\(\int \frac {x}{(a+b x) (c+d x)} \, dx\) [236]

Optimal result

Integrand size = 16, antiderivative size = 44 \[ \int \frac {x}{(a+b x) (c+d x)} \, dx=-\frac {a \log (a+b x)}{b (b c-a d)}+\frac {c \log (c+d x)}{d (b c-a d)} \]

[Out]

-a*ln(b*x+a)/b/(-a*d+b*c)+c*ln(d*x+c)/d/(-a*d+b*c)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {x}{(a+b x) (c+d x)} \, dx=\frac {c \log (c+d x)}{d (b c-a d)}-\frac {a \log (a+b x)}{b (b c-a d)} \]

[In]

Int[x/((a + b*x)*(c + d*x)),x]

[Out]

-((a*Log[a + b*x])/(b*(b*c - a*d))) + (c*Log[c + d*x])/(d*(b*c - a*d))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {x}{(a+b x) (c+d x)} \, dx=-\frac {a d \log (a+b x)-b c \log (c+d x)}{b^2 c d-a b d^2} \]

[In]

Integrate[x/((a + b*x)*(c + d*x)),x]

[Out]

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

Maple [A] (verified)

Time = 1.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86

method	result	size
parallelrisch	\(\frac {\ln \left (b x +a \right ) a d -c \ln \left (d x +c \right ) b}{\left (a d -b c \right ) b d}\)	\(38\)
default	\(-\frac {c \ln \left (d x +c \right )}{\left (a d -b c \right ) d}+\frac {a \ln \left (b x +a \right )}{\left (a d -b c \right ) b}\)	\(45\)
norman	\(-\frac {c \ln \left (d x +c \right )}{\left (a d -b c \right ) d}+\frac {a \ln \left (b x +a \right )}{\left (a d -b c \right ) b}\)	\(45\)
risch	\(\frac {a \ln \left (-b x -a \right )}{\left (a d -b c \right ) b}-\frac {c \ln \left (d x +c \right )}{\left (a d -b c \right ) d}\)	\(48\)

[In]

int(x/(b*x+a)/(d*x+c),x,method=_RETURNVERBOSE)

[Out]

(ln(b*x+a)*a*d-c*ln(d*x+c)*b)/(a*d-b*c)/b/d

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {x}{(a+b x) (c+d x)} \, dx=-\frac {a d \log \left (b x + a\right ) - b c \log \left (d x + c\right )}{b^{2} c d - a b d^{2}} \]

[In]

integrate(x/(b*x+a)/(d*x+c),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (32) = 64\).

Time = 0.39 (sec) , antiderivative size = 138, normalized size of antiderivative = 3.14 \[ \int \frac {x}{(a+b x) (c+d x)} \, dx=\frac {a \log {\left (x + \frac {\frac {a^{3} d^{2}}{b \left (a d - b c\right )} - \frac {2 a^{2} c d}{a d - b c} + \frac {a b c^{2}}{a d - b c} + 2 a c}{a d + b c} \right )}}{b \left (a d - b c\right )} - \frac {c \log {\left (x + \frac {- \frac {a^{2} c d}{a d - b c} + \frac {2 a b c^{2}}{a d - b c} + 2 a c - \frac {b^{2} c^{3}}{d \left (a d - b c\right )}}{a d + b c} \right )}}{d \left (a d - b c\right )} \]

[In]

integrate(x/(b*x+a)/(d*x+c),x)

[Out]

a*log(x + (a**3*d**2/(b*(a*d - b*c)) - 2*a**2*c*d/(a*d - b*c) + a*b*c**2/(a*d - b*c) + 2*a*c)/(a*d + b*c))/(b*
(a*d - b*c)) - c*log(x + (-a**2*c*d/(a*d - b*c) + 2*a*b*c**2/(a*d - b*c) + 2*a*c - b**2*c**3/(d*(a*d - b*c)))/
(a*d + b*c))/(d*(a*d - b*c))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {x}{(a+b x) (c+d x)} \, dx=-\frac {a \log \left (b x + a\right )}{b^{2} c - a b d} + \frac {c \log \left (d x + c\right )}{b c d - a d^{2}} \]

[In]

integrate(x/(b*x+a)/(d*x+c),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {x}{(a+b x) (c+d x)} \, dx=-\frac {a \log \left ({\left | b x + a \right |}\right )}{b^{2} c - a b d} + \frac {c \log \left ({\left | d x + c \right |}\right )}{b c d - a d^{2}} \]

[In]

integrate(x/(b*x+a)/(d*x+c),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {x}{(a+b x) (c+d x)} \, dx=\frac {a\,d\,\ln \left (a+b\,x\right )-b\,c\,\ln \left (c+d\,x\right )}{b\,d\,\left (a\,d-b\,c\right )} \]

[In]

int(x/((a + b*x)*(c + d*x)),x)

[Out]

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