\(\int \frac {\log ^2(c (a+b x^2)^p)}{x^6} \, dx\) [89]

Optimal result

Integrand size = 18, antiderivative size = 296 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=-\frac {8 b^2 p^2}{15 a^2 x}-\frac {32 b^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{15 a^{5/2}}+\frac {4 i b^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 a^{5/2}}+\frac {8 b^{5/2} p^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}}-\frac {4 b p \log \left (c \left (a+b x^2\right )^p\right )}{15 a x^3}+\frac {4 b^2 p \log \left (c \left (a+b x^2\right )^p\right )}{5 a^2 x}+\frac {4 b^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 a^{5/2}}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {4 i b^{5/2} p^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 a^{5/2}} \] Output:

-8/15*b^2*p^2/a^2/x-32/15*b^(5/2)*p^2*arctan(b^(1/2)*x/a^(1/2))/a^(5/2)+4/ 
5*I*b^(5/2)*p^2*arctan(b^(1/2)*x/a^(1/2))^2/a^(5/2)+8/5*b^(5/2)*p^2*arctan 
(b^(1/2)*x/a^(1/2))*ln(2*a^(1/2)/(a^(1/2)+I*b^(1/2)*x))/a^(5/2)-4/15*b*p*l 
n(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*arc 
tan(b^(1/2)*x/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+4/5*I*b^(5/2)*p^2*polylog(2,1-2*a^(1/2)/(a^(1/2)+I*b^(1/2)*x))/a^(5/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.21 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.99 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {4}{5} b p \left (-\frac {2 b^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2}}-\frac {2 b p \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {b x^2}{a}\right )}{3 a^2 x}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 a x^3}+\frac {b \log \left (c \left (a+b x^2\right )^p\right )}{a^2 x}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{a^{5/2}}+\frac {p \left (i b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2+2 b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 i \sqrt {a}}{i \sqrt {a}-\sqrt {b} x}\right )+i b^{3/2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a}+\sqrt {b} x}{i \sqrt {a}-\sqrt {b} x}\right )\right )}{a^{5/2}}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.92, 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.

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

\(\Big \downarrow \) 2907

\(\displaystyle \frac {4}{5} b p \int \frac {\log \left (c \left (b x^2+a\right )^p\right )}{x^4 \left (b x^2+a\right )}dx-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}\)

\(\Big \downarrow \) 2926

\(\displaystyle \frac {4}{5} b p \int \left (\frac {\log \left (c \left (b x^2+a\right )^p\right ) b^2}{a^2 \left (b x^2+a\right )}-\frac {\log \left (c \left (b x^2+a\right )^p\right ) b}{a^2 x^2}+\frac {\log \left (c \left (b x^2+a\right )^p\right )}{a x^4}\right )dx-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{5 x^5}+\frac {4}{5} b p \left (\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{a^{5/2}}+\frac {i b^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{a^{5/2}}-\frac {8 b^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 a^{5/2}}+\frac {2 b^{3/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{a^{5/2}}+\frac {i b^{3/2} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{a^{5/2}}+\frac {b \log \left (c \left (a+b x^2\right )^p\right )}{a^2 x}-\frac {2 b p}{3 a^2 x}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 a x^3}\right )\)

Input:

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

Output:

-1/5*Log[c*(a + b*x^2)^p]^2/x^5 + (4*b*p*((-2*b*p)/(3*a^2*x) - (8*b^(3/2)* 
p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(3*a^(5/2)) + (I*b^(3/2)*p*ArcTan[(Sqrt[b]* 
x)/Sqrt[a]]^2)/a^(5/2) + (2*b^(3/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[(2*S 
qrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/a^(5/2) - Log[c*(a + b*x^2)^p]/(3*a*x^3) 
 + (b*Log[c*(a + b*x^2)^p])/(a^2*x) + (b^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]] 
*Log[c*(a + b*x^2)^p])/a^(5/2) + (I*b^(3/2)*p*PolyLog[2, 1 - (2*Sqrt[a])/( 
Sqrt[a] + I*Sqrt[b]*x)])/a^(5/2)))/5
 

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 = 1.48 (sec) , antiderivative size = 568, normalized size of antiderivative = 1.92

method	result	size
risch	\(-\frac {{\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}^{2}}{5 x^{5}}-\frac {4 p^{2} b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (b \,x^{2}+a \right )}{5 a^{2} \sqrt {a b}}+\frac {4 p \,b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5 a^{2} \sqrt {a b}}-\frac {4 p b \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{15 a \,x^{3}}+\frac {4 p \,b^{2} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5 a^{2} x}-\frac {32 p^{2} b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{15 a^{2} \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 }\frac {\left (\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 )\right ) b}{2 a^{2} \underline {\hspace {1.25 ex}}\alpha }\right )}{5}-\frac {8 b^{2} p^{2}}{15 a^{2} x}+\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 )}{5 x^{5}}+\frac {2 p b \left (\frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}-\frac {1}{3 a \,x^{3}}+\frac {b}{a^{2} x}\right )}{5}\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}}{20 x^{5}}\)	\(568\)

Input:

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

Output:

-1/5*ln((b*x^2+a)^p)^2/x^5-4/5*p^2*b^3/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1 
/2))*ln(b*x^2+a)+4/5*p*b^3/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*ln((b*x 
^2+a)^p)-4/15*p*b*ln((b*x^2+a)^p)/a/x^3+4/5*p*b^2*ln((b*x^2+a)^p)/a^2/x-32 
/15*p^2*b^3/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))+4/5*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*_alpha/a*dilog(1/2*(x+_alpha)/_alpha) 
))*b/a^2/_alpha,_alpha=RootOf(_Z^2*b+a))-8/15*b^2*p^2/a^2/x+(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))*(-1/5/x^5*ln((b*x^2+a)^p)+2/5*p*b*(b^2/a^2/(a*b)^(1/ 
2)*arctan(b*x/(a*b)^(1/2))-1/3/a/x^3+b/a^2/x))-1/20*(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^5
 

Fricas [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{6}} \,d x } \] Input:

integrate(log(c*(b*x^2+a)^p)^2/x^6,x, algorithm="fricas")
 

Output:

integral(log((b*x^2 + a)^p*c)^2/x^6, x)
 

Sympy [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{x^{6}}\, dx \] Input:

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

Output:

Integral(log(c*(a + b*x**2)**p)**2/x**6, x)
 

Maxima [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{6}} \,d x } \] Input:

integrate(log(c*(b*x^2+a)^p)^2/x^6,x, algorithm="maxima")
 

Output:

-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)
 

Giac [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{6}} \,d x } \] Input:

integrate(log(c*(b*x^2+a)^p)^2/x^6,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2}{x^6} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\frac {24 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} p^{2} x^{5}-60 \left (\int \frac {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}{b \,x^{8}+a \,x^{6}}d x \right ) a^{4} p \,x^{5}-15 {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} a^{3}-12 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a^{3} p -8 a^{2} b \,p^{2} x^{2}+24 a \,b^{2} p^{2} x^{4}}{75 a^{3} x^{5}} \] Input:

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

Output:

(24*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**2*p**2*x**5 - 60*int( 
log((a + b*x**2)**p*c)/(a*x**6 + b*x**8),x)*a**4*p*x**5 - 15*log((a + b*x* 
*2)**p*c)**2*a**3 - 12*log((a + b*x**2)**p*c)*a**3*p - 8*a**2*b*p**2*x**2 
+ 24*a*b**2*p**2*x**4)/(75*a**3*x**5)