\(\int (A+B \log (\frac {e (a+b x)^2}{(c+d x)^2})) \, dx\) [266]

Optimal result

Integrand size = 21, antiderivative size = 54 \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=A x+\frac {B (a+b x) \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{b}-\frac {2 B (b c-a d) \log (c+d x)}{b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2536, 31} \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {B (a+b x) \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{b}-\frac {2 B (b c-a d) \log (c+d x)}{b d}+A x \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2536

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.), x_Symbol] :> Simp[
(a + b*x)*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])^p/b), x] - Dist[B*n*p*((b*c - a*d)/b), Int[(A + B*Log[e*((
a + b*x)^n/(c + d*x)^n)])^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && EqQ[n + mn, 0] &&
NeQ[b*c - a*d, 0] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = A x+B \int \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right ) \, dx \\ & = A x+\frac {B (a+b x) \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{b}-\frac {(2 B (b c-a d)) \int \frac {1}{c+d x} \, dx}{b} \\ & = A x+\frac {B (a+b x) \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{b}-\frac {2 B (b c-a d) \log (c+d x)}{b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=A x+\frac {B (a+b x) \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{b}-\frac {2 B (b c-a d) \log (c+d x)}{b d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00

method	result	size
risch	\(A x +B x \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )-\frac {2 B c \ln \left (d x +c \right )}{d}+\frac {2 B a \ln \left (-b x -a \right )}{b}\)	\(54\)
parallelrisch	\(\frac {B \left (2 x \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) b d +4 \ln \left (b x +a \right ) a d -4 \ln \left (b x +a \right ) b c +2 \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right ) b c \right )}{2 b d}+A x\)	\(78\)
default	\(A x -\frac {B \left (-\left (d x +c \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )-\left (-2 a d +2 c b \right ) \left (\frac {\ln \left (\frac {1}{d x +c}\right )}{b}+\frac {\left (-a d +c b \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b \left (a d -c b \right )}\right )\right )}{d}\)	\(120\)
parts	\(A x -\frac {B \left (-\left (d x +c \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )-\left (-2 a d +2 c b \right ) \left (\frac {\ln \left (\frac {1}{d x +c}\right )}{b}+\frac {\left (-a d +c b \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b \left (a d -c b \right )}\right )\right )}{d}\)	\(120\)
derivativedivides	\(-\frac {-A \left (d x +c \right )+B \left (-\left (d x +c \right ) \ln \left (\frac {e \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2}}{d^{2}}\right )-\left (-2 a d +2 c b \right ) \left (\frac {\ln \left (\frac {1}{d x +c}\right )}{b}+\frac {\left (-a d +c b \right ) \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b \left (a d -c b \right )}\right )\right )}{d}\)	\(126\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.48 \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=\frac {B b d x \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + A b d x + 2 \, B a d \log \left (b x + a\right ) - 2 \, B b c \log \left (d x + c\right )}{b d} \]

[In]

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

[Out]

(B*b*d*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*x + 2*B*a*d*log(b*x + a) - 2*B
*b*c*log(d*x + c))/(b*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (48) = 96\).

Time = 0.48 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.93 \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=A x + \frac {2 B a \log {\left (x + \frac {\frac {2 B a^{2} d}{b} + 2 B a c}{2 B a d + 2 B b c} \right )}}{b} - \frac {2 B c \log {\left (x + \frac {2 B a c + \frac {2 B b c^{2}}{d}}{2 B a d + 2 B b c} \right )}}{d} + B x \log {\left (\frac {e \left (a + b x\right )^{2}}{\left (c + d x\right )^{2}} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.06 \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx={\left (x \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + \frac {2 \, {\left (\frac {a e \log \left (b x + a\right )}{b} - \frac {c e \log \left (d x + c\right )}{d}\right )}}{e}\right )} B + A x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.52 \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx={\left (2 \, {\left (b c - a d\right )} {\left (\frac {a \log \left ({\left | b x + a \right |}\right )}{b^{2} c - a b d} - \frac {c \log \left ({\left | d x + c \right |}\right )}{b c d - a d^{2}}\right )} + x \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right )\right )} B + A x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.93 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \, dx=A\,x+B\,x\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )+\frac {2\,B\,a\,\ln \left (a+b\,x\right )}{b}-\frac {2\,B\,c\,\ln \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

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