Integrand size = 16, antiderivative size = 78 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=-\frac {b p}{12 a x^4}+\frac {b^2 p}{6 a^2 x^2}+\frac {b^3 p \log (x)}{3 a^3}-\frac {b^3 p \log \left (a+b x^2\right )}{6 a^3}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6} \] Output:
-1/12*b*p/a/x^4+1/6*b^2*p/a^2/x^2+1/3*b^3*p*ln(x)/a^3-1/6*b^3*p*ln(b*x^2+a )/a^3-1/6*ln(c*(b*x^2+a)^p)/x^6
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=\frac {1}{3} b p \left (-\frac {1}{4 a x^4}+\frac {b}{2 a^2 x^2}+\frac {b^2 \log (x)}{a^3}-\frac {b^2 \log \left (a+b x^2\right )}{2 a^3}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6} \] Input:
Integrate[Log[c*(a + b*x^2)^p]/x^7,x]
Output:
(b*p*(-1/4*1/(a*x^4) + b/(2*a^2*x^2) + (b^2*Log[x])/a^3 - (b^2*Log[a + b*x ^2])/(2*a^3)))/3 - Log[c*(a + b*x^2)^p]/(6*x^6)
Time = 0.43 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2904, 2842, 54, 2009}
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.
\(\displaystyle \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{2} \int \frac {\log \left (c \left (b x^2+a\right )^p\right )}{x^8}dx^2\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} b p \int \frac {1}{x^6 \left (b x^2+a\right )}dx^2-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 x^6}\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} b p \int \left (-\frac {b^3}{a^3 \left (b x^2+a\right )}+\frac {b^2}{a^3 x^2}-\frac {b}{a^2 x^4}+\frac {1}{a x^6}\right )dx^2-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 x^6}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} b p \left (\frac {b^2 \log \left (x^2\right )}{a^3}-\frac {b^2 \log \left (a+b x^2\right )}{a^3}+\frac {b}{a^2 x^2}-\frac {1}{2 a x^4}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{3 x^6}\right )\) |
Input:
Int[Log[c*(a + b*x^2)^p]/x^7,x]
Output:
((b*p*(-1/2*1/(a*x^4) + b/(a^2*x^2) + (b^2*Log[x^2])/a^3 - (b^2*Log[a + b* x^2])/a^3))/3 - Log[c*(a + b*x^2)^p]/(3*x^6))/2
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
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 = 1.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85
method | result | size |
parts | \(-\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{6 x^{6}}+\frac {p b \left (-\frac {b^{2} \ln \left (b \,x^{2}+a \right )}{2 a^{3}}-\frac {1}{4 a \,x^{4}}+\frac {b^{2} \ln \left (x \right )}{a^{3}}+\frac {b}{2 a^{2} x^{2}}\right )}{3}\) | \(66\) |
parallelrisch | \(\frac {4 b^{3} p^{2} \ln \left (x \right ) x^{6}-2 x^{6} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b^{3} p -2 x^{6} b^{3} p^{2}+2 x^{4} a \,b^{2} p^{2}-x^{2} a^{2} b \,p^{2}-2 \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{3} p}{12 x^{6} p \,a^{3}}\) | \(100\) |
risch | \(-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{6 x^{6}}-\frac {2 b^{3} p \ln \left (b \,x^{2}+a \right ) x^{6}-4 b^{3} p \ln \left (x \right ) x^{6}+i \pi \,a^{3} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,a^{3} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi \,a^{3} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi \,a^{3} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-2 a \,b^{2} p \,x^{4}+a^{2} b p \,x^{2}+2 \ln \left (c \right ) a^{3}}{12 a^{3} x^{6}}\) | \(206\) |
Input:
int(ln(c*(b*x^2+a)^p)/x^7,x,method=_RETURNVERBOSE)
Output:
-1/6*ln(c*(b*x^2+a)^p)/x^6+1/3*p*b*(-1/2*b^2/a^3*ln(b*x^2+a)-1/4/a/x^4+b^2 /a^3*ln(x)+1/2*b/a^2/x^2)
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=\frac {4 \, b^{3} p x^{6} \log \left (x\right ) + 2 \, a b^{2} p x^{4} - a^{2} b p x^{2} - 2 \, a^{3} \log \left (c\right ) - 2 \, {\left (b^{3} p x^{6} + a^{3} p\right )} \log \left (b x^{2} + a\right )}{12 \, a^{3} x^{6}} \] Input:
