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

Optimal. Leaf size=81 \[ -\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e (d+e x)}+\frac{a p \log (a x+b)}{e (a d-b e)}-\frac{b p \log (d+e x)}{d (a d-b e)}-\frac{p \log (x)}{d e} \]

[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))

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Rubi [A]  time = 0.0774764, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2463, 514, 72} \[ -\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e (d+e x)}+\frac{a p \log (a x+b)}{e (a d-b e)}-\frac{b p \log (d+e x)}{d (a d-b e)}-\frac{p \log (x)}{d e} \]

Antiderivative was successfully verified.

[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 2463

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]

Rule 514

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 72

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]

Rubi steps

\begin{align*} \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{(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]  time = 0.0658213, size = 81, normalized size = 1. \[ -\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e (d+e x)}+\frac{a p \log (a x+b)}{e (a d-b e)}-\frac{b p \log (d+e x)}{d (a d-b e)}-\frac{p \log (x)}{d e} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.543, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( ex+d \right ) ^{2}}\ln \left ( c \left ( a+{\frac{b}{x}} \right ) ^{p} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01932, size = 115, normalized size = 1.42 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.34873, size = 320, normalized size = 3.95 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.25694, size = 163, normalized size = 2.01 \begin{align*} \frac{a d p x e \log \left (a x + b\right ) - a d p x e \log \left (x\right ) + b d p e \log \left (a x + b\right ) - b p x e^{2} \log \left (x e + d\right ) - b d p e \log \left (x e + d\right ) + b p x e^{2} \log \left (x\right ) - a d^{2} \log \left (c\right ) + b d e \log \left (c\right )}{a d^{2} x e^{2} + a d^{3} e - b d x e^{3} - b d^{2} e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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