3.11.61 \(\int \frac {1}{(a+b x)^2 (a c-b c x)^2} \, dx\) [1061]

Optimal. Leaf size=46 \[ \frac {x}{2 a^2 c^2 \left (a^2-b^2 x^2\right )}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{2 a^3 b c^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {41, 205, 214} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{2 a^3 b c^2}+\frac {x}{2 a^2 c^2 \left (a^2-b^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(2*a^2*c^2*(a^2 - b^2*x^2)) + ArcTanh[(b*x)/a]/(2*a^3*b*c^2)

Rule 41

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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)^2} \, dx &=\int \frac {1}{\left (a^2 c-b^2 c x^2\right )^2} \, dx\\ &=\frac {x}{2 a^2 c^2 \left (a^2-b^2 x^2\right )}+\frac {\int \frac {1}{a^2 c-b^2 c x^2} \, dx}{2 a^2 c}\\ &=\frac {x}{2 a^2 c^2 \left (a^2-b^2 x^2\right )}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{2 a^3 b c^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 74, normalized size = 1.61 \begin {gather*} \frac {2 a b x+\left (-a^2+b^2 x^2\right ) \log (a-b x)+\left (a^2-b^2 x^2\right ) \log (a+b x)}{4 a^3 b c^2 (a-b x) (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*x + (-a^2 + b^2*x^2)*Log[a - b*x] + (a^2 - b^2*x^2)*Log[a + b*x])/(4*a^3*b*c^2*(a - b*x)*(a + b*x))

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Mathics [A]
time = 2.17, size = 69, normalized size = 1.50 \begin {gather*} \frac {2 a b x+\left (a^2-b^2 x^2\right ) \left (\text {Log}\left [\frac {a+b x}{b}\right ]-\text {Log}\left [\frac {-a+b x}{b}\right ]\right )}{4 a^3 b c^2 \left (a^2-b^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2 a b x + (a ^ 2 - b ^ 2 x ^ 2) (Log[(a + b x) / b] - Log[(-a + b x) / b])) / (4 a ^ 3 b c ^ 2 (a ^ 2 - b ^ 2
x ^ 2))

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Maple [A]
time = 0.16, size = 66, normalized size = 1.43

method	result	size
norman	\(\frac {x}{2 a^{2} c^{2} \left (b x +a \right ) \left (-b x +a \right )}-\frac {\ln \left (-b x +a \right )}{4 a^{3} b \,c^{2}}+\frac {\ln \left (b x +a \right )}{4 a^{3} b \,c^{2}}\)	\(61\)
risch	\(\frac {x}{2 a^{2} c^{2} \left (b x +a \right ) \left (-b x +a \right )}-\frac {\ln \left (-b x +a \right )}{4 a^{3} b \,c^{2}}+\frac {\ln \left (b x +a \right )}{4 a^{3} b \,c^{2}}\)	\(61\)
default	\(\frac {-\frac {\ln \left (-b x +a \right )}{4 a^{3} b}+\frac {1}{4 a^{2} b \left (-b x +a \right )}+\frac {\ln \left (b x +a \right )}{4 a^{3} b}-\frac {1}{4 a^{2} b \left (b x +a \right )}}{c^{2}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 64, normalized size = 1.39 \begin {gather*} -\frac {x}{2 \, {\left (a^{2} b^{2} c^{2} x^{2} - a^{4} c^{2}\right )}} + \frac {\log \left (b x + a\right )}{4 \, a^{3} b c^{2}} - \frac {\log \left (b x - a\right )}{4 \, a^{3} b c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.30, size = 76, normalized size = 1.65 \begin {gather*} -\frac {2 \, a b x - {\left (b^{2} x^{2} - a^{2}\right )} \log \left (b x + a\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \log \left (b x - a\right )}{4 \, {\left (a^{3} b^{3} c^{2} x^{2} - a^{5} b c^{2}\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)^2,x, algorithm="fricas")

[Out]

-1/4*(2*a*b*x - (b^2*x^2 - a^2)*log(b*x + a) + (b^2*x^2 - a^2)*log(b*x - a))/(a^3*b^3*c^2*x^2 - a^5*b*c^2)

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Sympy [A]
time = 0.13, size = 49, normalized size = 1.07 \begin {gather*} - \frac {x}{- 2 a^{4} c^{2} + 2 a^{2} b^{2} c^{2} x^{2}} + \frac {- \frac {\log {\left (- \frac {a}{b} + x \right )}}{4} + \frac {\log {\left (\frac {a}{b} + x \right )}}{4}}{a^{3} b c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 68, normalized size = 1.48 \begin {gather*} -\frac {b \ln \left |x b-a\right |}{4 b^{2} a^{3} c^{2}}+\frac {b \ln \left |x b+a\right |}{4 b^{2} a^{3} c^{2}}+\frac {x}{2 a^{2} c^{2} \left (-x^{2} b^{2}+a^{2}\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)^2,x)

[Out]

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

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Mupad [B]
time = 0.18, size = 46, normalized size = 1.00 \begin {gather*} \frac {x}{2\,a^2\,\left (a^2\,c^2-b^2\,c^2\,x^2\right )}+\frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{2\,a^3\,b\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*a^2*(a^2*c^2 - b^2*c^2*x^2)) + atanh((b*x)/a)/(2*a^3*b*c^2)

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