\(\int (a g+b g x) (A+B \log (\frac {e (c+d x)}{a+b x})) \, dx\) [176]

Optimal result

Integrand size = 28, antiderivative size = 81 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right ) \, dx=\frac {B (b c-a d) g x}{2 d}-\frac {B (b c-a d)^2 g \log (c+d x)}{2 b d^2}+\frac {g (a+b x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2548, 21, 45} \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right ) \, dx=\frac {g (a+b x)^2 \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )}{2 b}-\frac {B g (b c-a d)^2 \log (c+d x)}{2 b d^2}+\frac {B g x (b c-a d)}{2 d} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right ) \, dx=\frac {g \left (\frac {B (b c-a d) (b d x+(-b c+a d) \log (c+d x))}{d^2}+(a+b x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )\right )}{2 b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {g B x \left (b x +2 a \right ) \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )}{2}+\frac {g b A \,x^{2}}{2}+g A a x -\frac {B \,a^{2} g \ln \left (b x +a \right )}{2 b}+\frac {g B \ln \left (-d x -c \right ) a c}{d}-\frac {g b B \ln \left (-d x -c \right ) c^{2}}{2 d^{2}}-\frac {g B a x}{2}+\frac {g b B c x}{2 d}\)	\(111\)
parallelrisch	\(\frac {B \,x^{2} \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) b^{2} d^{2} g +A \,x^{2} b^{2} d^{2} g +2 B x \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) a b \,d^{2} g +2 A x a b \,d^{2} g -B \ln \left (b x +a \right ) a^{2} d^{2} g +2 B \ln \left (b x +a \right ) a b c d g -B \ln \left (b x +a \right ) b^{2} c^{2} g -B x a b \,d^{2} g +B x \,b^{2} c d g +2 B \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) a b c d g -B \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) b^{2} c^{2} g -2 A \,a^{2} d^{2} g -3 A a b c d g +B \,a^{2} d^{2} g -B \,b^{2} c^{2} g}{2 b \,d^{2}}\)	\(234\)
parts	\(A g \left (\frac {1}{2} b \,x^{2}+a x \right )-B g \,e^{2} \left (a d -c b \right )^{2} \left (-\frac {1}{2 d e b \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )}-\frac {\ln \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )}{2 d^{2} e^{2} b}+\frac {\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -2 d e \right )}{2 d^{2} e^{2} \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )^{2}}\right )\)	\(265\)
derivativedivides	\(\frac {e \left (a d -c b \right ) \left (\frac {A b e g \left (a d -c b \right )}{2 \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )^{2}}-B \,b^{2} e g \left (a d -c b \right ) \left (-\frac {1}{2 d e b \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )}-\frac {\ln \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )}{2 d^{2} e^{2} b}+\frac {\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -2 d e \right )}{2 d^{2} e^{2} \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )^{2}}\right )\right )}{b^{2}}\)	\(315\)
default	\(\frac {e \left (a d -c b \right ) \left (\frac {A b e g \left (a d -c b \right )}{2 \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )^{2}}-B \,b^{2} e g \left (a d -c b \right ) \left (-\frac {1}{2 d e b \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )}-\frac {\ln \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )}{2 d^{2} e^{2} b}+\frac {\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -2 d e \right )}{2 d^{2} e^{2} \left (\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right ) b -d e \right )^{2}}\right )\right )}{b^{2}}\)	\(315\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.57 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right ) \, dx=\frac {A b^{2} d^{2} g x^{2} - B a^{2} d^{2} g \log \left (b x + a\right ) + {\left (B b^{2} c d + {\left (2 \, A - B\right )} a b d^{2}\right )} g x - {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} g \log \left (d x + c\right ) + {\left (B b^{2} d^{2} g x^{2} + 2 \, B a b d^{2} g x\right )} \log \left (\frac {d e x + c e}{b x + a}\right )}{2 \, b d^{2}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (68) = 136\).

Time = 0.94 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.12 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right ) \, dx=\frac {A b g x^{2}}{2} - \frac {B a^{2} g \log {\left (x + \frac {\frac {B a^{3} d^{2} g}{b} + 2 B a^{2} c d g - B a b c^{2} g}{B a^{2} d^{2} g + 2 B a b c d g - B b^{2} c^{2} g} \right )}}{2 b} + \frac {B c g \left (2 a d - b c\right ) \log {\left (x + \frac {3 B a^{2} c d g - B a b c^{2} g - B a c g \left (2 a d - b c\right ) + \frac {B b c^{2} g \left (2 a d - b c\right )}{d}}{B a^{2} d^{2} g + 2 B a b c d g - B b^{2} c^{2} g} \right )}}{2 d^{2}} + x \left (A a g - \frac {B a g}{2} + \frac {B b c g}{2 d}\right ) + \left (B a g x + \frac {B b g x^{2}}{2}\right ) \log {\left (\frac {e \left (c + d x\right )}{a + b x} \right )} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.77 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right ) \, dx=\frac {1}{2} \, A b g x^{2} + {\left (x \log \left (\frac {d e x}{b x + a} + \frac {c e}{b x + a}\right ) - \frac {a \log \left (b x + a\right )}{b} + \frac {c \log \left (d x + c\right )}{d}\right )} B a g + \frac {1}{2} \, {\left (x^{2} \log \left (\frac {d e x}{b x + a} + \frac {c e}{b x + a}\right ) + \frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} B b g + A a g x \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 627 vs. \(2 (75) = 150\).

