3.1.58 \(\int \frac {(a+b \log (c (d+e x)^n))^3}{(f+g x)^3} \, dx\) [58]

Optimal. Leaf size=342 \[ -\frac {3 b e n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 (e f-d g)^2 (f+g x)}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2}+\frac {3 b^2 e^2 n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g (e f-d g)^2}-\frac {3 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (1+\frac {e f-d g}{g (d+e x)}\right )}{2 g (e f-d g)^2}+\frac {3 b^2 e^2 n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \text {Li}_2\left (-\frac {e f-d g}{g (d+e x)}\right )}{g (e f-d g)^2}+\frac {3 b^3 e^2 n^3 \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}+\frac {3 b^3 e^2 n^3 \text {Li}_3\left (-\frac {e f-d g}{g (d+e x)}\right )}{g (e f-d g)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.38, antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2445, 2458, 2389, 2379, 2421, 6724, 2355, 2354, 2438} \begin {gather*} \frac {3 b^2 e^2 n^2 \text {PolyLog}\left (2,-\frac {e f-d g}{g (d+e x)}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)^2}+\frac {3 b^3 e^2 n^3 \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}+\frac {3 b^3 e^2 n^3 \text {PolyLog}\left (3,-\frac {e f-d g}{g (d+e x)}\right )}{g (e f-d g)^2}+\frac {3 b^2 e^2 n^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 )}{g (e f-d g)^2}-\frac {3 b e^2 n \log \left (\frac {e f-d g}{g (d+e x)}+1\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 g (e f-d g)^2}-\frac {3 b e n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 (f+g x) (e f-d g)^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2354

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

Rule 2355

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_))^2, x_Symbol] :> Simp[x*((a + b*Log[c*x^n])
^p/(d*(d + e*x))), x] - Dist[b*n*(p/d), Int[(a + b*Log[c*x^n])^(p - 1)/(d + e*x), x], x] /; FreeQ[{a, b, c, d,
e, n, p}, x] && GtQ[p, 0]

Rule 2379

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

Rule 2389

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[(d
+ e*x)^(q + 1)*((a + b*Log[c*x^n])^p/x), x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c*x^n])^p/m), x] + Dist[b*n*(p/m), Int[PolyLog[2, (-d)*f*x^m]*((a + b*L
og[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2438

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]

