\(\int \frac {A+B x}{x^2 (a+b x)^2} \, dx\) [190]

Optimal result

Integrand size = 16, antiderivative size = 65 \[ \int \frac {A+B x}{x^2 (a+b x)^2} \, dx=-\frac {A}{a^2 x}-\frac {A b-a B}{a^2 (a+b x)}-\frac {(2 A b-a B) \log (x)}{a^3}+\frac {(2 A b-a B) \log (a+b x)}{a^3} \]

[Out]

-A/a^2/x+(-A*b+B*a)/a^2/(b*x+a)-(2*A*b-B*a)*ln(x)/a^3+(2*A*b-B*a)*ln(b*x+a)/a^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, 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} \[ \int \frac {A+B x}{x^2 (a+b x)^2} \, dx=-\frac {\log (x) (2 A b-a B)}{a^3}+\frac {(2 A b-a B) \log (a+b x)}{a^3}-\frac {A b-a B}{a^2 (a+b x)}-\frac {A}{a^2 x} \]

[In]

Int[(A + B*x)/(x^2*(a + b*x)^2),x]

[Out]

-(A/(a^2*x)) - (A*b - a*B)/(a^2*(a + b*x)) - ((2*A*b - a*B)*Log[x])/a^3 + ((2*A*b - a*B)*Log[a + b*x])/a^3

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}{a^2 x^2}+\frac {-2 A b+a B}{a^3 x}-\frac {b (-A b+a B)}{a^2 (a+b x)^2}-\frac {b (-2 A b+a B)}{a^3 (a+b x)}\right ) \, dx \\ & = -\frac {A}{a^2 x}-\frac {A b-a B}{a^2 (a+b x)}-\frac {(2 A b-a B) \log (x)}{a^3}+\frac {(2 A b-a B) \log (a+b x)}{a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{x^2 (a+b x)^2} \, dx=\frac {-\frac {a A}{x}+\frac {a (-A b+a B)}{a+b x}+(-2 A b+a B) \log (x)+(2 A b-a B) \log (a+b x)}{a^3} \]

[In]

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

[Out]

(-((a*A)/x) + (a*(-(A*b) + a*B))/(a + b*x) + (-2*A*b + a*B)*Log[x] + (2*A*b - a*B)*Log[a + b*x])/a^3

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {A}{a^{2} x}+\frac {\left (-2 A b +B a \right ) \ln \left (x \right )}{a^{3}}+\frac {\left (2 A b -B a \right ) \ln \left (b x +a \right )}{a^{3}}-\frac {A b -B a}{a^{2} \left (b x +a \right )}\)	\(64\)
norman	\(\frac {\frac {b \left (2 A b -B a \right ) x^{2}}{a^{3}}-\frac {A}{a}}{x \left (b x +a \right )}+\frac {\left (2 A b -B a \right ) \ln \left (b x +a \right )}{a^{3}}-\frac {\left (2 A b -B a \right ) \ln \left (x \right )}{a^{3}}\)	\(72\)
risch	\(\frac {-\frac {\left (2 A b -B a \right ) x}{a^{2}}-\frac {A}{a}}{x \left (b x +a \right )}-\frac {2 \ln \left (x \right ) A b}{a^{3}}+\frac {\ln \left (x \right ) B}{a^{2}}+\frac {2 \ln \left (-b x -a \right ) A b}{a^{3}}-\frac {\ln \left (-b x -a \right ) B}{a^{2}}\)	\(82\)
parallelrisch	\(-\frac {2 A \ln \left (x \right ) x^{2} b^{3}-2 A \ln \left (b x +a \right ) x^{2} b^{3}-B \ln \left (x \right ) x^{2} a \,b^{2}+B \ln \left (b x +a \right ) x^{2} a \,b^{2}+2 A \ln \left (x \right ) x a \,b^{2}-2 A \ln \left (b x +a \right ) x a \,b^{2}-B \ln \left (x \right ) x \,a^{2} b +B \ln \left (b x +a \right ) x \,a^{2} b +2 a \,b^{2} A x -a^{2} b B x +a^{2} b A}{a^{3} b x \left (b x +a \right )}\)	\(142\)

[In]

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

[Out]

