Integrand size = 31, antiderivative size = 86 \[ \int (c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=-\frac {B (b c-a d) i n x}{2 b}-\frac {B (b c-a d)^2 i n \log (a+b x)}{2 b^2 d}+\frac {i (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 d} \]
-1/2*B*(-a*d+b*c)*i*n*x/b-1/2*B*(-a*d+b*c)^2*i*n*ln(b*x+a)/b^2/d+1/2*i*(d* x+c)^2*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/d
Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int (c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {i \left (-\frac {B (b c-a d) n (b d x+(b c-a d) \log (a+b x))}{b^2}+(c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )\right )}{2 d} \]
(i*(-((B*(b*c - a*d)*n*(b*d*x + (b*c - a*d)*Log[a + b*x]))/b^2) + (c + d*x )^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n])))/(2*d)
Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2947, 27, 49, 2009}
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.
\(\displaystyle \int (c i+d i x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right ) \, dx\) |
\(\Big \downarrow \) 2947 |
\(\displaystyle \frac {i (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d}-\frac {B n (b c-a d) \int \frac {i^2 (c+d x)}{a+b x}dx}{2 d i}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d}-\frac {B i n (b c-a d) \int \frac {c+d x}{a+b x}dx}{2 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {i (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d}-\frac {B i n (b c-a d) \int \left (\frac {d}{b}+\frac {b c-a d}{b (a+b x)}\right )dx}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d}-\frac {B i n (b c-a d) \left (\frac {(b c-a d) \log (a+b x)}{b^2}+\frac {d x}{b}\right )}{2 d}\) |
-1/2*(B*(b*c - a*d)*i*n*((d*x)/b + ((b*c - a*d)*Log[a + b*x])/b^2))/d + (i *(c + d*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*d)
3.2.11.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(g*(m + 1))), x] - Simp[B*n*((b*c - a*d) /(g*(m + 1))) Int[(f + g*x)^(m + 1)/((a + b*x)*(c + d*x)), x], x] /; Free Q[{a, b, c, d, e, f, g, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, -2]
Leaf count of result is larger than twice the leaf count of optimal. \(249\) vs. \(2(80)=160\).
Time = 0.68 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.91
method | result | size |
parallelrisch | \(\frac {B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} d^{2} i n +A \,x^{2} b^{2} d^{2} i n -B \ln \left (b x +a \right ) a^{2} d^{2} i \,n^{2}+2 B \ln \left (b x +a \right ) a b c d i \,n^{2}-B \ln \left (b x +a \right ) b^{2} c^{2} i \,n^{2}+2 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} c d i n +B x a b \,d^{2} i \,n^{2}-B x \,b^{2} c d i \,n^{2}+2 A x \,b^{2} c d i n +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} c^{2} i n -B \,a^{2} d^{2} i \,n^{2}+B \,b^{2} c^{2} i \,n^{2}-3 A a b c d i n -2 A \,b^{2} c^{2} i n}{2 b^{2} n d}\) | \(250\) |
1/2*(B*x^2*ln(e*((b*x+a)/(d*x+c))^n)*b^2*d^2*i*n+A*x^2*b^2*d^2*i*n-B*ln(b* x+a)*a^2*d^2*i*n^2+2*B*ln(b*x+a)*a*b*c*d*i*n^2-B*ln(b*x+a)*b^2*c^2*i*n^2+2 *B*x*ln(e*((b*x+a)/(d*x+c))^n)*b^2*c*d*i*n+B*x*a*b*d^2*i*n^2-B*x*b^2*c*d*i *n^2+2*A*x*b^2*c*d*i*n+B*ln(e*((b*x+a)/(d*x+c))^n)*b^2*c^2*i*n-B*a^2*d^2*i *n^2+B*b^2*c^2*i*n^2-3*A*a*b*c*d*i*n-2*A*b^2*c^2*i*n)/b^2/n/d
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (80) = 160\).
Time = 0.35 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.88 \[ \int (c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {A b^{2} d^{2} i x^{2} - B b^{2} c^{2} i n \log \left (d x + c\right ) + {\left (2 \, B a b c d - B a^{2} d^{2}\right )} i n \log \left (b x + a\right ) + {\left (2 \, A b^{2} c d i - {\left (B b^{2} c d - B a b d^{2}\right )} i n\right )} x + {\left (B b^{2} d^{2} i x^{2} + 2 \, B b^{2} c d i x\right )} \log \left (e\right ) + {\left (B b^{2} d^{2} i n x^{2} + 2 \, B b^{2} c d i n x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{2 \, b^{2} d} \]
1/2*(A*b^2*d^2*i*x^2 - B*b^2*c^2*i*n*log(d*x + c) + (2*B*a*b*c*d - B*a^2*d ^2)*i*n*log(b*x + a) + (2*A*b^2*c*d*i - (B*b^2*c*d - B*a*b*d^2)*i*n)*x + ( B*b^2*d^2*i*x^2 + 2*B*b^2*c*d*i*x)*log(e) + (B*b^2*d^2*i*n*x^2 + 2*B*b^2*c *d*i*n*x)*log((b*x + a)/(d*x + c)))/(b^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 382 vs. \(2 (73) = 146\).
