Integrand size = 40, antiderivative size = 45 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3}{(a+b x) (c+d x)} \, dx=\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^4}{4 B (b c-a d) n} \] Output:
1/4*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^4/B/(-a*d+b*c)/n
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3}{(a+b x) (c+d x)} \, dx=\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^4}{4 (b B c n-a B d n)} \] Input:
Integrate[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^3/((a + b*x)*(c + d*x)) ,x]
Output:
(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^4/(4*(b*B*c*n - a*B*d*n))
Time = 0.46 (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.
\(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^3}{(a+b x) (c+d x)} \, dx\) |
\(\Big \downarrow \) 2973 |
\(\displaystyle \int \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^3}{(a+b x) (c+d x)}dx\) |
\(\Big \downarrow \) 2961 |
\(\displaystyle \frac {\int \frac {(c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3}{a+b x}d\frac {a+b x}{c+d x}}{b c-a d}\) |
\(\Big \downarrow \) 2739 |
\(\displaystyle \frac {\int \frac {(a+b x)^3}{(c+d x)^3}d\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{B n (b c-a d)}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {(a+b x)^4}{4 B n (c+d x)^4 (b c-a d)}\) |
Input:
Int[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^3/((a + b*x)*(c + d*x)),x]
Output:
(a + b*x)^4/(4*B*(b*c - a*d)*n*(c + d*x)^4)
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 = 39.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(-\frac {{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{4}}{4 n \left (d a -b c \right ) B}\) | \(44\) |
default | \(-\frac {{\left (A +B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )\right )}^{4}}{4 n \left (d a -b c \right ) B}\) | \(44\) |
parallelrisch | \(-\frac {B^{3} \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{4} b^{2} d^{2}+4 A \,B^{2} \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3} b^{2} d^{2}+6 A^{2} B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2} b^{2} d^{2}+4 A^{3} \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) b^{2} d^{2}}{4 n \,b^{2} d^{2} \left (d a -b c \right )}\) | \(150\) |
parts | \(\frac {A^{3} \ln \left (d x +c \right )}{d a -b c}-\frac {A^{3} \ln \left (b x +a \right )}{d a -b c}-\frac {B^{3} \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{4}}{4 n \left (d a -b c \right )}-\frac {A \,B^{2} \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3}}{n \left (d a -b c \right )}-\frac {3 A^{2} B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2}}{2 n \left (d a -b c \right )}\) | \(162\) |
risch | \(\text {Expression too large to display}\) | \(64289\) |
Input:
int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^3/(b*x+a)/(d*x+c),x,method=_RETURNVE RBOSE)
Output:
-1/4/n/(a*d-b*c)*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^4/B
Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (43) = 86\).
