\(\int x^2 \log (c (a+\frac {b}{x})^p) \, dx\) [28]

Optimal result

Integrand size = 16, antiderivative size = 61 \[ \int x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {b^2 p x}{3 a^2}+\frac {b p x^2}{6 a}+\frac {1}{3} x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b^3 p \log (b+a x)}{3 a^3} \]

[Out]

-1/3*b^2*p*x/a^2+1/6*b*p*x^2/a+1/3*x^3*ln(c*(a+b/x)^p)+1/3*b^3*p*ln(a*x+b)/a^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2505, 269, 45} \[ \int x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {b^3 p \log (a x+b)}{3 a^3}-\frac {b^2 p x}{3 a^2}+\frac {1}{3} x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p x^2}{6 a} \]

[In]

Int[x^2*Log[c*(a + b/x)^p],x]

[Out]

-1/3*(b^2*p*x)/a^2 + (b*p*x^2)/(6*a) + (x^3*Log[c*(a + b/x)^p])/3 + (b^3*p*Log[b + a*x])/(3*a^3)

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 269

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]

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}{3} x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{3} (b p) \int \frac {x}{a+\frac {b}{x}} \, dx \\ & = \frac {1}{3} x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{3} (b p) \int \frac {x^2}{b+a x} \, dx \\ & = \frac {1}{3} x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{3} (b p) \int \left (-\frac {b}{a^2}+\frac {x}{a}+\frac {b^2}{a^2 (b+a x)}\right ) \, dx \\ & = -\frac {b^2 p x}{3 a^2}+\frac {b p x^2}{6 a}+\frac {1}{3} x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b^3 p \log (b+a x)}{3 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {a b p x (-2 b+a x)+2 b^3 p \log \left (a+\frac {b}{x}\right )+2 a^3 x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+2 b^3 p \log (x)}{6 a^3} \]

[In]

Integrate[x^2*Log[c*(a + b/x)^p],x]

[Out]

(a*b*p*x*(-2*b + a*x) + 2*b^3*p*Log[a + b/x] + 2*a^3*x^3*Log[c*(a + b/x)^p] + 2*b^3*p*Log[x])/(6*a^3)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85

method	result	size
parts	\(\frac {x^{3} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{3}+\frac {p b \left (\frac {\frac {1}{2} x^{2} a -b x}{a^{2}}+\frac {b^{2} \ln \left (a x +b \right )}{a^{3}}\right )}{3}\)	\(52\)
parallelrisch	\(-\frac {-2 x^{3} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{3} p -x^{2} a^{2} b \,p^{2}-2 \ln \left (x \right ) b^{3} p^{2}+2 x a \,b^{2} p^{2}-2 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{3} p -2 b^{3} p^{2}}{6 a^{3} p}\)	\(93\)

[In]

int(x^2*ln(c*(a+b/x)^p),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05 \[ \int x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {2 \, a^{3} p x^{3} \log \left (\frac {a x + b}{x}\right ) + 2 \, a^{3} x^{3} \log \left (c\right ) + a^{2} b p x^{2} - 2 \, a b^{2} p x + 2 \, b^{3} p \log \left (a x + b\right )}{6 \, a^{3}} \]

[In]

integrate(x^2*log(c*(a+b/x)^p),x, algorithm="fricas")

[Out]

1/6*(2*a^3*p*x^3*log((a*x + b)/x) + 2*a^3*x^3*log(c) + a^2*b*p*x^2 - 2*a*b^2*p*x + 2*b^3*p*log(a*x + b))/a^3

Sympy [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\begin {cases} \frac {x^{3} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{3} + \frac {b p x^{2}}{6 a} - \frac {b^{2} p x}{3 a^{2}} + \frac {b^{3} p \log {\left (a x + b \right )}}{3 a^{3}} & \text {for}\: a \neq 0 \\\frac {p x^{3}}{9} + \frac {x^{3} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{3} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*ln(c*(a+b/x)**p),x)

[Out]

Piecewise((x**3*log(c*(a + b/x)**p)/3 + b*p*x**2/(6*a) - b**2*p*x/(3*a**2) + b**3*p*log(a*x + b)/(3*a**3), Ne(
a, 0)), (p*x**3/9 + x**3*log(c*(b/x)**p)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.84 \[ \int x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {1}{3} \, x^{3} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) + \frac {1}{6} \, b p {\left (\frac {2 \, b^{2} \log \left (a x + b\right )}{a^{3}} + \frac {a x^{2} - 2 \, b x}{a^{2}}\right )} \]

[In]

integrate(x^2*log(c*(a+b/x)^p),x, algorithm="maxima")

[Out]

1/3*x^3*log((a + b/x)^p*c) + 1/6*b*p*(2*b^2*log(a*x + b)/a^3 + (a*x^2 - 2*b*x)/a^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (53) = 106\).

Time = 0.33 (sec) , antiderivative size = 210, normalized size of antiderivative = 3.44 \[ \int x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {\frac {2 \, b^{4} p \log \left (\frac {a x + b}{x}\right )}{a^{3} - \frac {3 \, {\left (a x + b\right )} a^{2}}{x} + \frac {3 \, {\left (a x + b\right )}^{2} a}{x^{2}} - \frac {{\left (a x + b\right )}^{3}}{x^{3}}} + \frac {2 \, b^{4} p \log \left (-a + \frac {a x + b}{x}\right )}{a^{3}} - \frac {2 \, b^{4} p \log \left (\frac {a x + b}{x}\right )}{a^{3}} - \frac {3 \, a^{2} b^{4} p - 2 \, a^{2} b^{4} \log \left (c\right ) - \frac {5 \, {\left (a x + b\right )} a b^{4} p}{x} + \frac {2 \, {\left (a x + b\right )}^{2} b^{4} p}{x^{2}}}{a^{5} - \frac {3 \, {\left (a x + b\right )} a^{4}}{x} + \frac {3 \, {\left (a x + b\right )}^{2} a^{3}}{x^{2}} - \frac {{\left (a x + b\right )}^{3} a^{2}}{x^{3}}}}{6 \, b} \]

[In]

integrate(x^2*log(c*(a+b/x)^p),x, algorithm="giac")

[Out]

-1/6*(2*b^4*p*log((a*x + b)/x)/(a^3 - 3*(a*x + b)*a^2/x + 3*(a*x + b)^2*a/x^2 - (a*x + b)^3/x^3) + 2*b^4*p*log
(-a + (a*x + b)/x)/a^3 - 2*b^4*p*log((a*x + b)/x)/a^3 - (3*a^2*b^4*p - 2*a^2*b^4*log(c) - 5*(a*x + b)*a*b^4*p/
x + 2*(a*x + b)^2*b^4*p/x^2)/(a^5 - 3*(a*x + b)*a^4/x + 3*(a*x + b)^2*a^3/x^2 - (a*x + b)^3*a^2/x^3))/b

Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {x^3\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{3}+\frac {b^3\,p\,\ln \left (b+a\,x\right )}{3\,a^3}+\frac {b\,p\,x^2}{6\,a}-\frac {b^2\,p\,x}{3\,a^2} \]

[In]

int(x^2*log(c*(a + b/x)^p),x)

[Out]

(x^3*log(c*(a + b/x)^p))/3 + (b^3*p*log(b + a*x))/(3*a^3) + (b*p*x^2)/(6*a) - (b^2*p*x)/(3*a^2)