Time = 5.09 (sec) , antiderivative size = 382, normalized size of antiderivative = 4.44 \[ \int (c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\begin {cases} c i x \left (A + B \log {\left (e \left (\frac {a}{c}\right )^{n} \right )}\right ) & \text {for}\: b = 0 \wedge d = 0 \\A c i x + \frac {A d i x^{2}}{2} + \frac {B c^{2} i \log {\left (e \left (\frac {a}{c + d x}\right )^{n} \right )}}{2 d} + \frac {B c i n x}{2} + B c i x \log {\left (e \left (\frac {a}{c + d x}\right )^{n} \right )} + \frac {B d i n x^{2}}{4} + \frac {B d i x^{2} \log {\left (e \left (\frac {a}{c + d x}\right )^{n} \right )}}{2} & \text {for}\: b = 0 \\c i \left (A x + \frac {B a \log {\left (e \left (\frac {a}{c} + \frac {b x}{c}\right )^{n} \right )}}{b} - B n x + B x \log {\left (e \left (\frac {a}{c} + \frac {b x}{c}\right )^{n} \right )}\right ) & \text {for}\: d = 0 \\A c i x + \frac {A d i x^{2}}{2} - \frac {B a^{2} d i n \log {\left (\frac {c}{d} + x \right )}}{2 b^{2}} - \frac {B a^{2} d i \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{2 b^{2}} + \frac {B a c i n \log {\left (\frac {c}{d} + x \right )}}{b} + \frac {B a c i \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{b} + \frac {B a d i n x}{2 b} - \frac {B c^{2} i n \log {\left (\frac {c}{d} + x \right )}}{2 d} - \frac {B c i n x}{2} + B c i x \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + \frac {B d i x^{2} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{2} & \text {otherwise} \end {cases} \]
Piecewise((c*i*x*(A + B*log(e*(a/c)**n)), Eq(b, 0) & Eq(d, 0)), (A*c*i*x + A*d*i*x**2/2 + B*c**2*i*log(e*(a/(c + d*x))**n)/(2*d) + B*c*i*n*x/2 + B*c *i*x*log(e*(a/(c + d*x))**n) + B*d*i*n*x**2/4 + B*d*i*x**2*log(e*(a/(c + d *x))**n)/2, Eq(b, 0)), (c*i*(A*x + B*a*log(e*(a/c + b*x/c)**n)/b - B*n*x + B*x*log(e*(a/c + b*x/c)**n)), Eq(d, 0)), (A*c*i*x + A*d*i*x**2/2 - B*a**2 *d*i*n*log(c/d + x)/(2*b**2) - B*a**2*d*i*log(e*(a/(c + d*x) + b*x/(c + d* x))**n)/(2*b**2) + B*a*c*i*n*log(c/d + x)/b + B*a*c*i*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/b + B*a*d*i*n*x/(2*b) - B*c**2*i*n*log(c/d + x)/(2*d) - B*c*i*n*x/2 + B*c*i*x*log(e*(a/(c + d*x) + b*x/(c + d*x))**n) + B*d*i*x* *2*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/2, True))
Time = 0.20 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.81 \[ \int (c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {1}{2} \, B d i x^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {1}{2} \, A d i x^{2} - \frac {1}{2} \, B d i n {\left (\frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} + B c i n {\left (\frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} + B c i x \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A c i x \]
1/2*B*d*i*x^2*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 1/2*A*d*i*x^2 - 1/2 *B*d*i*n*(a^2*log(b*x + a)/b^2 - c^2*log(d*x + c)/d^2 + (b*c - a*d)*x/(b*d )) + B*c*i*n*(a*log(b*x + a)/b - c*log(d*x + c)/d) + B*c*i*x*log(e*(b*x/(d *x + c) + a/(d*x + c))^n) + A*c*i*x
Leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (80) = 160\).
