\(\int x \log (c (a+\frac {b}{x})^p) \, dx\) [29]

Optimal result

Integrand size = 14, antiderivative size = 47 \[ \int x \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {b p x}{2 a}+\frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )-\frac {b^2 p \log (b+a x)}{2 a^2} \]

[Out]

1/2*b*p*x/a+1/2*x^2*ln(c*(a+b/x)^p)-1/2*b^2*p*ln(a*x+b)/a^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2505, 199, 45} \[ \int x \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {b^2 p \log (a x+b)}{2 a^2}+\frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p x}{2 a} \]

[In]

Int[x*Log[c*(a + b/x)^p],x]

[Out]

(b*p*x)/(2*a) + (x^2*Log[c*(a + b/x)^p])/2 - (b^2*p*Log[b + a*x])/(2*a^2)

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

Rule 2505

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{2} (b p) \int \frac {1}{a+\frac {b}{x}} \, dx \\ & = \frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{2} (b p) \int \frac {x}{b+a x} \, dx \\ & = \frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{2} (b p) \int \left (\frac {1}{a}-\frac {b}{a (b+a x)}\right ) \, dx \\ & = \frac {b p x}{2 a}+\frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )-\frac {b^2 p \log (b+a x)}{2 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int x \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {1}{2} \left (x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p (a x-b \log (b+a x))}{a^2}\right ) \]

[In]

Integrate[x*Log[c*(a + b/x)^p],x]

[Out]

(x^2*Log[c*(a + b/x)^p] + (b*p*(a*x - b*Log[b + a*x]))/a^2)/2

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.87

method	result	size
parts	\(\frac {x^{2} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{2}+\frac {p b \left (\frac {x}{a}-\frac {b \ln \left (a x +b \right )}{a^{2}}\right )}{2}\)	\(41\)
parallelrisch	\(-\frac {-x^{2} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{2} p +\ln \left (x \right ) b^{2} p^{2}-x a b \,p^{2}+\ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{2} p +b^{2} p^{2}}{2 a^{2} p}\)	\(76\)

[In]

int(x*ln(c*(a+b/x)^p),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2*ln(c*(a+b/x)^p)+1/2*p*b*(x/a-1/a^2*b*ln(a*x+b))

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06 \[ \int x \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {a^{2} p x^{2} \log \left (\frac {a x + b}{x}\right ) + a^{2} x^{2} \log \left (c\right ) + a b p x - b^{2} p \log \left (a x + b\right )}{2 \, a^{2}} \]

[In]

integrate(x*log(c*(a+b/x)^p),x, algorithm="fricas")

[Out]

1/2*(a^2*p*x^2*log((a*x + b)/x) + a^2*x^2*log(c) + a*b*p*x - b^2*p*log(a*x + b))/a^2

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.28 \[ \int x \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\begin {cases} \frac {x^{2} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{2} + \frac {b p x}{2 a} - \frac {b^{2} p \log {\left (a x + b \right )}}{2 a^{2}} & \text {for}\: a \neq 0 \\\frac {p x^{2}}{4} + \frac {x^{2} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

integrate(x*ln(c*(a+b/x)**p),x)

[Out]

Piecewise((x**2*log(c*(a + b/x)**p)/2 + b*p*x/(2*a) - b**2*p*log(a*x + b)/(2*a**2), Ne(a, 0)), (p*x**2/4 + x**
2*log(c*(b/x)**p)/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int x \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {1}{2} \, b p {\left (\frac {x}{a} - \frac {b \log \left (a x + b\right )}{a^{2}}\right )} + \frac {1}{2} \, x^{2} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \]

[In]

integrate(x*log(c*(a+b/x)^p),x, algorithm="maxima")

[Out]

1/2*b*p*(x/a - b*log(a*x + b)/a^2) + 1/2*x^2*log((a + b/x)^p*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (41) = 82\).

Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 3.23 \[ \int x \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {\frac {b^{3} p \log \left (\frac {a x + b}{x}\right )}{a^{2} - \frac {2 \, {\left (a x + b\right )} a}{x} + \frac {{\left (a x + b\right )}^{2}}{x^{2}}} + \frac {b^{3} p \log \left (-a + \frac {a x + b}{x}\right )}{a^{2}} - \frac {b^{3} p \log \left (\frac {a x + b}{x}\right )}{a^{2}} - \frac {a b^{3} p - a b^{3} \log \left (c\right ) - \frac {{\left (a x + b\right )} b^{3} p}{x}}{a^{3} - \frac {2 \, {\left (a x + b\right )} a^{2}}{x} + \frac {{\left (a x + b\right )}^{2} a}{x^{2}}}}{2 \, b} \]

[In]

integrate(x*log(c*(a+b/x)^p),x, algorithm="giac")

[Out]

1/2*(b^3*p*log((a*x + b)/x)/(a^2 - 2*(a*x + b)*a/x + (a*x + b)^2/x^2) + b^3*p*log(-a + (a*x + b)/x)/a^2 - b^3*
p*log((a*x + b)/x)/a^2 - (a*b^3*p - a*b^3*log(c) - (a*x + b)*b^3*p/x)/(a^3 - 2*(a*x + b)*a^2/x + (a*x + b)^2*a
/x^2))/b

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.87 \[ \int x \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {x^2\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{2}+\frac {b\,p\,x}{2\,a}-\frac {b^2\,p\,\ln \left (b+a\,x\right )}{2\,a^2} \]

[In]

int(x*log(c*(a + b/x)^p),x)

[Out]

(x^2*log(c*(a + b/x)^p))/2 + (b*p*x)/(2*a) - (b^2*p*log(b + a*x))/(2*a^2)