integrate(log(c*(b*x^2+a)^p)/x^7,x, algorithm="fricas")
Output:
1/12*(4*b^3*p*x^6*log(x) + 2*a*b^2*p*x^4 - a^2*b*p*x^2 - 2*a^3*log(c) - 2* (b^3*p*x^6 + a^3*p)*log(b*x^2 + a))/(a^3*x^6)
Time = 7.01 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.24 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=\begin {cases} - \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{6 x^{6}} - \frac {b p}{12 a x^{4}} + \frac {b^{2} p}{6 a^{2} x^{2}} + \frac {b^{3} p \log {\left (x \right )}}{3 a^{3}} - \frac {b^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{6 a^{3}} & \text {for}\: a \neq 0 \\- \frac {p}{18 x^{6}} - \frac {\log {\left (c \left (b x^{2}\right )^{p} \right )}}{6 x^{6}} & \text {otherwise} \end {cases} \] Input:
integrate(ln(c*(b*x**2+a)**p)/x**7,x)
Output:
Piecewise((-log(c*(a + b*x**2)**p)/(6*x**6) - b*p/(12*a*x**4) + b**2*p/(6* a**2*x**2) + b**3*p*log(x)/(3*a**3) - b**3*log(c*(a + b*x**2)**p)/(6*a**3) , Ne(a, 0)), (-p/(18*x**6) - log(c*(b*x**2)**p)/(6*x**6), True))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=-\frac {1}{12} \, b p {\left (\frac {2 \, b^{2} \log \left (b x^{2} + a\right )}{a^{3}} - \frac {2 \, b^{2} \log \left (x^{2}\right )}{a^{3}} - \frac {2 \, b x^{2} - a}{a^{2} x^{4}}\right )} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{6 \, x^{6}} \] Input:
integrate(log(c*(b*x^2+a)^p)/x^7,x, algorithm="maxima")
Output:
-1/12*b*p*(2*b^2*log(b*x^2 + a)/a^3 - 2*b^2*log(x^2)/a^3 - (2*b*x^2 - a)/( a^2*x^4)) - 1/6*log((b*x^2 + a)^p*c)/x^6
Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (68) = 136\).
Time = 0.12 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.45 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=-\frac {\frac {2 \, b^{4} p \log \left (b x^{2} + a\right )}{{\left (b x^{2} + a\right )}^{3} - 3 \, {\left (b x^{2} + a\right )}^{2} a + 3 \, {\left (b x^{2} + a\right )} a^{2} - a^{3}} + \frac {2 \, b^{4} p \log \left (b x^{2} + a\right )}{a^{3}} - \frac {2 \, b^{4} p \log \left (b x^{2}\right )}{a^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{2} b^{4} p - 5 \, {\left (b x^{2} + a\right )} a b^{4} p + 3 \, a^{2} b^{4} p - 2 \, a^{2} b^{4} \log \left (c\right )}{{\left (b x^{2} + a\right )}^{3} a^{2} - 3 \, {\left (b x^{2} + a\right )}^{2} a^{3} + 3 \, {\left (b x^{2} + a\right )} a^{4} - a^{5}}}{12 \, b} \] Input:
integrate(log(c*(b*x^2+a)^p)/x^7,x, algorithm="giac")
Output:
-1/12*(2*b^4*p*log(b*x^2 + a)/((b*x^2 + a)^3 - 3*(b*x^2 + a)^2*a + 3*(b*x^ 2 + a)*a^2 - a^3) + 2*b^4*p*log(b*x^2 + a)/a^3 - 2*b^4*p*log(b*x^2)/a^3 - (2*(b*x^2 + a)^2*b^4*p - 5*(b*x^2 + a)*a*b^4*p + 3*a^2*b^4*p - 2*a^2*b^4*l og(c))/((b*x^2 + a)^3*a^2 - 3*(b*x^2 + a)^2*a^3 + 3*(b*x^2 + a)*a^4 - a^5) )/b
Time = 26.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=\frac {b^2\,p}{6\,a^2\,x^2}-\frac {b^3\,p\,\ln \left (b\,x^2+a\right )}{6\,a^3}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{6\,x^6}+\frac {b^3\,p\,\ln \left (x\right )}{3\,a^3}-\frac {b\,p}{12\,a\,x^4} \] Input:
int(log(c*(a + b*x^2)^p)/x^7,x)
Output:
(b^2*p)/(6*a^2*x^2) - (b^3*p*log(a + b*x^2))/(6*a^3) - log(c*(a + b*x^2)^p )/(6*x^6) + (b^3*p*log(x))/(3*a^3) - (b*p)/(12*a*x^4)
Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.99 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=\frac {-2 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a^{3}-2 \,\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b^{3} x^{6}+4 \,\mathrm {log}\left (x \right ) b^{3} p \,x^{6}-a^{2} b p \,x^{2}+2 a \,b^{2} p \,x^{4}}{12 a^{3} x^{6}} \] Input:
int(log(c*(b*x^2+a)^p)/x^7,x)
Output:
( - 2*log((a + b*x**2)**p*c)*a**3 - 2*log((a + b*x**2)**p*c)*b**3*x**6 + 4 *log(x)*b**3*p*x**6 - a**2*b*p*x**2 + 2*a*b**2*p*x**4)/(12*a**3*x**6)