\(\int (a g+b g x) (A+B \log (\frac {e (a+b x)^2}{(c+d x)^2})) \, dx\) [122]

Optimal result

Integrand size = 30, antiderivative size = 78 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=-\frac {B (b c-a d) g x}{d}+\frac {g (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{2 b}+\frac {B (b c-a d)^2 g \log (c+d x)}{b d^2} \]

[Out]

-B*(-a*d+b*c)*g*x/d+1/2*g*(b*x+a)^2*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))/b+B*(-a*d+b*c)^2*g*ln(d*x+c)/b/d^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2548, 21, 45} \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {g (a+b x)^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{2 b}+\frac {B g (b c-a d)^2 \log (c+d x)}{b d^2}-\frac {B g x (b c-a d)}{d} \]

[In]

Int[(a*g + b*g*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]),x]

[Out]

-((B*(b*c - a*d)*g*x)/d) + (g*(a + b*x)^2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]))/(2*b) + (B*(b*c - a*d)^2*g
*Log[c + d*x])/(b*d^2)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2548

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] - Dist[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])

Rubi steps \begin{align*} \text {integral}& = \frac {g (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{2 b}-\frac {(B (b c-a d)) \int \frac {(a g+b g x)^2}{(a+b x) (c+d x)} \, dx}{b g} \\ & = \frac {g (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{2 b}-\frac {(B (b c-a d) g) \int \frac {a+b x}{c+d x} \, dx}{b} \\ & = \frac {g (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{2 b}-\frac {(B (b c-a d) g) \int \left (\frac {b}{d}+\frac {-b c+a d}{d (c+d x)}\right ) \, dx}{b} \\ & = -\frac {B (b c-a d) g x}{d}+\frac {g (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}{2 b}+\frac {B (b c-a d)^2 g \log (c+d x)}{b d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {g \left ((a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )+\frac {2 B (-b c+a d) (b d x+(-b c+a d) \log (c+d x))}{d^2}\right )}{2 b} \]

[In]

Integrate[(a*g + b*g*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]),x]

[Out]

(g*((a + b*x)^2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2]) + (2*B*(-(b*c) + a*d)*(b*d*x + (-(b*c) + a*d)*Log[c +
d*x]))/d^2))/(2*b)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.38

method	result	size
risch	\(\frac {g B x \left (b x +2 a \right ) \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )}{2}+\frac {g b A \,x^{2}}{2}+g A a x -\frac {2 g B \ln \left (d x +c \right ) a c}{d}+\frac {g b B \ln \left (d x +c \right ) c^{2}}{d^{2}}+\frac {B \,a^{2} g \ln \left (-b x -a \right )}{b}+g B a x -\frac {g b B c x}{d}\)	\(108\)
parallelrisch	\(\frac {B \,x^{2} \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) b^{2} d^{2} g +A \,x^{2} b^{2} d^{2} g +2 B x \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) a b \,d^{2} g +2 A x a b \,d^{2} g +2 B \ln \left (b x +a \right ) a^{2} d^{2} g -4 B \ln \left (b x +a \right ) a b c d g +2 B \ln \left (b x +a \right ) b^{2} c^{2} g +2 B x a b \,d^{2} g -2 B x \,b^{2} c d g +2 B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) a b c d g -B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) b^{2} c^{2} g -2 A \,a^{2} d^{2} g -3 A a b c d g -2 B \,a^{2} d^{2} g +2 B \,b^{2} c^{2} g}{2 b \,d^{2}}\)	\(244\)
parts	\(A g \left (\frac {1}{2} b \,x^{2}+a x \right )-\frac {B g \left (\left (-\frac {\left (d x +c \right )^{2} \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{2}-\left (-a d +c b \right ) \left (\frac {\left (a d -c b \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b^{2}}-\frac {d x +c}{b}+\frac {\left (-a d +c b \right ) \ln \left (\frac {1}{d x +c}\right )}{b^{2}}\right )\right ) b +\left (-\left (d x +c \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )-\left (-2 a d +2 c b \right ) \left (\frac {\left (-a d +c b \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b \left (a d -c b \right )}+\frac {\ln \left (\frac {1}{d x +c}\right )}{b}\right )\right ) \left (a d -c b \right )\right )}{d^{2}}\)	\(261\)
derivativedivides	\(-\frac {\frac {A g \left (-\frac {b \left (d x +c \right )^{2}}{2}-\left (a d -c b \right ) \left (d x +c \right )\right )}{d}+\frac {B g \left (\left (-\frac {\left (d x +c \right )^{2} \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{2}-\left (-a d +c b \right ) \left (\frac {\left (a d -c b \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b^{2}}-\frac {d x +c}{b}+\frac {\left (-a d +c b \right ) \ln \left (\frac {1}{d x +c}\right )}{b^{2}}\right )\right ) b +\left (-\left (d x +c \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )-\left (-2 a d +2 c b \right ) \left (\frac {\left (-a d +c b \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b \left (a d -c b \right )}+\frac {\ln \left (\frac {1}{d x +c}\right )}{b}\right )\right ) \left (a d -c b \right )\right )}{d}}{d}\)	\(284\)
default	\(-\frac {\frac {A g \left (-\frac {b \left (d x +c \right )^{2}}{2}-\left (a d -c b \right ) \left (d x +c \right )\right )}{d}+\frac {B g \left (\left (-\frac {\left (d x +c \right )^{2} \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )}{2}-\left (-a d +c b \right ) \left (\frac {\left (a d -c b \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b^{2}}-\frac {d x +c}{b}+\frac {\left (-a d +c b \right ) \ln \left (\frac {1}{d x +c}\right )}{b^{2}}\right )\right ) b +\left (-\left (d x +c \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )-\left (-2 a d +2 c b \right ) \left (\frac {\left (-a d +c b \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b \left (a d -c b \right )}+\frac {\ln \left (\frac {1}{d x +c}\right )}{b}\right )\right ) \left (a d -c b \right )\right )}{d}}{d}\)	\(284\)

