\(\int \frac {x (a+b \log (c (d+e x)^n))}{(f+g x)^2} \, dx\) [251]

Optimal result

Integrand size = 23, antiderivative size = 138 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=-\frac {b e f n \log (d+e x)}{g^2 (e f-d g)}+\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}+\frac {b e f n \log (f+g x)}{g^2 (e f-d g)}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^2}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {45, 2463, 2442, 36, 31, 2441, 2440, 2438} \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}+\frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^2}-\frac {b e f n \log (d+e x)}{g^2 (e f-d g)}+\frac {b e f n \log (f+g x)}{g^2 (e f-d g)} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2442

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

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (f+g x)^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{g (f+g x)}\right ) \, dx \\ & = \frac {\int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{g}-\frac {f \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx}{g} \\ & = \frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}+\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 )}{g^2}-\frac {(b e n) \int \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right )}{d+e x} \, dx}{g^2}-\frac {(b e f n) \int \frac {1}{(d+e x) (f+g x)} \, dx}{g^2} \\ & = \frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}+\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 )}{g^2}-\frac {(b n) \text {Subst}\left (\int \frac {\log \left (1+\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g^2}-\frac {\left (b e^2 f n\right ) \int \frac {1}{d+e x} \, dx}{g^2 (e f-d g)}+\frac {(b e f n) \int \frac {1}{f+g x} \, dx}{g (e f-d g)} \\ & = -\frac {b e f n \log (d+e x)}{g^2 (e f-d g)}+\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}+\frac {b e f n \log (f+g x)}{g^2 (e f-d g)}+\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 )}{g^2}+\frac {b n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\frac {\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-\frac {b e f n (\log (d+e x)-\log (f+g x))}{e f-d g}+\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )+b n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{g^2} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.80 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.25

method	result	size
risch	\(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f}{g^{2} \left (g x +f \right )}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{g^{2}}-\frac {b n \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{2}}-\frac {b n \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{2}}-\frac {b e n f \ln \left (g x +f \right )}{g^{2} \left (d g -e f \right )}+\frac {b e n f \ln \left (\left (g x +f \right ) e +d g -e f \right )}{g^{2} \left (d g -e f \right )}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (\frac {f}{g^{2} \left (g x +f \right )}+\frac {\ln \left (g x +f \right )}{g^{2}}\right )\)	\(311\)

[In]

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

[Out]

b*ln((e*x+d)^n)/g^2*f/(g*x+f)+b*ln((e*x+d)^n)/g^2*ln(g*x+f)-b*n/g^2*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))-b*n/g
^2*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d*g-e*f))-b*e*n/g^2*f/(d*g-e*f)*ln(g*x+f)+b*e*n/g^2*f/(d*g-e*f)*ln((g*x+f
)*e+d*g-e*f)+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d
)^n)^2+1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+b*ln(c)+a)*(1/g^2*f
/(g*x+f)+1/g^2*ln(g*x+f))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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