\(\int \frac {(a+b \log (c (d+e x)^n))^2}{(f+g x)^2} \, dx\) [49]

Optimal result

Integrand size = 24, antiderivative size = 132 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(e f-d g) (f+g x)}-\frac {2 b e n \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g (e f-d g)}-\frac {2 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2444, 2441, 2440, 2438} \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=-\frac {2 b e n \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x) (e f-d g)}-\frac {2 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)} \]

[In]

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

[Out]

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

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 2440

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

Rule 2441

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

Rule 2444

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
]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=\frac {-\left (\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (a g (d+e x)+b g (d+e x) \log \left (c (d+e x)^n\right )-2 b e n (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )\right )\right )+2 b^2 e n^2 (f+g x) \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{g (-e f+d g) (f+g x)} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.91 (sec) , antiderivative size = 537, normalized size of antiderivative = 4.07

method	result	size
risch	\(-\frac {b^{2} \ln \left (\left (e x +d \right )^{n}\right )^{2}}{\left (g x +f \right ) g}-\frac {2 b^{2} n e \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (e x +d \right )}{g \left (d g -e f \right )}+\frac {2 b^{2} n e \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{g \left (d g -e f \right )}-\frac {2 b^{2} n^{2} e \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g \left (d g -e f \right )}-\frac {2 b^{2} n^{2} e \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g \left (d g -e f \right )}+\frac {b^{2} n^{2} e \ln \left (e x +d \right )^{2}}{g \left (d g -e f \right )}+\left (-i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )+i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b +i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b -i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b +2 b \ln \left (c \right )+2 a \right ) b \left (-\frac {\ln \left (\left (e x +d \right )^{n}\right )}{\left (g x +f \right ) g}+\frac {n e \left (-\frac {\ln \left (e x +d \right )}{d g -e f}+\frac {\ln \left (g x +f \right )}{d g -e f}\right )}{g}\right )-\frac {{\left (-i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )+i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b +i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b -i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b +2 b \ln \left (c \right )+2 a \right )}^{2}}{4 \left (g x +f \right ) g}\)	\(537\)

[In]

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

[Out]

-b^2*ln((e*x+d)^n)^2/(g*x+f)/g-2*b^2/g*n*e*ln((e*x+d)^n)/(d*g-e*f)*ln(e*x+d)+2*b^2/g*n*e*ln((e*x+d)^n)/(d*g-e*
f)*ln(g*x+f)-2*b^2/g*n^2*e/(d*g-e*f)*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))-2*b^2/g*n^2*e/(d*g-e*f)*ln(g*x+f)*ln
(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+b^2/g*n^2*e/(d*g-e*f)*ln(e*x+d)^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*(-ln((e*x+d)^n)/(g*x+f)/g+1/g*n*e*(-1/(d*g-e*f)*ln(e*x+d)+1/(d*g-e*f)*ln(g*
x+f)))-1/4*(-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*P
i*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/(g*x+f)/g

Fricas [F]

\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}}{{\left (g x + f\right )}^{2}} \,d x } \]

[In]

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

Sympy [F]

\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=\int \frac {\left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{2}}{\left (f + g x\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}}{{\left (g x + f\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}}{{\left (g x + f\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}{{\left (f+g\,x\right )}^2} \,d x \]

[In]

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

[Out]

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