3.677 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^4} \, dx\)

Optimal. Leaf size=144 \[ -\frac {11}{64 a c^4 (1-a x)}+\frac {99}{32 a c^4 (a x+1)}+\frac {1}{64 a c^4 (1-a x)^2}-\frac {35}{32 a c^4 (a x+1)^2}+\frac {13}{48 a c^4 (a x+1)^3}-\frac {1}{32 a c^4 (a x+1)^4}-\frac {47 \log (1-a x)}{128 a c^4}+\frac {303 \log (a x+1)}{128 a c^4}-\frac {x}{c^4} \]

[Out]

-x/c^4+1/64/a/c^4/(-a*x+1)^2-11/64/a/c^4/(-a*x+1)-1/32/a/c^4/(a*x+1)^4+13/48/a/c^4/(a*x+1)^3-35/32/a/c^4/(a*x+
1)^2+99/32/a/c^4/(a*x+1)-47/128*ln(-a*x+1)/a/c^4+303/128*ln(a*x+1)/a/c^4

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Rubi [A]  time = 0.20, antiderivative size = 144, 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 = {6157, 6150, 88} \[ -\frac {11}{64 a c^4 (1-a x)}+\frac {99}{32 a c^4 (a x+1)}+\frac {1}{64 a c^4 (1-a x)^2}-\frac {35}{32 a c^4 (a x+1)^2}+\frac {13}{48 a c^4 (a x+1)^3}-\frac {1}{32 a c^4 (a x+1)^4}-\frac {47 \log (1-a x)}{128 a c^4}+\frac {303 \log (a x+1)}{128 a c^4}-\frac {x}{c^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - c/(a^2*x^2))^4),x]

[Out]

-(x/c^4) + 1/(64*a*c^4*(1 - a*x)^2) - 11/(64*a*c^4*(1 - a*x)) - 1/(32*a*c^4*(1 + a*x)^4) + 13/(48*a*c^4*(1 + a
*x)^3) - 35/(32*a*c^4*(1 + a*x)^2) + 99/(32*a*c^4*(1 + a*x)) - (47*Log[1 - a*x])/(128*a*c^4) + (303*Log[1 + a*
x])/(128*a*c^4)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {e^{-2 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^4} \, dx &=\frac {a^8 \int \frac {e^{-2 \tanh ^{-1}(a x)} x^8}{\left (1-a^2 x^2\right )^4} \, dx}{c^4}\\ &=\frac {a^8 \int \frac {x^8}{(1-a x)^3 (1+a x)^5} \, dx}{c^4}\\ &=\frac {a^8 \int \left (-\frac {1}{a^8}-\frac {1}{32 a^8 (-1+a x)^3}-\frac {11}{64 a^8 (-1+a x)^2}-\frac {47}{128 a^8 (-1+a x)}+\frac {1}{8 a^8 (1+a x)^5}-\frac {13}{16 a^8 (1+a x)^4}+\frac {35}{16 a^8 (1+a x)^3}-\frac {99}{32 a^8 (1+a x)^2}+\frac {303}{128 a^8 (1+a x)}\right ) \, dx}{c^4}\\ &=-\frac {x}{c^4}+\frac {1}{64 a c^4 (1-a x)^2}-\frac {11}{64 a c^4 (1-a x)}-\frac {1}{32 a c^4 (1+a x)^4}+\frac {13}{48 a c^4 (1+a x)^3}-\frac {35}{32 a c^4 (1+a x)^2}+\frac {99}{32 a c^4 (1+a x)}-\frac {47 \log (1-a x)}{128 a c^4}+\frac {303 \log (1+a x)}{128 a c^4}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 121, normalized size = 0.84 \[ \frac {-384 a^7 x^7-768 a^6 x^6+1638 a^5 x^5+2508 a^4 x^4-1732 a^3 x^3-2516 a^2 x^2+550 a x-141 (a x-1)^2 (a x+1)^4 \log (1-a x)+909 (a x-1)^2 (a x+1)^4 \log (a x+1)+800}{384 a (a x-1)^2 (a c x+c)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - c/(a^2*x^2))^4),x]

[Out]

(800 + 550*a*x - 2516*a^2*x^2 - 1732*a^3*x^3 + 2508*a^4*x^4 + 1638*a^5*x^5 - 768*a^6*x^6 - 384*a^7*x^7 - 141*(
-1 + a*x)^2*(1 + a*x)^4*Log[1 - a*x] + 909*(-1 + a*x)^2*(1 + a*x)^4*Log[1 + a*x])/(384*a*(-1 + a*x)^2*(c + a*c
*x)^4)

