\(\int \frac {(a+b \log (c (d+e x)^n))^3}{f+g x} \, dx\) [56]

Optimal result

Integrand size = 24, antiderivative size = 158 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g}+\frac {3 b n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g}-\frac {6 b^2 n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \operatorname {PolyLog}\left (3,-\frac {g (d+e x)}{e f-d g}\right )}{g}+\frac {6 b^3 n^3 \operatorname {PolyLog}\left (4,-\frac {g (d+e x)}{e f-d g}\right )}{g} \]

[Out]

(a+b*ln(c*(e*x+d)^n))^3*ln(e*(g*x+f)/(-d*g+e*f))/g+3*b*n*(a+b*ln(c*(e*x+d)^n))^2*polylog(2,-g*(e*x+d)/(-d*g+e*
f))/g-6*b^2*n^2*(a+b*ln(c*(e*x+d)^n))*polylog(3,-g*(e*x+d)/(-d*g+e*f))/g+6*b^3*n^3*polylog(4,-g*(e*x+d)/(-d*g+
e*f))/g

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2443, 2481, 2421, 2430, 6724} \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=-\frac {6 b^2 n^2 \operatorname {PolyLog}\left (3,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}+\frac {3 b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g}+\frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{g}+\frac {6 b^3 n^3 \operatorname {PolyLog}\left (4,-\frac {g (d+e x)}{e f-d g}\right )}{g} \]

[In]

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

[Out]

((a + b*Log[c*(d + e*x)^n])^3*Log[(e*(f + g*x))/(e*f - d*g)])/g + (3*b*n*(a + b*Log[c*(d + e*x)^n])^2*PolyLog[
2, -((g*(d + e*x))/(e*f - d*g))])/g - (6*b^2*n^2*(a + b*Log[c*(d + e*x)^n])*PolyLog[3, -((g*(d + e*x))/(e*f -
d*g))])/g + (6*b^3*n^3*PolyLog[4, -((g*(d + e*x))/(e*f - d*g))])/g

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 2430

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

Rule 2443

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

Rule 2481

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(k*(x/d))^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + j*(x/e))^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

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*} \text {integral}& = \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g}-\frac {(3 b e n) \int \frac {\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 )}{d+e x} \, dx}{g} \\ & = \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g}-\frac {(3 b n) \text {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (\frac {e \left (\frac {e f-d g}{e}+\frac {g x}{e}\right )}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g} \\ & = \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g}+\frac {3 b n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g}-\frac {\left (6 b^2 n^2\right ) \text {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g} \\ & = \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g}+\frac {3 b n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g}-\frac {6 b^2 n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \text {Li}_3\left (-\frac {g (d+e x)}{e f-d g}\right )}{g}+\frac {\left (6 b^3 n^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g} \\ & = \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g}+\frac {3 b n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g}-\frac {6 b^2 n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \text {Li}_3\left (-\frac {g (d+e x)}{e f-d g}\right )}{g}+\frac {6 b^3 n^3 \text {Li}_4\left (-\frac {g (d+e x)}{e f-d g}\right )}{g} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(335\) vs. \(2(158)=316\).

Time = 0.18 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.12 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\frac {\left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^3 \log (f+g x)+3 b n \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2 \left (\log (d+e x) \log \left (\frac {e (f+g x)}{e f-d g}\right )+\operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )\right )+6 b^2 n^2 \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \left (\frac {1}{2} \log ^2(d+e x) \log \left (\frac {e (f+g x)}{e f-d g}\right )+\log (d+e x) \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )-\operatorname {PolyLog}\left (3,\frac {g (d+e x)}{-e f+d g}\right )\right )+b^3 n^3 \left (\log ^3(d+e x) \log \left (\frac {e (f+g x)}{e f-d g}\right )+3 \log ^2(d+e x) \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )-6 \log (d+e x) \operatorname {PolyLog}\left (3,\frac {g (d+e x)}{-e f+d g}\right )+6 \operatorname {PolyLog}\left (4,\frac {g (d+e x)}{-e f+d g}\right )\right )}{g} \]

[In]

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

[Out]

