3.308 \(\int \frac{a+b \log (c (d+e x)^n)}{(f+\frac{g}{x})^3 x^3} \, dx\)

Optimal. Leaf size=112 \[ -\frac{a+b \log \left (c (d+e x)^n\right )}{2 f (f x+g)^2}+\frac{b e^2 n \log (d+e x)}{2 f (d f-e g)^2}-\frac{b e^2 n \log (f x+g)}{2 f (d f-e g)^2}-\frac{b e n}{2 f (f x+g) (d f-e g)} \]

[Out]

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

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Rubi [A]  time = 0.11776, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2412, 2395, 44} \[ -\frac{a+b \log \left (c (d+e x)^n\right )}{2 f (f x+g)^2}+\frac{b e^2 n \log (d+e x)}{2 f (d f-e g)^2}-\frac{b e^2 n \log (f x+g)}{2 f (d f-e g)^2}-\frac{b e n}{2 f (f x+g) (d f-e g)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/((f + g/x)^3*x^3),x]

[Out]

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

Rule 2412

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol]
:> Int[(g + f*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q}, x] && EqQ[m,
q] && IntegerQ[q]

Rule 2395

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.102319, size = 83, normalized size = 0.74 \[ -\frac{a+b \log \left (c (d+e x)^n\right )-\frac{b e n (f x+g) (e (f x+g) \log (d+e x)-d f-e (f x+g) \log (f x+g)+e g)}{(d f-e g)^2}}{2 f (f x+g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.347, size = 633, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/f/(f*x+g)^2*ln((e*x+d)^n)-1/4*(2*I*Pi*b*d*e*f*g*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-I*Pi*b*
e^2*g^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+2*a*f^2*d^2+2*I*Pi*b*d*e*f*g*csgn(I*c*(e*x+d)^n)^3-4*l
n(c)*b*d*e*f*g-4*a*d*e*f*g+2*ln(c)*b*d^2*f^2+2*ln(c)*b*e^2*g^2+2*b*d*e*f^2*n*x-2*b*e^2*g^2*n+2*ln(f*x+g)*b*e^2
*g^2*n-2*ln(-e*x-d)*b*e^2*g^2*n+2*a*e^2*g^2-I*Pi*b*d^2*f^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-2*I
*Pi*b*d*e*f*g*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*Pi*b*e^2*g^2*csgn(I*c*(e*x+d)^n)^3-I*Pi*b*d^2*f^2*csgn
(I*c*(e*x+d)^n)^3-2*b*e^2*f*g*n*x+2*b*d*e*f*g*n-2*ln(-e*x-d)*b*e^2*f^2*n*x^2+I*Pi*b*e^2*g^2*csgn(I*c)*csgn(I*c
*(e*x+d)^n)^2+I*Pi*b*e^2*g^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+I*Pi*b*d^2*f^2*csgn(I*c)*csgn(I*c*(e*x+d)
^n)^2+2*ln(f*x+g)*b*e^2*f^2*n*x^2-4*ln(-e*x-d)*b*e^2*f*g*n*x-2*I*Pi*b*d*e*f*g*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+
I*Pi*b*d^2*f^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+4*ln(f*x+g)*b*e^2*f*g*n*x)/(f*x+g)^2/(d*f-e*g)^2/f

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Maxima [A]  time = 1.11387, size = 228, normalized size = 2.04 \begin{align*} \frac{1}{2} \, b e n{\left (\frac{e \log \left (e x + d\right )}{d^{2} f^{3} - 2 \, d e f^{2} g + e^{2} f g^{2}} - \frac{e \log \left (f x + g\right )}{d^{2} f^{3} - 2 \, d e f^{2} g + e^{2} f g^{2}} - \frac{1}{d f^{2} g - e f g^{2} +{\left (d f^{3} - e f^{2} g\right )} x}\right )} - \frac{b \log \left ({\left (e x + d\right )}^{n} c\right )}{2 \,{\left (f^{3} x^{2} + 2 \, f^{2} g x + f g^{2}\right )}} - \frac{a}{2 \,{\left (f^{3} x^{2} + 2 \, f^{2} g x + f g^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*e*n*(e*log(e*x + d)/(d^2*f^3 - 2*d*e*f^2*g + e^2*f*g^2) - e*log(f*x + g)/(d^2*f^3 - 2*d*e*f^2*g + e^2*f*
g^2) - 1/(d*f^2*g - e*f*g^2 + (d*f^3 - e*f^2*g)*x)) - 1/2*b*log((e*x + d)^n*c)/(f^3*x^2 + 2*f^2*g*x + f*g^2) -
1/2*a/(f^3*x^2 + 2*f^2*g*x + f*g^2)

