3.7.5 \(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^4} \, dx\) [605]

Optimal. Leaf size=119 \[ -\frac {1}{64 a c^4 (1-a x)^2}-\frac {5}{64 a c^4 (1-a x)}+\frac {1}{32 a c^4 (1+a x)^4}+\frac {1}{16 a c^4 (1+a x)^3}+\frac {3}{32 a c^4 (1+a x)^2}+\frac {5}{32 a c^4 (1+a x)}-\frac {15 \tanh ^{-1}(a x)}{64 a c^4} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6302, 6275, 46, 213} \begin {gather*} -\frac {5}{64 a c^4 (1-a x)}+\frac {5}{32 a c^4 (a x+1)}-\frac {1}{64 a c^4 (1-a x)^2}+\frac {3}{32 a c^4 (a x+1)^2}+\frac {1}{16 a c^4 (a x+1)^3}+\frac {1}{32 a c^4 (a x+1)^4}-\frac {15 \tanh ^{-1}(a x)}{64 a c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*1/(a*c^4*(1 - a*x)^2) - 5/(64*a*c^4*(1 - a*x)) + 1/(32*a*c^4*(1 + a*x)^4) + 1/(16*a*c^4*(1 + a*x)^3) + 3
/(32*a*c^4*(1 + a*x)^2) + 5/(32*a*c^4*(1 + a*x)) - (15*ArcTanh[a*x])/(64*a*c^4)

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 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6275

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

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx &=-\int \frac {e^{-2 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx\\ &=-\frac {\int \frac {1}{(1-a x)^3 (1+a x)^5} \, dx}{c^4}\\ &=-\frac {\int \left (-\frac {1}{32 (-1+a x)^3}+\frac {5}{64 (-1+a x)^2}+\frac {1}{8 (1+a x)^5}+\frac {3}{16 (1+a x)^4}+\frac {3}{16 (1+a x)^3}+\frac {5}{32 (1+a x)^2}-\frac {15}{64 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^4}\\ &=-\frac {1}{64 a c^4 (1-a x)^2}-\frac {5}{64 a c^4 (1-a x)}+\frac {1}{32 a c^4 (1+a x)^4}+\frac {1}{16 a c^4 (1+a x)^3}+\frac {3}{32 a c^4 (1+a x)^2}+\frac {5}{32 a c^4 (1+a x)}+\frac {15 \int \frac {1}{-1+a^2 x^2} \, dx}{64 c^4}\\ &=-\frac {1}{64 a c^4 (1-a x)^2}-\frac {5}{64 a c^4 (1-a x)}+\frac {1}{32 a c^4 (1+a x)^4}+\frac {1}{16 a c^4 (1+a x)^3}+\frac {3}{32 a c^4 (1+a x)^2}+\frac {5}{32 a c^4 (1+a x)}-\frac {15 \tanh ^{-1}(a x)}{64 a c^4}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 80, normalized size = 0.67 \begin {gather*} \frac {16-17 a x-50 a^2 x^2-10 a^3 x^3+30 a^4 x^4+15 a^5 x^5-15 (-1+a x)^2 (1+a x)^4 \tanh ^{-1}(a x)}{64 a (-1+a x)^2 (c+a c x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(16 - 17*a*x - 50*a^2*x^2 - 10*a^3*x^3 + 30*a^4*x^4 + 15*a^5*x^5 - 15*(-1 + a*x)^2*(1 + a*x)^4*ArcTanh[a*x])/(
64*a*(-1 + a*x)^2*(c + a*c*x)^4)

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Maple [A]
time = 0.16, size = 100, normalized size = 0.84

method	result	size
risch	\(\frac {\frac {15 a^{4} x^{5}}{64}+\frac {15 a^{3} x^{4}}{32}-\frac {5 a^{2} x^{3}}{32}-\frac {25 a \,x^{2}}{32}-\frac {17 x}{64}+\frac {1}{4 a}}{\left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right )^{2} c^{4}}-\frac {15 \ln \left (a x +1\right )}{128 a \,c^{4}}+\frac {15 \ln \left (-a x +1\right )}{128 a \,c^{4}}\)	\(92\)
default	\(\frac {\frac {1}{32 a \left (a x +1\right )^{4}}+\frac {1}{16 a \left (a x +1\right )^{3}}+\frac {3}{32 a \left (a x +1\right )^{2}}+\frac {5}{32 a \left (a x +1\right )}-\frac {15 \ln \left (a x +1\right )}{128 a}-\frac {1}{64 a \left (a x -1\right )^{2}}+\frac {5}{64 a \left (a x -1\right )}+\frac {15 \ln \left (a x -1\right )}{128 a}}{c^{4}}\)	\(100\)
norman	\(\frac {\frac {49 x}{64 c}-\frac {15 a \,x^{2}}{64 c}-\frac {11 a^{2} x^{3}}{8 c}+\frac {a^{3} x^{4}}{8 c}+\frac {63 a^{4} x^{5}}{64 c}-\frac {a^{5} x^{6}}{64 c}-\frac {a^{6} x^{7}}{4 c}}{\left (a x -1\right )^{3} \left (a x +1\right )^{4} c^{3}}+\frac {15 \ln \left (a x -1\right )}{128 c^{4} a}-\frac {15 \ln \left (a x +1\right )}{128 a \,c^{4}}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(1/32/a/(a*x+1)^4+1/16/a/(a*x+1)^3+3/32/a/(a*x+1)^2+5/32/a/(a*x+1)-15/128/a*ln(a*x+1)-1/64/a/(a*x-1)^2+5
/64/a/(a*x-1)+15/128/a*ln(a*x-1))