Time = 0.10 (sec) , antiderivative size = 375, normalized size of antiderivative = 8.33 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3}{(a+b x) (c+d x)} \, dx=\frac {B^{3} n^{3} \log \left (b x + a\right )^{4} + B^{3} n^{3} \log \left (d x + c\right )^{4} + 4 \, {\left (B^{3} n^{2} \log \left (e\right ) + A B^{2} n^{2}\right )} \log \left (b x + a\right )^{3} - 4 \, {\left (B^{3} n^{3} \log \left (b x + a\right ) + B^{3} n^{2} \log \left (e\right ) + A B^{2} n^{2}\right )} \log \left (d x + c\right )^{3} + 6 \, {\left (B^{3} n \log \left (e\right )^{2} + 2 \, A B^{2} n \log \left (e\right ) + A^{2} B n\right )} \log \left (b x + a\right )^{2} + 6 \, {\left (B^{3} n^{3} \log \left (b x + a\right )^{2} + B^{3} n \log \left (e\right )^{2} + 2 \, A B^{2} n \log \left (e\right ) + A^{2} B n + 2 \, {\left (B^{3} n^{2} \log \left (e\right ) + A B^{2} n^{2}\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )^{2} + 4 \, {\left (B^{3} \log \left (e\right )^{3} + 3 \, A B^{2} \log \left (e\right )^{2} + 3 \, A^{2} B \log \left (e\right ) + A^{3}\right )} \log \left (b x + a\right ) - 4 \, {\left (B^{3} n^{3} \log \left (b x + a\right )^{3} + B^{3} \log \left (e\right )^{3} + 3 \, A B^{2} \log \left (e\right )^{2} + 3 \, A^{2} B \log \left (e\right ) + A^{3} + 3 \, {\left (B^{3} n^{2} \log \left (e\right ) + A B^{2} n^{2}\right )} \log \left (b x + a\right )^{2} + 3 \, {\left (B^{3} n \log \left (e\right )^{2} + 2 \, A B^{2} n \log \left (e\right ) + A^{2} B n\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{4 \, {\left (b c - a d\right )}} \] Input:
integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3/(b*x+a)/(d*x+c),x, algorith m="fricas")
Output:
1/4*(B^3*n^3*log(b*x + a)^4 + B^3*n^3*log(d*x + c)^4 + 4*(B^3*n^2*log(e) + A*B^2*n^2)*log(b*x + a)^3 - 4*(B^3*n^3*log(b*x + a) + B^3*n^2*log(e) + A* B^2*n^2)*log(d*x + c)^3 + 6*(B^3*n*log(e)^2 + 2*A*B^2*n*log(e) + A^2*B*n)* log(b*x + a)^2 + 6*(B^3*n^3*log(b*x + a)^2 + B^3*n*log(e)^2 + 2*A*B^2*n*lo g(e) + A^2*B*n + 2*(B^3*n^2*log(e) + A*B^2*n^2)*log(b*x + a))*log(d*x + c) ^2 + 4*(B^3*log(e)^3 + 3*A*B^2*log(e)^2 + 3*A^2*B*log(e) + A^3)*log(b*x + a) - 4*(B^3*n^3*log(b*x + a)^3 + B^3*log(e)^3 + 3*A*B^2*log(e)^2 + 3*A^2*B *log(e) + A^3 + 3*(B^3*n^2*log(e) + A*B^2*n^2)*log(b*x + a)^2 + 3*(B^3*n*l og(e)^2 + 2*A*B^2*n*log(e) + A^2*B*n)*log(b*x + a))*log(d*x + c))/(b*c - a *d)
Timed out. \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3}{(a+b x) (c+d x)} \, dx=\text {Timed out} \] Input:
integrate((A+B*ln(e*(b*x+a)**n/((d*x+c)**n)))**3/(b*x+a)/(d*x+c),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 766 vs. \(2 (43) = 86\).
Time = 0.10 (sec) , antiderivative size = 766, normalized size of antiderivative = 17.02 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3}{(a+b x) (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3/(b*x+a)/(d*x+c),x, algorith m="maxima")
Output:
