Integrand size = 27, antiderivative size = 104 \[ \int (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=-\frac {B (b c-a d) g x}{b d}-\frac {B (b f-a g)^2 \log (a+b x)}{b^2 g}+\frac {(f+g x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{2 g}+\frac {B (d f-c g)^2 \log (c+d x)}{d^2 g} \] Output:
-B*(-a*d+b*c)*g*x/b/d-B*(-a*g+b*f)^2*ln(b*x+a)/b^2/g+1/2*(g*x+f)^2*(A+B*ln (e*(b*x+a)^2/(d*x+c)^2))/g+B*(-c*g+d*f)^2*ln(d*x+c)/d^2/g
Time = 0.13 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.13 \[ \int (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {-2 B d^2 (b f-a g)^2 \log (a+b x)+b \left (d \left (2 B (-b c+a d) g^2 x+A b d (f+g x)^2\right )+b B d^2 (f+g x)^2 \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+2 b B (d f-c g)^2 \log (c+d x)\right )}{2 b^2 d^2 g} \] Input:
Integrate[(f + g*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]),x]
Output:
(-2*B*d^2*(b*f - a*g)^2*Log[a + b*x] + b*(d*(2*B*(-(b*c) + a*d)*g^2*x + A* b*d*(f + g*x)^2) + b*B*d^2*(f + g*x)^2*Log[(e*(a + b*x)^2)/(c + d*x)^2] + 2*b*B*(d*f - c*g)^2*Log[c + d*x]))/(2*b^2*d^2*g)
Time = 0.33 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2948, 93, 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.
| \(\displaystyle \int (f+g x) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right ) \, dx\) |
| \(\Big \downarrow \) 2948 |
| \(\displaystyle \frac {(f+g x)^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{2 g}-\frac {B (b c-a d) \int \frac {(f+g x)^2}{(a+b x) (c+d x)}dx}{g}\) |
| \(\Big \downarrow \) 93 |
| \(\displaystyle \frac {(f+g x)^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{2 g}-\frac {B (b c-a d) \int \left (\frac {g^2}{b d}+\frac {(b f-a g)^2}{b (b c-a d) (a+b x)}+\frac {(d f-c g)^2}{d (a d-b c) (c+d x)}\right )dx}{g}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(f+g x)^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{2 g}-\frac {B (b c-a d) \left (\frac {(b f-a g)^2 \log (a+b x)}{b^2 (b c-a d)}-\frac {(d f-c g)^2 \log (c+d x)}{d^2 (b c-a d)}+\frac {g^2 x}{b d}\right )}{g}\) |
Input:
Int[(f + g*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]),x]
Output:
((f + g*x)^2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/(2*g) - (B*(b*c - a *d)*((g^2*x)/(b*d) + ((b*f - a*g)^2*Log[a + b*x])/(b^2*(b*c - a*d)) - ((d* f - c*g)^2*Log[c + d*x])/(d^2*(b*c - a*d))))/g
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ )]*(B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*( (A + B*Log[e*((a + b*x)^n/(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] / ; FreeQ[{a, b, c, d, e, f, g, A, B, m, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && !(EqQ[m, -2] && IntegerQ[n])
Time = 0.88 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18
| method | result | size |
| risch | \(\frac {B x \left (g x +2 f \right ) \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )}{2}+\frac {A \,x^{2} g}{2}+A f x -\frac {B \ln \left (b x +a \right ) a^{2} g}{b^{2}}+\frac {2 B \ln \left (b x +a \right ) a f}{b}+\frac {B \ln \left (-d x -c \right ) c^{2} g}{d^{2}}-\frac {2 B \ln \left (-d x -c \right ) c f}{d}+\frac {B x a g}{b}-\frac {B x c g}{d}\) | \(123\) |
| parts | \(A \left (\frac {1}{2} x^{2} g +f x \right )+\frac {B \left (-\left (-\frac {\left (d x +c \right )^{2} \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{2}-\left (-d a +b c \right ) \left (\frac {\left (d a -b c \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b^{2}}-\frac {d x +c}{b}+\frac {\left (-d a +b c \right ) \ln \left (\frac {1}{d x +c}\right )}{b^{2}}\right )\right ) g +\left (-\left (d x +c \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )-\left (-2 d a +2 b c \right ) \left (\frac {\left (-d a +b c \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b \left (d a -b c \right )}+\frac {\ln \left (\frac {1}{d x +c}\right )}{b}\right )\right ) \left (c g -d f \right )\right )}{d^{2}}\) | \(259\) |
