\(\int \frac {1}{(a+b x)^2 (a^2-b^2 x^2)} \, dx\) [754]

Optimal result

Integrand size = 22, antiderivative size = 52 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx=-\frac {1}{4 a b (a+b x)^2}-\frac {1}{4 a^2 b (a+b x)}+\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{4 a^3 b} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {641, 46, 214} \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx=\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{4 a^3 b}-\frac {1}{4 a^2 b (a+b x)}-\frac {1}{4 a b (a+b x)^2} \]

[In]

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

[Out]

-1/4*1/(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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx=\frac {-2 a (2 a+b x)-(a+b x)^2 \log (a-b x)+(a+b x)^2 \log (a+b x)}{8 a^3 b (a+b x)^2} \]

[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)

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {-\frac {x}{4 a^{2}}-\frac {1}{2 b a}}{\left (b x +a \right )^{2}}-\frac {\ln \left (-b x +a \right )}{8 a^{3} b}+\frac {\ln \left (b x +a \right )}{8 a^{3} b}\)	\(54\)
norman	\(\frac {\frac {3 x}{4 a^{2}}+\frac {b \,x^{2}}{2 a^{3}}}{\left (b x +a \right )^{2}}-\frac {\ln \left (-b x +a \right )}{8 a^{3} b}+\frac {\ln \left (b x +a \right )}{8 a^{3} b}\)	\(55\)
default	\(\frac {\ln \left (b x +a \right )}{8 a^{3} b}-\frac {1}{4 a^{2} b \left (b x +a \right )}-\frac {1}{4 a b \left (b x +a \right )^{2}}-\frac {\ln \left (-b x +a \right )}{8 a^{3} b}\)	\(61\)
parallelrisch	\(-\frac {\ln \left (b x -a \right ) x^{2} b^{2}-b^{2} \ln \left (b x +a \right ) x^{2}+2 \ln \left (b x -a \right ) x a b -2 \ln \left (b x +a \right ) x a b -4 b^{2} x^{2}+a^{2} \ln \left (b x -a \right )-a^{2} \ln \left (b x +a \right )-6 a b x}{8 a^{3} \left (b x +a \right )^{2} b}\)	\(106\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.71 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx=-\frac {2 \, a b x + 4 \, a^{2} - {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right ) + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x - a\right )}{8 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx=- \frac {2 a + b x}{4 a^{4} b + 8 a^{3} b^{2} x + 4 a^{2} b^{3} x^{2}} - \frac {\frac {\log {\left (- \frac {a}{b} + x \right )}}{8} - \frac {\log {\left (\frac {a}{b} + x \right )}}{8}}{a^{3} b} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx=-\frac {b x + 2 \, a}{4 \, {\left (a^{2} b^{3} x^{2} + 2 \, a^{3} b^{2} x + a^{4} b\right )}} + \frac {\log \left (b x + a\right )}{8 \, a^{3} b} - \frac {\log \left (b x - a\right )}{8 \, a^{3} b} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx=-\frac {\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}}{4 \, a^{2} b^{2}} - \frac {\log \left ({\left | -\frac {2 \, a}{b x + a} + 1 \right |}\right )}{8 \, a^{3} b} \]

[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)

Mupad [B] (verification not implemented)

Time = 9.71 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )} \, dx=\frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{4\,a^3\,b}-\frac {\frac {x}{4\,a^2}+\frac {1}{2\,a\,b}}{a^2+2\,a\,b\,x+b^2\,x^2} \]

[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)