[In]

int((b*g*x+a*g)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2)),x,method=_RETURNVERBOSE)

[Out]

1/2*g*B*x*(b*x+2*a)*ln(e*(b*x+a)^2/(d*x+c)^2)+1/2*g*b*A*x^2+g*A*a*x-2*g/d*B*ln(d*x+c)*a*c+g*b/d^2*B*ln(d*x+c)*
c^2+B*a^2*g/b*ln(-b*x-a)+g*B*a*x-g*b/d*B*c*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.90 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {A b^{2} d^{2} g x^{2} + 2 \, B a^{2} d^{2} g \log \left (b x + a\right ) - 2 \, {\left (B b^{2} c d - {\left (A + B\right )} a b d^{2}\right )} g x + 2 \, {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} g \log \left (d x + c\right ) + {\left (B b^{2} d^{2} g x^{2} + 2 \, B a b d^{2} g x\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \, b d^{2}} \]

[In]

integrate((b*g*x+a*g)*(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x, algorithm="fricas")

[Out]

1/2*(A*b^2*d^2*g*x^2 + 2*B*a^2*d^2*g*log(b*x + a) - 2*(B*b^2*c*d - (A + B)*a*b*d^2)*g*x + 2*(B*b^2*c^2 - 2*B*a
*b*c*d)*g*log(d*x + c) + (B*b^2*d^2*g*x^2 + 2*B*a*b*d^2*g*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*d^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (68) = 136\).

Time = 0.84 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.21 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {A b g x^{2}}{2} + \frac {B a^{2} g \log {\left (x + \frac {\frac {B a^{3} d^{2} g}{b} + 2 B a^{2} c d g - B a b c^{2} g}{B a^{2} d^{2} g + 2 B a b c d g - B b^{2} c^{2} g} \right )}}{b} - \frac {B c g \left (2 a d - b c\right ) \log {\left (x + \frac {3 B a^{2} c d g - B a b c^{2} g - B a c g \left (2 a d - b c\right ) + \frac {B b c^{2} g \left (2 a d - b c\right )}{d}}{B a^{2} d^{2} g + 2 B a b c d g - B b^{2} c^{2} g} \right )}}{d^{2}} + x \left (A a g + B a g - \frac {B b c g}{d}\right ) + \left (B a g x + \frac {B b g x^{2}}{2}\right ) \log {\left (\frac {e \left (a + b x\right )^{2}}{\left (c + d x\right )^{2}} \right )} \]

