3.203 \(\int \frac{\log (c (a+\frac{b}{x})^p)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=127 \[ \frac{a^2 p \log (a x+b)}{2 e (a d-b e)^2}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 e (d+e x)^2}-\frac{b p (2 a d-b e) \log (d+e x)}{2 d^2 (a d-b e)^2}+\frac{b p}{2 d (d+e x) (a d-b e)}-\frac{p \log (x)}{2 d^2 e} \]

[Out]

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

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Rubi [A]  time = 0.11854, antiderivative size = 127, 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} \[ \frac{a^2 p \log (a x+b)}{2 e (a d-b e)^2}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 e (d+e x)^2}-\frac{b p (2 a d-b e) \log (d+e x)}{2 d^2 (a d-b e)^2}+\frac{b p}{2 d (d+e x) (a d-b e)}-\frac{p \log (x)}{2 d^2 e} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x)^p]/(d + e*x)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.196341, size = 113, normalized size = 0.89 \[ \frac{\frac{a^2 p \log (a x+b)}{(a d-b e)^2}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{(d+e x)^2}+\frac{b e p (b e-2 a d) \log (d+e x)}{d^2 (a d-b e)^2}+\frac{b e p}{d (d+e x) (a d-b e)}-\frac{p \log (x)}{d^2}}{2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x)^p]/(d + e*x)^3,x]

[Out]

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

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Maple [F]  time = 0.544, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( ex+d \right ) ^{3}}\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)^3,x)

[Out]

int(ln(c*(a+b/x)^p)/(e*x+d)^3,x)

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Maxima [A]  time = 1.05119, size = 216, normalized size = 1.7 \begin{align*} \frac{{\left (\frac{a^{2} \log \left (a x + b\right )}{a^{2} b d^{2} - 2 \, a b^{2} d e + b^{3} e^{2}} - \frac{{\left (2 \, a d e - b e^{2}\right )} \log \left (e x + d\right )}{a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}} + \frac{e}{a d^{3} - b d^{2} e +{\left (a d^{2} e - b d e^{2}\right )} x} - \frac{\log \left (x\right )}{b d^{2}}\right )} b p}{2 \, e} - \frac{\log \left ({\left (a + \frac{b}{x}\right )}^{p} c\right )}{2 \,{\left (e x + d\right )}^{2} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 6.63349, size = 864, normalized size = 6.8 \begin{align*} \frac{{\left (a b d^{2} e^{2} - b^{2} d e^{3}\right )} p x -{\left (a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}\right )} p \log \left (\frac{a x + b}{x}\right ) +{\left (a b d^{3} e - b^{2} d^{2} e^{2}\right )} p +{\left (a^{2} d^{2} e^{2} p x^{2} + 2 \, a^{2} d^{3} e p x + a^{2} d^{4} p\right )} \log \left (a x + b\right ) -{\left ({\left (2 \, a b d e^{3} - b^{2} e^{4}\right )} p x^{2} + 2 \,{\left (2 \, a b d^{2} e^{2} - b^{2} d e^{3}\right )} p x +{\left (2 \, a b d^{3} e - b^{2} d^{2} e^{2}\right )} p\right )} \log \left (e x + d\right ) -{\left (a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}\right )} \log \left (c\right ) -{\left ({\left (a^{2} d^{2} e^{2} - 2 \, a b d e^{3} + b^{2} e^{4}\right )} p x^{2} + 2 \,{\left (a^{2} d^{3} e - 2 \, a b d^{2} e^{2} + b^{2} d e^{3}\right )} p x +{\left (a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2}\right )} p\right )} \log \left (x\right )}{2 \,{\left (a^{2} d^{6} e - 2 \, a b d^{5} e^{2} + b^{2} d^{4} e^{3} +{\left (a^{2} d^{4} e^{3} - 2 \, a b d^{3} e^{4} + b^{2} d^{2} e^{5}\right )} x^{2} + 2 \,{\left (a^{2} d^{5} e^{2} - 2 \, a b d^{4} e^{3} + b^{2} d^{3} e^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/(e*x+d)^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.2969, size = 579, normalized size = 4.56 \begin{align*} \frac{a^{2} d^{2} p x^{2} e^{2} \log \left (a x + b\right ) + 2 \, a^{2} d^{3} p x e \log \left (a x + b\right ) - a^{2} d^{2} p x^{2} e^{2} \log \left (x\right ) - 2 \, a^{2} d^{3} p x e \log \left (x\right ) + 2 \, a b d^{3} p e \log \left (a x + b\right ) - 2 \, a b d p x^{2} e^{3} \log \left (x e + d\right ) - 4 \, a b d^{2} p x e^{2} \log \left (x e + d\right ) - 2 \, a b d^{3} p e \log \left (x e + d\right ) + 2 \, a b d p x^{2} e^{3} \log \left (x\right ) + 4 \, a b d^{2} p x e^{2} \log \left (x\right ) + a b d^{2} p x e^{2} + a b d^{3} p e - b^{2} d^{2} p e^{2} \log \left (a x + b\right ) + b^{2} p x^{2} e^{4} \log \left (x e + d\right ) + 2 \, b^{2} d p x e^{3} \log \left (x e + d\right ) + b^{2} d^{2} p e^{2} \log \left (x e + d\right ) - a^{2} d^{4} \log \left (c\right ) + 2 \, a b d^{3} e \log \left (c\right ) - b^{2} p x^{2} e^{4} \log \left (x\right ) - 2 \, b^{2} d p x e^{3} \log \left (x\right ) - b^{2} d p x e^{3} - b^{2} d^{2} p e^{2} - b^{2} d^{2} e^{2} \log \left (c\right )}{2 \,{\left (a^{2} d^{4} x^{2} e^{3} + 2 \, a^{2} d^{5} x e^{2} + a^{2} d^{6} e - 2 \, a b d^{3} x^{2} e^{4} - 4 \, a b d^{4} x e^{3} - 2 \, a b d^{5} e^{2} + b^{2} d^{2} x^{2} e^{5} + 2 \, b^{2} d^{3} x e^{4} + b^{2} d^{4} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/(e*x+d)^3,x, algorithm="giac")

[Out]

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