\(\int \frac {\log (c (a+b x^2)^p)}{x^3} \, dx\) [8]

Optimal result

Integrand size = 16, antiderivative size = 38 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {b p \log (x)}{a}-\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 a x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2504, 2442, 36, 29, 31} \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 x^2}-\frac {b p \log \left (a+b x^2\right )}{2 a}+\frac {b p \log (x)}{a} \]

[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 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 2442

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 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 \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) \text {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) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 a}-\frac {\left (b^2 p\right ) \text {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] (verified)

Time = 0.00 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.18 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {b p \log (x)}{a}-\frac {b p \log \left (a+b x^2\right )}{2 a}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 x^2} \]

[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 [A] (verified)

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11

method	result	size
parts	\(-\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{2 x^{2}}+p b \left (\frac {\ln \left (x \right )}{a}-\frac {\ln \left (b \,x^{2}+a \right )}{2 a}\right )\)	\(42\)
parallelrisch	\(\frac {2 p^{2} b \ln \left (x \right ) x^{2}-x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b p -\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a p}{2 x^{2} a p}\)	\(59\)
risch	\(-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{2 x^{2}}-\frac {i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-4 b p \ln \left (x \right ) x^{2}+2 p b \ln \left (b \,x^{2}+a \right ) x^{2}+2 \ln \left (c \right ) a}{4 x^{2} a}\)	\(173\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.13 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {2 \, b p x^{2} \log \left (x\right ) - {\left (b p x^{2} + a p\right )} \log \left (b x^{2} + a\right ) - a \log \left (c\right )}{2 \, a x^{2}} \]

[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] (verification not implemented)

Time = 0.90 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.71 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\begin {cases} - \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2 x^{2}} + \frac {b p \log {\left (x \right )}}{a} - \frac {b \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2 a} & \text {for}\: a \neq 0 \\- \frac {p}{2 x^{2}} - \frac {\log {\left (c \left (b x^{2}\right )^{p} \right )}}{2 x^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=-\frac {1}{2} \, b p {\left (\frac {\log \left (b x^{2} + a\right )}{a} - \frac {\log \left (x^{2}\right )}{a}\right )} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{2 \, x^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.53 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=-\frac {\frac {b^{2} p \log \left (b x^{2} + a\right )}{a} - \frac {b^{2} p \log \left (b x^{2}\right )}{a} + \frac {b p \log \left (b x^{2} + a\right )}{x^{2}} + \frac {b \log \left (c\right )}{x^{2}}}{2 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {b\,p\,\ln \left (x\right )}{a}-\frac {b\,p\,\ln \left (b\,x^2+a\right )}{2\,a}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2\,x^2} \]

[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)