\(\int \frac {x}{(a+b x) (c+d x)^2} \, dx\) [246]

Optimal result

Integrand size = 16, antiderivative size = 61 \[ \int \frac {x}{(a+b x) (c+d x)^2} \, dx=-\frac {c}{d (b c-a d) (c+d x)}-\frac {a \log (a+b x)}{(b c-a d)^2}+\frac {a \log (c+d x)}{(b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, 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} \[ \int \frac {x}{(a+b x) (c+d x)^2} \, dx=-\frac {c}{d (c+d x) (b c-a d)}-\frac {a \log (a+b x)}{(b c-a d)^2}+\frac {a \log (c+d x)}{(b c-a d)^2} \]

[In]

Int[x/((a + b*x)*(c + d*x)^2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \frac {x}{(a+b x) (c+d x)^2} \, dx=\frac {c}{d (-b c+a d) (c+d x)}-\frac {a \log (a+b x)}{(b c-a d)^2}+\frac {a \log (c+d x)}{(b c-a d)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {a \ln \left (d x +c \right )}{\left (a d -b c \right )^{2}}+\frac {c}{\left (a d -b c \right ) d \left (d x +c \right )}-\frac {a \ln \left (b x +a \right )}{\left (a d -b c \right )^{2}}\)	\(61\)
norman	\(-\frac {x}{\left (a d -b c \right ) \left (d x +c \right )}+\frac {a \ln \left (d x +c \right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}-\frac {a \ln \left (b x +a \right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}\)	\(85\)
risch	\(\frac {c}{\left (a d -b c \right ) d \left (d x +c \right )}+\frac {a \ln \left (-d x -c \right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}-\frac {a \ln \left (b x +a \right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}\)	\(90\)
parallelrisch	\(-\frac {\ln \left (b x +a \right ) x a \,d^{2}-\ln \left (d x +c \right ) x a \,d^{2}+\ln \left (b x +a \right ) a c d -\ln \left (d x +c \right ) a c d -a d c +b \,c^{2}}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (d x +c \right ) d}\)	\(93\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.75 \[ \int \frac {x}{(a+b x) (c+d x)^2} \, dx=-\frac {b c^{2} - a c d + {\left (a d^{2} x + a c d\right )} \log \left (b x + a\right ) - {\left (a d^{2} x + a c d\right )} \log \left (d x + c\right )}{b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x} \]

[In]

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

[Out]

-(b*c^2 - a*c*d + (a*d^2*x + a*c*d)*log(b*x + a) - (a*d^2*x + a*c*d)*log(d*x + c))/(b^2*c^3*d - 2*a*b*c^2*d^2
+ a^2*c*d^3 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (48) = 96\).

Time = 0.36 (sec) , antiderivative size = 238, normalized size of antiderivative = 3.90 \[ \int \frac {x}{(a+b x) (c+d x)^2} \, dx=\frac {a \log {\left (x + \frac {- \frac {a^{4} d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 a^{3} b c d^{2}}{\left (a d - b c\right )^{2}} - \frac {3 a^{2} b^{2} c^{2} d}{\left (a d - b c\right )^{2}} + a^{2} d + \frac {a b^{3} c^{3}}{\left (a d - b c\right )^{2}} + a b c}{2 a b d} \right )}}{\left (a d - b c\right )^{2}} - \frac {a \log {\left (x + \frac {\frac {a^{4} d^{3}}{\left (a d - b c\right )^{2}} - \frac {3 a^{3} b c d^{2}}{\left (a d - b c\right )^{2}} + \frac {3 a^{2} b^{2} c^{2} d}{\left (a d - b c\right )^{2}} + a^{2} d - \frac {a b^{3} c^{3}}{\left (a d - b c\right )^{2}} + a b c}{2 a b d} \right )}}{\left (a d - b c\right )^{2}} + \frac {c}{a c d^{2} - b c^{2} d + x \left (a d^{3} - b c d^{2}\right )} \]

[In]

integrate(x/(b*x+a)/(d*x+c)**2,x)

[Out]

a*log(x + (-a**4*d**3/(a*d - b*c)**2 + 3*a**3*b*c*d**2/(a*d - b*c)**2 - 3*a**2*b**2*c**2*d/(a*d - b*c)**2 + a*
*2*d + a*b**3*c**3/(a*d - b*c)**2 + a*b*c)/(2*a*b*d))/(a*d - b*c)**2 - a*log(x + (a**4*d**3/(a*d - b*c)**2 - 3
*a**3*b*c*d**2/(a*d - b*c)**2 + 3*a**2*b**2*c**2*d/(a*d - b*c)**2 + a**2*d - a*b**3*c**3/(a*d - b*c)**2 + a*b*
c)/(2*a*b*d))/(a*d - b*c)**2 + c/(a*c*d**2 - b*c**2*d + x*(a*d**3 - b*c*d**2))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.61 \[ \int \frac {x}{(a+b x) (c+d x)^2} \, dx=-\frac {a \log \left (b x + a\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} + \frac {a \log \left (d x + c\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac {c}{b c^{2} d - a c d^{2} + {\left (b c d^{2} - a d^{3}\right )} x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.39 \[ \int \frac {x}{(a+b x) (c+d x)^2} \, dx=-\frac {\frac {a d^{2} \log \left ({\left | b - \frac {b c}{d x + c} + \frac {a d}{d x + c} \right |}\right )}{b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}} + \frac {c d}{{\left (b c d - a d^{2}\right )} {\left (d x + c\right )}}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int \frac {x}{(a+b x) (c+d x)^2} \, dx=\frac {a\,\ln \left (\frac {c+d\,x}{a+b\,x}\right )}{{\left (a\,d-b\,c\right )}^2}+\frac {c}{d\,\left (a\,d-b\,c\right )\,\left (c+d\,x\right )} \]

[In]

int(x/((a + b*x)*(c + d*x)^2),x)

[Out]

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