Time = 0.46 (sec) , antiderivative size = 627, normalized size of antiderivative = 7.74 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right ) \, dx=\frac {1}{2} \, {\left (\frac {{\left (B b^{3} c^{3} e^{3} g - 3 \, B a b^{2} c^{2} d e^{3} g + 3 \, B a^{2} b c d^{2} e^{3} g - B a^{3} d^{3} e^{3} g\right )} \log \left (\frac {d e x + c e}{b x + a}\right )}{b d^{2} e^{2} - \frac {2 \, {\left (d e x + c e\right )} b^{2} d e}{b x + a} + \frac {{\left (d e x + c e\right )}^{2} b^{3}}{{\left (b x + a\right )}^{2}}} + \frac {A b^{3} c^{3} d e^{3} g - B b^{3} c^{3} d e^{3} g - 3 \, A a b^{2} c^{2} d^{2} e^{3} g + 3 \, B a b^{2} c^{2} d^{2} e^{3} g + 3 \, A a^{2} b c d^{3} e^{3} g - 3 \, B a^{2} b c d^{3} e^{3} g - A a^{3} d^{4} e^{3} g + B a^{3} d^{4} e^{3} g + \frac {{\left (d e x + c e\right )} B b^{4} c^{3} e^{2} g}{b x + a} - \frac {3 \, {\left (d e x + c e\right )} B a b^{3} c^{2} d e^{2} g}{b x + a} + \frac {3 \, {\left (d e x + c e\right )} B a^{2} b^{2} c d^{2} e^{2} g}{b x + a} - \frac {{\left (d e x + c e\right )} B a^{3} b d^{3} e^{2} g}{b x + a}}{b d^{3} e^{2} - \frac {2 \, {\left (d e x + c e\right )} b^{2} d^{2} e}{b x + a} + \frac {{\left (d e x + c e\right )}^{2} b^{3} d}{{\left (b x + a\right )}^{2}}} + \frac {{\left (B b^{3} c^{3} e g - 3 \, B a b^{2} c^{2} d e g + 3 \, B a^{2} b c d^{2} e g - B a^{3} d^{3} e g\right )} \log \left (-d e + \frac {{\left (d e x + c e\right )} b}{b x + a}\right )}{b d^{2}} - \frac {{\left (B b^{3} c^{3} e g - 3 \, B a b^{2} c^{2} d e g + 3 \, B a^{2} b c d^{2} e g - B a^{3} d^{3} e g\right )} \log \left (\frac {d e x + c e}{b x + a}\right )}{b d^{2}}\right )} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )} \]

[In]

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

[Out]

1/2*((B*b^3*c^3*e^3*g - 3*B*a*b^2*c^2*d*e^3*g + 3*B*a^2*b*c*d^2*e^3*g - B*a^3*d^3*e^3*g)*log((d*e*x + c*e)/(b*
x + a))/(b*d^2*e^2 - 2*(d*e*x + c*e)*b^2*d*e/(b*x + a) + (d*e*x + c*e)^2*b^3/(b*x + a)^2) + (A*b^3*c^3*d*e^3*g
- B*b^3*c^3*d*e^3*g - 3*A*a*b^2*c^2*d^2*e^3*g + 3*B*a*b^2*c^2*d^2*e^3*g + 3*A*a^2*b*c*d^3*e^3*g - 3*B*a^2*b*c
*d^3*e^3*g - A*a^3*d^4*e^3*g + B*a^3*d^4*e^3*g + (d*e*x + c*e)*B*b^4*c^3*e^2*g/(b*x + a) - 3*(d*e*x + c*e)*B*a
*b^3*c^2*d*e^2*g/(b*x + a) + 3*(d*e*x + c*e)*B*a^2*b^2*c*d^2*e^2*g/(b*x + a) - (d*e*x + c*e)*B*a^3*b*d^3*e^2*g
/(b*x + a))/(b*d^3*e^2 - 2*(d*e*x + c*e)*b^2*d^2*e/(b*x + a) + (d*e*x + c*e)^2*b^3*d/(b*x + a)^2) + (B*b^3*c^3
*e*g - 3*B*a*b^2*c^2*d*e*g + 3*B*a^2*b*c*d^2*e*g - B*a^3*d^3*e*g)*log(-d*e + (d*e*x + c*e)*b/(b*x + a))/(b*d^2
) - (B*b^3*c^3*e*g - 3*B*a*b^2*c^2*d*e*g + 3*B*a^2*b*c*d^2*e*g - B*a^3*d^3*e*g)*log((d*e*x + c*e)/(b*x + a))/(
b*d^2))*(b*c/((b*c*e - a*d*e)*(b*c - a*d)) - a*d/((b*c*e - a*d*e)*(b*c - a*d)))

Mupad [B] (verification not implemented)

Time = 1.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.56 \[ \int (a g+b g x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right ) \, dx=x\,\left (\frac {g\,\left (4\,A\,a\,d+2\,A\,b\,c-B\,a\,d+B\,b\,c\right )}{2\,d}-\frac {A\,g\,\left (2\,a\,d+2\,b\,c\right )}{2\,d}\right )+\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )\,\left (\frac {B\,b\,g\,x^2}{2}+B\,a\,g\,x\right )-\frac {\ln \left (c+d\,x\right )\,\left (B\,b\,c^2\,g-2\,B\,a\,c\,d\,g\right )}{2\,d^2}+\frac {A\,b\,g\,x^2}{2}-\frac {B\,a^2\,g\,\ln \left (a+b\,x\right )}{2\,b} \]

[In]

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

[Out]

x*((g*(4*A*a*d + 2*A*b*c - B*a*d + B*b*c))/(2*d) - (A*g*(2*a*d + 2*b*c))/(2*d)) + log((e*(c + d*x))/(a + b*x))
*((B*b*g*x^2)/2 + B*a*g*x) - (log(c + d*x)*(B*b*c^2*g - 2*B*a*c*d*g))/(2*d^2) + (A*b*g*x^2)/2 - (B*a^2*g*log(a
+ b*x))/(2*b)