\(\int \frac {1}{(a+b x)^2 (a^2-b^2 x^2)^3} \, dx\) [776]

Optimal result

Integrand size = 22, antiderivative size = 122 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^3} \, dx=\frac {1}{64 a^5 b (a-b x)^2}+\frac {5}{64 a^6 b (a-b x)}-\frac {1}{32 a^3 b (a+b x)^4}-\frac {1}{16 a^4 b (a+b x)^3}-\frac {3}{32 a^5 b (a+b x)^2}-\frac {5}{32 a^6 b (a+b x)}+\frac {15 \text {arctanh}\left (\frac {b x}{a}\right )}{64 a^7 b} \]

[Out]

1/64/a^5/b/(-b*x+a)^2+5/64/a^6/b/(-b*x+a)-1/32/a^3/b/(b*x+a)^4-1/16/a^4/b/(b*x+a)^3-3/32/a^5/b/(b*x+a)^2-5/32/
a^6/b/(b*x+a)+15/64*arctanh(b*x/a)/a^7/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 122, 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} \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^3} \, dx=\frac {15 \text {arctanh}\left (\frac {b x}{a}\right )}{64 a^7 b}+\frac {5}{64 a^6 b (a-b x)}-\frac {5}{32 a^6 b (a+b x)}+\frac {1}{64 a^5 b (a-b x)^2}-\frac {3}{32 a^5 b (a+b x)^2}-\frac {1}{16 a^4 b (a+b x)^3}-\frac {1}{32 a^3 b (a+b x)^4} \]

[In]

Int[1/((a + b*x)^2*(a^2 - b^2*x^2)^3),x]

[Out]

1/(64*a^5*b*(a - b*x)^2) + 5/(64*a^6*b*(a - b*x)) - 1/(32*a^3*b*(a + b*x)^4) - 1/(16*a^4*b*(a + b*x)^3) - 3/(3
2*a^5*b*(a + b*x)^2) - 5/(32*a^6*b*(a + b*x)) + (15*ArcTanh[(b*x)/a])/(64*a^7*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)^5} \, dx \\ & = \int \left (\frac {1}{32 a^5 (a-b x)^3}+\frac {5}{64 a^6 (a-b x)^2}+\frac {1}{8 a^3 (a+b x)^5}+\frac {3}{16 a^4 (a+b x)^4}+\frac {3}{16 a^5 (a+b x)^3}+\frac {5}{32 a^6 (a+b x)^2}+\frac {15}{64 a^6 \left (a^2-b^2 x^2\right )}\right ) \, dx \\ & = \frac {1}{64 a^5 b (a-b x)^2}+\frac {5}{64 a^6 b (a-b x)}-\frac {1}{32 a^3 b (a+b x)^4}-\frac {1}{16 a^4 b (a+b x)^3}-\frac {3}{32 a^5 b (a+b x)^2}-\frac {5}{32 a^6 b (a+b x)}+\frac {15 \int \frac {1}{a^2-b^2 x^2} \, dx}{64 a^6} \\ & = \frac {1}{64 a^5 b (a-b x)^2}+\frac {5}{64 a^6 b (a-b x)}-\frac {1}{32 a^3 b (a+b x)^4}-\frac {1}{16 a^4 b (a+b x)^3}-\frac {3}{32 a^5 b (a+b x)^2}-\frac {5}{32 a^6 b (a+b x)}+\frac {15 \tanh ^{-1}\left (\frac {b x}{a}\right )}{64 a^7 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^3} \, dx=\frac {\frac {2 a \left (-16 a^5+17 a^4 b x+50 a^3 b^2 x^2+10 a^2 b^3 x^3-30 a b^4 x^4-15 b^5 x^5\right )}{(a-b x)^2 (a+b x)^4}-15 \log (a-b x)+15 \log (a+b x)}{128 a^7 b} \]

[In]

Integrate[1/((a + b*x)^2*(a^2 - b^2*x^2)^3),x]

[Out]