| parallelrisch | \(\frac {B \,x^{2} \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) b^{2} d^{2} g +A \,x^{2} b^{2} d^{2} g +2 B x \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) b^{2} d^{2} f +2 A \,b^{2} d^{2} f x -2 B \ln \left (b x +a \right ) a^{2} d^{2} g +4 B \ln \left (b x +a \right ) a b \,d^{2} f +2 B \ln \left (b x +a \right ) b^{2} c^{2} g -4 B \ln \left (b x +a \right ) b^{2} c d f +2 B x a b \,d^{2} g -2 B x \,b^{2} c d g -B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) b^{2} c^{2} g +2 B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) b^{2} c d f -A a b c d g -2 A a b \,d^{2} f -2 A \,b^{2} c d f -2 B \,a^{2} d^{2} g +2 B \,b^{2} c^{2} g}{2 b^{2} d^{2}}\) | \(271\) |
| derivativedivides | \(-\frac {-\frac {A \left (\frac {g \left (d x +c \right )^{2}}{2}-\left (c g -d f \right ) \left (d x +c \right )\right )}{d}-\frac {B \left (-\left (-\frac {\left (d x +c \right )^{2} \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{2}-\left (-d a +b c \right ) \left (\frac {\left (d a -b c \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b^{2}}-\frac {d x +c}{b}+\frac {\left (-d a +b c \right ) \ln \left (\frac {1}{d x +c}\right )}{b^{2}}\right )\right ) g +\left (-\left (d x +c \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )-\left (-2 d a +2 b c \right ) \left (\frac {\left (-d a +b c \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b \left (d a -b c \right )}+\frac {\ln \left (\frac {1}{d x +c}\right )}{b}\right )\right ) \left (c g -d f \right )\right )}{d}}{d}\) | \(285\) |
| default | \(-\frac {-\frac {A \left (\frac {g \left (d x +c \right )^{2}}{2}-\left (c g -d f \right ) \left (d x +c \right )\right )}{d}-\frac {B \left (-\left (-\frac {\left (d x +c \right )^{2} \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{2}-\left (-d a +b c \right ) \left (\frac {\left (d a -b c \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b^{2}}-\frac {d x +c}{b}+\frac {\left (-d a +b c \right ) \ln \left (\frac {1}{d x +c}\right )}{b^{2}}\right )\right ) g +\left (-\left (d x +c \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )-\left (-2 d a +2 b c \right ) \left (\frac {\left (-d a +b c \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b \left (d a -b c \right )}+\frac {\ln \left (\frac {1}{d x +c}\right )}{b}\right )\right ) \left (c g -d f \right )\right )}{d}}{d}\) | \(285\) |
Input:
int((g*x+f)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2)),x,method=_RETURNVERBOSE)
Output:
1/2*B*x*(g*x+2*f)*ln(e*(b*x+a)^2/(d*x+c)^2)+1/2*A*x^2*g+A*f*x-1/b^2*B*ln(b *x+a)*a^2*g+2/b*B*ln(b*x+a)*a*f+1/d^2*B*ln(-d*x-c)*c^2*g-2/d*B*ln(-d*x-c)* c*f+1/b*B*x*a*g-1/d*B*x*c*g
Time = 0.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.67 \[ \int (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {A b^{2} d^{2} g x^{2} + 2 \, {\left (A b^{2} d^{2} f - {\left (B b^{2} c d - B a b d^{2}\right )} g\right )} x + 2 \, {\left (2 \, B a b d^{2} f - B a^{2} d^{2} g\right )} \log \left (b x + a\right ) - 2 \, {\left (2 \, B b^{2} c d f - B b^{2} c^{2} g\right )} \log \left (d x + c\right ) + {\left (B b^{2} d^{2} g x^{2} + 2 \, B b^{2} d^{2} f x\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \, b^{2} d^{2}} \] Input:
integrate((g*x+f)*(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x, algorithm="fricas")
Output:
1/2*(A*b^2*d^2*g*x^2 + 2*(A*b^2*d^2*f - (B*b^2*c*d - B*a*b*d^2)*g)*x + 2*( 2*B*a*b*d^2*f - B*a^2*d^2*g)*log(b*x + a) - 2*(2*B*b^2*c*d*f - B*b^2*c^2*g )*log(d*x + c) + (B*b^2*d^2*g*x^2 + 2*B*b^2*d^2*f*x)*log((b^2*e*x^2 + 2*a* b*e*x + a^2*e)/(d^2*x^2 + 2*c*d*x + c^2)))/(b^2*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (88) = 176\).
