3.255 \(\int \frac{a+b \log (c (d+e x)^n)}{x^3 (f+g x)^2} \, dx\)

Optimal. Leaf size=335 \[ -\frac{3 b g^2 n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{f^4}+\frac{3 b g^2 n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{f^4}+\frac{3 g^2 \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac{g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}-\frac{3 g^2 \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac{2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}-\frac{a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}-\frac{b e^2 n \log (x)}{2 d^2 f^2}+\frac{b e^2 n \log (d+e x)}{2 d^2 f^2}-\frac{b e g^2 n \log (d+e x)}{f^3 (e f-d g)}+\frac{b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac{2 b e g n \log (x)}{d f^3}+\frac{2 b e g n \log (d+e x)}{d f^3}-\frac{b e n}{2 d f^2 x} \]

[Out]

-(b*e*n)/(2*d*f^2*x) - (b*e^2*n*Log[x])/(2*d^2*f^2) - (2*b*e*g*n*Log[x])/(d*f^3) + (b*e^2*n*Log[d + e*x])/(2*d
^2*f^2) + (2*b*e*g*n*Log[d + e*x])/(d*f^3) - (b*e*g^2*n*Log[d + e*x])/(f^3*(e*f - d*g)) - (a + b*Log[c*(d + e*
x)^n])/(2*f^2*x^2) + (2*g*(a + b*Log[c*(d + e*x)^n]))/(f^3*x) + (g^2*(a + b*Log[c*(d + e*x)^n]))/(f^3*(f + g*x
)) + (3*g^2*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^4 + (b*e*g^2*n*Log[f + g*x])/(f^3*(e*f - d*g)) - (3*
g^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/f^4 - (3*b*g^2*n*PolyLog[2, -((g*(d + e*x))/(e*
f - d*g))])/f^4 + (3*b*g^2*n*PolyLog[2, 1 + (e*x)/d])/f^4

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Rubi [A]  time = 0.314122, antiderivative size = 335, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {44, 2416, 2395, 36, 29, 31, 2394, 2315, 2393, 2391} \[ -\frac{3 b g^2 n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{f^4}+\frac{3 b g^2 n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{f^4}+\frac{3 g^2 \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac{g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}-\frac{3 g^2 \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac{2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}-\frac{a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}-\frac{b e^2 n \log (x)}{2 d^2 f^2}+\frac{b e^2 n \log (d+e x)}{2 d^2 f^2}-\frac{b e g^2 n \log (d+e x)}{f^3 (e f-d g)}+\frac{b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac{2 b e g n \log (x)}{d f^3}+\frac{2 b e g n \log (d+e x)}{d f^3}-\frac{b e n}{2 d f^2 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/(x^3*(f + g*x)^2),x]

[Out]

-(b*e*n)/(2*d*f^2*x) - (b*e^2*n*Log[x])/(2*d^2*f^2) - (2*b*e*g*n*Log[x])/(d*f^3) + (b*e^2*n*Log[d + e*x])/(2*d
^2*f^2) + (2*b*e*g*n*Log[d + e*x])/(d*f^3) - (b*e*g^2*n*Log[d + e*x])/(f^3*(e*f - d*g)) - (a + b*Log[c*(d + e*
x)^n])/(2*f^2*x^2) + (2*g*(a + b*Log[c*(d + e*x)^n]))/(f^3*x) + (g^2*(a + b*Log[c*(d + e*x)^n]))/(f^3*(f + g*x
)) + (3*g^2*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^4 + (b*e*g^2*n*Log[f + g*x])/(f^3*(e*f - d*g)) - (3*
g^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/f^4 - (3*b*g^2*n*PolyLog[2, -((g*(d + e*x))/(e*
f - d*g))])/f^4 + (3*b*g^2*n*PolyLog[2, 1 + (e*x)/d])/f^4

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])

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && 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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2394