((2*a*(-16*a^5 + 17*a^4*b*x + 50*a^3*b^2*x^2 + 10*a^2*b^3*x^3 - 30*a*b^4*x^4 - 15*b^5*x^5))/((a - b*x)^2*(a +
b*x)^4) - 15*Log[a - b*x] + 15*Log[a + b*x])/(128*a^7*b)

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.85

method	result	size
norman	\(\frac {-\frac {1}{4 b a}+\frac {25 b \,x^{2}}{32 a^{3}}-\frac {15 b^{3} x^{4}}{32 a^{5}}-\frac {15 b^{4} x^{5}}{64 a^{6}}+\frac {5 b^{2} x^{3}}{32 a^{4}}+\frac {17 x}{64 a^{2}}}{\left (b x +a \right )^{4} \left (-b x +a \right )^{2}}-\frac {15 \ln \left (-b x +a \right )}{128 a^{7} b}+\frac {15 \ln \left (b x +a \right )}{128 a^{7} b}\)	\(104\)
risch	\(\frac {-\frac {1}{4 b a}+\frac {25 b \,x^{2}}{32 a^{3}}-\frac {15 b^{3} x^{4}}{32 a^{5}}-\frac {15 b^{4} x^{5}}{64 a^{6}}+\frac {5 b^{2} x^{3}}{32 a^{4}}+\frac {17 x}{64 a^{2}}}{\left (b x +a \right )^{2} \left (-b^{2} x^{2}+a^{2}\right )^{2}}-\frac {15 \ln \left (-b x +a \right )}{128 a^{7} b}+\frac {15 \ln \left (b x +a \right )}{128 a^{7} b}\)	\(110\)
default	\(\frac {15 \ln \left (b x +a \right )}{128 a^{7} b}-\frac {5}{32 a^{6} b \left (b x +a \right )}-\frac {3}{32 a^{5} b \left (b x +a \right )^{2}}-\frac {1}{16 a^{4} b \left (b x +a \right )^{3}}-\frac {1}{32 a^{3} b \left (b x +a \right )^{4}}-\frac {15 \ln \left (-b x +a \right )}{128 a^{7} b}+\frac {5}{64 a^{6} b \left (-b x +a \right )}+\frac {1}{64 a^{5} b \left (-b x +a \right )^{2}}\)	\(123\)
parallelrisch	\(-\frac {32 a^{6} b^{5}+60 \ln \left (b x +a \right ) x^{3} a^{3} b^{8}-15 \ln \left (b x -a \right ) x^{2} a^{4} b^{7}+15 \ln \left (b x +a \right ) x^{2} a^{4} b^{7}+30 \ln \left (b x -a \right ) x \,a^{5} b^{6}-30 \ln \left (b x +a \right ) x \,a^{5} b^{6}-15 \ln \left (b x +a \right ) x^{6} b^{11}+15 \ln \left (b x -a \right ) a^{6} b^{5}-15 \ln \left (b x +a \right ) a^{6} b^{5}+30 x^{5} a \,b^{10}+60 x^{4} a^{2} b^{9}-20 x^{3} a^{3} b^{8}-100 x^{2} a^{4} b^{7}-34 x \,a^{5} b^{6}+15 \ln \left (b x -a \right ) x^{6} b^{11}+30 \ln \left (b x -a \right ) x^{5} a \,b^{10}-30 \ln \left (b x +a \right ) x^{5} a \,b^{10}-15 \ln \left (b x -a \right ) x^{4} a^{2} b^{9}+15 \ln \left (b x +a \right ) x^{4} a^{2} b^{9}-60 \ln \left (b x -a \right ) x^{3} a^{3} b^{8}}{128 a^{7} b^{6} \left (b x +a \right )^{2} \left (b^{2} x^{2}-a^{2}\right )^{2}}\)	\(323\)

[In]

int(1/(b*x+a)^2/(-b^2*x^2+a^2)^3,x,method=_RETURNVERBOSE)

[Out]

(-1/4/b/a+25/32/a^3*b*x^2-15/32*b^3/a^5*x^4-15/64*b^4/a^6*x^5+5/32*b^2*x^3/a^4+17/64/a^2*x)/(b*x+a)^4/(-b*x+a)
^2-15/128/a^7/b*ln(-b*x+a)+15/128/a^7/b*ln(b*x+a)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (110) = 220\).

