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

Optimal. Leaf size=80 \[ \frac{b p^2 \text{PolyLog}\left (2,\frac{b x^2}{a}+1\right )}{a}-\frac{\left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 a x^2}+\frac{b p \log \left (-\frac{b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.0838589, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2454, 2397, 2394, 2315} \[ \frac{b p^2 \text{PolyLog}\left (2,\frac{b x^2}{a}+1\right )}{a}-\frac{\left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 a x^2}+\frac{b p \log \left (-\frac{b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]^2/x^3,x]

[Out]

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

Rule 2454

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])

Rule 2397

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] - Dist[(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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0239508, size = 93, normalized size = 1.16 \[ \frac{b p^2 \text{PolyLog}\left (2,\frac{a+b x^2}{a}\right )}{a}-\frac{b \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 a}-\frac{\log ^2\left (c \left (a+b x^2\right )^p\right )}{2 x^2}+\frac{b p \log \left (-\frac{b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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

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Maple [C]  time = 0.494, size = 841, normalized size = 10.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x^2*ln((b*x^2+a)^p)^2-b*p*ln((b*x^2+a)^p)/a*ln(b*x^2+a)+2*b*p*ln((b*x^2+a)^p)/a*ln(x)-2*b*p^2/a*ln(x)*ln(
(-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))-2*b*p^2/a*ln(x)*ln((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))-2*b*p^2/a*dilog((-b*x+(-
a*b)^(1/2))/(-a*b)^(1/2))-2*b*p^2/a*dilog((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))+1/2*b*p^2/a*ln(b*x^2+a)^2+1/2*I*b*p
/a*ln(b*x^2+a)*Pi*csgn(I*c*(b*x^2+a)^p)^3-1/2*I*b*p/a*ln(b*x^2+a)*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-I*b*p/a
*ln(x)*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)+1/2*I*b*p/a*ln(b*x^2+a)*Pi*csgn(I*(b*x^2+a)^p)*c
sgn(I*c*(b*x^2+a)^p)*csgn(I*c)-1/x^2*ln((b*x^2+a)^p)*ln(c)+1/2*I/x^2*ln((b*x^2+a)^p)*Pi*csgn(I*(b*x^2+a)^p)*cs
gn(I*c*(b*x^2+a)^p)*csgn(I*c)+I*b*p/a*ln(x)*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-1/2*I/x^2*ln((b*x^2
+a)^p)*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-I*b*p/a*ln(x)*Pi*csgn(I*c*(b*x^2+a)^p)^3-b*p/a*ln(b*x^2+a)*ln(c)-1
/2*I*b*p/a*ln(b*x^2+a)*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-1/2*I/x^2*ln((b*x^2+a)^p)*Pi*csgn(I*(b*x
^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2+I*b*p/a*ln(x)*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+1/2*I/x^2*ln((b*x^2+a)^p)*
Pi*csgn(I*c*(b*x^2+a)^p)^3+2*b*p/a*ln(x)*ln(c)-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

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Maxima [A]  time = 1.06904, size = 159, normalized size = 1.99 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log{\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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