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2393

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx &=\int \left (\frac{a+b \log \left (c (d+e x)^n\right )}{f^2 x^3}-\frac{2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x^2}+\frac{3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 x}-\frac{g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)^2}-\frac{3 g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 (f+g x)}\right ) \, dx\\ &=\frac{\int \frac{a+b \log \left (c (d+e x)^n\right )}{x^3} \, dx}{f^2}-\frac{(2 g) \int \frac{a+b \log \left (c (d+e x)^n\right )}{x^2} \, dx}{f^3}+\frac{\left (3 g^2\right ) \int \frac{a+b \log \left (c (d+e x)^n\right )}{x} \, dx}{f^4}-\frac{\left (3 g^3\right ) \int \frac{a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{f^4}-\frac{g^3 \int \frac{a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx}{f^3}\\ &=-\frac{a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac{2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac{g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac{3 g^2 \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac{3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e (f+g x)}{e f-d g}\right )}{f^4}+\frac{(b e n) \int \frac{1}{x^2 (d+e x)} \, dx}{2 f^2}-\frac{(2 b e g n) \int \frac{1}{x (d+e x)} \, dx}{f^3}-\frac{\left (3 b e g^2 n\right ) \int \frac{\log \left (-\frac{e x}{d}\right )}{d+e x} \, dx}{f^4}+\frac{\left (3 b e g^2 n\right ) \int \frac{\log \left (\frac{e (f+g x)}{e f-d g}\right )}{d+e x} \, dx}{f^4}-\frac{\left (b e g^2 n\right ) \int \frac{1}{(d+e x) (f+g x)} \, dx}{f^3}\\ &=-\frac{a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac{2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac{g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac{3 g^2 \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac{3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e (f+g x)}{e f-d g}\right )}{f^4}+\frac{3 b g^2 n \text{Li}_2\left (1+\frac{e x}{d}\right )}{f^4}+\frac{(b e n) \int \left (\frac{1}{d x^2}-\frac{e}{d^2 x}+\frac{e^2}{d^2 (d+e x)}\right ) \, dx}{2 f^2}-\frac{(2 b e g n) \int \frac{1}{x} \, dx}{d f^3}+\frac{\left (2 b e^2 g n\right ) \int \frac{1}{d+e x} \, dx}{d f^3}+\frac{\left (3 b g^2 n\right ) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{f^4}-\frac{\left (b e^2 g^2 n\right ) \int \frac{1}{d+e x} \, dx}{f^3 (e f-d g)}+\frac{\left (b e g^3 n\right ) \int \frac{1}{f+g x} \, dx}{f^3 (e f-d g)}\\ &=-\frac{b e n}{2 d f^2 x}-\frac{b e^2 n \log (x)}{2 d^2 f^2}-\frac{2 b e g n \log (x)}{d f^3}+\frac{b e^2 n \log (d+e x)}{2 d^2 f^2}+\frac{2 b e g n \log (d+e x)}{d f^3}-\frac{b e g^2 n \log (d+e x)}{f^3 (e f-d g)}-\frac{a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac{2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac{g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac{3 g^2 \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac{b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac{3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e (f+g x)}{e f-d g}\right )}{f^4}-\frac{3 b g^2 n \text{Li}_2\left (-\frac{g (d+e x)}{e f-d g}\right )}{f^4}+\frac{3 b g^2 n \text{Li}_2\left (1+\frac{e x}{d}\right )}{f^4}\\ \end{align*}

Mathematica [A]  time = 0.399546, size = 269, normalized size = 0.8 \[ -\frac{6 b g^2 n \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )-6 b g^2 n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )+\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^2}-\frac{2 f g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}+6 g^2 \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac{4 f g \left (a+b \log \left (c (d+e x)^n\right )\right )}{x}-6 g^2 \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{b e f^2 n (-e x \log (d+e x)+d+e x \log (x))}{d^2 x}+\frac{2 b e f g^2 n (\log (d+e x)-\log (f+g x))}{e f-d g}+\frac{4 b e f g n (\log (x)-\log (d+e x))}{d}}{2 f^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])/(x^3*(f + g*x)^2),x]

[Out]

-((4*b*e*f*g*n*(Log[x] - Log[d + e*x]))/d + (b*e*f^2*n*(d + e*x*Log[x] - e*x*Log[d + e*x]))/(d^2*x) + (f^2*(a
+ b*Log[c*(d + e*x)^n]))/x^2 - (4*f*g*(a + b*Log[c*(d + e*x)^n]))/x - (2*f*g^2*(a + b*Log[c*(d + e*x)^n]))/(f
+ g*x) - 6*g^2*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]) + (2*b*e*f*g^2*n*(Log[d + e*x] - Log[f + g*x]))/(e*f
- d*g) + 6*g^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)] + 6*b*g^2*n*PolyLog[2, (g*(d + e*x))
/(-(e*f) + d*g)] - 6*b*g^2*n*PolyLog[2, 1 + (e*x)/d])/(2*f^4)

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Maple [C]  time = 0.536, size = 1224, normalized size = 3.7 \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))/x^3/(g*x+f)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a{\left (\frac{6 \, g^{2} x^{2} + 3 \, f g x - f^{2}}{f^{3} g x^{3} + f^{4} x^{2}} - \frac{6 \, g^{2} \log \left (g x + f\right )}{f^{4}} + \frac{6 \, g^{2} \log \left (x\right )}{f^{4}}\right )} + b \int \frac{\log \left ({\left (e x + d\right )}^{n}\right ) + \log \left (c\right )}{g^{2} x^{5} + 2 \, f g x^{4} + f^{2} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^3/(g*x+f)^2,x, algorithm="maxima")

[Out]

1/2*a*((6*g^2*x^2 + 3*f*g*x - f^2)/(f^3*g*x^3 + f^4*x^2) - 6*g^2*log(g*x + f)/f^4 + 6*g^2*log(x)/f^4) + b*inte
grate((log((e*x + d)^n) + log(c))/(g^2*x^5 + 2*f*g*x^4 + f^2*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{g^{2} x^{5} + 2 \, f g x^{4} + f^{2} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^3/(g*x+f)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^3/(g*x+f)^2,x, algorithm="giac")

[Out]

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