Time = 0.28 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.18 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^3} \, dx=-\frac {30 \, a b^{5} x^{5} + 60 \, a^{2} b^{4} x^{4} - 20 \, a^{3} b^{3} x^{3} - 100 \, a^{4} b^{2} x^{2} - 34 \, a^{5} b x + 32 \, a^{6} - 15 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{5} - a^{2} b^{4} x^{4} - 4 \, a^{3} b^{3} x^{3} - a^{4} b^{2} x^{2} + 2 \, a^{5} b x + a^{6}\right )} \log \left (b x + a\right ) + 15 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{5} - a^{2} b^{4} x^{4} - 4 \, a^{3} b^{3} x^{3} - a^{4} b^{2} x^{2} + 2 \, a^{5} b x + a^{6}\right )} \log \left (b x - a\right )}{128 \, {\left (a^{7} b^{7} x^{6} + 2 \, a^{8} b^{6} x^{5} - a^{9} b^{5} x^{4} - 4 \, a^{10} b^{4} x^{3} - a^{11} b^{3} x^{2} + 2 \, a^{12} b^{2} x + a^{13} b\right )}} \]

[In]

integrate(1/(b*x+a)^2/(-b^2*x^2+a^2)^3,x, algorithm="fricas")

[Out]

-1/128*(30*a*b^5*x^5 + 60*a^2*b^4*x^4 - 20*a^3*b^3*x^3 - 100*a^4*b^2*x^2 - 34*a^5*b*x + 32*a^6 - 15*(b^6*x^6 +
2*a*b^5*x^5 - a^2*b^4*x^4 - 4*a^3*b^3*x^3 - a^4*b^2*x^2 + 2*a^5*b*x + a^6)*log(b*x + a) + 15*(b^6*x^6 + 2*a*b
^5*x^5 - a^2*b^4*x^4 - 4*a^3*b^3*x^3 - a^4*b^2*x^2 + 2*a^5*b*x + a^6)*log(b*x - a))/(a^7*b^7*x^6 + 2*a^8*b^6*x
^5 - a^9*b^5*x^4 - 4*a^10*b^4*x^3 - a^11*b^3*x^2 + 2*a^12*b^2*x + a^13*b)

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^3} \, dx=- \frac {16 a^{5} - 17 a^{4} b x - 50 a^{3} b^{2} x^{2} - 10 a^{2} b^{3} x^{3} + 30 a b^{4} x^{4} + 15 b^{5} x^{5}}{64 a^{12} b + 128 a^{11} b^{2} x - 64 a^{10} b^{3} x^{2} - 256 a^{9} b^{4} x^{3} - 64 a^{8} b^{5} x^{4} + 128 a^{7} b^{6} x^{5} + 64 a^{6} b^{7} x^{6}} - \frac {\frac {15 \log {\left (- \frac {a}{b} + x \right )}}{128} - \frac {15 \log {\left (\frac {a}{b} + x \right )}}{128}}{a^{7} b} \]

[In]

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

[Out]

-(16*a**5 - 17*a**4*b*x - 50*a**3*b**2*x**2 - 10*a**2*b**3*x**3 + 30*a*b**4*x**4 + 15*b**5*x**5)/(64*a**12*b +
128*a**11*b**2*x - 64*a**10*b**3*x**2 - 256*a**9*b**4*x**3 - 64*a**8*b**5*x**4 + 128*a**7*b**6*x**5 + 64*a**6
*b**7*x**6) - (15*log(-a/b + x)/128 - 15*log(a/b + x)/128)/(a**7*b)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^3} \, dx=-\frac {15 \, b^{5} x^{5} + 30 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 50 \, a^{3} b^{2} x^{2} - 17 \, a^{4} b x + 16 \, a^{5}}{64 \, {\left (a^{6} b^{7} x^{6} + 2 \, a^{7} b^{6} x^{5} - a^{8} b^{5} x^{4} - 4 \, a^{9} b^{4} x^{3} - a^{10} b^{3} x^{2} + 2 \, a^{11} b^{2} x + a^{12} b\right )}} + \frac {15 \, \log \left (b x + a\right )}{128 \, a^{7} b} - \frac {15 \, \log \left (b x - a\right )}{128 \, a^{7} b} \]