-A/a^2/x+(-2*A*b+B*a)/a^3*ln(x)+(2*A*b-B*a)*ln(b*x+a)/a^3-(A*b-B*a)/a^2/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.65 \[ \int \frac {A+B x}{x^2 (a+b x)^2} \, dx=-\frac {A a^{2} - {\left (B a^{2} - 2 \, A a b\right )} x + {\left ({\left (B a b - 2 \, A b^{2}\right )} x^{2} + {\left (B a^{2} - 2 \, A a b\right )} x\right )} \log \left (b x + a\right ) - {\left ({\left (B a b - 2 \, A b^{2}\right )} x^{2} + {\left (B a^{2} - 2 \, A a b\right )} x\right )} \log \left (x\right )}{a^{3} b x^{2} + a^{4} x} \]

[In]

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

[Out]

-(A*a^2 - (B*a^2 - 2*A*a*b)*x + ((B*a*b - 2*A*b^2)*x^2 + (B*a^2 - 2*A*a*b)*x)*log(b*x + a) - ((B*a*b - 2*A*b^2
)*x^2 + (B*a^2 - 2*A*a*b)*x)*log(x))/(a^3*b*x^2 + a^4*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (54) = 108\).

Time = 0.26 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.97 \[ \int \frac {A+B x}{x^2 (a+b x)^2} \, dx=\frac {- A a + x \left (- 2 A b + B a\right )}{a^{3} x + a^{2} b x^{2}} + \frac {\left (- 2 A b + B a\right ) \log {\left (x + \frac {- 2 A a b + B a^{2} - a \left (- 2 A b + B a\right )}{- 4 A b^{2} + 2 B a b} \right )}}{a^{3}} - \frac {\left (- 2 A b + B a\right ) \log {\left (x + \frac {- 2 A a b + B a^{2} + a \left (- 2 A b + B a\right )}{- 4 A b^{2} + 2 B a b} \right )}}{a^{3}} \]

[In]

integrate((B*x+A)/x**2/(b*x+a)**2,x)

[Out]

(-A*a + x*(-2*A*b + B*a))/(a**3*x + a**2*b*x**2) + (-2*A*b + B*a)*log(x + (-2*A*a*b + B*a**2 - a*(-2*A*b + B*a
))/(-4*A*b**2 + 2*B*a*b))/a**3 - (-2*A*b + B*a)*log(x + (-2*A*a*b + B*a**2 + a*(-2*A*b + B*a))/(-4*A*b**2 + 2*
B*a*b))/a**3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x}{x^2 (a+b x)^2} \, dx=-\frac {A a - {\left (B a - 2 \, A b\right )} x}{a^{2} b x^{2} + a^{3} x} - \frac {{\left (B a - 2 \, A b\right )} \log \left (b x + a\right )}{a^{3}} + \frac {{\left (B a - 2 \, A b\right )} \log \left (x\right )}{a^{3}} \]

[In]

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

[Out]

-(A*a - (B*a - 2*A*b)*x)/(a^2*b*x^2 + a^3*x) - (B*a - 2*A*b)*log(b*x + a)/a^3 + (B*a - 2*A*b)*log(x)/a^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.32 \[ \int \frac {A+B x}{x^2 (a+b x)^2} \, dx=\frac {A b}{a^{3} {\left (\frac {a}{b x + a} - 1\right )}} + \frac {{\left (B a b - 2 \, A b^{2}\right )} \log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{3} b} + \frac {\frac {B a b^{2}}{b x + a} - \frac {A b^{3}}{b x + a}}{a^{2} b^{2}} \]

[In]

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

[Out]

A*b/(a^3*(a/(b*x + a) - 1)) + (B*a*b - 2*A*b^2)*log(abs(-a/(b*x + a) + 1))/(a^3*b) + (B*a*b^2/(b*x + a) - A*b^
3/(b*x + a))/(a^2*b^2)

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{x^2 (a+b x)^2} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )\,\left (2\,A\,b-B\,a\right )}{a^3}-\frac {\frac {A}{a}+\frac {x\,\left (2\,A\,b-B\,a\right )}{a^2}}{b\,x^2+a\,x} \]

[In]

int((A + B*x)/(x^2*(a + b*x)^2),x)

[Out]

(2*atanh((2*b*x)/a + 1)*(2*A*b - B*a))/a^3 - (A/a + (x*(2*A*b - B*a))/a^2)/(a*x + b*x^2)