3.328 \(\int \frac {(f+g x^2)^2 \log (c (d+e x^2)^p)}{x^5} \, dx\)

Optimal. Leaf size=172 \[ -\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 x^4}-\frac {f g \log \left (c \left (d+e x^2\right )^p\right )}{x^2}+\frac {1}{2} g^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {e^2 f^2 p \log \left (d+e x^2\right )}{4 d^2}-\frac {e^2 f^2 p \log (x)}{2 d^2}-\frac {e f^2 p}{4 d x^2}-\frac {e f g p \log \left (d+e x^2\right )}{d}+\frac {2 e f g p \log (x)}{d}+\frac {1}{2} g^2 p \text {Li}_2\left (\frac {e x^2}{d}+1\right ) \]

[Out]

-1/4*e*f^2*p/d/x^2-1/2*e^2*f^2*p*ln(x)/d^2+2*e*f*g*p*ln(x)/d+1/4*e^2*f^2*p*ln(e*x^2+d)/d^2-e*f*g*p*ln(e*x^2+d)
/d-1/4*f^2*ln(c*(e*x^2+d)^p)/x^4-f*g*ln(c*(e*x^2+d)^p)/x^2+1/2*g^2*ln(-e*x^2/d)*ln(c*(e*x^2+d)^p)+1/2*g^2*p*po
lylog(2,1+e*x^2/d)

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Rubi [A]  time = 0.22, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2475, 43, 2416, 2395, 44, 36, 29, 31, 2394, 2315} \[ \frac {1}{2} g^2 p \text {PolyLog}\left (2,\frac {e x^2}{d}+1\right )-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 x^4}-\frac {f g \log \left (c \left (d+e x^2\right )^p\right )}{x^2}+\frac {1}{2} g^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {e^2 f^2 p \log \left (d+e x^2\right )}{4 d^2}-\frac {e^2 f^2 p \log (x)}{2 d^2}-\frac {e f^2 p}{4 d x^2}-\frac {e f g p \log \left (d+e x^2\right )}{d}+\frac {2 e f g p \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^5,x]

[Out]

-(e*f^2*p)/(4*d*x^2) - (e^2*f^2*p*Log[x])/(2*d^2) + (2*e*f*g*p*Log[x])/d + (e^2*f^2*p*Log[d + e*x^2])/(4*d^2)
- (e*f*g*p*Log[d + e*x^2])/d - (f^2*Log[c*(d + e*x^2)^p])/(4*x^4) - (f*g*Log[c*(d + e*x^2)^p])/x^2 + (g^2*Log[
-((e*x^2)/d)]*Log[c*(d + e*x^2)^p])/2 + (g^2*p*PolyLog[2, 1 + (e*x^2)/d])/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2394

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] - Dist[(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 2395