Rule 2445

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f
+ g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])^p/(g*(q + 1))), x] - Dist[b*e*n*(p/(g*(q + 1))), Int[(f + g*x)^(q
+ 1)*((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{(f+g x)^3} \, dx &=-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2}+\frac {(3 b e n) \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(d+e x) (f+g x)^2} \, dx}{2 g}\\ &=-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2}+\frac {(3 b n) \text {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x \left (\frac {e f-d g}{e}+\frac {g x}{e}\right )^2} \, dx,x,d+e x\right )}{2 g}\\ &=-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2}-\frac {(3 b n) \text {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{\left (\frac {e f-d g}{e}+\frac {g x}{e}\right )^2} \, dx,x,d+e x\right )}{2 (e f-d g)}+\frac {(3 b e n) \text {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x \left (\frac {e f-d g}{e}+\frac {g x}{e}\right )} \, dx,x,d+e x\right )}{2 g (e f-d g)}\\ &=-\frac {3 b e n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 (e f-d g)^2 (f+g x)}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2}-\frac {(3 b e n) \text {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{\frac {e f-d g}{e}+\frac {g x}{e}} \, dx,x,d+e x\right )}{2 (e f-d g)^2}+\frac {\left (3 b e^2 n\right ) \text {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x} \, dx,x,d+e x\right )}{2 g (e f-d g)^2}+\frac {\left (3 b^2 e n^2\right ) \text {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{\frac {e f-d g}{e}+\frac {g x}{e}} \, dx,x,d+e x\right )}{(e f-d g)^2}\\ &=-\frac {3 b e n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 (e f-d g)^2 (f+g x)}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2}+\frac {3 b^2 e^2 n^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 )}{g (e f-d g)^2}-\frac {3 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{2 g (e f-d g)^2}+\frac {\left (3 e^2\right ) \text {Subst}\left (\int x^2 \, dx,x,a+b \log \left (c (d+e x)^n\right )\right )}{2 g (e f-d g)^2}+\frac {\left (3 b^2 e^2 n^2\right ) \text {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g (e f-d g)^2}-\frac {\left (3 b^3 e^2 n^3\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g (e f-d g)^2}\\ &=-\frac {3 b e n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 (e f-d g)^2 (f+g x)}+\frac {e^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (e f-d g)^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2}+\frac {3 b^2 e^2 n^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 )}{g (e f-d g)^2}-\frac {3 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{2 g (e f-d g)^2}+\frac {3 b^3 e^2 n^3 \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}-\frac {3 b^2 e^2 n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}+\frac {\left (3 b^3 e^2 n^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g (e f-d g)^2}\\ &=-\frac {3 b e n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 (e f-d g)^2 (f+g x)}+\frac {e^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (e f-d g)^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2}+\frac {3 b^2 e^2 n^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 )}{g (e f-d g)^2}-\frac {3 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{2 g (e f-d g)^2}+\frac {3 b^3 e^2 n^3 \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}-\frac {3 b^2 e^2 n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}+\frac {3 b^3 e^2 n^3 \text {Li}_3\left (-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.48, size = 620, normalized size = 1.81 \begin {gather*} -\frac {-3 b e (e f-d g) n (f+g x) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2+3 b (e f-d g)^2 n \log (d+e x) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2-3 b e^2 n (f+g x)^2 \log (d+e x) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2+(e f-d g)^2 \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^3+3 b e^2 n (f+g x)^2 \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2 \log (f+g x)+3 b^2 n^2 \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \left (g (d+e x) (d g-e (2 f+g x)) \log ^2(d+e x)-2 e^2 (f+g x)^2 \log \left (\frac {e (f+g x)}{e f-d g}\right )+2 e (f+g x) \log (d+e x) \left (g (d+e x)+e (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )\right )+2 e^2 (f+g x)^2 \text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )\right )+b^3 n^3 \left (g (d+e x) (d g-e (2 f+g x)) \log ^3(d+e x)+3 e (f+g x) \log ^2(d+e x) \left (g (d+e x)+e (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )\right )-6 e^2 (f+g x)^2 \log (d+e x) \left (\log \left (\frac {e (f+g x)}{e f-d g}\right )-\text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )\right )-6 e^2 (f+g x)^2 \text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )-6 e^2 (f+g x)^2 \text {Li}_3\left (\frac {g (d+e x)}{-e f+d g}\right )\right )}{2 g (e f-d g)^2 (f+g x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(-3*b*e*(e*f - d*g)*n*(f + g*x)*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 + 3*b*(e*f - d*g)^2*n*Log
[d + e*x]*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 - 3*b*e^2*n*(f + g*x)^2*Log[d + e*x]*(a - b*n*Log[d
+ e*x] + b*Log[c*(d + e*x)^n])^2 + (e*f - d*g)^2*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^3 + 3*b*e^2*n*(
f + g*x)^2*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2*Log[f + g*x] + 3*b^2*n^2*(a - b*n*Log[d + e*x] + b*
Log[c*(d + e*x)^n])*(g*(d + e*x)*(d*g - e*(2*f + g*x))*Log[d + e*x]^2 - 2*e^2*(f + g*x)^2*Log[(e*(f + g*x))/(e
*f - d*g)] + 2*e*(f + g*x)*Log[d + e*x]*(g*(d + e*x) + e*(f + g*x)*Log[(e*(f + g*x))/(e*f - d*g)]) + 2*e^2*(f
+ g*x)^2*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)]) + b^3*n^3*(g*(d + e*x)*(d*g - e*(2*f + g*x))*Log[d + e*x]^3
+ 3*e*(f + g*x)*Log[d + e*x]^2*(g*(d + e*x) + e*(f + g*x)*Log[(e*(f + g*x))/(e*f - d*g)]) - 6*e^2*(f + g*x)^2
*Log[d + e*x]*(Log[(e*(f + g*x))/(e*f - d*g)] - PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)]) - 6*e^2*(f + g*x)^2*
PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)] - 6*e^2*(f + g*x)^2*PolyLog[3, (g*(d + e*x))/(-(e*f) + d*g)]))/(g*(e*
f - d*g)^2*(f + g*x)^2)

________________________________________________________________________________________

Maple [F]
time = 0.17, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )^{3}}{\left (g x +f \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{3}}{\left (f + g x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^3}{{\left (f+g\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________