3.1060 \(\int \frac{1}{(a+b x)^2 (a c-b c x)} \, dx\)

Optimal. Leaf size=41 \[ \frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{2 a^2 b c}-\frac{1}{2 a b c (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0279136, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {44, 208} \[ \frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{2 a^2 b c}-\frac{1}{2 a b c (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^2*(a*c - b*c*x)),x]

[Out]

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

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.0136945, size = 50, normalized size = 1.22 \[ \frac{-(a+b x) \log (a-b x)+(a+b x) \log (a+b x)-2 a}{4 a^2 b c (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 56, normalized size = 1.4 \begin{align*}{\frac{\ln \left ( bx+a \right ) }{4\,c{a}^{2}b}}-{\frac{1}{2\,abc \left ( bx+a \right ) }}-{\frac{\ln \left ( bx-a \right ) }{4\,c{a}^{2}b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01729, size = 74, normalized size = 1.8 \begin{align*} -\frac{1}{2 \,{\left (a b^{2} c x + a^{2} b c\right )}} + \frac{\log \left (b x + a\right )}{4 \, a^{2} b c} - \frac{\log \left (b x - a\right )}{4 \, a^{2} b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.53883, size = 115, normalized size = 2.8 \begin{align*} \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 x + a^{3} b c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^2/(-b*c*x+a*c),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*x + a^3*b*c)

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Sympy [A]  time = 0.421321, size = 44, normalized size = 1.07 \begin{align*} - \frac{1}{2 a^{2} b c + 2 a b^{2} c x} - \frac{\frac{\log{\left (- \frac{a}{b} + x \right )}}{4} - \frac{\log{\left (\frac{a}{b} + x \right )}}{4}}{a^{2} b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.06203, size = 59, normalized size = 1.44 \begin{align*} -\frac{\log \left ({\left | -\frac{2 \, a}{b x + a} + 1 \right |}\right )}{4 \, a^{2} b c} - \frac{1}{2 \,{\left (b x + a\right )} a b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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