Integrand size = 16, antiderivative size = 74 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=-\frac {2 b p}{15 a x^3}+\frac {2 b^2 p}{5 a^2 x}+\frac {2 b^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 a^{5/2}}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5} \]
-2/15*b*p/a/x^3+2/5*b^2*p/a^2/x+2/5*b^(5/2)*p*arctan(x*b^(1/2)/a^(1/2))/a^ (5/2)-1/5*ln(c*(b*x^2+a)^p)/x^5
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.00 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.66 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=-\frac {2 b p \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {b x^2}{a}\right )}{15 a x^3}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5} \]
(-2*b*p*Hypergeometric2F1[-3/2, 1, -1/2, -((b*x^2)/a)])/(15*a*x^3) - Log[c *(a + b*x^2)^p]/(5*x^5)
Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2905, 264, 264, 218}
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.
\(\displaystyle \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {2}{5} b p \int \frac {1}{x^4 \left (b x^2+a\right )}dx-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{5} b p \left (-\frac {b \int \frac {1}{x^2 \left (b x^2+a\right )}dx}{a}-\frac {1}{3 a x^3}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{5} b p \left (-\frac {b \left (-\frac {b \int \frac {1}{b x^2+a}dx}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2}{5} b p \left (-\frac {b \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{5 x^5}\) |
(2*b*p*(-1/3*1/(a*x^3) - (b*(-(1/(a*x)) - (Sqrt[b]*ArcTan[(Sqrt[b]*x)/Sqrt [a]])/a^(3/2)))/a))/5 - Log[c*(a + b*x^2)^p]/(5*x^5)
3.1.11.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^ (m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Simp[b*e*n*(p/(f*(m + 1))) Int[x^(n - 1)*((f*x)^(m + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]
Time = 0.94 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82
method | result | size |
parts | \(-\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{5 x^{5}}+\frac {2 p b \left (-\frac {1}{3 a \,x^{3}}+\frac {b}{a^{2} x}+\frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}\right )}{5}\) | \(61\) |
risch | \(-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5 x^{5}}-\frac {-6 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{5} \textit {\_Z}^{2}+b^{5} p^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{5}+2 b^{5} p^{2}\right ) x -a^{3} b^{2} p \textit {\_R} \right )\right ) a^{2} x^{5}+3 i \pi \,a^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-3 i \pi \,a^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-3 i \pi \,a^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+3 i \pi \,a^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-12 b^{2} p \,x^{4}+4 a b p \,x^{2}+6 \ln \left (c \right ) a^{2}}{30 a^{2} x^{5}}\) | \(236\) |
-1/5*ln(c*(b*x^2+a)^p)/x^5+2/5*p*b*(-1/3/a/x^3+b/a^2/x+b^2/a^2/(a*b)^(1/2) *arctan(b*x/(a*b)^(1/2)))
Time = 0.32 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.30 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\left [\frac {3 \, b^{2} p x^{5} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 6 \, b^{2} p x^{4} - 2 \, a b p x^{2} - 3 \, a^{2} p \log \left (b x^{2} + a\right ) - 3 \, a^{2} \log \left (c\right )}{15 \, a^{2} x^{5}}, \frac {6 \, b^{2} p x^{5} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 6 \, b^{2} p x^{4} - 2 \, a b p x^{2} - 3 \, a^{2} p \log \left (b x^{2} + a\right ) - 3 \, a^{2} \log \left (c\right )}{15 \, a^{2} x^{5}}\right ] \]
[1/15*(3*b^2*p*x^5*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 6*b^2*p*x^4 - 2*a*b*p*x^2 - 3*a^2*p*log(b*x^2 + a) - 3*a^2*log(c))/( a^2*x^5), 1/15*(6*b^2*p*x^5*sqrt(b/a)*arctan(x*sqrt(b/a)) + 6*b^2*p*x^4 - 2*a*b*p*x^2 - 3*a^2*p*log(b*x^2 + a) - 3*a^2*log(c))/(a^2*x^5)]
Leaf count of result is larger than twice the leaf count of optimal. 583 vs. \(2 (70) = 140\).