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

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rubi steps

\begin {align*} \int \frac {\left (f+g x^2\right )^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^5} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(f+g x)^2 \log \left (c (d+e x)^p\right )}{x^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {f^2 \log \left (c (d+e x)^p\right )}{x^3}+\frac {2 f g \log \left (c (d+e x)^p\right )}{x^2}+\frac {g^2 \log \left (c (d+e x)^p\right )}{x}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{2} f^2 \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x^3} \, dx,x,x^2\right )+(f g) \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x^2} \, dx,x,x^2\right )+\frac {1}{2} g^2 \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^2\right )\\ &=-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 x^4}-\frac {f g \log \left (c \left (d+e x^2\right )^p\right )}{x^2}+\frac {1}{2} g^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{4} \left (e f^2 p\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 (d+e x)} \, dx,x,x^2\right )+(e f g p) \operatorname {Subst}\left (\int \frac {1}{x (d+e x)} \, dx,x,x^2\right )-\frac {1}{2} \left (e g^2 p\right ) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^2\right )\\ &=-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 x^4}-\frac {f g \log \left (c \left (d+e x^2\right )^p\right )}{x^2}+\frac {1}{2} g^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} g^2 p \text {Li}_2\left (1+\frac {e x^2}{d}\right )+\frac {1}{4} \left (e f^2 p\right ) \operatorname {Subst}\left (\int \left (\frac {1}{d x^2}-\frac {e}{d^2 x}+\frac {e^2}{d^2 (d+e x)}\right ) \, dx,x,x^2\right )+\frac {(e f g p) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{d}-\frac {\left (e^2 f g p\right ) \operatorname {Subst}\left (\int \frac {1}{d+e x} \, dx,x,x^2\right )}{d}\\ &=-\frac {e f^2 p}{4 d x^2}-\frac {e^2 f^2 p \log (x)}{2 d^2}+\frac {2 e f g p \log (x)}{d}+\frac {e^2 f^2 p \log \left (d+e x^2\right )}{4 d^2}-\frac {e f g p \log \left (d+e x^2\right )}{d}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{4 x^4}-\frac {f g \log \left (c \left (d+e x^2\right )^p\right )}{x^2}+\frac {1}{2} g^2 \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} g^2 p \text {Li}_2\left (1+\frac {e x^2}{d}\right )\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 148, normalized size = 0.86 \[ \frac {1}{4} \left (-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^4}-\frac {4 f g \log \left (c \left (d+e x^2\right )^p\right )}{x^2}+2 g^2 \left (\log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+p \text {Li}_2\left (\frac {e x^2}{d}+1\right )\right )-\frac {e f^2 p \left (-e x^2 \log \left (d+e x^2\right )+d+2 e x^2 \log (x)\right )}{d^2 x^2}+\frac {4 e f g p \left (2 \log (x)-\log \left (d+e x^2\right )\right )}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x^2)^2*Log[c*(d + e*x^2)^p])/x^5,x]

[Out]

((4*e*f*g*p*(2*Log[x] - Log[d + e*x^2]))/d - (e*f^2*p*(d + 2*e*x^2*Log[x] - e*x^2*Log[d + e*x^2]))/(d^2*x^2) -
(f^2*Log[c*(d + e*x^2)^p])/x^4 - (4*f*g*Log[c*(d + e*x^2)^p])/x^2 + 2*g^2*(Log[-((e*x^2)/d)]*Log[c*(d + e*x^2
)^p] + p*PolyLog[2, 1 + (e*x^2)/d]))/4

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (g^{2} x^{4} + 2 \, f g x^{2} + f^{2}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^5,x, algorithm="fricas")

[Out]

integral((g^2*x^4 + 2*f*g*x^2 + f^2)*log((e*x^2 + d)^p*c)/x^5, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (g x^{2} + f\right )}^{2} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^5,x, algorithm="giac")

[Out]

integrate((g*x^2 + f)^2*log((e*x^2 + d)^p*c)/x^5, x)

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maple [C]  time = 0.26, size = 663, normalized size = 3.85 \[ -g^{2} p \dilog \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )-g^{2} p \dilog \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )-\frac {f g \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{x^{2}}+g^{2} \ln \relax (c ) \ln \relax (x )-\frac {f^{2} \ln \relax (c )}{4 x^{4}}+g^{2} \ln \relax (x ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )-\frac {f^{2} \ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{4 x^{4}}+\frac {i \pi f g \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}{2 x^{2}}-g^{2} p \ln \relax (x ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )-g^{2} p \ln \relax (x ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )-\frac {f g \ln \relax (c )}{x^{2}}+\frac {i \pi \,f^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{8 x^{4}}-\frac {i \pi \,g^{2} \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3} \ln \relax (x )}{2}+\frac {i \pi \,g^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \ln \relax (x )}{2}+\frac {i \pi \,g^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2} \ln \relax (x )}{2}+\frac {i \pi f g \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{3}}{2 x^{2}}-\frac {i \pi \,f^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{8 x^{4}}-\frac {e^{2} f^{2} p \ln \relax (x )}{2 d^{2}}+\frac {e^{2} f^{2} p \ln \left (e \,x^{2}+d \right )}{4 d^{2}}-\frac {i \pi \,f^{2} \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{8 x^{4}}+\frac {2 e f g p \ln \relax (x )}{d}-\frac {e f g p \ln \left (e \,x^{2}+d \right )}{d}-\frac {e \,f^{2} p}{4 d \,x^{2}}-\frac {i \pi \,g^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \ln \relax (x )}{2}-\frac {i \pi f g \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{2 x^{2}}-\frac {i \pi f g \,\mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )^{2}}{2 x^{2}}+\frac {i \pi \,f^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}{8 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x^2+f)^2*ln(c*(e*x^2+d)^p)/x^5,x)

