Integrand size = 16, antiderivative size = 80 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {2 a^2 p x}{5 b^2}+\frac {2 a p x^3}{15 b}-\frac {2 p x^5}{25}+\frac {2 a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right ) \]
-2/5*a^2*p*x/b^2+2/15*a*p*x^3/b-2/25*p*x^5+2/5*a^(5/2)*p*arctan(x*b^(1/2)/ a^(1/2))/b^(5/2)+1/5*x^5*ln(c*(b*x^2+a)^p)
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{75} \left (-\frac {30 a^2 p x}{b^2}+\frac {10 a p x^3}{b}-6 p x^5+\frac {30 a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}+15 x^5 \log \left (c \left (a+b x^2\right )^p\right )\right ) \]
((-30*a^2*p*x)/b^2 + (10*a*p*x^3)/b - 6*p*x^5 + (30*a^(5/2)*p*ArcTan[(Sqrt [b]*x)/Sqrt[a]])/b^(5/2) + 15*x^5*Log[c*(a + b*x^2)^p])/75
Time = 0.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2905, 254, 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.
\(\displaystyle \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )-\frac {2}{5} b p \int \frac {x^6}{b x^2+a}dx\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )-\frac {2}{5} b p \int \left (\frac {x^4}{b}-\frac {a x^2}{b^2}-\frac {a^3}{b^3 \left (b x^2+a\right )}+\frac {a^2}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )-\frac {2}{5} b p \left (-\frac {a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {a^2 x}{b^3}-\frac {a x^3}{3 b^2}+\frac {x^5}{5 b}\right )\) |
(-2*b*p*((a^2*x)/b^3 - (a*x^3)/(3*b^2) + x^5/(5*b) - (a^(5/2)*ArcTan[(Sqrt [b]*x)/Sqrt[a]])/b^(7/2)))/5 + (x^5*Log[c*(a + b*x^2)^p])/5
3.1.1.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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 = 0.65 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89
method | result | size |
parts | \(\frac {x^{5} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{5}-\frac {2 p b \left (\frac {\frac {1}{5} x^{5} b^{2}-\frac {1}{3} a b \,x^{3}+a^{2} x}{b^{3}}-\frac {a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\right )}{5}\) | \(71\) |
risch | \(\frac {x^{5} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5}-\frac {i \pi \,x^{5} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{10}+\frac {i \pi \,x^{5} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{10}+\frac {i \pi \,x^{5} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{10}-\frac {i \pi \,x^{5} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{10}+\frac {\ln \left (c \right ) x^{5}}{5}-\frac {2 p \,x^{5}}{25}+\frac {2 a p \,x^{3}}{15 b}+\frac {\sqrt {-a b}\, a^{2} p \ln \left (-\sqrt {-a b}\, x +a \right )}{5 b^{3}}-\frac {\sqrt {-a b}\, a^{2} p \ln \left (\sqrt {-a b}\, x +a \right )}{5 b^{3}}-\frac {2 a^{2} p x}{5 b^{2}}\) | \(229\) |
1/5*x^5*ln(c*(b*x^2+a)^p)-2/5*p*b*(1/b^3*(1/5*x^5*b^2-1/3*a*b*x^3+a^2*x)-a ^3/b^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.32 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.35 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\left [\frac {15 \, b^{2} p x^{5} \log \left (b x^{2} + a\right ) - 6 \, b^{2} p x^{5} + 15 \, b^{2} x^{5} \log \left (c\right ) + 10 \, a b p x^{3} + 15 \, a^{2} p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 30 \, a^{2} p x}{75 \, b^{2}}, \frac {15 \, b^{2} p x^{5} \log \left (b x^{2} + a\right ) - 6 \, b^{2} p x^{5} + 15 \, b^{2} x^{5} \log \left (c\right ) + 10 \, a b p x^{3} + 30 \, a^{2} p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 30 \, a^{2} p x}{75 \, b^{2}}\right ] \]
[1/75*(15*b^2*p*x^5*log(b*x^2 + a) - 6*b^2*p*x^5 + 15*b^2*x^5*log(c) + 10* a*b*p*x^3 + 15*a^2*p*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) - 30*a^2*p*x)/b^2, 1/75*(15*b^2*p*x^5*log(b*x^2 + a) - 6*b^2*p*x^5 + 15*b^2*x^5*log(c) + 10*a*b*p*x^3 + 30*a^2*p*sqrt(a/b)*arctan(b*x*sqrt(a/b )/a) - 30*a^2*p*x)/b^2]
Time = 31.80 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.95 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \frac {x^{5} \log {\left (0^{p} c \right )}}{5} & \text {for}\: a = 0 \wedge b = 0 \\\frac {x^{5} \log {\left (a^{p} c \right )}}{5} & \text {for}\: b = 0 \\- \frac {2 p x^{5}}{25} + \frac {x^{5} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{5} & \text {for}\: a = 0 \\\frac {2 a^{3} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{5 b^{3} \sqrt {- \frac {a}{b}}} - \frac {a^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{5 b^{3} \sqrt {- \frac {a}{b}}} - \frac {2 a^{2} p x}{5 b^{2}} + \frac {2 a p x^{3}}{15 b} - \frac {2 p x^{5}}{25} + \frac {x^{5} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{5} & \text {otherwise} \end {cases} \]
Piecewise((x**5*log(0**p*c)/5, Eq(a, 0) & Eq(b, 0)), (x**5*log(a**p*c)/5, Eq(b, 0)), (-2*p*x**5/25 + x**5*log(c*(b*x**2)**p)/5, Eq(a, 0)), (2*a**3*p *log(x - sqrt(-a/b))/(5*b**3*sqrt(-a/b)) - a**3*log(c*(a + b*x**2)**p)/(5* b**3*sqrt(-a/b)) - 2*a**2*p*x/(5*b**2) + 2*a*p*x**3/(15*b) - 2*p*x**5/25 + x**5*log(c*(a + b*x**2)**p)/5, True))
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{5} \, x^{5} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) + \frac {2}{75} \, b p {\left (\frac {15 \, a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {3 \, b^{2} x^{5} - 5 \, a b x^{3} + 15 \, a^{2} x}{b^{3}}\right )} \]
1/5*x^5*log((b*x^2 + a)^p*c) + 2/75*b*p*(15*a^3*arctan(b*x/sqrt(a*b))/(sqr t(a*b)*b^3) - (3*b^2*x^5 - 5*a*b*x^3 + 15*a^2*x)/b^3)
Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{5} \, p x^{5} \log \left (b x^{2} + a\right ) - \frac {1}{25} \, {\left (2 \, p - 5 \, \log \left (c\right )\right )} x^{5} + \frac {2 \, a p x^{3}}{15 \, b} + \frac {2 \, a^{3} p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{5 \, \sqrt {a b} b^{2}} - \frac {2 \, a^{2} p x}{5 \, b^{2}} \]
1/5*p*x^5*log(b*x^2 + a) - 1/25*(2*p - 5*log(c))*x^5 + 2/15*a*p*x^3/b + 2/ 5*a^3*p*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2) - 2/5*a^2*p*x/b^2
Time = 1.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {x^5\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{5}-\frac {2\,p\,x^5}{25}+\frac {2\,a^{5/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{5\,b^{5/2}}+\frac {2\,a\,p\,x^3}{15\,b}-\frac {2\,a^2\,p\,x}{5\,b^2} \]