Integrand size = 16, antiderivative size = 75 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {b^3 p x}{4 a^3}-\frac {b^2 p x^2}{8 a^2}+\frac {b p x^3}{12 a}+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )-\frac {b^4 p \log (b+a x)}{4 a^4} \] Output:
1/4*b^3*p*x/a^3-1/8*b^2*p*x^2/a^2+1/12*b*p*x^3/a+1/4*x^4*ln(c*(a+b/x)^p)-1 /4*b^4*p*ln(a*x+b)/a^4
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {a b p x \left (6 b^2-3 a b x+2 a^2 x^2\right )-6 b^4 p \log \left (a+\frac {b}{x}\right )+6 a^4 x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )-6 b^4 p \log (x)}{24 a^4} \] Input:
Integrate[x^3*Log[c*(a + b/x)^p],x]
Output:
(a*b*p*x*(6*b^2 - 3*a*b*x + 2*a^2*x^2) - 6*b^4*p*Log[a + b/x] + 6*a^4*x^4* Log[c*(a + b/x)^p] - 6*b^4*p*Log[x])/(24*a^4)
Time = 0.39 (sec) , antiderivative size = 70, 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.
\(\displaystyle \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {1}{4} b p \int \frac {x^2}{a+\frac {b}{x}}dx+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {1}{4} b p \int \frac {x^3}{b+a x}dx+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{4} b p \int \left (-\frac {b^3}{a^3 (b+a x)}+\frac {b^2}{a^3}-\frac {x b}{a^2}+\frac {x^2}{a}\right )dx+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} b p \left (-\frac {b^3 \log (a x+b)}{a^4}+\frac {b^2 x}{a^3}-\frac {b x^2}{2 a^2}+\frac {x^3}{3 a}\right )+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )\) |
Input:
Int[x^3*Log[c*(a + b/x)^p],x]
Output:
(x^4*Log[c*(a + b/x)^p])/4 + (b*p*((b^2*x)/a^3 - (b*x^2)/(2*a^2) + x^3/(3* a) - (b^3*Log[b + a*x])/a^4))/4
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 = 0.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84
method | result | size |
parts | \(\frac {x^{4} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{4}+\frac {p b \left (\frac {\frac {1}{3} a^{2} x^{3}-\frac {1}{2} a b \,x^{2}+b^{2} x}{a^{3}}-\frac {b^{3} \ln \left (a x +b \right )}{a^{4}}\right )}{4}\) | \(63\) |
parallelrisch | \(-\frac {-6 x^{4} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{4} p -2 x^{3} a^{3} b \,p^{2}+3 x^{2} a^{2} b^{2} p^{2}+6 \ln \left (x \right ) b^{4} p^{2}-6 x a \,b^{3} p^{2}+6 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{4} p +6 b^{4} p^{2}}{24 a^{4} p}\) | \(107\) |
Input:
int(x^3*ln(c*(a+b/x)^p),x,method=_RETURNVERBOSE)
Output:
1/4*x^4*ln(c*(a+b/x)^p)+1/4*p*b*(1/a^3*(1/3*a^2*x^3-1/2*a*b*x^2+b^2*x)-b^3 /a^4*ln(a*x+b))
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {6 \, a^{4} p x^{4} \log \left (\frac {a x + b}{x}\right ) + 6 \, a^{4} x^{4} \log \left (c\right ) + 2 \, a^{3} b p x^{3} - 3 \, a^{2} b^{2} p x^{2} + 6 \, a b^{3} p x - 6 \, b^{4} p \log \left (a x + b\right )}{24 \, a^{4}} \] Input:
integrate(x^3*log(c*(a+b/x)^p),x, algorithm="fricas")
Output:
1/24*(6*a^4*p*x^4*log((a*x + b)/x) + 6*a^4*x^4*log(c) + 2*a^3*b*p*x^3 - 3* a^2*b^2*p*x^2 + 6*a*b^3*p*x - 6*b^4*p*log(a*x + b))/a^4
Time = 1.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.16 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\begin {cases} \frac {x^{4} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{4} + \frac {b p x^{3}}{12 a} - \frac {b^{2} p x^{2}}{8 a^{2}} + \frac {b^{3} p x}{4 a^{3}} - \frac {b^{4} p \log {\left (a x + b \right )}}{4 a^{4}} & \text {for}\: a \neq 0 \\\frac {p x^{4}}{16} + \frac {x^{4} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*ln(c*(a+b/x)**p),x)
Output:
Piecewise((x**4*log(c*(a + b/x)**p)/4 + b*p*x**3/(12*a) - b**2*p*x**2/(8*a **2) + b**3*p*x/(4*a**3) - b**4*p*log(a*x + b)/(4*a**4), Ne(a, 0)), (p*x** 4/16 + x**4*log(c*(b/x)**p)/4, True))
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.85 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {1}{4} \, x^{4} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) - \frac {1}{24} \, b p {\left (\frac {6 \, b^{3} \log \left (a x + b\right )}{a^{4}} - \frac {2 \, a^{2} x^{3} - 3 \, a b x^{2} + 6 \, b^{2} x}{a^{3}}\right )} \] Input:
integrate(x^3*log(c*(a+b/x)^p),x, algorithm="maxima")
Output:
1/4*x^4*log((a + b/x)^p*c) - 1/24*b*p*(6*b^3*log(a*x + b)/a^4 - (2*a^2*x^3 - 3*a*b*x^2 + 6*b^2*x)/a^3)
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (65) = 130\).
