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

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

[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

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Rubi [A]  time = 0.62, antiderivative size = 370, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 11, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {2398, 2411, 2347, 2344, 2302, 30, 2317, 2374, 6589, 2318, 2391} \[ -\frac {3 b^2 e^2 n^2 \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)^2}+\frac {3 b^3 e^2 n^3 \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}+\frac {3 b^3 e^2 n^3 \text {PolyLog}\left (3,-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}+\frac {3 b^2 e^2 n^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)^2}-\frac {3 b e^2 n \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 g (e f-d g)^2}+\frac {e^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (e f-d g)^2}-\frac {3 b e n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 (f+g x) (e f-d g)^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{2 g (f+g x)^2} \]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2317

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 2318

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 2344

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

Rule 2347

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 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(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*Log
[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 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]

Rule 2398

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 2411

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 6589

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) \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.75, size = 620, normalized size = 1.81 \[ -\frac {3 b^2 n^2 \left (2 e^2 (f+g x)^2 \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )-2 e^2 (f+g x)^2 \log \left (\frac {e (f+g x)}{e f-d g}\right )+g (d+e x) \log ^2(d+e x) (d g-e (2 f+g x))+2 e (f+g x) \log (d+e x) \left (e (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )+g (d+e x)\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )-3 b e^2 n (f+g x)^2 \log (d+e x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2+3 b e^2 n (f+g x)^2 \log (f+g x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2-3 b e n (f+g x) (e f-d g) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2+3 b n (e f-d g)^2 \log (d+e x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2+(e f-d g)^2 \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^3+b^3 n^3 \left (-6 e^2 (f+g x)^2 \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )-6 e^2 (f+g x)^2 \text {Li}_3\left (\frac {g (d+e x)}{d g-e f}\right )-6 e^2 (f+g x)^2 \log (d+e x) \left (\log \left (\frac {e (f+g x)}{e f-d g}\right )-\text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )\right )+g (d+e x) \log ^3(d+e x) (d g-e (2 f+g x))+3 e (f+g x) \log ^2(d+e x) \left (e (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )+g (d+e x)\right )\right )}{2 g (f+g x)^2 (e f-d g)^2} \]

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)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \log \left ({\left (e x + d\right )}^{n} c\right )^{3} + 3 \, a b^{2} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 3 \, a^{2} b \log \left ({\left (e x + d\right )}^{n} c\right ) + a^{3}}{g^{3} x^{3} + 3 \, f g^{2} x^{2} + 3 \, f^{2} g x + f^{3}}, x\right ) \]

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

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

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

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maple [F]  time = 2.13, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )^{3}}{\left (g x +f \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {3}{2} \, a^{2} b e n {\left (\frac {e \log \left (e x + d\right )}{e^{2} f^{2} g - 2 \, d e f g^{2} + d^{2} g^{3}} - \frac {e \log \left (g x + f\right )}{e^{2} f^{2} g - 2 \, d e f g^{2} + d^{2} g^{3}} + \frac {1}{e f^{2} g - d f g^{2} + {\left (e f g^{2} - d g^{3}\right )} x}\right )} - \frac {b^{3} \log \left ({\left (e x + d\right )}^{n}\right )^{3}}{2 \, {\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} - \frac {3 \, a^{2} b \log \left ({\left (e x + d\right )}^{n} c\right )}{2 \, {\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} - \frac {a^{3}}{2 \, {\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} + \int \frac {2 \, b^{3} d g \log \relax (c)^{3} + 6 \, a b^{2} d g \log \relax (c)^{2} + 3 \, {\left (2 \, a b^{2} d g + {\left (e f n + 2 \, d g \log \relax (c)\right )} b^{3} + {\left (2 \, a b^{2} e g + {\left (e g n + 2 \, e g \log \relax (c)\right )} b^{3}\right )} x\right )} \log \left ({\left (e x + d\right )}^{n}\right )^{2} + 2 \, {\left (b^{3} e g \log \relax (c)^{3} + 3 \, a b^{2} e g \log \relax (c)^{2}\right )} x + 6 \, {\left (b^{3} d g \log \relax (c)^{2} + 2 \, a b^{2} d g \log \relax (c) + {\left (b^{3} e g \log \relax (c)^{2} + 2 \, a b^{2} e g \log \relax (c)\right )} x\right )} \log \left ({\left (e x + d\right )}^{n}\right )}{2 \, {\left (e g^{4} x^{4} + d f^{3} g + {\left (3 \, e f g^{3} + d g^{4}\right )} x^{3} + 3 \, {\left (e f^{2} g^{2} + d f g^{3}\right )} x^{2} + {\left (e f^{3} g + 3 \, d f^{2} g^{2}\right )} x\right )}}\,{d x} \]

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

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

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)

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

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)

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