3.11.60 \(\int \frac {1}{(a+b x)^2 (a c-b c x)} \, dx\) [1060]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, 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 = {46, 214} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{2 a^2 b c}-\frac {1}{2 a b c (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^2*(a*c - b*c*x)),x]

[Out]

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

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*} \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*}

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Mathematica [A]
time = 0.01, size = 50, normalized size = 1.22 \begin {gather*} \frac {-2 a-(a+b x) \log (a-b x)+(a+b x) \log (a+b x)}{4 a^2 b c (a+b x)} \end {gather*}

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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Mathics [A]
time = 2.09, size = 53, normalized size = 1.29 \begin {gather*} \frac {-2 a+\left (a+b x\right ) \left (\text {Log}\left [\frac {a+b x}{b}\right ]-\text {Log}\left [\frac {-a+b x}{b}\right ]\right )}{4 a^2 b c \left (a+b x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.15, size = 50, normalized size = 1.22

method	result	size
default	\(\frac {-\frac {\ln \left (-b x +a \right )}{4 a^{2} b}+\frac {\ln \left (b x +a \right )}{4 a^{2} b}-\frac {1}{2 a b \left (b x +a \right )}}{c}\)	\(50\)
norman	\(-\frac {1}{2 a b c \left (b x +a \right )}-\frac {\ln \left (-b x +a \right )}{4 a^{2} b c}+\frac {\ln \left (b x +a \right )}{4 a^{2} b c}\)	\(55\)
risch	\(-\frac {1}{2 a b c \left (b x +a \right )}-\frac {\ln \left (-b x +a \right )}{4 a^{2} b c}+\frac {\ln \left (b x +a \right )}{4 a^{2} b c}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 55, normalized size = 1.34 \begin {gather*} -\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 {gather*}

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 = 0.30, size = 51, normalized size = 1.24 \begin {gather*} \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 {gather*}

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 = 0.14, size = 44, normalized size = 1.07 \begin {gather*} - \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 {gather*}

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 = 0.00, size = 56, normalized size = 1.37 \begin {gather*} -\frac {\ln \left |x b-a\right |}{4 b a^{2} c}+\frac {\ln \left |x b+a\right |}{4 b a^{2} c}-\frac {\frac {1}{4}\cdot 2 a}{a^{2} c b \left (x b+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.18, size = 37, normalized size = 0.90 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{2\,a^2\,b\,c}-\frac {1}{2\,a\,b\,\left (a\,c+b\,c\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*c - b*c*x)*(a + b*x)^2),x)

[Out]

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

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