\(\int \frac {\log ^2(c (a+b x^2)^p)}{x^3} \, dx\) [81]

Optimal result

Integrand size = 18, antiderivative size = 80 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {b p \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{a}-\frac {\left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 a x^2}+\frac {b p^2 \operatorname {PolyLog}\left (2,1+\frac {b x^2}{a}\right )}{a} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.16 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {b p \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{a}-\frac {b \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 a}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{2 x^2}+\frac {b p^2 \operatorname {PolyLog}\left (2,\frac {a+b x^2}{a}\right )}{a} \] Input:

Integrate[Log[c*(a + b*x^2)^p]^2/x^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2904, 2844, 2841, 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.

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

\(\Big \downarrow \) 2904

\(\displaystyle \frac {1}{2} \int \frac {\log ^2\left (c \left (b x^2+a\right )^p\right )}{x^4}dx^2\)

\(\Big \downarrow \) 2844

\(\displaystyle \frac {1}{2} \left (\frac {2 b p \int \frac {\log \left (c \left (b x^2+a\right )^p\right )}{x^2}dx^2}{a}-\frac {\left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{a x^2}\right )\)

\(\Big \downarrow \) 2841

\(\displaystyle \frac {1}{2} \left (\frac {2 b p \left (\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )-b p \int \frac {\log \left (-\frac {b x^2}{a}\right )}{b x^2+a}dx^2\right )}{a}-\frac {\left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{a x^2}\right )\)

\(\Big \downarrow \) 2752

\(\displaystyle \frac {1}{2} \left (\frac {2 b p \left (\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+p \operatorname {PolyLog}\left (2,\frac {b x^2}{a}+1\right )\right )}{a}-\frac {\left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{a x^2}\right )\)

Input:

Int[Log[c*(a + b*x^2)^p]^2/x^3,x]
 

Output:

(-(((a + b*x^2)*Log[c*(a + b*x^2)^p]^2)/(a*x^2)) + (2*b*p*(Log[-((b*x^2)/a 
)]*Log[c*(a + b*x^2)^p] + p*PolyLog[2, 1 + (b*x^2)/a]))/a)/2
 

Defintions of rubi rules used

rule 2752

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]
 

rule 2841

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_ 
)), x_Symbol] :> Simp[Log[e*((f + g*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x 
)^n])/g), x] - Simp[b*e*(n/g)   Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d + e*x 
), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]
 

rule 2844

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_. 
)*(x_))^2, x_Symbol] :> Simp[(d + e*x)*((a + b*Log[c*(d + e*x)^n])^p/((e*f 
- d*g)*(f + g*x))), x] - Simp[b*e*n*(p/(e*f - d*g))   Int[(a + b*Log[c*(d + 
 e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] & 
& NeQ[e*f - d*g, 0] && GtQ[p, 0]
 

rule 2904

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L 
og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, 
 x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & 
&  !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
 

Maple [C] (warning: unable to verify)

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

Time = 0.60 (sec) , antiderivative size = 481, normalized size of antiderivative = 6.01

method	result	size
risch	\(-\frac {{\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}^{2}}{2 x^{2}}-\frac {p b \ln \left (\left (b \,x^{2}+a \right )^{p}\right ) \ln \left (b \,x^{2}+a \right )}{a}+\frac {2 p b \ln \left (\left (b \,x^{2}+a \right )^{p}\right ) \ln \left (x \right )}{a}-\frac {2 p^{2} b \ln \left (x \right ) \ln \left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{a}-\frac {2 p^{2} b \ln \left (x \right ) \ln \left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{a}-\frac {2 p^{2} b \operatorname {dilog}\left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{a}-\frac {2 p^{2} b \operatorname {dilog}\left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{a}+\frac {p^{2} b \ln \left (b \,x^{2}+a \right )^{2}}{2 a}+\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 )}{2 x^{2}}+p b \left (-\frac {\ln \left (b \,x^{2}+a \right )}{2 a}+\frac {\ln \left (x \right )}{a}\right )\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}}{8 x^{2}}\)	\(481\)

Input:

int(ln(c*(b*x^2+a)^p)^2/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*ln((b*x^2+a)^p)^2/x^2-p*b*ln((b*x^2+a)^p)/a*ln(b*x^2+a)+2*p*b*ln((b*x 
^2+a)^p)/a*ln(x)-2*p^2*b/a*ln(x)*ln((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))-2*p^ 
2*b/a*ln(x)*ln((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))-2*p^2*b/a*dilog((-b*x+(-a* 
b)^(1/2))/(-a*b)^(1/2))-2*p^2*b/a*dilog((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))+1 
/2*p^2*b/a*ln(b*x^2+a)^2+(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))*(-1/2/x^2*ln( 
(b*x^2+a)^p)+p*b*(-1/2/a*ln(b*x^2+a)+ln(x)/a))-1/8*(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^2
 

Fricas [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{3}} \,d x } \] Input:

integrate(log(c*(b*x^2+a)^p)^2/x^3,x, algorithm="fricas")
 

Output:

integral(log((b*x^2 + a)^p*c)^2/x^3, x)
 

Sympy [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{x^{3}}\, dx \] Input:

integrate(ln(c*(b*x**2+a)**p)**2/x**3,x)
 

Output:

Integral(log(c*(a + b*x**2)**p)**2/x**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.48 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {1}{2} \, b^{2} p^{2} {\left (\frac {\log \left (b x^{2} + a\right )^{2}}{a b} - \frac {2 \, {\left (2 \, \log \left (\frac {b x^{2}}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x^{2}}{a}\right )\right )}}{a b}\right )} - b p {\left (\frac {\log \left (b x^{2} + a\right )}{a} - \frac {\log \left (x^{2}\right )}{a}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{2 \, x^{2}} \] Input:

integrate(log(c*(b*x^2+a)^p)^2/x^3,x, algorithm="maxima")
 

Output:

1/2*b^2*p^2*(log(b*x^2 + a)^2/(a*b) - 2*(2*log(b*x^2/a + 1)*log(x) + dilog 
(-b*x^2/a))/(a*b)) - b*p*(log(b*x^2 + a)/a - log(x^2)/a)*log((b*x^2 + a)^p 
*c) - 1/2*log((b*x^2 + a)^p*c)^2/x^2
 

Giac [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{3}} \,d x } \] Input:

integrate(log(c*(b*x^2+a)^p)^2/x^3,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2}{x^3} \,d x \] Input:

int(log(c*(a + b*x^2)^p)^2/x^3,x)
 

Output:

int(log(c*(a + b*x^2)^p)^2/x^3, x)
 

Reduce [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {-4 \left (\int \frac {\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}{b \,x^{5}+a \,x^{3}}d x \right ) a^{2} p \,x^{2}-{\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )}^{2} a -2 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a p -2 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b p \,x^{2}+4 \,\mathrm {log}\left (x \right ) b \,p^{2} x^{2}}{2 a \,x^{2}} \] Input:

int(log(c*(b*x^2+a)^p)^2/x^3,x)
 

Output:

( - 4*int(log((a + b*x**2)**p*c)/(a*x**3 + b*x**5),x)*a**2*p*x**2 - log((a 
 + b*x**2)**p*c)**2*a - 2*log((a + b*x**2)**p*c)*a*p - 2*log((a + b*x**2)* 
*p*c)*b*p*x**2 + 4*log(x)*b*p**2*x**2)/(2*a*x**2)