\(\int \frac {1}{(a+b x) (a c-b c x)} \, dx\) [1053]

Optimal result

Integrand size = 19, antiderivative size = 17 \[ \int \frac {1}{(a+b x) (a c-b c x)} \, dx=\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{a b c} \]

[Out]

arctanh(b*x/a)/a/b/c

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {35, 214} \[ \int \frac {1}{(a+b x) (a c-b c x)} \, dx=\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{a b c} \]

[In]

Int[1/((a + b*x)*(a*c - b*c*x)),x]

[Out]

ArcTanh[(b*x)/a]/(a*b*c)

Rule 35

Int[1/(((a_) + (b_.)*(x_))*((c_) + (d_.)*(x_))), x_Symbol] :> Int[1/(a*c + b*d*x^2), x] /; FreeQ[{a, b, c, d},
x] && EqQ[b*c + a*d, 0]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{a^2 c-b^2 c x^2} \, dx \\ & = \frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{a b c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x) (a c-b c x)} \, dx=\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{a b c} \]

[In]

Integrate[1/((a + b*x)*(a*c - b*c*x)),x]

[Out]

ArcTanh[(b*x)/a]/(a*b*c)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.71

method	result	size
parallelrisch	\(\frac {-\ln \left (b x -a \right )+\ln \left (b x +a \right )}{2 a c b}\)	\(29\)
default	\(\frac {-\frac {\ln \left (-b x +a \right )}{2 b a}+\frac {\ln \left (b x +a \right )}{2 b a}}{c}\)	\(35\)
norman	\(-\frac {\ln \left (-b x +a \right )}{2 a c b}+\frac {\ln \left (b x +a \right )}{2 a c b}\)	\(37\)
risch	\(-\frac {\ln \left (-b x +a \right )}{2 a c b}+\frac {\ln \left (b x +a \right )}{2 a c b}\)	\(37\)

[In]

int(1/(b*x+a)/(-b*c*x+a*c),x,method=_RETURNVERBOSE)

[Out]

1/2*(-ln(b*x-a)+ln(b*x+a))/a/c/b

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.65 \[ \int \frac {1}{(a+b x) (a c-b c x)} \, dx=\frac {\log \left (b x + a\right ) - \log \left (b x - a\right )}{2 \, a b c} \]

[In]

integrate(1/(b*x+a)/(-b*c*x+a*c),x, algorithm="fricas")

[Out]

1/2*(log(b*x + a) - log(b*x - a))/(a*b*c)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 22 vs. \(2 (10) = 20\).

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(a+b x) (a c-b c x)} \, dx=- \frac {\frac {\log {\left (- \frac {a}{b} + x \right )}}{2} - \frac {\log {\left (\frac {a}{b} + x \right )}}{2}}{a b c} \]

[In]

integrate(1/(b*x+a)/(-b*c*x+a*c),x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.18 \[ \int \frac {1}{(a+b x) (a c-b c x)} \, dx=\frac {\log \left (b x + a\right )}{2 \, a b c} - \frac {\log \left (b x - a\right )}{2 \, a b c} \]

[In]

integrate(1/(b*x+a)/(-b*c*x+a*c),x, algorithm="maxima")

[Out]

1/2*log(b*x + a)/(a*b*c) - 1/2*log(b*x - a)/(a*b*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (17) = 34\).

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.29 \[ \int \frac {1}{(a+b x) (a c-b c x)} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{2 \, a b c} - \frac {\log \left ({\left | b x - a \right |}\right )}{2 \, a b c} \]

[In]

integrate(1/(b*x+a)/(-b*c*x+a*c),x, algorithm="giac")

[Out]

1/2*log(abs(b*x + a))/(a*b*c) - 1/2*log(abs(b*x - a))/(a*b*c)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x) (a c-b c x)} \, dx=\frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{a\,b\,c} \]

[In]

int(1/((a*c - b*c*x)*(a + b*x)),x)

[Out]

atanh((b*x)/a)/(a*b*c)