3.226 \(\int \frac{(a+b \log (c (d+e x)^n))^2}{f+g x} \, dx\)
Optimal. Leaf size=111 \[ \frac{2 b n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac{2 b^2 n^2 \text{PolyLog}\left (3,-\frac{g (d+e x)}{e f-d g}\right )}{g}+\frac{\log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g} \]
[Out]
((a + b*Log[c*(d + e*x)^n])^2*Log[(e*(f + g*x))/(e*f - d*g)])/g + (2*b*n*(a + b*Log[c*(d + e*x)^n])*PolyLog[2,
-((g*(d + e*x))/(e*f - d*g))])/g - (2*b^2*n^2*PolyLog[3, -((g*(d + e*x))/(e*f - d*g))])/g
________________________________________________________________________________________
Rubi [A] time = 0.113661, antiderivative size = 111, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used =
{2396, 2433, 2374, 6589} \[ \frac{2 b n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac{2 b^2 n^2 \text{PolyLog}\left (3,-\frac{g (d+e x)}{e f-d g}\right )}{g}+\frac{\log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Log[c*(d + e*x)^n])^2/(f + g*x),x]
[Out]
((a + b*Log[c*(d + e*x)^n])^2*Log[(e*(f + g*x))/(e*f - d*g)])/g + (2*b*n*(a + b*Log[c*(d + e*x)^n])*PolyLog[2,
-((g*(d + e*x))/(e*f - d*g))])/g - (2*b^2*n^2*PolyLog[3, -((g*(d + e*x))/(e*f - d*g))])/g
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 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 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 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 )^2}{f+g x} \, dx &=\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 )}{g}-\frac{(2 b e n) \int \frac{\left (a+b \log \left (c (d+e x)^n\right )\right ) \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 )^2 \log \left (\frac{e (f+g x)}{e f-d g}\right )}{g}-\frac{(2 b n) \operatorname{Subst}\left (\int \frac{\left (a+b \log \left (c x^n\right )\right ) \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 )^2 \log \left (\frac{e (f+g x)}{e f-d g}\right )}{g}+\frac{2 b n \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}-\frac{\left (2 b^2 n^2\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}\\ &=\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 )}{g}+\frac{2 b n \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}-\frac{2 b^2 n^2 \text{Li}_3\left (-\frac{g (d+e x)}{e f-d g}\right )}{g}\\ \end{align*}
Mathematica [B] time = 0.235963, size = 226, normalized size = 2.04 \[ \frac{b \left (2 n \left (\text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )+\log (d+e x) \log \left (\frac{e (f+g x)}{e f-d g}\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )+b n^2 \left (-2 \text{PolyLog}\left (3,\frac{g (d+e x)}{d g-e f}\right )+2 \log (d+e x) \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )+\log ^2(d+e x) \log \left (\frac{e (f+g x)}{e f-d g}\right )\right )+\log (f+g x) \left (\log \left (c (d+e x)^n\right )-n \log (d+e x)\right ) \left (2 a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )\right )}{g}+\frac{a^2 \log (f+g x)}{g} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Log[c*(d + e*x)^n])^2/(f + g*x),x]
[Out]
(a^2*Log[f + g*x])/g + (b*((-(n*Log[d + e*x]) + Log[c*(d + e*x)^n])*(2*a - b*n*Log[d + e*x] + b*Log[c*(d + e*x
)^n])*Log[f + g*x] + 2*n*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])*(Log[d + e*x]*Log[(e*(f + g*x))/(e*f -
d*g)] + PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)]) + b*n^2*(Log[d + e*x]^2*Log[(e*(f + g*x))/(e*f - d*g)] + 2*L
og[d + e*x]*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)] - 2*PolyLog[3, (g*(d + e*x))/(-(e*f) + d*g)])))/g
________________________________________________________________________________________
Maple [C] time = 0.075, size = 2018, normalized size = 18.2 \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))^2/(g*x+f),x)
[Out]
I/g*n*ln(g*x+f)*ln(((g*x+f)*e+d*g-f*e)/(d*g-e*f))*b^2*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+b^2*l
n(g*(e*x+d)-d*g+f*e)/g*ln((e*x+d)^n)^2-1/4*ln(g*x+f)/g*Pi^2*b^2*csgn(I*c*(e*x+d)^n)^6+I*ln(g*x+f)/g*ln(c)*Pi*b
^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-I/g*n*dilog(((g*x+f)*e+d*g-f*e)/(d*g-e*f))*b^2*Pi*csgn(I*(e*x+d)^n)*csgn(I*
c*(e*x+d)^n)^2+I*ln(g*x+f)/g*Pi*a*b*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+I/g*n*ln(g*x+f)*ln(((g*x+f)*e+d*g-
