\(\int \frac {\log (c (a+\frac {b}{x})^p)}{(d+e x)^2} \, dx\) [202]

Optimal result

Integrand size = 20, antiderivative size = 81 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}-\frac {p \log (x)}{d e}+\frac {a p \log (b+a x)}{e (a d-b e)}-\frac {b p \log (d+e x)}{d (a d-b e)} \]

[Out]

-ln(c*(a+b/x)^p)/e/(e*x+d)-p*ln(x)/d/e+a*p*ln(a*x+b)/e/(a*d-b*e)-b*p*ln(e*x+d)/d/(a*d-b*e)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2513, 528, 84} \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}+\frac {a p \log (a x+b)}{e (a d-b e)}-\frac {b p \log (d+e x)}{d (a d-b e)}-\frac {p \log (x)}{d e} \]

[In]

Int[Log[c*(a + b/x)^p]/(d + e*x)^2,x]

[Out]

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

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 528

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 2513

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Simp[(f
+ g*x)^(r + 1)*((a + b*Log[c*(d + e*x^n)^p])/(g*(r + 1))), x] - Dist[b*e*n*(p/(g*(r + 1))), Int[x^(n - 1)*((f
+ g*x)^(r + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x] && (IGtQ[r, 0] || RationalQ[n
]) && NeQ[r, -1]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}-\frac {p \log (x)}{d e}+\frac {a p \log (b+a x)}{e (a d-b e)}-\frac {b p \log (d+e x)}{d (a d-b e)} \]

[In]

Integrate[Log[c*(a + b/x)^p]/(d + e*x)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06

method	result	size
parts	\(-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{e \left (e x +d \right )}-\frac {p b \left (\frac {\ln \left (x \right )}{b d}+\frac {e \ln \left (e x +d \right )}{d \left (a d -b e \right )}-\frac {a \ln \left (a x +b \right )}{b \left (a d -b e \right )}\right )}{e}\)	\(86\)
parallelrisch	\(-\frac {-\ln \left (x \right ) x \,b^{2} e \,p^{2}+\ln \left (e x +d \right ) x \,b^{2} e \,p^{2}-\ln \left (x \right ) b^{2} d \,p^{2}+\ln \left (e x +d \right ) b^{2} d \,p^{2}-x \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a b d p -\ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{2} d p}{\left (e x +d \right ) p b d \left (a d -b e \right )}\)	\(124\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.83 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {{\left (a d^{2} - b d e\right )} p \log \left (\frac {a x + b}{x}\right ) - {\left (a d e p x + a d^{2} p\right )} \log \left (a x + b\right ) + {\left (b e^{2} p x + b d e p\right )} \log \left (e x + d\right ) + {\left (a d^{2} - b d e\right )} \log \left (c\right ) + {\left ({\left (a d e - b e^{2}\right )} p x + {\left (a d^{2} - b d e\right )} p\right )} \log \left (x\right )}{a d^{3} e - b d^{2} e^{2} + {\left (a d^{2} e^{2} - b d e^{3}\right )} x} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (61) = 122\).

Time = 3.72 (sec) , antiderivative size = 425, normalized size of antiderivative = 5.25 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=\begin {cases} \frac {d p \log {\left (\frac {d}{e} + x \right )}}{d^{2} e + d e^{2} x} + \frac {e p x \log {\left (\frac {d}{e} + x \right )}}{d^{2} e + d e^{2} x} + \frac {e x \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{d^{2} e + d e^{2} x} & \text {for}\: a = 0 \\\frac {x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )} + \frac {b p \log {\left (a x + b \right )}}{a}}{d^{2}} & \text {for}\: e = 0 \\- \frac {d p}{d^{2} e + d e^{2} x} + \frac {e x \log {\left (c \left (\frac {b}{x} + \frac {b e}{d}\right )^{p} \right )}}{d^{2} e + d e^{2} x} & \text {for}\: a = \frac {b e}{d} \\\frac {- \frac {a \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{b} + \frac {p}{x} - \frac {\log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{x}}{e^{2}} & \text {for}\: d = 0 \\\frac {a d x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} + \frac {b d p \log {\left (x + \frac {b}{a} \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} - \frac {b d p \log {\left (\frac {d}{e} + x \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} + \frac {b e p x \log {\left (x + \frac {b}{a} \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} - \frac {b e p x \log {\left (\frac {d}{e} + x \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} - \frac {b e x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((d*p*log(d/e + x)/(d**2*e + d*e**2*x) + e*p*x*log(d/e + x)/(d**2*e + d*e**2*x) + e*x*log(c*(b/x)**p)
/(d**2*e + d*e**2*x), Eq(a, 0)), ((x*log(c*(a + b/x)**p) + b*p*log(a*x + b)/a)/d**2, Eq(e, 0)), (-d*p/(d**2*e
+ d*e**2*x) + e*x*log(c*(b/x + b*e/d)**p)/(d**2*e + d*e**2*x), Eq(a, b*e/d)), ((-a*log(c*(a + b/x)**p)/b + p/x
- log(c*(a + b/x)**p)/x)/e**2, Eq(d, 0)), (a*d*x*log(c*(a + b/x)**p)/(a*d**3 + a*d**2*e*x - b*d**2*e - b*d*e*
*2*x) + b*d*p*log(x + b/a)/(a*d**3 + a*d**2*e*x - b*d**2*e - b*d*e**2*x) - b*d*p*log(d/e + x)/(a*d**3 + a*d**2
*e*x - b*d**2*e - b*d*e**2*x) + b*e*p*x*log(x + b/a)/(a*d**3 + a*d**2*e*x - b*d**2*e - b*d*e**2*x) - b*e*p*x*l
og(d/e + x)/(a*d**3 + a*d**2*e*x - b*d**2*e - b*d*e**2*x) - b*e*x*log(c*(a + b/x)**p)/(a*d**3 + a*d**2*e*x - b
*d**2*e - b*d*e**2*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=\frac {b p {\left (\frac {a \log \left (a x + b\right )}{a b d - b^{2} e} - \frac {e \log \left (e x + d\right )}{a d^{2} - b d e} - \frac {\log \left (x\right )}{b d}\right )}}{e} - \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{{\left (e x + d\right )} e} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.79 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\frac {b^{2} p \log \left (-a d + b e + \frac {{\left (a x + b\right )} d}{x}\right )}{a d^{2} - b d e} + \frac {b^{2} p \log \left (\frac {a x + b}{x}\right )}{a d^{2} - b d e - \frac {{\left (a x + b\right )} d^{2}}{x}} - \frac {b^{2} p \log \left (\frac {a x + b}{x}\right )}{a d^{2} - b d e} + \frac {b^{2} \log \left (c\right )}{a d^{2} - b d e - \frac {{\left (a x + b\right )} d^{2}}{x}}}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\ln \left (c\,{\left (\frac {b+a\,x}{x}\right )}^p\right )}{x\,e^2+d\,e}-\frac {p\,\ln \left (x\right )}{d\,e}-\frac {a\,p\,\ln \left (b+a\,x\right )}{b\,e^2-a\,d\,e}-\frac {b\,p\,\ln \left (d+e\,x\right )}{a\,d^2-b\,d\,e} \]

[In]

int(log(c*(a + b/x)^p)/(d + e*x)^2,x)

[Out]

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