Integrand size = 16, antiderivative size = 38 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {b p \log (x)}{a}-\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 a x^2} \] Output:
b*p*ln(x)/a-1/2*(b*x^2+a)*ln(c*(b*x^2+a)^p)/a/x^2
Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.18 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {b p \log (x)}{a}-\frac {b p \log \left (a+b x^2\right )}{2 a}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{2 x^2} \] Input:
Integrate[Log[c*(a + b*x^2)^p]/x^3,x]
Output:
(b*p*Log[x])/a - (b*p*Log[a + b*x^2])/(2*a) - Log[c*(a + b*x^2)^p]/(2*x^2)
Time = 0.35 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2904, 2842, 47, 14, 16}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{2} \int \frac {\log \left (c \left (b x^2+a\right )^p\right )}{x^4}dx^2\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{2} \left (b p \int \frac {1}{x^2 \left (b x^2+a\right )}dx^2-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^2}\right )\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {1}{2} \left (b p \left (\frac {\int \frac {1}{x^2}dx^2}{a}-\frac {b \int \frac {1}{b x^2+a}dx^2}{a}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^2}\right )\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {1}{2} \left (b p \left (\frac {\log \left (x^2\right )}{a}-\frac {b \int \frac {1}{b x^2+a}dx^2}{a}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^2}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (b p \left (\frac {\log \left (x^2\right )}{a}-\frac {\log \left (a+b x^2\right )}{a}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^2}\right )\) |
Input:
Int[Log[c*(a + b*x^2)^p]/x^3,x]
Output:
(b*p*(Log[x^2]/a - Log[a + b*x^2]/a) - Log[c*(a + b*x^2)^p]/x^2)/2
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
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] - Simp[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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[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])
Time = 0.44 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11
method | result | size |
parts | \(-\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{2 x^{2}}+p b \left (-\frac {\ln \left (b \,x^{2}+a \right )}{2 a}+\frac {\ln \left (x \right )}{a}\right )\) | \(42\) |
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 | \(-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{2 x^{2}}-\frac {i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-4 b p \ln \left (x \right ) x^{2}+2 p b \ln \left (b \,x^{2}+a \right ) x^{2}+2 \ln \left (c \right ) a}{4 a \,x^{2}}\) | \(173\) |
Input:
int(ln(c*(b*x^2+a)^p)/x^3,x,method=_RETURNVERBOSE)
Output:
-1/2*ln(c*(b*x^2+a)^p)/x^2+p*b*(-1/2/a*ln(b*x^2+a)+ln(x)/a)
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.13 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {2 \, b p x^{2} \log \left (x\right ) - {\left (b p x^{2} + a p\right )} \log \left (b x^{2} + a\right ) - a \log \left (c\right )}{2 \, a x^{2}} \] Input:
integrate(log(c*(b*x^2+a)^p)/x^3,x, algorithm="fricas")
Output:
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)
Time = 0.93 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.71 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\begin {cases} - \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2 x^{2}} + \frac {b p \log {\left (x \right )}}{a} - \frac {b \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2 a} & \text {for}\: a \neq 0 \\- \frac {p}{2 x^{2}} - \frac {\log {\left (c \left (b x^{2}\right )^{p} \right )}}{2 x^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(ln(c*(b*x**2+a)**p)/x**3,x)
Output:
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))
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=-\frac {1}{2} \, b p {\left (\frac {\log \left (b x^{2} + a\right )}{a} - \frac {\log \left (x^{2}\right )}{a}\right )} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{2 \, x^{2}} \] Input:
integrate(log(c*(b*x^2+a)^p)/x^3,x, algorithm="maxima")
Output:
-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
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.53 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=-\frac {\frac {b^{2} p \log \left (b x^{2} + a\right )}{a} - \frac {b^{2} p \log \left (b x^{2}\right )}{a} + \frac {b p \log \left (b x^{2} + a\right )}{x^{2}} + \frac {b \log \left (c\right )}{x^{2}}}{2 \, b} \] Input:
integrate(log(c*(b*x^2+a)^p)/x^3,x, algorithm="giac")
Output:
-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
Time = 26.35 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {b\,p\,\ln \left (x\right )}{a}-\frac {b\,p\,\ln \left (b\,x^2+a\right )}{2\,a}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2\,x^2} \] Input:
int(log(c*(a + b*x^2)^p)/x^3,x)
Output:
(b*p*log(x))/a - (b*p*log(a + b*x^2))/(2*a) - log(c*(a + b*x^2)^p)/(2*x^2)
Time = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.34 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^3} \, dx=\frac {-\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a -\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b \,x^{2}+2 \,\mathrm {log}\left (x \right ) b p \,x^{2}}{2 a \,x^{2}} \] Input:
int(log(c*(b*x^2+a)^p)/x^3,x)
Output:
( - log((a + b*x**2)**p*c)*a - log((a + b*x**2)**p*c)*b*x**2 + 2*log(x)*b* p*x**2)/(2*a*x**2)