3.8.73 \(\int \frac {(a+b x)^2}{(a^2-b^2 x^2)^3} \, dx\) [773]

Optimal. Leaf size=54 \[ \frac {1}{4 a b (a-b x)^2}+\frac {1}{4 a^2 b (a-b x)}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{4 a^3 b} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 54, 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 = {641, 46, 214} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(a^2 - b^2*x^2)^3,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 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]

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 {(a+b x)^2}{\left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac {1}{(a-b x)^3 (a+b x)} \, 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 = 0.01, size = 62, normalized size = 1.15 \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[(a + b*x)^2/(a^2 - b^2*x^2)^3,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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Maple [A]
time = 0.46, size = 63, normalized size = 1.17

method	result	size
risch	\(\frac {-\frac {x}{4 a^{2}}+\frac {1}{2 b a}}{\left (-b x +a \right )^{2}}-\frac {\ln \left (-b x +a \right )}{8 a^{3} b}+\frac {\ln \left (b x +a \right )}{8 a^{3} b}\)	\(55\)
default	\(\frac {\ln \left (b x +a \right )}{8 a^{3} b}-\frac {\ln \left (-b x +a \right )}{8 a^{3} b}+\frac {1}{4 a^{2} b \left (-b x +a \right )}+\frac {1}{4 a b \left (-b x +a \right )^{2}}\)	\(63\)
norman	\(\frac {\frac {3 x}{4}-\frac {x^{3} b^{2}}{4 a^{2}}+\frac {a}{2 b}}{\left (-b^{2} x^{2}+a^{2}\right )^{2}}-\frac {\ln \left (-b x +a \right )}{8 a^{3} b}+\frac {\ln \left (b x +a \right )}{8 a^{3} b}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/(-b^2*x^2+a^2)^3,x,method=_RETURNVERBOSE)

[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 = 0.30, size = 67, normalized size = 1.24 \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((b*x+a)^2/(-b^2*x^2+a^2)^3,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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Fricas [A]
time = 1.95, size = 89, normalized size = 1.65 \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((b*x+a)^2/(-b^2*x^2+a^2)^3,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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Sympy [A]
time = 0.15, size = 58, normalized size = 1.07 \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((b*x+a)**2/(-b**2*x**2+a**2)**3,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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Giac [A]
time = 0.93, size = 60, normalized size = 1.11 \begin {gather*} \frac {\log \left ({\left | b x + a \right |}\right )}{8 \, a^{3} b} - \frac {\log \left ({\left | b x - a \right |}\right )}{8 \, a^{3} b} - \frac {a b x - 2 \, a^{2}}{4 \, {\left (b x - a\right )}^{2} a^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(-b^2*x^2+a^2)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.06, size = 51, normalized size = 0.94 \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((a + b*x)^2/(a^2 - b^2*x^2)^3,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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