\(\int \frac {a+b x}{(a^2-b^2 x^2)^3} \, dx\) [774]

Optimal result

Integrand size = 20, antiderivative size = 71 \[ \int \frac {a+b x}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {1}{8 a^2 b (a-b x)^2}+\frac {1}{4 a^3 b (a-b x)}-\frac {1}{8 a^3 b (a+b x)}+\frac {3 \text {arctanh}\left (\frac {b x}{a}\right )}{8 a^4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {641, 46, 214} \[ \int \frac {a+b x}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {3 \text {arctanh}\left (\frac {b x}{a}\right )}{8 a^4 b}+\frac {1}{4 a^3 b (a-b x)}-\frac {1}{8 a^3 b (a+b x)}+\frac {1}{8 a^2 b (a-b x)^2} \]

[In]

Int[(a + b*x)/(a^2 - b^2*x^2)^3,x]

[Out]

1/(8*a^2*b*(a - b*x)^2) + 1/(4*a^3*b*(a - b*x)) - 1/(8*a^3*b*(a + b*x)) + (3*ArcTanh[(b*x)/a])/(8*a^4*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)^3 (a+b x)^2} \, dx \\ & = \int \left (\frac {1}{4 a^2 (a-b x)^3}+\frac {1}{4 a^3 (a-b x)^2}+\frac {1}{8 a^3 (a+b x)^2}+\frac {3}{8 a^3 \left (a^2-b^2 x^2\right )}\right ) \, dx \\ & = \frac {1}{8 a^2 b (a-b x)^2}+\frac {1}{4 a^3 b (a-b x)}-\frac {1}{8 a^3 b (a+b x)}+\frac {3 \int \frac {1}{a^2-b^2 x^2} \, dx}{8 a^3} \\ & = \frac {1}{8 a^2 b (a-b x)^2}+\frac {1}{4 a^3 b (a-b x)}-\frac {1}{8 a^3 b (a+b x)}+\frac {3 \tanh ^{-1}\left (\frac {b x}{a}\right )}{8 a^4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {a+b x}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {\frac {2 a \left (2 a^2+3 a b x-3 b^2 x^2\right )}{(a-b x)^2 (a+b x)}-3 \log (a-b x)+3 \log (a+b x)}{16 a^4 b} \]

[In]

Integrate[(a + b*x)/(a^2 - b^2*x^2)^3,x]

[Out]

((2*a*(2*a^2 + 3*a*b*x - 3*b^2*x^2))/((a - b*x)^2*(a + b*x)) - 3*Log[a - b*x] + 3*Log[a + b*x])/(16*a^4*b)

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97

method	result	size
norman	\(\frac {\frac {5 x}{8 a}+\frac {1}{4 b}-\frac {3 b^{2} x^{3}}{8 a^{3}}}{\left (-b^{2} x^{2}+a^{2}\right )^{2}}-\frac {3 \ln \left (-b x +a \right )}{16 a^{4} b}+\frac {3 \ln \left (b x +a \right )}{16 a^{4} b}\)	\(69\)
default	\(\frac {3 \ln \left (b x +a \right )}{16 a^{4} b}-\frac {1}{8 a^{3} b \left (b x +a \right )}-\frac {3 \ln \left (-b x +a \right )}{16 a^{4} b}+\frac {1}{4 a^{3} b \left (-b x +a \right )}+\frac {1}{8 a^{2} b \left (-b x +a \right )^{2}}\)	\(78\)
risch	\(\frac {-\frac {3 b \,x^{2}}{8 a^{3}}+\frac {3 x}{8 a^{2}}+\frac {1}{4 b a}}{\left (-b x +a \right ) \left (-b^{2} x^{2}+a^{2}\right )}-\frac {3 \ln \left (-b x +a \right )}{16 a^{4} b}+\frac {3 \ln \left (b x +a \right )}{16 a^{4} b}\)	\(78\)
parallelrisch	\(-\frac {3 \ln \left (b x -a \right ) x^{3} b^{5}-3 \ln \left (b x +a \right ) x^{3} b^{5}-3 \ln \left (b x -a \right ) x^{2} a \,b^{4}+3 \ln \left (b x +a \right ) x^{2} a \,b^{4}-3 \ln \left (b x -a \right ) x \,a^{2} b^{3}+3 \ln \left (b x +a \right ) x \,a^{2} b^{3}+6 b^{4} x^{2} a +3 \ln \left (b x -a \right ) a^{3} b^{2}-3 \ln \left (b x +a \right ) a^{3} b^{2}-6 a^{2} b^{3} x -4 a^{3} b^{2}}{16 a^{4} b^{3} \left (b x -a \right ) \left (b^{2} x^{2}-a^{2}\right )}\)	\(184\)

[In]

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

[Out]

(5/8*x/a+1/4/b-3/8*b^2*x^3/a^3)/(-b^2*x^2+a^2)^2-3/16/a^4/b*ln(-b*x+a)+3/16/a^4/b*ln(b*x+a)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (65) = 130\).

Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.89 \[ \int \frac {a+b x}{\left (a^2-b^2 x^2\right )^3} \, dx=-\frac {6 \, a b^{2} x^{2} - 6 \, a^{2} b x - 4 \, a^{3} - 3 \, {\left (b^{3} x^{3} - a b^{2} x^{2} - a^{2} b x + a^{3}\right )} \log \left (b x + a\right ) + 3 \, {\left (b^{3} x^{3} - a b^{2} x^{2} - a^{2} b x + a^{3}\right )} \log \left (b x - a\right )}{16 \, {\left (a^{4} b^{4} x^{3} - a^{5} b^{3} x^{2} - a^{6} b^{2} x + a^{7} b\right )}} \]

[In]

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

[Out]

-1/16*(6*a*b^2*x^2 - 6*a^2*b*x - 4*a^3 - 3*(b^3*x^3 - a*b^2*x^2 - a^2*b*x + a^3)*log(b*x + a) + 3*(b^3*x^3 - a
*b^2*x^2 - a^2*b*x + a^3)*log(b*x - a))/(a^4*b^4*x^3 - a^5*b^3*x^2 - a^6*b^2*x + a^7*b)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.23 \[ \int \frac {a+b x}{\left (a^2-b^2 x^2\right )^3} \, dx=- \frac {- 2 a^{2} - 3 a b x + 3 b^{2} x^{2}}{8 a^{6} b - 8 a^{5} b^{2} x - 8 a^{4} b^{3} x^{2} + 8 a^{3} b^{4} x^{3}} - \frac {\frac {3 \log {\left (- \frac {a}{b} + x \right )}}{16} - \frac {3 \log {\left (\frac {a}{b} + x \right )}}{16}}{a^{4} b} \]

[In]

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

[Out]

-(-2*a**2 - 3*a*b*x + 3*b**2*x**2)/(8*a**6*b - 8*a**5*b**2*x - 8*a**4*b**3*x**2 + 8*a**3*b**4*x**3) - (3*log(-
a/b + x)/16 - 3*log(a/b + x)/16)/(a**4*b)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.27 \[ \int \frac {a+b x}{\left (a^2-b^2 x^2\right )^3} \, dx=-\frac {3 \, b^{2} x^{2} - 3 \, a b x - 2 \, a^{2}}{8 \, {\left (a^{3} b^{4} x^{3} - a^{4} b^{3} x^{2} - a^{5} b^{2} x + a^{6} b\right )}} + \frac {3 \, \log \left (b x + a\right )}{16 \, a^{4} b} - \frac {3 \, \log \left (b x - a\right )}{16 \, a^{4} b} \]

[In]

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

[Out]

-1/8*(3*b^2*x^2 - 3*a*b*x - 2*a^2)/(a^3*b^4*x^3 - a^4*b^3*x^2 - a^5*b^2*x + a^6*b) + 3/16*log(b*x + a)/(a^4*b)
- 3/16*log(b*x - a)/(a^4*b)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.11 \[ \int \frac {a+b x}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {3 \, \log \left ({\left | b x + a \right |}\right )}{16 \, a^{4} b} - \frac {3 \, \log \left ({\left | b x - a \right |}\right )}{16 \, a^{4} b} - \frac {3 \, a b^{2} x^{2} - 3 \, a^{2} b x - 2 \, a^{3}}{8 \, {\left (b x + a\right )} {\left (b x - a\right )}^{2} a^{4} b} \]

[In]

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

[Out]

3/16*log(abs(b*x + a))/(a^4*b) - 3/16*log(abs(b*x - a))/(a^4*b) - 1/8*(3*a*b^2*x^2 - 3*a^2*b*x - 2*a^3)/((b*x
+ a)*(b*x - a)^2*a^4*b)

Mupad [B] (verification not implemented)

Time = 10.00 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99 \[ \int \frac {a+b x}{\left (a^2-b^2 x^2\right )^3} \, dx=\frac {\frac {3\,x}{8\,a^2}+\frac {1}{4\,a\,b}-\frac {3\,b\,x^2}{8\,a^3}}{a^3-a^2\,b\,x-a\,b^2\,x^2+b^3\,x^3}+\frac {3\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{8\,a^4\,b} \]

[In]

int((a + b*x)/(a^2 - b^2*x^2)^3,x)

[Out]

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