\(\int \frac {\log ^2(c (a+b x^2)^p)}{x^5} \, dx\) [82]

Optimal result

Integrand size = 18, antiderivative size = 129 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=\frac {b^2 p^2 \log (x)}{a^2}-\frac {b p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 a^2 x^2}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{4 x^4}-\frac {b^2 p \log \left (c \left (a+b x^2\right )^p\right ) \log \left (1-\frac {a}{a+b x^2}\right )}{2 a^2}+\frac {b^2 p^2 \operatorname {PolyLog}\left (2,\frac {a}{a+b x^2}\right )}{2 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2504, 2445, 2458, 2389, 2379, 2438, 2351, 31} \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=-\frac {b^2 p \log \left (1-\frac {a}{a+b x^2}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 a^2}+\frac {b^2 p^2 \operatorname {PolyLog}\left (2,\frac {a}{b x^2+a}\right )}{2 a^2}+\frac {b^2 p^2 \log (x)}{a^2}-\frac {b p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 a^2 x^2}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{4 x^4} \]

[In]

Int[Log[c*(a + b*x^2)^p]^2/x^5,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2379

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[(-Log[1 +
d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)), x] + Dist[b*n*(p/(d*r)), Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^
(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]

Rule 2389

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[(d
+ e*x)^(q + 1)*((a + b*Log[c*x^n])^p/x), x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2445

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

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[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])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\log ^2\left (c (a+b x)^p\right )}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{4 x^4}+\frac {1}{2} (b p) \text {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x^2 (a+b x)} \, dx,x,x^2\right ) \\ & = -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{4 x^4}+\frac {1}{2} p \text {Subst}\left (\int \frac {\log \left (c x^p\right )}{x \left (-\frac {a}{b}+\frac {x}{b}\right )^2} \, dx,x,a+b x^2\right ) \\ & = -\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{4 x^4}+\frac {p \text {Subst}\left (\int \frac {\log \left (c x^p\right )}{\left (-\frac {a}{b}+\frac {x}{b}\right )^2} \, dx,x,a+b x^2\right )}{2 a}-\frac {(b p) \text {Subst}\left (\int \frac {\log \left (c x^p\right )}{x \left (-\frac {a}{b}+\frac {x}{b}\right )} \, dx,x,a+b x^2\right )}{2 a} \\ & = -\frac {b p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 a^2 x^2}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{4 x^4}-\frac {b^2 p \log \left (c \left (a+b x^2\right )^p\right ) \log \left (1-\frac {a}{a+b x^2}\right )}{2 a^2}+\frac {\left (b p^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x^2\right )}{2 a^2}+\frac {\left (b^2 p^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {a}{x}\right )}{x} \, dx,x,a+b x^2\right )}{2 a^2} \\ & = \frac {b^2 p^2 \log (x)}{a^2}-\frac {b p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 a^2 x^2}-\frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{4 x^4}-\frac {b^2 p \log \left (c \left (a+b x^2\right )^p\right ) \log \left (1-\frac {a}{a+b x^2}\right )}{2 a^2}+\frac {b^2 p^2 \text {Li}_2\left (\frac {a}{a+b x^2}\right )}{2 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.09 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=\frac {-\log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {b x^2 \left (4 b p^2 x^2 \log (x)-2 b p^2 x^2 \log \left (a+b x^2\right )-2 a p \log \left (c \left (a+b x^2\right )^p\right )+b x^2 \log ^2\left (c \left (a+b x^2\right )^p\right )-2 b p x^2 \left (\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+p \operatorname {PolyLog}\left (2,1+\frac {b x^2}{a}\right )\right )\right )}{a^2}}{4 x^4} \]

[In]

