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

Optimal. Leaf size=52 \[ \frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{4 a^3 b}-\frac {1}{4 a^2 b (a+b x)}-\frac {1}{4 a b (a+b x)^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {627, 44, 208} \begin {gather*} -\frac {1}{4 a^2 b (a+b x)}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{4 a^3 b}-\frac {1}{4 a b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(4*a*b*(a + b*x)^2) - 1/(4*a^2*b*(a + b*x)) + ArcTanh[(b*x)/a]/(4*a^3*b)

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]

Rule 627

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

Rubi steps

\begin {align*} \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx &=\int \frac {1}{(a-b x) (a+b x)^3} \, dx\\ &=\int \left (\frac {1}{2 a (a+b x)^3}+\frac {1}{4 a^2 (a+b x)^2}+\frac {1}{4 a^2 \left (a^2-b^2 x^2\right )}\right ) \, dx\\ &=-\frac {1}{4 a b (a+b x)^2}-\frac {1}{4 a^2 b (a+b x)}+\frac {\int \frac {1}{a^2-b^2 x^2} \, dx}{4 a^2}\\ &=-\frac {1}{4 a b (a+b x)^2}-\frac {1}{4 a^2 b (a+b x)}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{4 a^3 b}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 58, normalized size = 1.12 \begin {gather*} \frac {-2 a (2 a+b x)+(a+b x)^2 (-\log (a-b x))+(a+b x)^2 \log (a+b x)}{8 a^3 b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(2*a + b*x) - (a + b*x)^2*Log[a - b*x] + (a + b*x)^2*Log[a + b*x])/(8*a^3*b*(a + b*x)^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 89, normalized size = 1.71 \begin {gather*} -\frac {2 \, a b x + 4 \, a^{2} - {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right ) + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x - a\right )}{8 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*a*b*x + 4*a^2 - (b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a) + (b^2*x^2 + 2*a*b*x + a^2)*log(b*x - a))/(a^3*
b^3*x^2 + 2*a^4*b^2*x + a^5*b)

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giac [A]  time = 0.16, size = 51, normalized size = 0.98 \begin {gather*} -\frac {\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}}{4 \, a^{2} b^{2}} - \frac {\log \left ({\left | -\frac {2 \, a}{b x + a} + 1 \right |}\right )}{8 \, a^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b/(b*x + a) + a*b/(b*x + a)^2)/(a^2*b^2) - 1/8*log(abs(-2*a/(b*x + a) + 1))/(a^3*b)

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maple [A]  time = 0.05, size = 62, normalized size = 1.19 \begin {gather*} -\frac {1}{4 \left (b x +a \right )^{2} a b}-\frac {1}{4 \left (b x +a \right ) a^{2} b}-\frac {\ln \left (b x -a \right )}{8 a^{3} b}+\frac {\ln \left (b x +a \right )}{8 a^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/a^3/b*ln(b*x-a)+1/8/a^3/b*ln(b*x+a)-1/4/a^2/b/(b*x+a)-1/4/a/b/(b*x+a)^2

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maxima [A]  time = 1.37, size = 67, normalized size = 1.29 \begin {gather*} -\frac {b x + 2 \, a}{4 \, {\left (a^{2} b^{3} x^{2} + 2 \, a^{3} b^{2} x + a^{4} b\right )}} + \frac {\log \left (b x + a\right )}{8 \, a^{3} b} - \frac {\log \left (b x - a\right )}{8 \, a^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*x + 2*a)/(a^2*b^3*x^2 + 2*a^3*b^2*x + a^4*b) + 1/8*log(b*x + a)/(a^3*b) - 1/8*log(b*x - a)/(a^3*b)

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mupad [B]  time = 0.07, size = 51, normalized size = 0.98 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{4\,a^3\,b}-\frac {\frac {x}{4\,a^2}+\frac {1}{2\,a\,b}}{a^2+2\,a\,b\,x+b^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.43, size = 58, normalized size = 1.12 \begin {gather*} - \frac {2 a + b x}{4 a^{4} b + 8 a^{3} b^{2} x + 4 a^{2} b^{3} x^{2}} - \frac {\frac {\log {\left (- \frac {a}{b} + x \right )}}{8} - \frac {\log {\left (\frac {a}{b} + x \right )}}{8}}{a^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a + b*x)/(4*a**4*b + 8*a**3*b**2*x + 4*a**2*b**3*x**2) - (log(-a/b + x)/8 - log(a/b + x)/8)/(a**3*b)

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