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

Optimal. Leaf size=38 \[ \frac{b p \log (x)}{a}-\frac{\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 a x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.037124, antiderivative size = 45, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {2454, 2395, 36, 29, 31} \[ -\frac{\log \left (c \left (a+b x^2\right )^p\right )}{2 x^2}-\frac{b p \log \left (a+b x^2\right )}{2 a}+\frac{b p \log (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]/x^3,x]

[Out]

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

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0026732, size = 45, normalized size = 1.18 \[ -\frac{\log \left (c \left (a+b x^2\right )^p\right )}{2 x^2}-\frac{b p \log \left (a+b x^2\right )}{2 a}+\frac{b p \log (x)}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]/x^3,x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.279, size = 173, normalized size = 4.6 \begin{align*} -{\frac{\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{2\,{x}^{2}}}-{\frac{i\pi \,a{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}-i\pi \,a{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) -i\pi \,a \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+i\pi \,a \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -4\,bp\ln \left ( x \right ){x}^{2}+2\,bp\ln \left ( b{x}^{2}+a \right ){x}^{2}+2\,\ln \left ( c \right ) a}{4\,a{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x^2*ln((b*x^2+a)^p)-1/4*(I*Pi*a*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I*Pi*a*csgn(I*(b*x^2+a)^p)*cs
gn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*a*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*a*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-4*b*p*ln
(x)*x^2+2*b*p*ln(b*x^2+a)*x^2+2*ln(c)*a)/a/x^2

________________________________________________________________________________________

Maxima [A]  time = 1.12891, size = 59, normalized size = 1.55 \begin{align*} -\frac{1}{2} \, b p{\left (\frac{\log \left (b x^{2} + a\right )}{a} - \frac{\log \left (x^{2}\right )}{a}\right )} - \frac{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.02935, size = 103, normalized size = 2.71 \begin{align*} \frac{2 \, b p x^{2} \log \left (x\right ) -{\left (b p x^{2} + a p\right )} \log \left (b x^{2} + a\right ) - a \log \left (c\right )}{2 \, a x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 6.96032, size = 82, normalized size = 2.16 \begin{align*} \begin{cases} - \frac{p \log{\left (a + b x^{2} \right )}}{2 x^{2}} - \frac{\log{\left (c \right )}}{2 x^{2}} + \frac{b p \log{\left (x \right )}}{a} - \frac{b p \log{\left (a + b x^{2} \right )}}{2 a} & \text{for}\: a \neq 0 \\- \frac{p \log{\left (b \right )}}{2 x^{2}} - \frac{p \log{\left (x \right )}}{x^{2}} - \frac{p}{2 x^{2}} - \frac{\log{\left (c \right )}}{2 x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-p*log(a + b*x**2)/(2*x**2) - log(c)/(2*x**2) + b*p*log(x)/a - b*p*log(a + b*x**2)/(2*a), Ne(a, 0))
, (-p*log(b)/(2*x**2) - p*log(x)/x**2 - p/(2*x**2) - log(c)/(2*x**2), True))

________________________________________________________________________________________

Giac [A]  time = 1.23713, size = 78, normalized size = 2.05 \begin{align*} -\frac{\frac{b^{2} p \log \left (b x^{2} + a\right )}{a} - \frac{b^{2} p \log \left (b x^{2}\right )}{a} + \frac{b p \log \left (b x^{2} + a\right )}{x^{2}} + \frac{b \log \left (c\right )}{x^{2}}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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