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Maxima [A]
time = 0.28, size = 140, normalized size = 1.18 \begin {gather*} \frac {15 \, a^{5} x^{5} + 30 \, a^{4} x^{4} - 10 \, a^{3} x^{3} - 50 \, a^{2} x^{2} - 17 \, a x + 16}{64 \, {\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 {15 \, \log \left (a x + 1\right )}{128 \, a c^{4}} + \frac {15 \, \log \left (a x - 1\right )}{128 \, a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(15*a^5*x^5 + 30*a^4*x^4 - 10*a^3*x^3 - 50*a^2*x^2 - 17*a*x + 16)/(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) - 15/128*log(a*x + 1)/(a*c^4) + 15/128*log(a*x - 1)/(
a*c^4)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (103) = 206\).
time = 0.37, size = 217, normalized size = 1.82 \begin {gather*} \frac {30 \, a^{5} x^{5} + 60 \, a^{4} x^{4} - 20 \, a^{3} x^{3} - 100 \, a^{2} x^{2} - 34 \, a x - 15 \, {\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 ) + 15 \, {\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 ) + 32}{128 \, {\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(30*a^5*x^5 + 60*a^4*x^4 - 20*a^3*x^3 - 100*a^2*x^2 - 34*a*x - 15*(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) + 15*(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) + 32)/(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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Sympy [A]
time = 0.32, size = 141, normalized size = 1.18 \begin {gather*} \frac {15 a^{5} x^{5} + 30 a^{4} x^{4} - 10 a^{3} x^{3} - 50 a^{2} x^{2} - 17 a x + 16}{64 a^{7} c^{4} x^{6} + 128 a^{6} c^{4} x^{5} - 64 a^{5} c^{4} x^{4} - 256 a^{4} c^{4} x^{3} - 64 a^{3} c^{4} x^{2} + 128 a^{2} c^{4} x + 64 a c^{4}} + \frac {\frac {15 \log {\left (x - \frac {1}{a} \right )}}{128} - \frac {15 \log {\left (x + \frac {1}{a} \right )}}{128}}{a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*a**5*x**5 + 30*a**4*x**4 - 10*a**3*x**3 - 50*a**2*x**2 - 17*a*x + 16)/(64*a**7*c**4*x**6 + 128*a**6*c**4*x
**5 - 64*a**5*c**4*x**4 - 256*a**4*c**4*x**3 - 64*a**3*c**4*x**2 + 128*a**2*c**4*x + 64*a*c**4) + (15*log(x -
1/a)/128 - 15*log(x + 1/a)/128)/(a*c**4)

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Giac [A]
time = 0.40, size = 91, normalized size = 0.76 \begin {gather*} -\frac {15 \, \log \left ({\left | a x + 1 \right |}\right )}{128 \, a c^{4}} + \frac {15 \, \log \left ({\left | a x - 1 \right |}\right )}{128 \, a c^{4}} + \frac {15 \, a^{5} x^{5} + 30 \, a^{4} x^{4} - 10 \, a^{3} x^{3} - 50 \, a^{2} x^{2} - 17 \, a x + 16}{64 \, {\left (a x + 1\right )}^{4} {\left (a x - 1\right )}^{2} a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/128*log(abs(a*x + 1))/(a*c^4) + 15/128*log(abs(a*x - 1))/(a*c^4) + 1/64*(15*a^5*x^5 + 30*a^4*x^4 - 10*a^3*
x^3 - 50*a^2*x^2 - 17*a*x + 16)/((a*x + 1)^4*(a*x - 1)^2*a*c^4)

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Mupad [B]
time = 1.28, size = 121, normalized size = 1.02 \begin {gather*} -\frac {\frac {17\,x}{64}+\frac {25\,a\,x^2}{32}-\frac {1}{4\,a}+\frac {5\,a^2\,x^3}{32}-\frac {15\,a^3\,x^4}{32}-\frac {15\,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 {15\,\mathrm {atanh}\left (a\,x\right )}{64\,a\,c^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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