[Out]

-ln((e*x^2+d)^p)*f*g/x^2-p*g^2*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-p*g^2*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(
1/2))+ln(c)*g^2*ln(x)-1/4*ln(c)*f^2/x^4+1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*f*g/x^2+l
n((e*x^2+d)^p)*g^2*ln(x)-1/4*ln((e*x^2+d)^p)*f^2/x^4-1/8*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*f^2/x^4+1/2*I*
Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*g^2*ln(x)-1/8*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*
f^2/x^4+1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*f*g/x^2+1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*g^2*ln(x)-p*g^2*ln
(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-p*g^2*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-ln(c)*f*g/x^2-1/2*I*P
i*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*g^2*ln(x)-1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d
)^p)^2*f*g/x^2-1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*f*g/x^2+1/8*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2
+d)^p)*csgn(I*c)*f^2/x^4+1/8*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*f^2/x^4-1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*g^2*ln(x)-1
/2*e^2*f^2*p*ln(x)/d^2+1/4*e^2*f^2*p*ln(e*x^2+d)/d^2+2*e*f*g*p*ln(x)/d-e*f*g*p*ln(e*x^2+d)/d-1/4*e*f^2*p/d/x^2

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maxima [A]  time = 1.30, size = 166, normalized size = 0.97 \[ \frac {1}{2} \, {\left (\log \left (e x^{2} + d\right ) \log \left (-\frac {e x^{2} + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x^{2} + d}{d}\right )\right )} g^{2} p - \frac {{\left (e^{2} f^{2} p - 4 \, d e f g p - 2 \, d^{2} g^{2} \log \relax (c)\right )} \log \relax (x)}{2 \, d^{2}} - \frac {d^{2} f^{2} \log \relax (c) + {\left (d e f^{2} p + 4 \, d^{2} f g \log \relax (c)\right )} x^{2} + {\left (4 \, d^{2} f g p x^{2} + d^{2} f^{2} p - {\left (e^{2} f^{2} p - 4 \, d e f g p\right )} x^{4}\right )} \log \left (e x^{2} + d\right )}{4 \, d^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^2+f)^2*log(c*(e*x^2+d)^p)/x^5,x, algorithm="maxima")

[Out]

1/2*(log(e*x^2 + d)*log(-(e*x^2 + d)/d + 1) + dilog((e*x^2 + d)/d))*g^2*p - 1/2*(e^2*f^2*p - 4*d*e*f*g*p - 2*d
^2*g^2*log(c))*log(x)/d^2 - 1/4*(d^2*f^2*log(c) + (d*e*f^2*p + 4*d^2*f*g*log(c))*x^2 + (4*d^2*f*g*p*x^2 + d^2*
f^2*p - (e^2*f^2*p - 4*d*e*f*g*p)*x^4)*log(e*x^2 + d))/(d^2*x^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,{\left (g\,x^2+f\right )}^2}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(c*(d + e*x^2)^p)*(f + g*x^2)^2)/x^5,x)

[Out]

int((log(c*(d + e*x^2)^p)*(f + g*x^2)^2)/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (f + g x^{2}\right )^{2} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x**2+f)**2*ln(c*(e*x**2+d)**p)/x**5,x)

[Out]

Integral((f + g*x**2)**2*log(c*(d + e*x**2)**p)/x**5, x)

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