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Fricas [B]  time = 2.01476, size = 579, normalized size = 5.17 \begin{align*} -\frac{a d^{2} f^{2} - 2 \, a d e f g + a e^{2} g^{2} +{\left (b d e f^{2} - b e^{2} f g\right )} n x +{\left (b d e f g - b e^{2} g^{2}\right )} n -{\left (b e^{2} f^{2} n x^{2} + 2 \, b e^{2} f g n x -{\left (b d^{2} f^{2} - 2 \, b d e f g\right )} n\right )} \log \left (e x + d\right ) +{\left (b e^{2} f^{2} n x^{2} + 2 \, b e^{2} f g n x + b e^{2} g^{2} n\right )} \log \left (f x + g\right ) +{\left (b d^{2} f^{2} - 2 \, b d e f g + b e^{2} g^{2}\right )} \log \left (c\right )}{2 \,{\left (d^{2} f^{3} g^{2} - 2 \, d e f^{2} g^{3} + e^{2} f g^{4} +{\left (d^{2} f^{5} - 2 \, d e f^{4} g + e^{2} f^{3} g^{2}\right )} x^{2} + 2 \,{\left (d^{2} f^{4} g - 2 \, d e f^{3} g^{2} + e^{2} f^{2} g^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*d^2*f^2 - 2*a*d*e*f*g + a*e^2*g^2 + (b*d*e*f^2 - b*e^2*f*g)*n*x + (b*d*e*f*g - b*e^2*g^2)*n - (b*e^2*f
^2*n*x^2 + 2*b*e^2*f*g*n*x - (b*d^2*f^2 - 2*b*d*e*f*g)*n)*log(e*x + d) + (b*e^2*f^2*n*x^2 + 2*b*e^2*f*g*n*x +
b*e^2*g^2*n)*log(f*x + g) + (b*d^2*f^2 - 2*b*d*e*f*g + b*e^2*g^2)*log(c))/(d^2*f^3*g^2 - 2*d*e*f^2*g^3 + e^2*f
*g^4 + (d^2*f^5 - 2*d*e*f^4*g + e^2*f^3*g^2)*x^2 + 2*(d^2*f^4*g - 2*d*e*f^3*g^2 + e^2*f^2*g^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((a+b*ln(c*(e*x+d)**n))/(f+g/x)**3/x**3,x)

[Out]

Timed out

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Giac [B]  time = 1.17938, size = 408, normalized size = 3.64 \begin{align*} -\frac{b f^{2} n x^{2} e^{2} \log \left (f x + g\right ) - b f^{2} n x^{2} e^{2} \log \left (x e + d\right ) + b d f^{2} n x e + 2 \, b f g n x e^{2} \log \left (f x + g\right ) + b d^{2} f^{2} n \log \left (x e + d\right ) - 2 \, b f g n x e^{2} \log \left (x e + d\right ) - 2 \, b d f g n e \log \left (x e + d\right ) - b f g n x e^{2} + b d f g n e + b g^{2} n e^{2} \log \left (f x + g\right ) + b d^{2} f^{2} \log \left (c\right ) - 2 \, b d f g e \log \left (c\right ) + a d^{2} f^{2} - b g^{2} n e^{2} - 2 \, a d f g e + b g^{2} e^{2} \log \left (c\right ) + a g^{2} e^{2}}{2 \,{\left (d^{2} f^{5} x^{2} - 2 \, d f^{4} g x^{2} e + 2 \, d^{2} f^{4} g x + f^{3} g^{2} x^{2} e^{2} - 4 \, d f^{3} g^{2} x e + d^{2} f^{3} g^{2} + 2 \, f^{2} g^{3} x e^{2} - 2 \, d f^{2} g^{3} e + f g^{4} e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*f^2*n*x^2*e^2*log(f*x + g) - b*f^2*n*x^2*e^2*log(x*e + d) + b*d*f^2*n*x*e + 2*b*f*g*n*x*e^2*log(f*x +
g) + b*d^2*f^2*n*log(x*e + d) - 2*b*f*g*n*x*e^2*log(x*e + d) - 2*b*d*f*g*n*e*log(x*e + d) - b*f*g*n*x*e^2 + b*
d*f*g*n*e + b*g^2*n*e^2*log(f*x + g) + b*d^2*f^2*log(c) - 2*b*d*f*g*e*log(c) + a*d^2*f^2 - b*g^2*n*e^2 - 2*a*d
*f*g*e + b*g^2*e^2*log(c) + a*g^2*e^2)/(d^2*f^5*x^2 - 2*d*f^4*g*x^2*e + 2*d^2*f^4*g*x + f^3*g^2*x^2*e^2 - 4*d*
f^3*g^2*x*e + d^2*f^3*g^2 + 2*f^2*g^3*x*e^2 - 2*d*f^2*g^3*e + f*g^4*e^2)