3.3.33 \(\int \frac {1}{(a+b x) (c+d x) (A+B \log (e (a+b x)^n (c+d x)^{-n}))^3} \, dx\) [233]

3.3.33.1 Optimal result
3.3.33.2 Mathematica [A] (verified)
3.3.33.3 Rubi [A] (warning: unable to verify)
3.3.33.4 Maple [A] (verified)
3.3.33.5 Fricas [B] (verification not implemented)
3.3.33.6 Sympy [F(-1)]
3.3.33.7 Maxima [B] (verification not implemented)
3.3.33.8 Giac [B] (verification not implemented)
3.3.33.9 Mupad [B] (verification not implemented)

3.3.33.1 Optimal result

Integrand size = 40, antiderivative size = 45 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3} \, dx=-\frac {1}{2 B (b c-a d) n \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \]

output
-1/2/B/(-a*d+b*c)/n/(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2
 
3.3.33.2 Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3} \, dx=-\frac {1}{2 (b B c n-a B d n) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \]

input
Integrate[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^ 
3),x]
 
output
-1/2*1/((b*B*c*n - a*B*d*n)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2)
 
3.3.33.3 Rubi [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.76, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2973, 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.

\(\displaystyle \int \frac {1}{(a+b x) (c+d x) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^3} \, dx\)

\(\Big \downarrow \) 2973

\(\displaystyle \int \frac {1}{(a+b x) (c+d x) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^3}dx\)

\(\Big \downarrow \) 2961

\(\displaystyle \frac {\int \frac {c+d x}{(a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3}d\frac {a+b x}{c+d x}}{b c-a d}\)

\(\Big \downarrow \) 2739

\(\displaystyle \frac {\int \frac {(c+d x)^3}{(a+b x)^3}d\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n (b c-a d)}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {(c+d x)^2}{2 B n (a+b x)^2 (b c-a d)}\)

input
Int[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^3),x]
 
output
-1/2*(c + d*x)^2/(B*(b*c - a*d)*n*(a + b*x)^2)
 

3.3.33.3.1 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]
 
3.3.33.4 Maple [A] (verified)

Time = 163.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98

method result size
derivativedivides \(\frac {1}{2 {\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{2} B n \left (a d -c b \right )}\) \(44\)
default \(\frac {1}{2 {\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{2} B n \left (a d -c b \right )}\) \(44\)
parallelrisch \(\frac {1}{2 {\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{2} B n \left (a d -c b \right )}\) \(44\)
risch \(\frac {2}{B n \left (a d -c b \right ) {\left (2 A +2 B \ln \left (e \right )+2 B \ln \left (\left (b x +a \right )^{n}\right )-2 B \ln \left (\left (d x +c \right )^{n}\right )-i B \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )+i B \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2}+i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2}-i B \pi \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3}-i B \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )+i B \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}+i B \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}-i B \pi \operatorname {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{3}\right )}^{2}}\) \(366\)

input
int(1/(b*x+a)/(d*x+c)/(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^3,x,method=_RETURN 
VERBOSE)
 
output
1/2/(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2/B/n/(a*d-b*c)
 
3.3.33.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (43) = 86\).

Time = 0.33 (sec) , antiderivative size = 238, normalized size of antiderivative = 5.29 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3} \, dx=-\frac {1}{2 \, {\left ({\left (B^{3} b c - B^{3} a d\right )} n^{3} \log \left (b x + a\right )^{2} + {\left (B^{3} b c - B^{3} a d\right )} n^{3} \log \left (d x + c\right )^{2} + {\left (B^{3} b c - B^{3} a d\right )} n \log \left (e\right )^{2} + 2 \, {\left (A B^{2} b c - A B^{2} a d\right )} n \log \left (e\right ) + {\left (A^{2} B b c - A^{2} B a d\right )} n + 2 \, {\left ({\left (B^{3} b c - B^{3} a d\right )} n^{2} \log \left (e\right ) + {\left (A B^{2} b c - A B^{2} a d\right )} n^{2}\right )} \log \left (b x + a\right ) - 2 \, {\left ({\left (B^{3} b c - B^{3} a d\right )} n^{3} \log \left (b x + a\right ) + {\left (B^{3} b c - B^{3} a d\right )} n^{2} \log \left (e\right ) + {\left (A B^{2} b c - A B^{2} a d\right )} n^{2}\right )} \log \left (d x + c\right )\right )}} \]

input
integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3,x, algori 
thm="fricas")
 
output
-1/2/((B^3*b*c - B^3*a*d)*n^3*log(b*x + a)^2 + (B^3*b*c - B^3*a*d)*n^3*log 
(d*x + c)^2 + (B^3*b*c - B^3*a*d)*n*log(e)^2 + 2*(A*B^2*b*c - A*B^2*a*d)*n 
*log(e) + (A^2*B*b*c - A^2*B*a*d)*n + 2*((B^3*b*c - B^3*a*d)*n^2*log(e) + 
(A*B^2*b*c - A*B^2*a*d)*n^2)*log(b*x + a) - 2*((B^3*b*c - B^3*a*d)*n^3*log 
(b*x + a) + (B^3*b*c - B^3*a*d)*n^2*log(e) + (A*B^2*b*c - A*B^2*a*d)*n^2)* 
log(d*x + c))
 
3.3.33.6 Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3} \, dx=\text {Timed out} \]

input
integrate(1/(b*x+a)/(d*x+c)/(A+B*ln(e*(b*x+a)**n/((d*x+c)**n)))**3,x)
 
output
Timed out
 
3.3.33.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (43) = 86\).

