\(\int \frac {(A+B \log (e (a+b x)^n (c+d x)^{-n}))^p}{(a+b x) (c+d x)} \, dx\) [234]

Optimal result

Integrand size = 40, antiderivative size = 49 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{(a+b x) (c+d x)} \, dx=\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^{1+p}}{B (b c-a d) n (1+p)} \]

[Out]

(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^(p+1)/B/(-a*d+b*c)/n/(p+1)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2573, 2561, 2339, 30} \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{(a+b x) (c+d x)} \, dx=\frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^{p+1}}{B n (p+1) (b c-a d)} \]

[In]

Int[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^p/((a + b*x)*(c + d*x)),x]

[Out]

(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^(1 + p)/(B*(b*c - a*d)*n*(1 + p))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2561

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q, Subst[Int[x^m*((A +
B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i,
A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]

Rule 2573

Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] :> Subst[Int[w*(A + B*Log[e*(u/v)^
n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; FreeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !I
ntegerQ[n]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.96 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{(a+b x) (c+d x)} \, dx=\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^{1+p}}{(b B c n-a B d n) (1+p)} \]

[In]

Integrate[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^p/((a + b*x)*(c + d*x)),x]

[Out]

(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^(1 + p)/((b*B*c*n - a*B*d*n)*(1 + p))

Maple [A] (verified)

Time = 234.62 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(-\frac {{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{p +1}}{n \left (a d -c b \right ) B \left (p +1\right )}\)	\(51\)
default	\(-\frac {{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{p +1}}{n \left (a d -c b \right ) B \left (p +1\right )}\)	\(51\)
parallelrisch	\(-\frac {B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) {\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{p} a^{2} c^{2}+A {\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{p} a^{2} c^{2}}{\left (a d p -b c p +a d -c b \right ) B \,a^{2} c^{2} n}\)	\(120\)

[In]

int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^p/(b*x+a)/(d*x+c),x,method=_RETURNVERBOSE)

[Out]

-1/n/(a*d-b*c)*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^(p+1)/B/(p+1)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.65 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{(a+b x) (c+d x)} \, dx=\frac {{\left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + B \log \left (e\right ) + A\right )} {\left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + B \log \left (e\right ) + A\right )}^{p}}{{\left (B b c - B a d\right )} n p + {\left (B b c - B a d\right )} n} \]

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^p/(b*x+a)/(d*x+c),x, algorithm="fricas")

[Out]

(B*n*log(b*x + a) - B*n*log(d*x + c) + B*log(e) + A)*(B*n*log(b*x + a) - B*n*log(d*x + c) + B*log(e) + A)^p/((
B*b*c - B*a*d)*n*p + (B*b*c - B*a*d)*n)

Sympy [F(-1)]

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

[In]

integrate((A+B*ln(e*(b*x+a)**n/((d*x+c)**n)))**p/(b*x+a)/(d*x+c),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{(a+b x) (c+d x)} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{p}}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^p/(b*x+a)/(d*x+c),x, algorithm="maxima")

[Out]

integrate((B*log((b*x + a)^n*e/(d*x + c)^n) + A)^p/((b*x + a)*(d*x + c)), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{(a+b x) (c+d x)} \, dx=\frac {{\left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + B \log \left (e\right ) + A\right )}^{p + 1}}{{\left (B b c n - B a d n\right )} {\left (p + 1\right )}} \]

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^p/(b*x+a)/(d*x+c),x, algorithm="giac")

[Out]

(B*n*log(b*x + a) - B*n*log(d*x + c) + B*log(e) + A)^(p + 1)/((B*b*c*n - B*a*d*n)*(p + 1))

Mupad [F(-1)]

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

[In]

int((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^p/((a + b*x)*(c + d*x)),x)

[Out]

int((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^p/((a + b*x)*(c + d*x)), x)