[In]

integrate((b*g*x+a*g)*(A+B*ln(e*(b*x+a)**2/(d*x+c)**2)),x)

[Out]

A*b*g*x**2/2 + B*a**2*g*log(x + (B*a**3*d**2*g/b + 2*B*a**2*c*d*g - B*a*b*c**2*g)/(B*a**2*d**2*g + 2*B*a*b*c*d
*g - B*b**2*c**2*g))/b - B*c*g*(2*a*d - b*c)*log(x + (3*B*a**2*c*d*g - B*a*b*c**2*g - B*a*c*g*(2*a*d - b*c) +
B*b*c**2*g*(2*a*d - b*c)/d)/(B*a**2*d**2*g + 2*B*a*b*c*d*g - B*b**2*c**2*g))/d**2 + x*(A*a*g + B*a*g - B*b*c*g
/d) + (B*a*g*x + B*b*g*x**2/2)*log(e*(a + b*x)**2/(c + d*x)**2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (76) = 152\).

Time = 0.28 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.21 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {1}{2} \, A b g x^{2} + {\left (x \log \left (\frac {b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac {2 \, a \log \left (b x + a\right )}{b} - \frac {2 \, c \log \left (d x + c\right )}{d}\right )} B a g + \frac {1}{2} \, {\left (x^{2} \log \left (\frac {b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - \frac {2 \, a^{2} \log \left (b x + a\right )}{b^{2}} + \frac {2 \, c^{2} \log \left (d x + c\right )}{d^{2}} - \frac {2 \, {\left (b c - a d\right )} x}{b d}\right )} B b g + A a g x \]

[In]

integrate((b*g*x+a*g)*(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x, algorithm="maxima")

[Out]

1/2*A*b*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*a*g + 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*b*g + A*a*g*x

Giac [A] (verification not implemented)

none

Time = 0.54 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.65 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {1}{2} \, A b g x^{2} + \frac {B a^{2} g \log \left (b x + a\right )}{b} + \frac {1}{2} \, {\left (B b g x^{2} + 2 \, B a g x\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - \frac {{\left (B b c g - A a d g - B a d g\right )} x}{d} + \frac {{\left (B b c^{2} g - 2 \, B a c d g\right )} \log \left (d x + c\right )}{d^{2}} \]

[In]

integrate((b*g*x+a*g)*(A+B*log(e*(b*x+a)^2/(d*x+c)^2)),x, algorithm="giac")

[Out]

1/2*A*b*g*x^2 + B*a^2*g*log(b*x + a)/b + 1/2*(B*b*g*x^2 + 2*B*a*g*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*b*c*g - A*a*d*g - B*a*d*g)*x/d + (B*b*c^2*g - 2*B*a*c*d*g)*log(d*x + c)/d^2

Mupad [B] (verification not implemented)

Time = 1.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.54 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=x\,\left (\frac {g\,\left (2\,A\,a\,d+A\,b\,c+B\,a\,d-B\,b\,c\right )}{d}-\frac {A\,g\,\left (a\,d+b\,c\right )}{d}\right )+\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\,\left (\frac {B\,b\,g\,x^2}{2}+B\,a\,g\,x\right )+\frac {A\,b\,g\,x^2}{2}+\frac {B\,a^2\,g\,\ln \left (a+b\,x\right )}{b}-\frac {B\,c\,g\,\ln \left (c+d\,x\right )\,\left (2\,a\,d-b\,c\right )}{d^2} \]

[In]

int((a*g + b*g*x)*(A + B*log((e*(a + b*x)^2)/(c + d*x)^2)),x)

[Out]

x*((g*(2*A*a*d + A*b*c + B*a*d - B*b*c))/d - (A*g*(a*d + b*c))/d) + log((e*(a + b*x)^2)/(c + d*x)^2)*((B*b*g*x
^2)/2 + B*a*g*x) + (A*b*g*x^2)/2 + (B*a^2*g*log(a + b*x))/b - (B*c*g*log(c + d*x)*(2*a*d - b*c))/d^2