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fricas [A]  time = 0.49, size = 233, normalized size = 1.62 \[ -\frac {384 \, a^{7} x^{7} + 768 \, a^{6} x^{6} - 1638 \, a^{5} x^{5} - 2508 \, a^{4} x^{4} + 1732 \, a^{3} x^{3} + 2516 \, a^{2} x^{2} - 550 \, a x - 909 \, {\left (a^{6} x^{6} + 2 \, a^{5} x^{5} - a^{4} x^{4} - 4 \, a^{3} x^{3} - a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 141 \, {\left (a^{6} x^{6} + 2 \, a^{5} x^{5} - a^{4} x^{4} - 4 \, a^{3} x^{3} - a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x - 1\right ) - 800}{384 \, {\left (a^{7} c^{4} x^{6} + 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} + 2 \, a^{2} c^{4} x + a c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2)^4,x, algorithm="fricas")

[Out]

-1/384*(384*a^7*x^7 + 768*a^6*x^6 - 1638*a^5*x^5 - 2508*a^4*x^4 + 1732*a^3*x^3 + 2516*a^2*x^2 - 550*a*x - 909*
(a^6*x^6 + 2*a^5*x^5 - a^4*x^4 - 4*a^3*x^3 - a^2*x^2 + 2*a*x + 1)*log(a*x + 1) + 141*(a^6*x^6 + 2*a^5*x^5 - a^
4*x^4 - 4*a^3*x^3 - a^2*x^2 + 2*a*x + 1)*log(a*x - 1) - 800)/(a^7*c^4*x^6 + 2*a^6*c^4*x^5 - a^5*c^4*x^4 - 4*a^
4*c^4*x^3 - a^3*c^4*x^2 + 2*a^2*c^4*x + a*c^4)

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giac [A]  time = 0.17, size = 164, normalized size = 1.14 \[ -\frac {2 \, \log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a c^{4}} - \frac {47 \, \log \left ({\left | -\frac {2}{a x + 1} + 1 \right |}\right )}{128 \, a c^{4}} + \frac {{\left (a x + 1\right )} {\left (\frac {1045}{a x + 1} - \frac {1064}{{\left (a x + 1\right )}^{2}} - 256\right )}}{256 \, a c^{4} {\left (\frac {2}{a x + 1} - 1\right )}^{2}} + \frac {\frac {297 \, a^{19} c^{12}}{a x + 1} - \frac {105 \, a^{19} c^{12}}{{\left (a x + 1\right )}^{2}} + \frac {26 \, a^{19} c^{12}}{{\left (a x + 1\right )}^{3}} - \frac {3 \, a^{19} c^{12}}{{\left (a x + 1\right )}^{4}}}{96 \, a^{20} c^{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2)^4,x, algorithm="giac")

[Out]

-2*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/(a*c^4) - 47/128*log(abs(-2/(a*x + 1) + 1))/(a*c^4) + 1/256*(a*x + 1
)*(1045/(a*x + 1) - 1064/(a*x + 1)^2 - 256)/(a*c^4*(2/(a*x + 1) - 1)^2) + 1/96*(297*a^19*c^12/(a*x + 1) - 105*
a^19*c^12/(a*x + 1)^2 + 26*a^19*c^12/(a*x + 1)^3 - 3*a^19*c^12/(a*x + 1)^4)/(a^20*c^16)

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maple [A]  time = 0.04, size = 126, normalized size = 0.88 \[ -\frac {x}{c^{4}}+\frac {1}{64 a \,c^{4} \left (a x -1\right )^{2}}+\frac {11}{64 a \,c^{4} \left (a x -1\right )}-\frac {47 \ln \left (a x -1\right )}{128 a \,c^{4}}-\frac {1}{32 a \,c^{4} \left (a x +1\right )^{4}}+\frac {13}{48 a \,c^{4} \left (a x +1\right )^{3}}-\frac {35}{32 a \,c^{4} \left (a x +1\right )^{2}}+\frac {99}{32 a \,c^{4} \left (a x +1\right )}+\frac {303 \ln \left (a x +1\right )}{128 a \,c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2)^4,x)

[Out]