[In]

integrate(1/(b*x+a)^2/(-b^2*x^2+a^2)^3,x, algorithm="maxima")

[Out]

-1/64*(15*b^5*x^5 + 30*a*b^4*x^4 - 10*a^2*b^3*x^3 - 50*a^3*b^2*x^2 - 17*a^4*b*x + 16*a^5)/(a^6*b^7*x^6 + 2*a^7
*b^6*x^5 - a^8*b^5*x^4 - 4*a^9*b^4*x^3 - a^10*b^3*x^2 + 2*a^11*b^2*x + a^12*b) + 15/128*log(b*x + a)/(a^7*b) -
15/128*log(b*x - a)/(a^7*b)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^3} \, dx=-\frac {15 \, \log \left ({\left | -\frac {2 \, a}{b x + a} + 1 \right |}\right )}{128 \, a^{7} b} + \frac {\frac {24 \, a}{b x + a} - 11}{256 \, a^{7} b {\left (\frac {2 \, a}{b x + a} - 1\right )}^{2}} - \frac {\frac {5 \, a^{6} b^{11}}{b x + a} + \frac {3 \, a^{7} b^{11}}{{\left (b x + a\right )}^{2}} + \frac {2 \, a^{8} b^{11}}{{\left (b x + a\right )}^{3}} + \frac {a^{9} b^{11}}{{\left (b x + a\right )}^{4}}}{32 \, a^{12} b^{12}} \]

[In]

integrate(1/(b*x+a)^2/(-b^2*x^2+a^2)^3,x, algorithm="giac")

[Out]

-15/128*log(abs(-2*a/(b*x + a) + 1))/(a^7*b) + 1/256*(24*a/(b*x + a) - 11)/(a^7*b*(2*a/(b*x + a) - 1)^2) - 1/3
2*(5*a^6*b^11/(b*x + a) + 3*a^7*b^11/(b*x + a)^2 + 2*a^8*b^11/(b*x + a)^3 + a^9*b^11/(b*x + a)^4)/(a^12*b^12)

Mupad [B] (verification not implemented)

Time = 9.94 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(a+b x)^2 \left (a^2-b^2 x^2\right )^3} \, dx=\frac {\frac {17\,x}{64\,a^2}-\frac {1}{4\,a\,b}+\frac {25\,b\,x^2}{32\,a^3}+\frac {5\,b^2\,x^3}{32\,a^4}-\frac {15\,b^3\,x^4}{32\,a^5}-\frac {15\,b^4\,x^5}{64\,a^6}}{a^6+2\,a^5\,b\,x-a^4\,b^2\,x^2-4\,a^3\,b^3\,x^3-a^2\,b^4\,x^4+2\,a\,b^5\,x^5+b^6\,x^6}+\frac {15\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{64\,a^7\,b} \]

[In]

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

[Out]

((17*x)/(64*a^2) - 1/(4*a*b) + (25*b*x^2)/(32*a^3) + (5*b^2*x^3)/(32*a^4) - (15*b^3*x^4)/(32*a^5) - (15*b^4*x^
5)/(64*a^6))/(a^6 + b^6*x^6 + 2*a*b^5*x^5 - a^4*b^2*x^2 - 4*a^3*b^3*x^3 - a^2*b^4*x^4 + 2*a^5*b*x) + (15*atanh
((b*x)/a))/(64*a^7*b)