((a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^3*Log[f + g*x] + 3*b*n*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x
)^n])^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*b^2*n^2*(
a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])*((Log[d + e*x]^2*Log[(e*(f + g*x))/(e*f - d*g)])/2 + Log[d + e*x]
*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)] - PolyLog[3, (g*(d + e*x))/(-(e*f) + d*g)]) + b^3*n^3*(Log[d + e*x]^
3*Log[(e*(f + g*x))/(e*f - d*g)] + 3*Log[d + e*x]^2*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)] - 6*Log[d + e*x]*
PolyLog[3, (g*(d + e*x))/(-(e*f) + d*g)] + 6*PolyLog[4, (g*(d + e*x))/(-(e*f) + d*g)]))/g

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.99 (sec) , antiderivative size = 1396, normalized size of antiderivative = 8.84

method	result	size
risch	\(\text {Expression too large to display}\)	\(1396\)

[In]

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

[Out]

-b^3*ln(g*(e*x+d)-d*g+e*f)/g*ln(e*x+d)^3*n^3+3*b^3*ln(g*(e*x+d)-d*g+e*f)/g*ln((e*x+d)^n)*ln(e*x+d)^2*n^2-3*b^3
*ln(g*(e*x+d)-d*g+e*f)/g*ln((e*x+d)^n)^2*ln(e*x+d)*n+b^3*ln(g*(e*x+d)-d*g+e*f)/g*ln((e*x+d)^n)^3-2*b^3*n^3/g*l
n(e*x+d)^3*ln(1-g*(e*x+d)/(d*g-e*f))-3*b^3*n^3/g*ln(e*x+d)^2*polylog(2,g*(e*x+d)/(d*g-e*f))+6*b^3*n^3/g*polylo
g(4,g*(e*x+d)/(d*g-e*f))+3*b^3*n^3*dilog((g*(e*x+d)-d*g+e*f)/(-d*g+e*f))/g*ln(e*x+d)^2-6*b^3*n^2*dilog((g*(e*x
+d)-d*g+e*f)/(-d*g+e*f))/g*ln((e*x+d)^n)*ln(e*x+d)+3*b^3*n*dilog((g*(e*x+d)-d*g+e*f)/(-d*g+e*f))/g*ln((e*x+d)^
n)^2+3*b^3*n^3*ln(e*x+d)^3*ln((g*(e*x+d)-d*g+e*f)/(-d*g+e*f))/g-6*b^3*n^2*ln(e*x+d)^2*ln((g*(e*x+d)-d*g+e*f)/(
-d*g+e*f))/g*ln((e*x+d)^n)+3*b^3*n*ln(e*x+d)*ln((g*(e*x+d)-d*g+e*f)/(-d*g+e*f))/g*ln((e*x+d)^n)^2+3*b^3*n^2/g*
ln((e*x+d)^n)*ln(e*x+d)^2*ln(1-g*(e*x+d)/(d*g-e*f))+6*b^3*n^2/g*ln((e*x+d)^n)*ln(e*x+d)*polylog(2,g*(e*x+d)/(d
*g-e*f))-6*b^3*n^2/g*ln((e*x+d)^n)*polylog(3,g*(e*x+d)/(d*g-e*f))+1/8*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*c
sgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*cs
gn(I*c*(e*x+d)^n)^3*b+2*b*ln(c)+2*a)^3*ln(g*x+f)/g+3/2*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n
)+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^
n)^3*b+2*b*ln(c)+2*a)*b^2*((ln((e*x+d)^n)-n*ln(e*x+d))^2*ln(g*(e*x+d)-d*g+e*f)/g+n^2/g*ln(e*x+d)^2*ln(1-g*(e*x
+d)/(d*g-e*f))+2*n^2/g*ln(e*x+d)*polylog(2,g*(e*x+d)/(d*g-e*f))-2*n^2/g*polylog(3,g*(e*x+d)/(d*g-e*f))+2*n*(ln
((e*x+d)^n)-n*ln(e*x+d))*dilog((g*(e*x+d)-d*g+e*f)/(-d*g+e*f))/g+2*n*(ln((e*x+d)^n)-n*ln(e*x+d))*ln(e*x+d)*ln(
(g*(e*x+d)-d*g+e*f)/(-d*g+e*f))/g)+3/4*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)+I*Pi*csgn(I*c)
*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n)^3*b+2*b*ln(c)
+2*a)^2*b*(ln((e*x+d)^n)*ln(g*x+f)/g-1/g*n*e*(dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))/e+ln(g*x+f)*ln(((g*x+f)*e+d
*g-e*f)/(d*g-e*f))/e))

Fricas [F]

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

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^3}{f+g\,x} \,d x \]

[In]

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

[Out]

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