Integrand size = 40, antiderivative size = 43 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=-\frac {1}{B (b c-a d) n \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \] Output:
-1/B/(-a*d+b*c)/n/(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=-\frac {1}{(b B c n-a B d n) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )} \] Input:
Integrate[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^ 2),x]
Output:
-(1/((b*B*c*n - a*B*d*n)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])))
Time = 0.48 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.70, 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.
\(\displaystyle \int \frac {1}{(a+b x) (c+d x) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2} \, dx\) |
\(\Big \downarrow \) 2973 |
\(\displaystyle \int \frac {1}{(a+b x) (c+d x) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}dx\) |
\(\Big \downarrow \) 2961 |
\(\displaystyle \frac {\int \frac {c+d x}{(a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}d\frac {a+b x}{c+d x}}{b c-a d}\) |
\(\Big \downarrow \) 2739 |
\(\displaystyle \frac {\int \frac {(c+d x)^2}{(a+b x)^2}d\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n (b c-a d)}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {c+d x}{B n (a+b x) (b c-a d)}\) |
Input:
Int[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2),x]
Output:
-((c + d*x)/(B*(b*c - a*d)*n*(a + b*x)))
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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]
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]
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]
Time = 72.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {1}{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right ) B n \left (d a -b c \right )}\) | \(43\) |
default | \(\frac {1}{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right ) B n \left (d a -b c \right )}\) | \(43\) |
parallelrisch | \(\frac {1}{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right ) B n \left (d a -b c \right )}\) | \(43\) |
risch | \(\frac {2}{B n \left (d a -b c \right ) \left (2 A +2 B \ln \left (e \right )-2 B \ln \left (\left (d x +c \right )^{n}\right )+2 B \ln \left (\left (b x +a \right )^{n}\right )-i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )+i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}+i B \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}-i B \pi \operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{3}+i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2}-i B \pi \,\operatorname {csgn}\left (i \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \operatorname {csgn}\left (i e \right )-i B \pi \operatorname {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3}+i B \pi \operatorname {csgn}\left (i e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} \operatorname {csgn}\left (i e \right )\right )}\) | \(366\) |
Input:
int(1/(b*x+a)/(d*x+c)/(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2,x,method=_RETURN VERBOSE)
Output:
1/(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))/B/n/(a*d-b*c)
Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.00 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=-\frac {1}{{\left (B^{2} b c - B^{2} a d\right )} n^{2} \log \left (b x + a\right ) - {\left (B^{2} b c - B^{2} a d\right )} n^{2} \log \left (d x + c\right ) + {\left (B^{2} b c - B^{2} a d\right )} n \log \left (e\right ) + {\left (A B b c - A B a d\right )} n} \] Input:
integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2,x, algori thm="fricas")
Output:
-1/((B^2*b*c - B^2*a*d)*n^2*log(b*x + a) - (B^2*b*c - B^2*a*d)*n^2*log(d*x + c) + (B^2*b*c - B^2*a*d)*n*log(e) + (A*B*b*c - A*B*a*d)*n)
Timed out. \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x+a)/(d*x+c)/(A+B*ln(e*(b*x+a)**n/((d*x+c)**n)))**2,x)
Output:
Timed out
Time = 0.17 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.88 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=-\frac {1}{{\left (b c n - a d n\right )} B^{2} \log \left ({\left (b x + a\right )}^{n}\right ) - {\left (b c n - a d n\right )} B^{2} \log \left ({\left (d x + c\right )}^{n}\right ) + {\left (b c n - a d n\right )} A B + {\left (b c n \log \left (e\right ) - a d n \log \left (e\right )\right )} B^{2}} \] Input:
integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2,x, algori thm="maxima")
Output:
-1/((b*c*n - a*d*n)*B^2*log((b*x + a)^n) - (b*c*n - a*d*n)*B^2*log((d*x + c)^n) + (b*c*n - a*d*n)*A*B + (b*c*n*log(e) - a*d*n*log(e))*B^2)
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (43) = 86\).
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.30 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=-\frac {1}{B^{2} b c n^{2} \log \left (b x + a\right ) - B^{2} a d n^{2} \log \left (b x + a\right ) - B^{2} b c n^{2} \log \left (d x + c\right ) + B^{2} a d n^{2} \log \left (d x + c\right ) + B^{2} b c n \log \left (e\right ) - B^{2} a d n \log \left (e\right ) + A B b c n - A B a d n} \] Input:
integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2,x, algori thm="giac")
Output:
-1/(B^2*b*c*n^2*log(b*x + a) - B^2*a*d*n^2*log(b*x + a) - B^2*b*c*n^2*log( d*x + c) + B^2*a*d*n^2*log(d*x + c) + B^2*b*c*n*log(e) - B^2*a*d*n*log(e) + A*B*b*c*n - A*B*a*d*n)
Time = 25.79 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=\frac {1}{B\,n\,\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\right )\,\left (a\,d-b\,c\right )} \] Input:
int(1/((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^2*(a + b*x)*(c + d*x)),x)
Output:
1/(B*n*(A + B*log((e*(a + b*x)^n)/(c + d*x)^n))*(a*d - b*c))
Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.05 \[ \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \, dx=-\frac {\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right )}{a n \left (\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) a b d -\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right ) b^{2} c +a^{2} d -a b c \right )} \] Input:
int(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2,x)
Output:
( - log(((a + b*x)**n*e)/(c + d*x)**n))/(a*n*(log(((a + b*x)**n*e)/(c + d* x)**n)*a*b*d - log(((a + b*x)**n*e)/(c + d*x)**n)*b**2*c + a**2*d - a*b*c) )