\right )dx-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{3 x^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{3 x^3}+\frac {4}{3} b p \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{a^{3/2}}-\frac {i \sqrt {b} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{a^{3/2}}+\frac {2 \sqrt {b} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2 \sqrt {b} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{a^{3/2}}-\frac {i \sqrt {b} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{a^{3/2}}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{a x}\right )\)

Input:

Int[Log[c*(a + b*x^2)^p]^2/x^4,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2907

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]
 

rule 2926

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]
 

Maple [C] (warning: unable to verify)

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

Time = 1.02 (sec) , antiderivative size = 522, normalized size of antiderivative = 2.06

method	result	size
risch	\(-\frac {{\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}^{2}}{3 x^{3}}-\frac {4 p b \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3 a x}+\frac {4 p^{2} b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (b \,x^{2}+a \right )}{3 a \sqrt {a b}}-\frac {4 p \,b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3 a \sqrt {a b}}+\frac {8 p^{2} b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{3 a \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 }\left (-\frac {\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 )}{2 a \underline {\hspace {1.25 ex}}\alpha }\right )\right )}{3}+\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 )}{3 x^{3}}+\frac {2 p b \left (-\frac {1}{a x}-\frac {b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{a \sqrt {a b}}\right )}{3}\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}}{12 x^{3}}\)	\(522\)

Input:

int(ln(c*(b*x^2+a)^p)^2/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3*ln((b*x^2+a)^p)^2/x^3-4/3*p*b*ln((b*x^2+a)^p)/a/x+4/3*p^2*b^2/a/(a*b) 
^(1/2)*arctan(b*x/(a*b)^(1/2))*ln(b*x^2+a)-4/3*p*b^2/a/(a*b)^(1/2)*arctan( 
b*x/(a*b)^(1/2))*ln((b*x^2+a)^p)+8/3*p^2*b^2/a/(a*b)^(1/2)*arctan(b*x/(a*b 
)^(1/2))+4/3*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*_alph 
a/a*dilog(1/2*(x+_alpha)/_alpha)))/a/_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))*(-1/3/x^3*ln((b*x^2+a)^p)+2/3*p*b*(-1/a/x-1 
/a*b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))-1/12*(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)*csg 
n(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^3
 

Fricas [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{4}} \,d x } \] Input:

integrate(log(c*(b*x^2+a)^p)^2/x^4,x, algorithm="fricas")
 

Output:

integral(log((b*x^2 + a)^p*c)^2/x^4, x)
 

Sympy [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx=\int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{x^{4}}\, dx \] Input:

integrate(ln(c*(b*x**2+a)**p)**2/x**4,x)
 

Output:

Integral(log(c*(a + b*x**2)**p)**2/x**4, x)
 

Maxima [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{4}} \,d x } \] Input:

integrate(log(c*(b*x^2+a)^p)^2/x^4,x, algorithm="maxima")
 

Output:

-1/3*p^2*log(b*x^2 + a)^2/x^3 + integrate(1/3*(3*b*x^2*log(c)^2 + 3*a*log( 
c)^2 + 2*((2*p^2 + 3*p*log(c))*b*x^2 + 3*a*p*log(c))*log(b*x^2 + a))/(b*x^ 
6 + a*x^4), x)
 

Giac [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{4}} \,d x } \] Input:

integrate(log(c*(b*x^2+a)^p)^2/x^4,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx=\int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2}{x^4} \,d x \] Input:

int(log(c*(a + b*x^2)^p)^2/x^4,x)
 

Output:

int(log(c*(a + b*x^2)^p)^2/x^4, x)
 

Reduce [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^4} \, dx=\frac {-8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,p^{2} x^{3}-12 \left (\int \frac {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}{b \,x^{6}+a \,x^{4}}d x \right ) a^{3} p \,x^{3}-3 {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} a^{2}-4 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a^{2} p -8 a b \,p^{2} x^{2}}{9 a^{2} x^{3}} \] Input:

int(log(c*(b*x^2+a)^p)^2/x^4,x)
 

Output:

( - 8*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b*p**2*x**3 - 12*int(l 
og((a + b*x**2)**p*c)/(a*x**4 + b*x**6),x)*a**3*p*x**3 - 3*log((a + b*x**2 
)**p*c)**2*a**2 - 4*log((a + b*x**2)**p*c)*a**2*p - 8*a*b*p**2*x**2)/(9*a* 
*2*x**3)