Time = 1.38 (sec) , antiderivative size = 314, normalized size of antiderivative = 3.02 \[ \int (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {A g x^{2}}{2} - \frac {B a \left (a g - 2 b f\right ) \log {\left (x + \frac {B a^{2} c d g + \frac {B a^{2} d^{2} \left (a g - 2 b f\right )}{b} + B a b c^{2} g - 4 B a b c d f - B a c d \left (a g - 2 b f\right )}{B a^{2} d^{2} g - 2 B a b d^{2} f + B b^{2} c^{2} g - 2 B b^{2} c d f} \right )}}{b^{2}} + \frac {B c \left (c g - 2 d f\right ) \log {\left (x + \frac {B a^{2} c d g + B a b c^{2} g - 4 B a b c d f - B a b c \left (c g - 2 d f\right ) + \frac {B b^{2} c^{2} \left (c g - 2 d f\right )}{d}}{B a^{2} d^{2} g - 2 B a b d^{2} f + B b^{2} c^{2} g - 2 B b^{2} c d f} \right )}}{d^{2}} + x \left (A f + \frac {B a g}{b} - \frac {B c g}{d}\right ) + \left (B f x + \frac {B g x^{2}}{2}\right ) \log {\left (\frac {e \left (a + b x\right )^{2}}{\left (c + d x\right )^{2}} \right )} \] Input:
integrate((g*x+f)*(A+B*ln(e*(b*x+a)**2/(d*x+c)**2)),x)
Output:
A*g*x**2/2 - B*a*(a*g - 2*b*f)*log(x + (B*a**2*c*d*g + B*a**2*d**2*(a*g - 2*b*f)/b + B*a*b*c**2*g - 4*B*a*b*c*d*f - B*a*c*d*(a*g - 2*b*f))/(B*a**2*d **2*g - 2*B*a*b*d**2*f + B*b**2*c**2*g - 2*B*b**2*c*d*f))/b**2 + B*c*(c*g - 2*d*f)*log(x + (B*a**2*c*d*g + B*a*b*c**2*g - 4*B*a*b*c*d*f - B*a*b*c*(c *g - 2*d*f) + B*b**2*c**2*(c*g - 2*d*f)/d)/(B*a**2*d**2*g - 2*B*a*b*d**2*f + B*b**2*c**2*g - 2*B*b**2*c*d*f))/d**2 + x*(A*f + B*a*g/b - B*c*g/d) + ( B*f*x + B*g*x**2/2)*log(e*(a + b*x)**2/(c + d*x)**2)
Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (102) = 204\).
Time = 0.04 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.37 \[ \int (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {1}{2} \, A g x^{2} + {\left (x \log \left (\frac {b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac {2 \, a \log \left (b x + a\right )}{b} - \frac {2 \, c \log \left (d x + c\right )}{d}\right )} B f + \frac {1}{2} \, {\left (x^{2} \log \left (\frac {b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - \frac {2 \, a^{2} \log \left (b x + a\right )}{b^{2}} + \frac {2 \, c^{2} \log \left (d x + c\right )}{d^{2}} - \frac {2 \, {\left (b c - a d\right )} x}{b d}\right )} B g + A f x \] Input:
integrate((g*x+f)*(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x, algorithm="maxima")
Output:
1/2*A*g*x^2 + (x*log(b^2*e*x^2/(d^2*x^2 + 2*c*d*x + c^2) + 2*a*b*e*x/(d^2* x^2 + 2*c*d*x + c^2) + a^2*e/(d^2*x^2 + 2*c*d*x + c^2)) + 2*a*log(b*x + a) /b - 2*c*log(d*x + c)/d)*B*f + 1/2*(x^2*log(b^2*e*x^2/(d^2*x^2 + 2*c*d*x + c^2) + 2*a*b*e*x/(d^2*x^2 + 2*c*d*x + c^2) + a^2*e/(d^2*x^2 + 2*c*d*x + c ^2)) - 2*a^2*log(b*x + a)/b^2 + 2*c^2*log(d*x + c)/d^2 - 2*(b*c - a*d)*x/( b*d))*B*g + A*f*x