Time = 0.13 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.43 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {\frac {6 \, b^{5} p \log \left (\frac {a x + b}{x}\right )}{a^{4} - \frac {4 \, {\left (a x + b\right )} a^{3}}{x} + \frac {6 \, {\left (a x + b\right )}^{2} a^{2}}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3} a}{x^{3}} + \frac {{\left (a x + b\right )}^{4}}{x^{4}}} + \frac {6 \, b^{5} p \log \left (-a + \frac {a x + b}{x}\right )}{a^{4}} - \frac {6 \, b^{5} p \log \left (\frac {a x + b}{x}\right )}{a^{4}} - \frac {11 \, a^{3} b^{5} p - 6 \, a^{3} b^{5} \log \left (c\right ) - \frac {26 \, {\left (a x + b\right )} a^{2} b^{5} p}{x} + \frac {21 \, {\left (a x + b\right )}^{2} a b^{5} p}{x^{2}} - \frac {6 \, {\left (a x + b\right )}^{3} b^{5} p}{x^{3}}}{a^{7} - \frac {4 \, {\left (a x + b\right )} a^{6}}{x} + \frac {6 \, {\left (a x + b\right )}^{2} a^{5}}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3} a^{4}}{x^{3}} + \frac {{\left (a x + b\right )}^{4} a^{3}}{x^{4}}}}{24 \, b} \] Input:
integrate(x^3*log(c*(a+b/x)^p),x, algorithm="giac")
Output:
1/24*(6*b^5*p*log((a*x + b)/x)/(a^4 - 4*(a*x + b)*a^3/x + 6*(a*x + b)^2*a^ 2/x^2 - 4*(a*x + b)^3*a/x^3 + (a*x + b)^4/x^4) + 6*b^5*p*log(-a + (a*x + b )/x)/a^4 - 6*b^5*p*log((a*x + b)/x)/a^4 - (11*a^3*b^5*p - 6*a^3*b^5*log(c) - 26*(a*x + b)*a^2*b^5*p/x + 21*(a*x + b)^2*a*b^5*p/x^2 - 6*(a*x + b)^3*b ^5*p/x^3)/(a^7 - 4*(a*x + b)*a^6/x + 6*(a*x + b)^2*a^5/x^2 - 4*(a*x + b)^3 *a^4/x^3 + (a*x + b)^4*a^3/x^4))/b
Time = 26.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {x^4\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{4}-\frac {b^2\,p\,x^2}{8\,a^2}-\frac {b^4\,p\,\ln \left (b+a\,x\right )}{4\,a^4}+\frac {b\,p\,x^3}{12\,a}+\frac {b^3\,p\,x}{4\,a^3} \] Input:
int(x^3*log(c*(a + b/x)^p),x)
Output:
(x^4*log(c*(a + b/x)^p))/4 - (b^2*p*x^2)/(8*a^2) - (b^4*p*log(b + a*x))/(4 *a^4) + (b*p*x^3)/(12*a) + (b^3*p*x)/(4*a^3)
Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.16 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {6 \,\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right ) a^{4} x^{4}-6 \,\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right ) b^{4}-6 \,\mathrm {log}\left (x \right ) b^{4} p +2 a^{3} b p \,x^{3}-3 a^{2} b^{2} p \,x^{2}+6 a \,b^{3} p x}{24 a^{4}} \] Input:
int(x^3*log(c*(a+b/x)^p),x)
Output:
(6*log(((a*x + b)**p*c)/x**p)*a**4*x**4 - 6*log(((a*x + b)**p*c)/x**p)*b** 4 - 6*log(x)*b**4*p + 2*a**3*b*p*x**3 - 3*a**2*b**2*p*x**2 + 6*a*b**3*p*x) /(24*a**4)