\(\int \frac {1}{(a+b x) (a c-b c x)^2} \, dx\) [1054]

Optimal result

Integrand size = 19, antiderivative size = 42 \[ \int \frac {1}{(a+b x) (a c-b c x)^2} \, dx=\frac {1}{2 a b c^2 (a-b x)}+\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{2 a^2 b c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {46, 214} \[ \int \frac {1}{(a+b x) (a c-b c x)^2} \, dx=\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{2 a^2 b c^2}+\frac {1}{2 a b c^2 (a-b x)} \]

[In]

Int[1/((a + b*x)*(a*c - b*c*x)^2),x]

[Out]

1/(2*a*b*c^2*(a - b*x)) + ArcTanh[(b*x)/a]/(2*a^2*b*c^2)

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 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 \left (\frac {1}{2 a c^2 (a-b x)^2}+\frac {1}{2 a c^2 \left (a^2-b^2 x^2\right )}\right ) \, dx \\ & = \frac {1}{2 a b c^2 (a-b x)}+\frac {\int \frac {1}{a^2-b^2 x^2} \, dx}{2 a c^2} \\ & = \frac {1}{2 a b c^2 (a-b x)}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{2 a^2 b c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(a+b x) (a c-b c x)^2} \, dx=\frac {2 a+(-a+b x) \log (a-b x)+(a-b x) \log (a+b x)}{4 a^2 b c^2 (a-b x)} \]

[In]

Integrate[1/((a + b*x)*(a*c - b*c*x)^2),x]

[Out]

(2*a + (-a + b*x)*Log[a - b*x] + (a - b*x)*Log[a + b*x])/(4*a^2*b*c^2*(a - b*x))

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {\frac {\ln \left (b x +a \right )}{4 a^{2} b}-\frac {\ln \left (-b x +a \right )}{4 a^{2} b}+\frac {1}{2 b a \left (-b x +a \right )}}{c^{2}}\)	\(51\)
norman	\(\frac {1}{2 a b \,c^{2} \left (-b x +a \right )}-\frac {\ln \left (-b x +a \right )}{4 a^{2} c^{2} b}+\frac {\ln \left (b x +a \right )}{4 a^{2} c^{2} b}\)	\(56\)
risch	\(\frac {1}{2 a b \,c^{2} \left (-b x +a \right )}-\frac {\ln \left (-b x +a \right )}{4 a^{2} c^{2} b}+\frac {\ln \left (b x +a \right )}{4 a^{2} c^{2} b}\)	\(56\)
parallelrisch	\(\frac {-\ln \left (b x -a \right ) x b +b \ln \left (b x +a \right ) x +a \ln \left (b x -a \right )-a \ln \left (b x +a \right )-2 a}{4 a^{2} b \,c^{2} \left (b x -a \right )}\)	\(65\)

[In]

int(1/(b*x+a)/(-b*c*x+a*c)^2,x,method=_RETURNVERBOSE)

[Out]

1/c^2*(1/4/a^2/b*ln(b*x+a)-1/4/a^2/b*ln(-b*x+a)+1/2/b/a/(-b*x+a))

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(a+b x) (a c-b c x)^2} \, dx=\frac {{\left (b x - a\right )} \log \left (b x + a\right ) - {\left (b x - a\right )} \log \left (b x - a\right ) - 2 \, a}{4 \, {\left (a^{2} b^{2} c^{2} x - a^{3} b c^{2}\right )}} \]

[In]

integrate(1/(b*x+a)/(-b*c*x+a*c)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(a+b x) (a c-b c x)^2} \, dx=- \frac {1}{- 2 a^{2} b c^{2} + 2 a b^{2} c^{2} x} + \frac {- \frac {\log {\left (- \frac {a}{b} + x \right )}}{4} + \frac {\log {\left (\frac {a}{b} + x \right )}}{4}}{a^{2} b c^{2}} \]

[In]

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

[Out]

-1/(-2*a**2*b*c**2 + 2*a*b**2*c**2*x) + (-log(-a/b + x)/4 + log(a/b + x)/4)/(a**2*b*c**2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(a+b x) (a c-b c x)^2} \, dx=-\frac {1}{2 \, {\left (a b^{2} c^{2} x - a^{2} b c^{2}\right )}} + \frac {\log \left (b x + a\right )}{4 \, a^{2} b c^{2}} - \frac {\log \left (b x - a\right )}{4 \, a^{2} b c^{2}} \]

[In]

integrate(1/(b*x+a)/(-b*c*x+a*c)^2,x, algorithm="maxima")

[Out]

-1/2/(a*b^2*c^2*x - a^2*b*c^2) + 1/4*log(b*x + a)/(a^2*b*c^2) - 1/4*log(b*x - a)/(a^2*b*c^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(a+b x) (a c-b c x)^2} \, dx=-\frac {1}{2 \, {\left (b c x - a c\right )} a b c} + \frac {\log \left ({\left | -\frac {2 \, a c}{b c x - a c} - 1 \right |}\right )}{4 \, a^{2} b c^{2}} \]

[In]

integrate(1/(b*x+a)/(-b*c*x+a*c)^2,x, algorithm="giac")

[Out]

-1/2/((b*c*x - a*c)*a*b*c) + 1/4*log(abs(-2*a*c/(b*c*x - a*c) - 1))/(a^2*b*c^2)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x) (a c-b c x)^2} \, dx=\frac {1}{2\,a\,b\,\left (a\,c^2-b\,c^2\,x\right )}+\frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{2\,a^2\,b\,c^2} \]

[In]

int(1/((a*c - b*c*x)^2*(a + b*x)),x)

[Out]

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