-x/c^4+1/64/a/c^4/(a*x-1)^2+11/64/a/c^4/(a*x-1)-47/128/a/c^4*ln(a*x-1)-1/32/a/c^4/(a*x+1)^4+13/48/a/c^4/(a*x+1
)^3-35/32/a/c^4/(a*x+1)^2+99/32/a/c^4/(a*x+1)+303/128*ln(a*x+1)/a/c^4

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maxima [A]  time = 0.31, size = 146, normalized size = 1.01 \[ \frac {627 \, a^{5} x^{5} + 486 \, a^{4} x^{4} - 1058 \, a^{3} x^{3} - 874 \, a^{2} x^{2} + 467 \, a x + 400}{192 \, {\left (a^{7} c^{4} x^{6} + 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} + 2 \, a^{2} c^{4} x + a c^{4}\right )}} - \frac {x}{c^{4}} + \frac {303 \, \log \left (a x + 1\right )}{128 \, a c^{4}} - \frac {47 \, \log \left (a x - 1\right )}{128 \, a c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2)^4,x, algorithm="maxima")

[Out]

1/192*(627*a^5*x^5 + 486*a^4*x^4 - 1058*a^3*x^3 - 874*a^2*x^2 + 467*a*x + 400)/(a^7*c^4*x^6 + 2*a^6*c^4*x^5 -
a^5*c^4*x^4 - 4*a^4*c^4*x^3 - a^3*c^4*x^2 + 2*a^2*c^4*x + a*c^4) - x/c^4 + 303/128*log(a*x + 1)/(a*c^4) - 47/1
28*log(a*x - 1)/(a*c^4)

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mupad [B]  time = 0.95, size = 142, normalized size = 0.99 \[ \frac {\frac {467\,x}{192}-\frac {437\,a\,x^2}{96}+\frac {25}{12\,a}-\frac {529\,a^2\,x^3}{96}+\frac {81\,a^3\,x^4}{32}+\frac {209\,a^4\,x^5}{64}}{a^6\,c^4\,x^6+2\,a^5\,c^4\,x^5-a^4\,c^4\,x^4-4\,a^3\,c^4\,x^3-a^2\,c^4\,x^2+2\,a\,c^4\,x+c^4}-\frac {x}{c^4}-\frac {47\,\ln \left (a\,x-1\right )}{128\,a\,c^4}+\frac {303\,\ln \left (a\,x+1\right )}{128\,a\,c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^2*x^2 - 1)/((c - c/(a^2*x^2))^4*(a*x + 1)^2),x)

[Out]

((467*x)/192 - (437*a*x^2)/96 + 25/(12*a) - (529*a^2*x^3)/96 + (81*a^3*x^4)/32 + (209*a^4*x^5)/64)/(c^4 - a^2*
c^4*x^2 - 4*a^3*c^4*x^3 - a^4*c^4*x^4 + 2*a^5*c^4*x^5 + a^6*c^4*x^6 + 2*a*c^4*x) - x/c^4 - (47*log(a*x - 1))/(
128*a*c^4) + (303*log(a*x + 1))/(128*a*c^4)

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sympy [A]  time = 0.88, size = 158, normalized size = 1.10 \[ - a^{8} \left (\frac {- 627 a^{5} x^{5} - 486 a^{4} x^{4} + 1058 a^{3} x^{3} + 874 a^{2} x^{2} - 467 a x - 400}{192 a^{15} c^{4} x^{6} + 384 a^{14} c^{4} x^{5} - 192 a^{13} c^{4} x^{4} - 768 a^{12} c^{4} x^{3} - 192 a^{11} c^{4} x^{2} + 384 a^{10} c^{4} x + 192 a^{9} c^{4}} + \frac {x}{a^{8} c^{4}} + \frac {\frac {47 \log {\left (x - \frac {1}{a} \right )}}{128} - \frac {303 \log {\left (x + \frac {1}{a} \right )}}{128}}{a^{9} c^{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)**2*(-a**2*x**2+1)/(c-c/a**2/x**2)**4,x)

[Out]

-a**8*((-627*a**5*x**5 - 486*a**4*x**4 + 1058*a**3*x**3 + 874*a**2*x**2 - 467*a*x - 400)/(192*a**15*c**4*x**6
+ 384*a**14*c**4*x**5 - 192*a**13*c**4*x**4 - 768*a**12*c**4*x**3 - 192*a**11*c**4*x**2 + 384*a**10*c**4*x + 1
92*a**9*c**4) + x/(a**8*c**4) + (47*log(x - 1/a)/128 - 303*log(x + 1/a)/128)/(a**9*c**4))

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