Time = 0.46 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.34 \[ \int (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {1}{2} \, A g x^{2} + \frac {1}{2} \, {\left (B g x^{2} + 2 \, B f x\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac {{\left (A b d f - B b c g + B a d g\right )} x}{b d} + \frac {{\left (2 \, B a b f - B a^{2} g\right )} \log \left (b x + a\right )}{b^{2}} - \frac {{\left (2 \, B c d f - B c^{2} g\right )} \log \left (-d x - c\right )}{d^{2}} \] Input:
integrate((g*x+f)*(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x, algorithm="giac")
Output:
1/2*A*g*x^2 + 1/2*(B*g*x^2 + 2*B*f*x)*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)/ (d^2*x^2 + 2*c*d*x + c^2)) + (A*b*d*f - B*b*c*g + B*a*d*g)*x/(b*d) + (2*B* a*b*f - B*a^2*g)*log(b*x + a)/b^2 - (2*B*c*d*f - B*c^2*g)*log(-d*x - c)/d^ 2
Time = 25.51 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.28 \[ \int (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\,\left (\frac {B\,g\,x^2}{2}+B\,f\,x\right )+x\,\left (\frac {A\,a\,d\,g+A\,b\,c\,g+A\,b\,d\,f+B\,a\,d\,g-B\,b\,c\,g}{b\,d}-\frac {A\,g\,\left (a\,d+b\,c\right )}{b\,d}\right )+\frac {A\,g\,x^2}{2}-\frac {B\,a\,\ln \left (a+b\,x\right )\,\left (a\,g-2\,b\,f\right )}{b^2}+\frac {B\,c\,\ln \left (c+d\,x\right )\,\left (c\,g-2\,d\,f\right )}{d^2} \] Input:
int((f + g*x)*(A + B*log((e*(a + b*x)^2)/(c + d*x)^2)),x)
Output:
log((e*(a + b*x)^2)/(c + d*x)^2)*(B*f*x + (B*g*x^2)/2) + x*((A*a*d*g + A*b *c*g + A*b*d*f + B*a*d*g - B*b*c*g)/(b*d) - (A*g*(a*d + b*c))/(b*d)) + (A* g*x^2)/2 - (B*a*log(a + b*x)*(a*g - 2*b*f))/b^2 + (B*c*log(c + d*x)*(c*g - 2*d*f))/d^2
Time = 0.17 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.90 \[ \int (f+g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {-2 \,\mathrm {log}\left (d x +c \right ) a^{2} d^{2} g +4 \,\mathrm {log}\left (d x +c \right ) a b \,d^{2} f +2 \,\mathrm {log}\left (d x +c \right ) b^{2} c^{2} g -4 \,\mathrm {log}\left (d x +c \right ) b^{2} c d f -\mathrm {log}\left (\frac {b^{2} e \,x^{2}+2 a b e x +a^{2} e}{d^{2} x^{2}+2 c d x +c^{2}}\right ) a^{2} d^{2} g +2 \,\mathrm {log}\left (\frac {b^{2} e \,x^{2}+2 a b e x +a^{2} e}{d^{2} x^{2}+2 c d x +c^{2}}\right ) a b \,d^{2} f +2 \,\mathrm {log}\left (\frac {b^{2} e \,x^{2}+2 a b e x +a^{2} e}{d^{2} x^{2}+2 c d x +c^{2}}\right ) b^{2} d^{2} f x +\mathrm {log}\left (\frac {b^{2} e \,x^{2}+2 a b e x +a^{2} e}{d^{2} x^{2}+2 c d x +c^{2}}\right ) b^{2} d^{2} g \,x^{2}+2 a b \,d^{2} f x +a b \,d^{2} g \,x^{2}+2 a b \,d^{2} g x -2 b^{2} c d g x}{2 b \,d^{2}} \] Input:
int((g*x+f)*(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x)
Output:
( - 2*log(c + d*x)*a**2*d**2*g + 4*log(c + d*x)*a*b*d**2*f + 2*log(c + d*x )*b**2*c**2*g - 4*log(c + d*x)*b**2*c*d*f - log((a**2*e + 2*a*b*e*x + b**2 *e*x**2)/(c**2 + 2*c*d*x + d**2*x**2))*a**2*d**2*g + 2*log((a**2*e + 2*a*b *e*x + b**2*e*x**2)/(c**2 + 2*c*d*x + d**2*x**2))*a*b*d**2*f + 2*log((a**2 *e + 2*a*b*e*x + b**2*e*x**2)/(c**2 + 2*c*d*x + d**2*x**2))*b**2*d**2*f*x + log((a**2*e + 2*a*b*e*x + b**2*e*x**2)/(c**2 + 2*c*d*x + d**2*x**2))*b** 2*d**2*g*x**2 + 2*a*b*d**2*f*x + a*b*d**2*g*x**2 + 2*a*b*d**2*g*x - 2*b**2 *c*d*g*x)/(2*b*d**2)