B^3*(log(b*x + a)/(b*c - a*d) - log(d*x + c)/(b*c - a*d))*log((b*x + a)^n* e/(d*x + c)^n)^3 + 3*A*B^2*(log(b*x + a)/(b*c - a*d) - log(d*x + c)/(b*c - a*d))*log((b*x + a)^n*e/(d*x + c)^n)^2 + 3*A^2*B*(log(b*x + a)/(b*c - a*d ) - log(d*x + c)/(b*c - a*d))*log((b*x + a)^n*e/(d*x + c)^n) - 1/4*B^3*(6* (e*n*log(b*x + a)^2 - 2*e*n*log(b*x + a)*log(d*x + c) + e*n*log(d*x + c)^2 )*log((b*x + a)^n*e/(d*x + c)^n)^2/((b*c - a*d)*e) - (4*(e^2*n^2*log(b*x + a)^3 - 3*e^2*n^2*log(b*x + a)^2*log(d*x + c) + 3*e^2*n^2*log(b*x + a)*log (d*x + c)^2 - e^2*n^2*log(d*x + c)^3)*log((b*x + a)^n*e/(d*x + c)^n)/((b*c - a*d)*e) - (e^3*n^3*log(b*x + a)^4 - 4*e^3*n^3*log(b*x + a)^3*log(d*x + c) + 6*e^3*n^3*log(b*x + a)^2*log(d*x + c)^2 - 4*e^3*n^3*log(b*x + a)*log( d*x + c)^3 + e^3*n^3*log(d*x + c)^4)/((b*c - a*d)*e^2))/e) + A^3*(log(b*x + a)/(b*c - a*d) - log(d*x + c)/(b*c - a*d)) - A*B^2*(3*(e*n*log(b*x + a)^ 2 - 2*e*n*log(b*x + a)*log(d*x + c) + e*n*log(d*x + c)^2)*log((b*x + a)^n* e/(d*x + c)^n)/((b*c - a*d)*e) - (e^2*n^2*log(b*x + a)^3 - 3*e^2*n^2*log(b *x + a)^2*log(d*x + c) + 3*e^2*n^2*log(b*x + a)*log(d*x + c)^2 - e^2*n^2*l og(d*x + c)^3)/((b*c - a*d)*e^2)) - 3/2*(e*n*log(b*x + a)^2 - 2*e*n*log(b* x + a)*log(d*x + c) + e*n*log(d*x + c)^2)*A^2*B/((b*c - a*d)*e)
\[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3}{(a+b x) (c+d x)} \, dx=\int { \frac {{\left (B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A\right )}^{3}}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3/(b*x+a)/(d*x+c),x, algorith m="giac")
Output:
integrate((B*log((b*x + a)^n*e/(d*x + c)^n) + A)^3/((b*x + a)*(d*x + c)), x)
Time = 28.67 (sec) , antiderivative size = 141, normalized size of antiderivative = 3.13 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3}{(a+b x) (c+d x)} \, dx=-\frac {\frac {3\,A^2\,B\,{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^2}{2}+A\,B^2\,{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^3+\frac {B^3\,{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^4}{4}}{n\,\left (a\,d-b\,c\right )}+\frac {A^3\,\mathrm {atan}\left (\frac {a\,d\,1{}\mathrm {i}+b\,c\,1{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}\right )\,2{}\mathrm {i}}{a\,d-b\,c} \] Input:
int((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^3/((a + b*x)*(c + d*x)),x)
Output:
(A^3*atan((a*d*1i + b*c*1i + b*d*x*2i)/(a*d - b*c))*2i)/(a*d - b*c) - ((B^ 3*log((e*(a + b*x)^n)/(c + d*x)^n)^4)/4 + (3*A^2*B*log((e*(a + b*x)^n)/(c + d*x)^n)^2)/2 + A*B^2*log((e*(a + b*x)^n)/(c + d*x)^n)^3)/(n*(a*d - b*c))
Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.67 \[ \int \frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3}{(a+b x) (c+d x)} \, dx=\frac {-4 \,\mathrm {log}\left (b x +a \right ) a^{3} n +4 \,\mathrm {log}\left (d x +c \right ) a^{3} n -\mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right )^{4} b^{3}-4 \mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right )^{3} a \,b^{2}-6 \mathrm {log}\left (\frac {\left (b x +a \right )^{n} e}{\left (d x +c \right )^{n}}\right )^{2} a^{2} b}{4 n \left (a d -b c \right )} \] Input:
int((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3/(b*x+a)/(d*x+c),x)
Output:
( - 4*log(a + b*x)*a**3*n + 4*log(c + d*x)*a**3*n - log(((a + b*x)**n*e)/( c + d*x)**n)**4*b**3 - 4*log(((a + b*x)**n*e)/(c + d*x)**n)**3*a*b**2 - 6* log(((a + b*x)**n*e)/(c + d*x)**n)**2*a**2*b)/(4*n*(a*d - b*c))