Integrand size = 16, antiderivative size = 89 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {b^4 p x}{5 a^4}+\frac {b^3 p x^2}{10 a^3}-\frac {b^2 p x^3}{15 a^2}+\frac {b p x^4}{20 a}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b^5 p \log (b+a x)}{5 a^5} \] Output:
-1/5*b^4*p*x/a^4+1/10*b^3*p*x^2/a^3-1/15*b^2*p*x^3/a^2+1/20*b*p*x^4/a+1/5* x^5*ln(c*(a+b/x)^p)+1/5*b^5*p*ln(a*x+b)/a^5
Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {a b p x \left (-12 b^3+6 a b^2 x-4 a^2 b x^2+3 a^3 x^3\right )+12 b^5 p \log \left (a+\frac {b}{x}\right )+12 a^5 x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+12 b^5 p \log (x)}{60 a^5} \] Input:
Integrate[x^4*Log[c*(a + b/x)^p],x]
Output:
(a*b*p*x*(-12*b^3 + 6*a*b^2*x - 4*a^2*b*x^2 + 3*a^3*x^3) + 12*b^5*p*Log[a + b/x] + 12*a^5*x^5*Log[c*(a + b/x)^p] + 12*b^5*p*Log[x])/(60*a^5)
Time = 0.41 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2905, 795, 49, 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.
\(\displaystyle \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {1}{5} b p \int \frac {x^3}{a+\frac {b}{x}}dx+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {1}{5} b p \int \frac {x^4}{b+a x}dx+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{5} b p \int \left (\frac {b^4}{a^4 (b+a x)}-\frac {b^3}{a^4}+\frac {x b^2}{a^3}-\frac {x^2 b}{a^2}+\frac {x^3}{a}\right )dx+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} b p \left (\frac {b^4 \log (a x+b)}{a^5}-\frac {b^3 x}{a^4}+\frac {b^2 x^2}{2 a^3}-\frac {b x^3}{3 a^2}+\frac {x^4}{4 a}\right )+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )\) |
Input:
Int[x^4*Log[c*(a + b/x)^p],x]
Output:
(x^5*Log[c*(a + b/x)^p])/5 + (b*p*(-((b^3*x)/a^4) + (b^2*x^2)/(2*a^3) - (b *x^3)/(3*a^2) + x^4/(4*a) + (b^4*Log[b + a*x])/a^5))/5
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
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 = 0.66 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83
method | result | size |
parts | \(\frac {x^{5} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{5}+\frac {p b \left (\frac {\frac {1}{4} a^{3} x^{4}-\frac {1}{3} a^{2} b \,x^{3}+\frac {1}{2} a \,b^{2} x^{2}-b^{3} x}{a^{4}}+\frac {b^{4} \ln \left (a x +b \right )}{a^{5}}\right )}{5}\) | \(74\) |
parallelrisch | \(-\frac {-12 x^{5} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{5} p -3 x^{4} a^{4} b \,p^{2}+4 x^{3} a^{3} b^{2} p^{2}-6 x^{2} a^{2} b^{3} p^{2}-12 \ln \left (x \right ) b^{5} p^{2}+12 x a \,b^{4} p^{2}-12 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{5} p -12 b^{5} p^{2}}{60 a^{5} p}\) | \(121\) |
Input:
int(x^4*ln(c*(a+b/x)^p),x,method=_RETURNVERBOSE)
Output:
1/5*x^5*ln(c*(a+b/x)^p)+1/5*p*b*(1/a^4*(1/4*a^3*x^4-1/3*a^2*b*x^3+1/2*a*b^ 2*x^2-b^3*x)+b^4/a^5*ln(a*x+b))
Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {12 \, a^{5} p x^{5} \log \left (\frac {a x + b}{x}\right ) + 12 \, a^{5} x^{5} \log \left (c\right ) + 3 \, a^{4} b p x^{4} - 4 \, a^{3} b^{2} p x^{3} + 6 \, a^{2} b^{3} p x^{2} - 12 \, a b^{4} p x + 12 \, b^{5} p \log \left (a x + b\right )}{60 \, a^{5}} \] Input:
integrate(x^4*log(c*(a+b/x)^p),x, algorithm="fricas")
Output:
1/60*(12*a^5*p*x^5*log((a*x + b)/x) + 12*a^5*x^5*log(c) + 3*a^4*b*p*x^4 - 4*a^3*b^2*p*x^3 + 6*a^2*b^3*p*x^2 - 12*a*b^4*p*x + 12*b^5*p*log(a*x + b))/ a^5
Time = 2.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.12 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\begin {cases} \frac {x^{5} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{5} + \frac {b p x^{4}}{20 a} - \frac {b^{2} p x^{3}}{15 a^{2}} + \frac {b^{3} p x^{2}}{10 a^{3}} - \frac {b^{4} p x}{5 a^{4}} + \frac {b^{5} p \log {\left (a x + b \right )}}{5 a^{5}} & \text {for}\: a \neq 0 \\\frac {p x^{5}}{25} + \frac {x^{5} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{5} & \text {otherwise} \end {cases} \] Input:
integrate(x**4*ln(c*(a+b/x)**p),x)
Output:
Piecewise((x**5*log(c*(a + b/x)**p)/5 + b*p*x**4/(20*a) - b**2*p*x**3/(15* a**2) + b**3*p*x**2/(10*a**3) - b**4*p*x/(5*a**4) + b**5*p*log(a*x + b)/(5 *a**5), Ne(a, 0)), (p*x**5/25 + x**5*log(c*(b/x)**p)/5, True))
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {1}{5} \, x^{5} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) + \frac {1}{60} \, b p {\left (\frac {12 \, b^{4} \log \left (a x + b\right )}{a^{5}} + \frac {3 \, a^{3} x^{4} - 4 \, a^{2} b x^{3} + 6 \, a b^{2} x^{2} - 12 \, b^{3} x}{a^{4}}\right )} \] Input:
integrate(x^4*log(c*(a+b/x)^p),x, algorithm="maxima")
Output:
1/5*x^5*log((a + b/x)^p*c) + 1/60*b*p*(12*b^4*log(a*x + b)/a^5 + (3*a^3*x^ 4 - 4*a^2*b*x^3 + 6*a*b^2*x^2 - 12*b^3*x)/a^4)
Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (77) = 154\).