Time = 0.36 (sec) , antiderivative size = 220, normalized size of antiderivative = 4.89 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3} \, dx=-\frac {1}{2 \, {\left ({\left (b c n - a d n\right )} B^{3} \log \left ({\left (b x + a\right )}^{n}\right )^{2} + {\left (b c n - a d n\right )} B^{3} \log \left ({\left (d x + c\right )}^{n}\right )^{2} + {\left (b c n - a d n\right )} A^{2} B + 2 \, {\left (b c n \log \left (e\right ) - a d n \log \left (e\right )\right )} A B^{2} + {\left (b c n \log \left (e\right )^{2} - a d n \log \left (e\right )^{2}\right )} B^{3} + 2 \, {\left ({\left (b c n - a d n\right )} A B^{2} + {\left (b c n \log \left (e\right ) - a d n \log \left (e\right )\right )} B^{3}\right )} \log \left ({\left (b x + a\right )}^{n}\right ) - 2 \, {\left ({\left (b c n - a d n\right )} B^{3} \log \left ({\left (b x + a\right )}^{n}\right ) + {\left (b c n - a d n\right )} A B^{2} + {\left (b c n \log \left (e\right ) - a d n \log \left (e\right )\right )} B^{3}\right )} \log \left ({\left (d x + c\right )}^{n}\right )\right )}} \]

input
integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3,x, algori 
thm="maxima")
 
output
-1/2/((b*c*n - a*d*n)*B^3*log((b*x + a)^n)^2 + (b*c*n - a*d*n)*B^3*log((d* 
x + c)^n)^2 + (b*c*n - a*d*n)*A^2*B + 2*(b*c*n*log(e) - a*d*n*log(e))*A*B^ 
2 + (b*c*n*log(e)^2 - a*d*n*log(e)^2)*B^3 + 2*((b*c*n - a*d*n)*A*B^2 + (b* 
c*n*log(e) - a*d*n*log(e))*B^3)*log((b*x + a)^n) - 2*((b*c*n - a*d*n)*B^3* 
log((b*x + a)^n) + (b*c*n - a*d*n)*A*B^2 + (b*c*n*log(e) - a*d*n*log(e))*B 
^3)*log((d*x + c)^n))
 
3.3.33.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (43) = 86\).

Time = 0.29 (sec) , antiderivative size = 321, normalized size of antiderivative = 7.13 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3} \, dx=-\frac {1}{2 \, {\left (B^{3} b c n^{3} \log \left (b x + a\right )^{2} - B^{3} a d n^{3} \log \left (b x + a\right )^{2} - 2 \, B^{3} b c n^{3} \log \left (b x + a\right ) \log \left (d x + c\right ) + 2 \, B^{3} a d n^{3} \log \left (b x + a\right ) \log \left (d x + c\right ) + B^{3} b c n^{3} \log \left (d x + c\right )^{2} - B^{3} a d n^{3} \log \left (d x + c\right )^{2} + 2 \, B^{3} b c n^{2} \log \left (b x + a\right ) \log \left (e\right ) - 2 \, B^{3} a d n^{2} \log \left (b x + a\right ) \log \left (e\right ) - 2 \, B^{3} b c n^{2} \log \left (d x + c\right ) \log \left (e\right ) + 2 \, B^{3} a d n^{2} \log \left (d x + c\right ) \log \left (e\right ) + 2 \, A B^{2} b c n^{2} \log \left (b x + a\right ) - 2 \, A B^{2} a d n^{2} \log \left (b x + a\right ) - 2 \, A B^{2} b c n^{2} \log \left (d x + c\right ) + 2 \, A B^{2} a d n^{2} \log \left (d x + c\right ) + B^{3} b c n \log \left (e\right )^{2} - B^{3} a d n \log \left (e\right )^{2} + 2 \, A B^{2} b c n \log \left (e\right ) - 2 \, A B^{2} a d n \log \left (e\right ) + A^{2} B b c n - A^{2} B a d n\right )}} \]

input
integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3,x, algori 
thm="giac")
 
output
-1/2/(B^3*b*c*n^3*log(b*x + a)^2 - B^3*a*d*n^3*log(b*x + a)^2 - 2*B^3*b*c* 
n^3*log(b*x + a)*log(d*x + c) + 2*B^3*a*d*n^3*log(b*x + a)*log(d*x + c) + 
B^3*b*c*n^3*log(d*x + c)^2 - B^3*a*d*n^3*log(d*x + c)^2 + 2*B^3*b*c*n^2*lo 
g(b*x + a)*log(e) - 2*B^3*a*d*n^2*log(b*x + a)*log(e) - 2*B^3*b*c*n^2*log( 
d*x + c)*log(e) + 2*B^3*a*d*n^2*log(d*x + c)*log(e) + 2*A*B^2*b*c*n^2*log( 
b*x + a) - 2*A*B^2*a*d*n^2*log(b*x + a) - 2*A*B^2*b*c*n^2*log(d*x + c) + 2 
*A*B^2*a*d*n^2*log(d*x + c) + B^3*b*c*n*log(e)^2 - B^3*a*d*n*log(e)^2 + 2* 
A*B^2*b*c*n*log(e) - 2*A*B^2*a*d*n*log(e) + A^2*B*b*c*n - A^2*B*a*d*n)
 
3.3.33.9 Mupad [B] (verification not implemented)

Time = 1.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.60 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3} \, dx=\frac {1}{2\,B\,n\,\left (a\,d-b\,c\right )\,\left (A^2+2\,A\,B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )+B^2\,{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^2\right )} \]

input
int(1/((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^3*(a + b*x)*(c + d*x)),x)
 
output
1/(2*B*n*(a*d - b*c)*(B^2*log((e*(a + b*x)^n)/(c + d*x)^n)^2 + A^2 + 2*A*B 
*log((e*(a + b*x)^n)/(c + d*x)^n)))