\(\int \frac {A+B x}{x (a+b x)^3} \, dx\) [198]

Optimal result

Integrand size = 16, antiderivative size = 57 \[ \int \frac {A+B x}{x (a+b x)^3} \, dx=\frac {A b-a B}{2 a b (a+b x)^2}+\frac {A}{a^2 (a+b x)}+\frac {A \log (x)}{a^3}-\frac {A \log (a+b x)}{a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, 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} \[ \int \frac {A+B x}{x (a+b x)^3} \, dx=-\frac {A \log (a+b x)}{a^3}+\frac {A \log (x)}{a^3}+\frac {A}{a^2 (a+b x)}+\frac {A b-a B}{2 a b (a+b x)^2} \]

[In]

Int[(A + B*x)/(x*(a + b*x)^3),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{x (a+b x)^3} \, dx=\frac {\frac {a \left (3 a A b-a^2 B+2 A b^2 x\right )}{b (a+b x)^2}+2 A \log (x)-2 A \log (a+b x)}{2 a^3} \]

[In]

Integrate[(A + B*x)/(x*(a + b*x)^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {A \ln \left (x \right )}{a^{3}}-\frac {-A b +B a}{2 a b \left (b x +a \right )^{2}}-\frac {A \ln \left (b x +a \right )}{a^{3}}+\frac {A}{a^{2} \left (b x +a \right )}\)	\(56\)
risch	\(\frac {\frac {A b x}{a^{2}}+\frac {3 A b -B a}{2 a b}}{\left (b x +a \right )^{2}}+\frac {A \ln \left (-x \right )}{a^{3}}-\frac {A \ln \left (b x +a \right )}{a^{3}}\)	\(56\)
norman	\(\frac {-\frac {\left (2 A b -B a \right ) x}{a^{2}}-\frac {b \left (3 A b -B a \right ) x^{2}}{2 a^{3}}}{\left (b x +a \right )^{2}}+\frac {A \ln \left (x \right )}{a^{3}}-\frac {A \ln \left (b x +a \right )}{a^{3}}\)	\(63\)
parallelrisch	\(\frac {2 A \ln \left (x \right ) x^{2} b^{2}-2 A \ln \left (b x +a \right ) x^{2} b^{2}+4 A \ln \left (x \right ) x a b -4 A \ln \left (b x +a \right ) x a b -3 A \,b^{2} x^{2}+B a b \,x^{2}+2 a^{2} A \ln \left (x \right )-2 A \ln \left (b x +a \right ) a^{2}-4 a A b x +2 a^{2} B x}{2 a^{3} \left (b x +a \right )^{2}}\)	\(109\)

[In]

int((B*x+A)/x/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.91 \[ \int \frac {A+B x}{x (a+b x)^3} \, dx=\frac {2 \, A a b^{2} x - B a^{3} + 3 \, A a^{2} b - 2 \, {\left (A b^{3} x^{2} + 2 \, A a b^{2} x + A a^{2} b\right )} \log \left (b x + a\right ) + 2 \, {\left (A b^{3} x^{2} + 2 \, A a b^{2} x + A a^{2} b\right )} \log \left (x\right )}{2 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}} \]

[In]

integrate((B*x+A)/x/(b*x+a)^3,x, algorithm="fricas")

[Out]

1/2*(2*A*a*b^2*x - B*a^3 + 3*A*a^2*b - 2*(A*b^3*x^2 + 2*A*a*b^2*x + A*a^2*b)*log(b*x + a) + 2*(A*b^3*x^2 + 2*A
*a*b^2*x + A*a^2*b)*log(x))/(a^3*b^3*x^2 + 2*a^4*b^2*x + a^5*b)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x}{x (a+b x)^3} \, dx=\frac {A \left (\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}\right )}{a^{3}} + \frac {3 A a b + 2 A b^{2} x - B a^{2}}{2 a^{4} b + 4 a^{3} b^{2} x + 2 a^{2} b^{3} x^{2}} \]

[In]

integrate((B*x+A)/x/(b*x+a)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x}{x (a+b x)^3} \, dx=\frac {2 \, A b^{2} x - B a^{2} + 3 \, A a b}{2 \, {\left (a^{2} b^{3} x^{2} + 2 \, a^{3} b^{2} x + a^{4} b\right )}} - \frac {A \log \left (b x + a\right )}{a^{3}} + \frac {A \log \left (x\right )}{a^{3}} \]

[In]

integrate((B*x+A)/x/(b*x+a)^3,x, algorithm="maxima")

[Out]

1/2*(2*A*b^2*x - B*a^2 + 3*A*a*b)/(a^2*b^3*x^2 + 2*a^3*b^2*x + a^4*b) - A*log(b*x + a)/a^3 + A*log(x)/a^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x}{x (a+b x)^3} \, dx=-\frac {A \log \left ({\left | b x + a \right |}\right )}{a^{3}} + \frac {A \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {2 \, A a b^{2} x - B a^{3} + 3 \, A a^{2} b}{2 \, {\left (b x + a\right )}^{2} a^{3} b} \]

[In]

integrate((B*x+A)/x/(b*x+a)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x}{x (a+b x)^3} \, dx=\frac {\frac {3\,A\,b-B\,a}{2\,a\,b}+\frac {A\,b\,x}{a^2}}{a^2+2\,a\,b\,x+b^2\,x^2}-\frac {2\,A\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^3} \]

[In]

int((A + B*x)/(x*(a + b*x)^3),x)

[Out]

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