\(\int \frac {1}{(a+b x) (a^2-b^2 x^2)} \, dx\) [753]

Optimal result

Integrand size = 22, antiderivative size = 35 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )} \, dx=-\frac {1}{2 a b (a+b x)}+\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{2 a^2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, 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} \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )} \, dx=\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{2 a^2 b}-\frac {1}{2 a b (a+b x)} \]

[In]

Int[1/((a + b*x)*(a^2 - b^2*x^2)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.34 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )} \, dx=\frac {-2 a-(a+b x) \log (a-b x)+(a+b x) \log (a+b x)}{4 a^2 b (a+b x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26

method	result	size
norman	\(\frac {x}{2 a^{2} \left (b x +a \right )}-\frac {\ln \left (-b x +a \right )}{4 a^{2} b}+\frac {\ln \left (b x +a \right )}{4 a^{2} b}\)	\(44\)
default	\(\frac {\ln \left (b x +a \right )}{4 a^{2} b}-\frac {1}{2 a b \left (b x +a \right )}-\frac {\ln \left (-b x +a \right )}{4 a^{2} b}\)	\(46\)
risch	\(\frac {\ln \left (b x +a \right )}{4 a^{2} b}-\frac {1}{2 a b \left (b x +a \right )}-\frac {\ln \left (-b x +a \right )}{4 a^{2} b}\)	\(46\)
parallelrisch	\(-\frac {\ln \left (b x -a \right ) x b -b \ln \left (b x +a \right ) x +a \ln \left (b x -a \right )-a \ln \left (b x +a \right )-2 b x}{4 a^{2} \left (b x +a \right ) b}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )} \, dx=\frac {{\left (b x + a\right )} \log \left (b x + a\right ) - {\left (b x + a\right )} \log \left (b x - a\right ) - 2 \, a}{4 \, {\left (a^{2} b^{2} x + a^{3} b\right )}} \]

[In]

integrate(1/(b*x+a)/(-b^2*x^2+a^2),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*x + a^3*b)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )} \, dx=- \frac {1}{2 a^{2} b + 2 a b^{2} x} - \frac {\frac {\log {\left (- \frac {a}{b} + x \right )}}{4} - \frac {\log {\left (\frac {a}{b} + x \right )}}{4}}{a^{2} b} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.34 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )} \, dx=-\frac {1}{2 \, {\left (a b^{2} x + a^{2} b\right )}} + \frac {\log \left (b x + a\right )}{4 \, a^{2} b} - \frac {\log \left (b x - a\right )}{4 \, a^{2} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{4 \, a^{2} b} - \frac {\log \left ({\left | b x - a \right |}\right )}{4 \, a^{2} b} - \frac {1}{2 \, {\left (b x + a\right )} a b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(a+b x) \left (a^2-b^2 x^2\right )} \, dx=\frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{2\,a^2\,b}-\frac {1}{2\,a\,b\,\left (a+b\,x\right )} \]

[In]

int(1/((a^2 - b^2*x^2)*(a + b*x)),x)

[Out]

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