3.3.36 \(\int \frac {(A+B \log (e (a+b x)^n (c+d x)^{-n}))^p}{a c f+(b c+a d) f x+b d f x^2} \, dx\) [236]

3.3.36.1 Optimal result

Integrand size = 50, antiderivative size = 52 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{a c f+(b c+a d) f x+b d f x^2} \, dx=\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^{1+p}}{B (b c-a d) f n (1+p)} \]

output

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

3.3.36.2 Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{a c f+(b c+a d) f x+b d f x^2} \, dx=\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^{1+p}}{f (b B c n-a B d n) (1+p)} \]

input

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

output

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

3.3.36.3 Rubi [A] (warning: unable to verify)

Time = 0.55 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2973, 2974, 2961, 2739, 15}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^p}{f x (a d+b c)+a c f+b d f x^2} \, dx\)

\(\Big \downarrow \) 2973

\(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^p}{f x (a d+b c)+a c f+b d f x^2}dx\)

\(\Big \downarrow \) 2974

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

\(\Big \downarrow \) 2961

\(\displaystyle \frac {\int \frac {(c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^p}{a+b x}d\frac {a+b x}{c+d x}}{f (b c-a d)}\)

\(\Big \downarrow \) 2739

\(\displaystyle \frac {\int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^pd\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B f n (b c-a d)}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {\left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^{p+1}}{B f n (p+1) (b c-a d)}\)

input

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

output

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

3.3.36.3.1 Defintions of rubi rules used

rule 15

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

rule 2739

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

rule 2961

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol 
] :> Simp[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q   Subst[Int[x^m*((A + B*L 
og[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; Fre 
eQ[{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 2973

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)] /; Fr 
eeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !Intege 
rQ[n]
 

rule 2974

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))^(p_.)*((f_.) + (g_.)*(x_) + (h_.)*(x_)^2)^(m_.), x_Symbol] :> Simp[h^ 
m/(b^m*d^m)   Int[(a + b*x)^m*(c + d*x)^m*(A + B*Log[e*((a + b*x)/(c + d*x) 
)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, A, B, n, p}, x] && EqQ[b* 
d*f - a*c*h, 0] && EqQ[b*d*g - h*(b*c + a*d), 0] && IntegerQ[m]
 

3.3.36.4 Maple [F]

\[\int \frac {{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{p}}{a c f +\left (a d +c b \right ) f x +b d f \,x^{2}}d x\]

input

int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^p/(a*c*f+(a*d+b*c)*f*x+b*d*f*x^2),x)
 

output

int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^p/(a*c*f+(a*d+b*c)*f*x+b*d*f*x^2),x)
 

3.3.36.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.60 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{a c f+(b c+a d) f x+b d f x^2} \, dx=\frac {{\left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + B \log \left (e\right ) + A\right )} {\left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + B \log \left (e\right ) + A\right )}^{p}}{{\left (B b c - B a d\right )} f n p + {\left (B b c - B a d\right )} f n} \]

input

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

output

(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)*f*n*p + (B*b*c - B*a*d) 
*f*n)
 

3.3.36.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.3.36.7 Maxima [F]

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

input

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

output

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

3.3.36.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^p}{a c f+(b c+a d) f x+b d f x^2} \, dx=\frac {{\left (B n \log \left (b x + a\right ) - B n \log \left (d x + c\right ) + B \log \left (e\right ) + A\right )}^{p + 1}}{{\left (B b c f n - B a d f n\right )} {\left (p + 1\right )}} \]

input

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

output

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

3.3.36.9 Mupad [F(-1)]

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

input

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

output

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