\(\int \frac {A+B x}{x^2 (a+b x)} \, dx\) [180]

Optimal result

Integrand size = 16, antiderivative size = 43 \[ \int \frac {A+B x}{x^2 (a+b x)} \, dx=-\frac {A}{a x}-\frac {(A b-a B) \log (x)}{a^2}+\frac {(A b-a B) \log (a+b x)}{a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, 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} \[ \int \frac {A+B x}{x^2 (a+b x)} \, dx=-\frac {\log (x) (A b-a B)}{a^2}+\frac {(A b-a B) \log (a+b x)}{a^2}-\frac {A}{a x} \]

[In]

Int[(A + B*x)/(x^2*(a + b*x)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{x^2 (a+b x)} \, dx=-\frac {A}{a x}+\frac {(-A b+a B) \log (x)}{a^2}+\frac {(A b-a B) \log (a+b x)}{a^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {A}{a x}+\frac {\left (-A b +B a \right ) \ln \left (x \right )}{a^{2}}+\frac {\left (A b -B a \right ) \ln \left (b x +a \right )}{a^{2}}\)	\(43\)
norman	\(-\frac {A}{a x}-\frac {\left (A b -B a \right ) \ln \left (x \right )}{a^{2}}+\frac {\left (A b -B a \right ) \ln \left (b x +a \right )}{a^{2}}\)	\(44\)
parallelrisch	\(-\frac {A \ln \left (x \right ) x b -A \ln \left (b x +a \right ) x b -B \ln \left (x \right ) x a +B \ln \left (b x +a \right ) x a +A a}{a^{2} x}\)	\(47\)
risch	\(-\frac {A}{a x}-\frac {\ln \left (x \right ) A b}{a^{2}}+\frac {\ln \left (x \right ) B}{a}+\frac {\ln \left (-b x -a \right ) A b}{a^{2}}-\frac {\ln \left (-b x -a \right ) B}{a}\)	\(57\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x}{x^2 (a+b x)} \, dx=-\frac {{\left (B a - A b\right )} x \log \left (b x + a\right ) - {\left (B a - A b\right )} x \log \left (x\right ) + A a}{a^{2} x} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (34) = 68\).

Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.21 \[ \int \frac {A+B x}{x^2 (a+b x)} \, dx=- \frac {A}{a x} + \frac {\left (- A b + B a\right ) \log {\left (x + \frac {- A a b + B a^{2} - a \left (- A b + B a\right )}{- 2 A b^{2} + 2 B a b} \right )}}{a^{2}} - \frac {\left (- A b + B a\right ) \log {\left (x + \frac {- A a b + B a^{2} + a \left (- A b + B a\right )}{- 2 A b^{2} + 2 B a b} \right )}}{a^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x}{x^2 (a+b x)} \, dx=-\frac {{\left (B a - A b\right )} \log \left (b x + a\right )}{a^{2}} + \frac {{\left (B a - A b\right )} \log \left (x\right )}{a^{2}} - \frac {A}{a x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x}{x^2 (a+b x)} \, dx=\frac {{\left (B a - A b\right )} \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {A}{a x} - \frac {{\left (B a b - A b^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{2} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {A+B x}{x^2 (a+b x)} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )\,\left (A\,b-B\,a\right )}{a^2}-\frac {A}{a\,x} \]

[In]

int((A + B*x)/(x^2*(a + b*x)),x)

[Out]

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