Time = 0.46 (sec) , antiderivative size = 580, normalized size of antiderivative = 6.74 \[ \int (c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {1}{2} \, {\left (\frac {{\left (B b^{3} c^{3} i n - 3 \, B a b^{2} c^{2} d i n + 3 \, B a^{2} b c d^{2} i n - B a^{3} d^{3} i n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d - \frac {2 \, {\left (b x + a\right )} b d^{2}}{d x + c} + \frac {{\left (b x + a\right )}^{2} d^{3}}{{\left (d x + c\right )}^{2}}} - \frac {B b^{4} c^{3} i n - 3 \, B a b^{3} c^{2} d i n - \frac {{\left (b x + a\right )} B b^{3} c^{3} d i n}{d x + c} + 3 \, B a^{2} b^{2} c d^{2} i n + \frac {3 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} i n}{d x + c} - B a^{3} b d^{3} i n - \frac {3 \, {\left (b x + a\right )} B a^{2} b c d^{3} i n}{d x + c} + \frac {{\left (b x + a\right )} B a^{3} d^{4} i n}{d x + c} - B b^{4} c^{3} i \log \left (e\right ) + 3 \, B a b^{3} c^{2} d i \log \left (e\right ) - 3 \, B a^{2} b^{2} c d^{2} i \log \left (e\right ) + B a^{3} b d^{3} i \log \left (e\right ) - A b^{4} c^{3} i + 3 \, A a b^{3} c^{2} d i - 3 \, A a^{2} b^{2} c d^{2} i + A a^{3} b d^{3} i}{b^{3} d - \frac {2 \, {\left (b x + a\right )} b^{2} d^{2}}{d x + c} + \frac {{\left (b x + a\right )}^{2} b d^{3}}{{\left (d x + c\right )}^{2}}} + \frac {{\left (B b^{3} c^{3} i n - 3 \, B a b^{2} c^{2} d i n + 3 \, B a^{2} b c d^{2} i n - B a^{3} d^{3} i n\right )} \log \left (-b + \frac {{\left (b x + a\right )} d}{d x + c}\right )}{b^{2} d} - \frac {{\left (B b^{3} c^{3} i n - 3 \, B a b^{2} c^{2} d i n + 3 \, B a^{2} b c d^{2} i n - B a^{3} d^{3} i n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]
1/2*((B*b^3*c^3*i*n - 3*B*a*b^2*c^2*d*i*n + 3*B*a^2*b*c*d^2*i*n - B*a^3*d^ 3*i*n)*log((b*x + a)/(d*x + c))/(b^2*d - 2*(b*x + a)*b*d^2/(d*x + c) + (b* x + a)^2*d^3/(d*x + c)^2) - (B*b^4*c^3*i*n - 3*B*a*b^3*c^2*d*i*n - (b*x + a)*B*b^3*c^3*d*i*n/(d*x + c) + 3*B*a^2*b^2*c*d^2*i*n + 3*(b*x + a)*B*a*b^2 *c^2*d^2*i*n/(d*x + c) - B*a^3*b*d^3*i*n - 3*(b*x + a)*B*a^2*b*c*d^3*i*n/( d*x + c) + (b*x + a)*B*a^3*d^4*i*n/(d*x + c) - B*b^4*c^3*i*log(e) + 3*B*a* b^3*c^2*d*i*log(e) - 3*B*a^2*b^2*c*d^2*i*log(e) + B*a^3*b*d^3*i*log(e) - A *b^4*c^3*i + 3*A*a*b^3*c^2*d*i - 3*A*a^2*b^2*c*d^2*i + A*a^3*b*d^3*i)/(b^3 *d - 2*(b*x + a)*b^2*d^2/(d*x + c) + (b*x + a)^2*b*d^3/(d*x + c)^2) + (B*b ^3*c^3*i*n - 3*B*a*b^2*c^2*d*i*n + 3*B*a^2*b*c*d^2*i*n - B*a^3*d^3*i*n)*lo g(-b + (b*x + a)*d/(d*x + c))/(b^2*d) - (B*b^3*c^3*i*n - 3*B*a*b^2*c^2*d*i *n + 3*B*a^2*b*c*d^2*i*n - B*a^3*d^3*i*n)*log((b*x + a)/(d*x + c))/(b^2*d) )*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)
Time = 1.29 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.56 \[ \int (c i+d i x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=x\,\left (\frac {i\,\left (2\,A\,a\,d+4\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{2\,b}-\frac {A\,i\,\left (2\,a\,d+2\,b\,c\right )}{2\,b}\right )+\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {B\,d\,i\,x^2}{2}+B\,c\,i\,x\right )-\frac {\ln \left (a+b\,x\right )\,\left (B\,a^2\,d\,i\,n-2\,B\,a\,b\,c\,i\,n\right )}{2\,b^2}+\frac {A\,d\,i\,x^2}{2}-\frac {B\,c^2\,i\,n\,\ln \left (c+d\,x\right )}{2\,d} \]