Time = 0.13 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.46 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {\frac {12 \, b^{6} p \log \left (\frac {a x + b}{x}\right )}{a^{5} - \frac {5 \, {\left (a x + b\right )} a^{4}}{x} + \frac {10 \, {\left (a x + b\right )}^{2} a^{3}}{x^{2}} - \frac {10 \, {\left (a x + b\right )}^{3} a^{2}}{x^{3}} + \frac {5 \, {\left (a x + b\right )}^{4} a}{x^{4}} - \frac {{\left (a x + b\right )}^{5}}{x^{5}}} + \frac {12 \, b^{6} p \log \left (-a + \frac {a x + b}{x}\right )}{a^{5}} - \frac {12 \, b^{6} p \log \left (\frac {a x + b}{x}\right )}{a^{5}} - \frac {25 \, a^{4} b^{6} p - 12 \, a^{4} b^{6} \log \left (c\right ) - \frac {77 \, {\left (a x + b\right )} a^{3} b^{6} p}{x} + \frac {94 \, {\left (a x + b\right )}^{2} a^{2} b^{6} p}{x^{2}} - \frac {54 \, {\left (a x + b\right )}^{3} a b^{6} p}{x^{3}} + \frac {12 \, {\left (a x + b\right )}^{4} b^{6} p}{x^{4}}}{a^{9} - \frac {5 \, {\left (a x + b\right )} a^{8}}{x} + \frac {10 \, {\left (a x + b\right )}^{2} a^{7}}{x^{2}} - \frac {10 \, {\left (a x + b\right )}^{3} a^{6}}{x^{3}} + \frac {5 \, {\left (a x + b\right )}^{4} a^{5}}{x^{4}} - \frac {{\left (a x + b\right )}^{5} a^{4}}{x^{5}}}}{60 \, b} \] Input:
integrate(x^4*log(c*(a+b/x)^p),x, algorithm="giac")
Output:
-1/60*(12*b^6*p*log((a*x + b)/x)/(a^5 - 5*(a*x + b)*a^4/x + 10*(a*x + b)^2 *a^3/x^2 - 10*(a*x + b)^3*a^2/x^3 + 5*(a*x + b)^4*a/x^4 - (a*x + b)^5/x^5) + 12*b^6*p*log(-a + (a*x + b)/x)/a^5 - 12*b^6*p*log((a*x + b)/x)/a^5 - (2 5*a^4*b^6*p - 12*a^4*b^6*log(c) - 77*(a*x + b)*a^3*b^6*p/x + 94*(a*x + b)^ 2*a^2*b^6*p/x^2 - 54*(a*x + b)^3*a*b^6*p/x^3 + 12*(a*x + b)^4*b^6*p/x^4)/( a^9 - 5*(a*x + b)*a^8/x + 10*(a*x + b)^2*a^7/x^2 - 10*(a*x + b)^3*a^6/x^3 + 5*(a*x + b)^4*a^5/x^4 - (a*x + b)^5*a^4/x^5))/b
Time = 26.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.87 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {x^5\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{5}-\frac {b^2\,p\,x^3}{15\,a^2}+\frac {b^3\,p\,x^2}{10\,a^3}+\frac {b^5\,p\,\ln \left (b+a\,x\right )}{5\,a^5}+\frac {b\,p\,x^4}{20\,a}-\frac {b^4\,p\,x}{5\,a^4} \] Input:
int(x^4*log(c*(a + b/x)^p),x)
Output:
(x^5*log(c*(a + b/x)^p))/5 - (b^2*p*x^3)/(15*a^2) + (b^3*p*x^2)/(10*a^3) + (b^5*p*log(b + a*x))/(5*a^5) + (b*p*x^4)/(20*a) - (b^4*p*x)/(5*a^4)
Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {12 \,\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right ) a^{5} x^{5}+12 \,\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right ) b^{5}+12 \,\mathrm {log}\left (x \right ) b^{5} p +3 a^{4} b p \,x^{4}-4 a^{3} b^{2} p \,x^{3}+6 a^{2} b^{3} p \,x^{2}-12 a \,b^{4} p x}{60 a^{5}} \] Input:
int(x^4*log(c*(a+b/x)^p),x)
Output:
(12*log(((a*x + b)**p*c)/x**p)*a**5*x**5 + 12*log(((a*x + b)**p*c)/x**p)*b **5 + 12*log(x)*b**5*p + 3*a**4*b*p*x**4 - 4*a**3*b**2*p*x**3 + 6*a**2*b** 3*p*x**2 - 12*a*b**4*p*x)/(60*a**5)