\(\int \frac {c+d x}{x^2 (a+b x)} \, dx\) [210]

Optimal result

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

[Out]

-c/a/x-(-a*d+b*c)*ln(x)/a^2+(-a*d+b*c)*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} \[ \int \frac {c+d x}{x^2 (a+b x)} \, dx=-\frac {\log (x) (b c-a d)}{a^2}+\frac {(b c-a d) \log (a+b x)}{a^2}-\frac {c}{a x} \]

[In]

Int[(c + d*x)/(x^2*(a + b*x)),x]

[Out]

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

Mathematica [A] (verified)

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

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {c}{a x}+\frac {\left (a d -b c \right ) \ln \left (x \right )}{a^{2}}-\frac {\left (a d -b c \right ) \ln \left (b x +a \right )}{a^{2}}\)	\(44\)
norman	\(-\frac {c}{a x}+\frac {\left (a d -b c \right ) \ln \left (x \right )}{a^{2}}-\frac {\left (a d -b c \right ) \ln \left (b x +a \right )}{a^{2}}\)	\(44\)
parallelrisch	\(\frac {\ln \left (x \right ) x a d -\ln \left (x \right ) x b c -\ln \left (b x +a \right ) x a d +\ln \left (b x +a \right ) x b c -a c}{a^{2} x}\)	\(47\)
risch	\(-\frac {c}{a x}+\frac {\ln \left (-x \right ) d}{a}-\frac {\ln \left (-x \right ) b c}{a^{2}}-\frac {\ln \left (b x +a \right ) d}{a}+\frac {\ln \left (b x +a \right ) b c}{a^{2}}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {c+d x}{x^2 (a+b x)} \, dx=\frac {{\left (b c - a d\right )} x \log \left (b x + a\right ) - {\left (b c - a d\right )} x \log \left (x\right ) - a c}{a^{2} x} \]

[In]

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

[Out]

((b*c - a*d)*x*log(b*x + a) - (b*c - a*d)*x*log(x) - a*c)/(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 = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.21 \[ \int \frac {c+d x}{x^2 (a+b x)} \, dx=- \frac {c}{a x} + \frac {\left (a d - b c\right ) \log {\left (x + \frac {a^{2} d - a b c - a \left (a d - b c\right )}{2 a b d - 2 b^{2} c} \right )}}{a^{2}} - \frac {\left (a d - b c\right ) \log {\left (x + \frac {a^{2} d - a b c + a \left (a d - b c\right )}{2 a b d - 2 b^{2} c} \right )}}{a^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.19 \[ \int \frac {c+d x}{x^2 (a+b x)} \, dx=-\frac {{\left (b c - a d\right )} \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {c}{a x} + \frac {{\left (b^{2} c - a b d\right )} \log \left ({\left | b x + a \right |}\right )}{a^{2} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {c+d x}{x^2 (a+b x)} \, dx=-\frac {c}{a\,x}-\frac {2\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )\,\left (a\,d-b\,c\right )}{a^2} \]

[In]

int((c + d*x)/(x^2*(a + b*x)),x)

[Out]

- c/(a*x) - (2*atanh((2*b*x)/a + 1)*(a*d - b*c))/a^2