3.63 \(\int \frac {(a+b \log (c (d+e x)^n))^4}{(f+g x)^2} \, dx\)

Optimal. Leaf size=248 \[ \frac {24 b^3 e n^3 \text {Li}_3\left (-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)}-\frac {12 b^2 e n^2 \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g (e f-d g)}-\frac {4 b e n \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{g (e f-d g)}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^4}{(f+g x) (e f-d g)}-\frac {24 b^4 e n^4 \text {Li}_4\left (-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)} \]

[Out]

(e*x+d)*(a+b*ln(c*(e*x+d)^n))^4/(-d*g+e*f)/(g*x+f)-4*b*e*n*(a+b*ln(c*(e*x+d)^n))^3*ln(e*(g*x+f)/(-d*g+e*f))/g/
(-d*g+e*f)-12*b^2*e*n^2*(a+b*ln(c*(e*x+d)^n))^2*polylog(2,-g*(e*x+d)/(-d*g+e*f))/g/(-d*g+e*f)+24*b^3*e*n^3*(a+
b*ln(c*(e*x+d)^n))*polylog(3,-g*(e*x+d)/(-d*g+e*f))/g/(-d*g+e*f)-24*b^4*e*n^4*polylog(4,-g*(e*x+d)/(-d*g+e*f))
/g/(-d*g+e*f)

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Rubi [A]  time = 0.23, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2397, 2396, 2433, 2374, 2383, 6589} \[ \frac {24 b^3 e n^3 \text {PolyLog}\left (3,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)}-\frac {12 b^2 e n^2 \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g (e f-d g)}-\frac {24 b^4 e n^4 \text {PolyLog}\left (4,-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)}-\frac {4 b e n \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{g (e f-d g)}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^4}{(f+g x) (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)*(a + b*Log[c*(d + e*x)^n])^4)/((e*f - d*g)*(f + g*x)) - (4*b*e*n*(a + b*Log[c*(d + e*x)^n])^3*Log[(
e*(f + g*x))/(e*f - d*g)])/(g*(e*f - d*g)) - (12*b^2*e*n^2*(a + b*Log[c*(d + e*x)^n])^2*PolyLog[2, -((g*(d + e
*x))/(e*f - d*g))])/(g*(e*f - d*g)) + (24*b^3*e*n^3*(a + b*Log[c*(d + e*x)^n])*PolyLog[3, -((g*(d + e*x))/(e*f
- d*g))])/(g*(e*f - d*g)) - (24*b^4*e*n^4*PolyLog[4, -((g*(d + e*x))/(e*f - d*g))])/(g*(e*f - d*g))

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 2383

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[(PolyL
og[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 2396

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 2397

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

Rule 2433

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 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 )^4}{(f+g x)^2} \, dx &=\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^4}{(e f-d g) (f+g x)}-\frac {(4 b e n) \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{f+g x} \, dx}{e f-d g}\\ &=\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^4}{(e f-d g) (f+g x)}-\frac {4 b e n \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 (e f-d g)}+\frac {\left (12 b^2 e^2 n^2\right ) \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 (e f-d g)}\\ &=\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^4}{(e f-d g) (f+g x)}-\frac {4 b e n \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 (e f-d g)}+\frac {\left (12 b^2 e n^2\right ) \operatorname {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 (e f-d g)}\\ &=\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^4}{(e f-d g) (f+g x)}-\frac {4 b e n \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 (e f-d g)}-\frac {12 b^2 e n^2 \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 (e f-d g)}+\frac {\left (24 b^3 e n^3\right ) \operatorname {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 (e f-d g)}\\ &=\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^4}{(e f-d g) (f+g x)}-\frac {4 b e n \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 (e f-d g)}-\frac {12 b^2 e n^2 \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 (e f-d g)}+\frac {24 b^3 e n^3 \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 (e f-d g)}-\frac {\left (24 b^4 e n^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g (e f-d g)}\\ &=\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^4}{(e f-d g) (f+g x)}-\frac {4 b e n \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 (e f-d g)}-\frac {12 b^2 e n^2 \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 (e f-d g)}+\frac {24 b^3 e n^3 \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 (e f-d g)}-\frac {24 b^4 e n^4 \text {Li}_4\left (-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)}\\ \end {align*}