Time = 162.33 (sec) , antiderivative size = 583, normalized size of antiderivative = 7.88 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\begin {cases} - \frac {\log {\left (0^{p} c \right )}}{5 x^{5}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {\log {\left (a^{p} c \right )}}{5 x^{5}} & \text {for}\: b = 0 \\- \frac {2 p}{25 x^{5}} - \frac {\log {\left (c \left (b x^{2}\right )^{p} \right )}}{5 x^{5}} & \text {for}\: a = 0 \\- \frac {\log {\left (0^{p} c \right )}}{5 x^{5}} & \text {for}\: a = - b x^{2} \\- \frac {3 a^{3} \sqrt {- \frac {a}{b}} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{15 a^{3} x^{5} \sqrt {- \frac {a}{b}} + 15 a^{2} b x^{7} \sqrt {- \frac {a}{b}}} - \frac {2 a^{2} p x^{2} \sqrt {- \frac {a}{b}}}{\frac {15 a^{3} x^{5} \sqrt {- \frac {a}{b}}}{b} + 15 a^{2} x^{7} \sqrt {- \frac {a}{b}}} - \frac {3 a^{2} x^{2} \sqrt {- \frac {a}{b}} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{\frac {15 a^{3} x^{5} \sqrt {- \frac {a}{b}}}{b} + 15 a^{2} x^{7} \sqrt {- \frac {a}{b}}} + \frac {6 a b p x^{5} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{\frac {15 a^{3} x^{5} \sqrt {- \frac {a}{b}}}{b} + 15 a^{2} x^{7} \sqrt {- \frac {a}{b}}} + \frac {4 a b p x^{4} \sqrt {- \frac {a}{b}}}{\frac {15 a^{3} x^{5} \sqrt {- \frac {a}{b}}}{b} + 15 a^{2} x^{7} \sqrt {- \frac {a}{b}}} - \frac {3 a b x^{5} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{\frac {15 a^{3} x^{5} \sqrt {- \frac {a}{b}}}{b} + 15 a^{2} x^{7} \sqrt {- \frac {a}{b}}} + \frac {6 b^{2} p x^{7} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{\frac {15 a^{3} x^{5} \sqrt {- \frac {a}{b}}}{b} + 15 a^{2} x^{7} \sqrt {- \frac {a}{b}}} + \frac {6 b^{2} p x^{6} \sqrt {- \frac {a}{b}}}{\frac {15 a^{3} x^{5} \sqrt {- \frac {a}{b}}}{b} + 15 a^{2} x^{7} \sqrt {- \frac {a}{b}}} - \frac {3 b^{2} x^{7} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{\frac {15 a^{3} x^{5} \sqrt {- \frac {a}{b}}}{b} + 15 a^{2} x^{7} \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]
Piecewise((-log(0**p*c)/(5*x**5), Eq(a, 0) & Eq(b, 0)), (-log(a**p*c)/(5*x **5), Eq(b, 0)), (-2*p/(25*x**5) - log(c*(b*x**2)**p)/(5*x**5), Eq(a, 0)), (-log(0**p*c)/(5*x**5), Eq(a, -b*x**2)), (-3*a**3*sqrt(-a/b)*log(c*(a + b *x**2)**p)/(15*a**3*x**5*sqrt(-a/b) + 15*a**2*b*x**7*sqrt(-a/b)) - 2*a**2* p*x**2*sqrt(-a/b)/(15*a**3*x**5*sqrt(-a/b)/b + 15*a**2*x**7*sqrt(-a/b)) - 3*a**2*x**2*sqrt(-a/b)*log(c*(a + b*x**2)**p)/(15*a**3*x**5*sqrt(-a/b)/b + 15*a**2*x**7*sqrt(-a/b)) + 6*a*b*p*x**5*log(x - sqrt(-a/b))/(15*a**3*x**5 *sqrt(-a/b)/b + 15*a**2*x**7*sqrt(-a/b)) + 4*a*b*p*x**4*sqrt(-a/b)/(15*a** 3*x**5*sqrt(-a/b)/b + 15*a**2*x**7*sqrt(-a/b)) - 3*a*b*x**5*log(c*(a + b*x **2)**p)/(15*a**3*x**5*sqrt(-a/b)/b + 15*a**2*x**7*sqrt(-a/b)) + 6*b**2*p* x**7*log(x - sqrt(-a/b))/(15*a**3*x**5*sqrt(-a/b)/b + 15*a**2*x**7*sqrt(-a /b)) + 6*b**2*p*x**6*sqrt(-a/b)/(15*a**3*x**5*sqrt(-a/b)/b + 15*a**2*x**7* sqrt(-a/b)) - 3*b**2*x**7*log(c*(a + b*x**2)**p)/(15*a**3*x**5*sqrt(-a/b)/ b + 15*a**2*x**7*sqrt(-a/b)), True))
Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\frac {2}{15} \, b p {\left (\frac {3 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {3 \, b x^{2} - a}{a^{2} x^{3}}\right )} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{5 \, x^{5}} \]
2/15*b*p*(3*b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2) + (3*b*x^2 - a)/(a^2 *x^3)) - 1/5*log((b*x^2 + a)^p*c)/x^5
Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\frac {2 \, b^{3} p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{5 \, \sqrt {a b} a^{2}} - \frac {p \log \left (b x^{2} + a\right )}{5 \, x^{5}} + \frac {6 \, b^{2} p x^{4} - 2 \, a b p x^{2} - 3 \, a^{2} \log \left (c\right )}{15 \, a^{2} x^{5}} \]
2/5*b^3*p*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2) - 1/5*p*log(b*x^2 + a)/x^5 + 1/15*(6*b^2*p*x^4 - 2*a*b*p*x^2 - 3*a^2*log(c))/(a^2*x^5)
Time = 1.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^6} \, dx=\frac {2\,b^{5/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{5\,a^{5/2}}-\frac {\frac {2\,b\,p}{3\,a}-\frac {2\,b^2\,p\,x^2}{a^2}}{5\,x^3}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{5\,x^5} \]