Integrate[Log[c*(a + b*x^2)^p]^2/x^5,x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.64 (sec) , antiderivative size = 554, normalized size of antiderivative = 4.29

method	result	size
risch	\(-\frac {{\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}^{2}}{4 x^{4}}-\frac {p b \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{2 a \,x^{2}}-\frac {p \,b^{2} \ln \left (\left (b \,x^{2}+a \right )^{p}\right ) \ln \left (x \right )}{a^{2}}+\frac {p \,b^{2} \ln \left (\left (b \,x^{2}+a \right )^{p}\right ) \ln \left (b \,x^{2}+a \right )}{2 a^{2}}-\frac {p^{2} b^{2} \ln \left (b \,x^{2}+a \right )^{2}}{4 a^{2}}+\frac {b^{2} p^{2} \ln \left (x \right )}{a^{2}}-\frac {p^{2} b^{2} \ln \left (b \,x^{2}+a \right )}{2 a^{2}}+\frac {p^{2} b^{2} \ln \left (x \right ) \ln \left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{a^{2}}+\frac {p^{2} b^{2} \ln \left (x \right ) \ln \left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{a^{2}}+\frac {p^{2} b^{2} \operatorname {dilog}\left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{a^{2}}+\frac {p^{2} b^{2} \operatorname {dilog}\left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{a^{2}}+\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 )}{4 x^{4}}+\frac {p b \left (-\frac {1}{2 a \,x^{2}}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {b \ln \left (b \,x^{2}+a \right )}{2 a^{2}}\right )}{2}\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}}{16 x^{4}}\)	\(554\)

[In]

int(ln(c*(b*x^2+a)^p)^2/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*ln((b*x^2+a)^p)^2/x^4-1/2*p*b*ln((b*x^2+a)^p)/a/x^2-p*b^2*ln((b*x^2+a)^p)/a^2*ln(x)+1/2*p*b^2*ln((b*x^2+a
)^p)/a^2*ln(b*x^2+a)-1/4*p^2*b^2/a^2*ln(b*x^2+a)^2+b^2*p^2*ln(x)/a^2-1/2*p^2*b^2/a^2*ln(b*x^2+a)+p^2*b^2/a^2*l
n(x)*ln((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))+p^2*b^2/a^2*ln(x)*ln((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))+p^2*b^2/a^2*di
log((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))+p^2*b^2/a^2*dilog((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))+(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))*(-1/4/x^4*ln((b*x^2+a)^p)+1/2*p*b*(-1/2/a/x^2-1/a^2*b*ln(
x)+1/2*b/a^2*ln(b*x^2+a)))-1/16*(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)*csg
n(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^
4

Fricas [F]

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

[In]

integrate(log(c*(b*x^2+a)^p)^2/x^5,x, algorithm="fricas")

[Out]

integral(log((b*x^2 + a)^p*c)^2/x^5, x)

Sympy [F]

\[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=\int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{x^{5}}\, dx \]

[In]

integrate(ln(c*(b*x**2+a)**p)**2/x**5,x)

[Out]

Integral(log(c*(a + b*x**2)**p)**2/x**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.10 \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=-\frac {1}{4} \, b^{2} p^{2} {\left (\frac {\log \left (b x^{2} + a\right )^{2}}{a^{2}} - \frac {2 \, {\left (2 \, \log \left (\frac {b x^{2}}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x^{2}}{a}\right )\right )}}{a^{2}} + \frac {2 \, \log \left (b x^{2} + a\right )}{a^{2}} - \frac {4 \, \log \left (x\right )}{a^{2}}\right )} + \frac {1}{2} \, b p {\left (\frac {b \log \left (b x^{2} + a\right )}{a^{2}} - \frac {b \log \left (x^{2}\right )}{a^{2}} - \frac {1}{a x^{2}}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{4 \, x^{4}} \]

[In]

integrate(log(c*(b*x^2+a)^p)^2/x^5,x, algorithm="maxima")

[Out]

-1/4*b^2*p^2*(log(b*x^2 + a)^2/a^2 - 2*(2*log(b*x^2/a + 1)*log(x) + dilog(-b*x^2/a))/a^2 + 2*log(b*x^2 + a)/a^
2 - 4*log(x)/a^2) + 1/2*b*p*(b*log(b*x^2 + a)/a^2 - b*log(x^2)/a^2 - 1/(a*x^2))*log((b*x^2 + a)^p*c) - 1/4*log
((b*x^2 + a)^p*c)^2/x^4

Giac [F]

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

[In]

integrate(log(c*(b*x^2+a)^p)^2/x^5,x, algorithm="giac")

[Out]

integrate(log((b*x^2 + a)^p*c)^2/x^5, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log ^2\left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx=\int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2}{x^5} \,d x \]

[In]

int(log(c*(a + b*x^2)^p)^2/x^5,x)

[Out]

int(log(c*(a + b*x^2)^p)^2/x^5, x)