f*e)/(d*g-e*f))*b^2*Pi*csgn(I*c*(e*x+d)^n)^3+I*ln(g*x+f)/g*Pi*a*b*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-2/g*n*dilog(
((g*x+f)*e+d*g-f*e)/(d*g-e*f))*b^2*ln(c)+2*ln(g*x+f)/g*ln(c)*a*b-2*b/g*n*dilog(((g*x+f)*e+d*g-f*e)/(d*g-e*f))*
a-I/g*n*dilog(((g*x+f)*e+d*g-f*e)/(d*g-e*f))*b^2*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+2*ln(g*x+f)/g*ln((e*x+d)^n
)*b^2*ln(c)+2*b*ln(g*x+f)/g*ln((e*x+d)^n)*a-2*b/g*n*ln(g*x+f)*ln(((g*x+f)*e+d*g-f*e)/(d*g-e*f))*a+I*ln(g*x+f)/
g*ln(c)*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+I*ln(g*x+f)/g*ln((e*x+d)^n)*b^2*Pi*csgn(I*c)*csgn(I*c*(
e*x+d)^n)^2+I*ln(g*x+f)/g*ln((e*x+d)^n)*b^2*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+b^2*n^2/g*ln(e*x+d)^2*l
n(1-g/(d*g-e*f)*(e*x+d))+2*b^2*n^2/g*ln(e*x+d)*polylog(2,g/(d*g-e*f)*(e*x+d))-2*b^2*n^2*dilog((g*(e*x+d)-d*g+f
*e)/(-d*g+e*f))/g*ln(e*x+d)-2*b^2*n^2*ln(e*x+d)^2*ln((g*(e*x+d)-d*g+f*e)/(-d*g+e*f))/g-I*ln(g*x+f)/g*ln((e*x+d
)^n)*b^2*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-I*ln(g*x+f)/g*Pi*a*b*csgn(I*c)*csgn(I*(e*x+d)^n)*c
sgn(I*c*(e*x+d)^n)-I/g*n*ln(g*x+f)*ln(((g*x+f)*e+d*g-f*e)/(d*g-e*f))*b^2*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)
^n)^2+a^2*ln(g*x+f)/g-I*ln(g*x+f)/g*Pi*a*b*csgn(I*c*(e*x+d)^n)^3-2*b^2*n^2/g*polylog(3,g/(d*g-e*f)*(e*x+d))+ln
(g*x+f)/g*ln(c)^2*b^2+I/g*n*dilog(((g*x+f)*e+d*g-f*e)/(d*g-e*f))*b^2*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(
e*x+d)^n)-I/g*n*ln(g*x+f)*ln(((g*x+f)*e+d*g-f*e)/(d*g-e*f))*b^2*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-I*ln(g*x+f)
/g*ln(c)*Pi*b^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+2*b^2*n*dilog((g*(e*x+d)-d*g+f*e)/(-d*g+e*f))/
g*ln((e*x+d)^n)+b^2*ln(g*(e*x+d)-d*g+f*e)/g*ln(e*x+d)^2*n^2-I*ln(g*x+f)/g*ln(c)*Pi*b^2*csgn(I*c*(e*x+d)^n)^3-l
n(g*x+f)/g*Pi^2*b^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^4+1/2*ln(g*x+f)/g*Pi^2*b^2*csgn(I*c)^2*csg
n(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^3-2/g*n*ln(g*x+f)*ln(((g*x+f)*e+d*g-f*e)/(d*g-e*f))*b^2*ln(c)+I/g*n*dilog((
(g*x+f)*e+d*g-f*e)/(d*g-e*f))*b^2*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*ln(g*x+f)/g*Pi^2*b^2*csgn(I*c)*csgn(I*(e*x+d)^n
)^2*csgn(I*c*(e*x+d)^n)^3-1/4*ln(g*x+f)/g*Pi^2*b^2*csgn(I*c)^2*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^2-2*b^2
*ln(g*(e*x+d)-d*g+f*e)/g*ln((e*x+d)^n)*ln(e*x+d)*n+2*b^2*n*ln(e*x+d)*ln((g*(e*x+d)-d*g+f*e)/(-d*g+e*f))/g*ln((
e*x+d)^n)-I*ln(g*x+f)/g*ln((e*x+d)^n)*b^2*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*ln(g*x+f)/g*Pi^2*b^2*csgn(I*(e*x+d)^n)*
csgn(I*c*(e*x+d)^n)^5-1/4*ln(g*x+f)/g*Pi^2*b^2*csgn(I*c)^2*csgn(I*c*(e*x+d)^n)^4+1/2*ln(g*x+f)/g*Pi^2*b^2*csgn
(I*c)*csgn(I*c*(e*x+d)^n)^5-1/4*ln(g*x+f)/g*Pi^2*b^2*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^4
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{2} \log \left (g x + f\right )}{g} + \int \frac{b^{2} \log \left ({\left (e x + d\right )}^{n}\right )^{2} + b^{2} \log \left (c\right )^{2} + 2 \, a b \log \left (c\right ) + 2 \,{\left (b^{2} \log \left (c\right ) + a b\right )} \log \left ({\left (e x + d\right )}^{n}\right )}{g x + f}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f),x, algorithm="maxima")
[Out]
a^2*log(g*x + f)/g + integrate((b^2*log((e*x + d)^n)^2 + b^2*log(c)^2 + 2*a*b*log(c) + 2*(b^2*log(c) + a*b)*lo
g((e*x + d)^n))/(g*x + f), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 2 \, a b \log \left ({\left (e x + d\right )}^{n} c\right ) + a^{2}}{g x + f}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f),x, algorithm="fricas")
[Out]
integral((b^2*log((e*x + d)^n*c)^2 + 2*a*b*log((e*x + d)^n*c) + a^2)/(g*x + f), x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \log{\left (c \left (d + e x\right )^{n} \right )}\right )^{2}}{f + g x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*ln(c*(e*x+d)**n))**2/(g*x+f),x)
[Out]
Integral((a + b*log(c*(d + e*x)**n))**2/(f + g*x), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}}{g x + f}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f),x, algorithm="giac")
[Out]
integrate((b*log((e*x + d)^n*c) + a)^2/(g*x + f), x)