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Mathematica [B]  time = 0.74, size = 531, normalized size = 2.14 \[ \frac {4 b^3 n^3 \left (6 e (f+g x) \text {Li}_3\left (\frac {g (d+e x)}{d g-e f}\right )-6 e (f+g x) \log (d+e x) \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )+\log ^2(d+e x) \left (g (d+e x) \log (d+e x)-3 e (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )+6 b^2 n^2 \left (\log (d+e x) \left (g (d+e x) \log (d+e x)-2 e (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )\right )-2 e (f+g x) \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2+4 b n \left (g (d+e x) \log (d+e x)-e (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^3-(e f-d g) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^4+b^4 n^4 \left (-24 e (f+g x) \text {Li}_4\left (\frac {g (d+e x)}{d g-e f}\right )-12 e (f+g x) \log ^2(d+e x) \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )+24 e (f+g x) \log (d+e x) \text {Li}_3\left (\frac {g (d+e x)}{d g-e f}\right )-4 e (f+g x) \log ^3(d+e x) \log \left (\frac {e (f+g x)}{e f-d g}\right )+g (d+e x) \log ^4(d+e x)\right )}{g (f+g x) (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((e*f - d*g)*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^4) + 4*b*n*(a - b*n*Log[d + e*x] + b*Log[c*(d + e
*x)^n])^3*(g*(d + e*x)*Log[d + e*x] - e*(f + g*x)*Log[(e*(f + g*x))/(e*f - d*g)]) + 6*b^2*n^2*(a - b*n*Log[d +
e*x] + b*Log[c*(d + e*x)^n])^2*(Log[d + e*x]*(g*(d + e*x)*Log[d + e*x] - 2*e*(f + g*x)*Log[(e*(f + g*x))/(e*f
- d*g)]) - 2*e*(f + g*x)*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)]) + 4*b^3*n^3*(a - b*n*Log[d + e*x] + b*Log[
c*(d + e*x)^n])*(Log[d + e*x]^2*(g*(d + e*x)*Log[d + e*x] - 3*e*(f + g*x)*Log[(e*(f + g*x))/(e*f - d*g)]) - 6*
e*(f + g*x)*Log[d + e*x]*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)] + 6*e*(f + g*x)*PolyLog[3, (g*(d + e*x))/(-(
e*f) + d*g)]) + b^4*n^4*(g*(d + e*x)*Log[d + e*x]^4 - 4*e*(f + g*x)*Log[d + e*x]^3*Log[(e*(f + g*x))/(e*f - d*
g)] - 12*e*(f + g*x)*Log[d + e*x]^2*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)] + 24*e*(f + g*x)*Log[d + e*x]*Pol
yLog[3, (g*(d + e*x))/(-(e*f) + d*g)] - 24*e*(f + g*x)*PolyLog[4, (g*(d + e*x))/(-(e*f) + d*g)]))/(g*(e*f - d*
g)*(f + g*x))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{4} \log \left ({\left (e x + d\right )}^{n} c\right )^{4} + 4 \, a b^{3} \log \left ({\left (e x + d\right )}^{n} c\right )^{3} + 6 \, a^{2} b^{2} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 4 \, a^{3} b \log \left ({\left (e x + d\right )}^{n} c\right ) + a^{4}}{g^{2} x^{2} + 2 \, f g x + f^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^4/(g*x+f)^2,x, algorithm="fricas")

[Out]

integral((b^4*log((e*x + d)^n*c)^4 + 4*a*b^3*log((e*x + d)^n*c)^3 + 6*a^2*b^2*log((e*x + d)^n*c)^2 + 4*a^3*b*l
og((e*x + d)^n*c) + a^4)/(g^2*x^2 + 2*f*g*x + f^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^4/(g*x+f)^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 1.47, size = 21740, normalized size = 87.66 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*(e*x+d)^n)+a)^4/(g*x+f)^2,x)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^4/(g*x+f)^2,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))**4/(g*x+f)**2,x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))**4/(f + g*x)**2, x)

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