3.7.55 \(\int e^{4 \tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^3 \, dx\) [655]

Optimal. Leaf size=63 \[ \frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+c^3 x+\frac {4 c^3 \log (x)}{a} \]

[Out]

1/5*c^3/a^6/x^5+c^3/a^5/x^4+5/3*c^3/a^4/x^3-5*c^3/a^2/x+c^3*x+4*c^3*ln(x)/a

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Rubi [A]
time = 0.08, antiderivative size = 63, 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 = {6292, 6285, 76} \begin {gather*} \frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+\frac {4 c^3 \log (x)}{a}+c^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])*(c - c/(a^2*x^2))^3,x]

[Out]

c^3/(5*a^6*x^5) + c^3/(a^5*x^4) + (5*c^3)/(3*a^4*x^3) - (5*c^3)/(a^2*x) + c^3*x + (4*c^3*Log[x])/a

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 6285

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 6292

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

Rubi steps

\begin {align*} \int e^{4 \tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx &=-\frac {c^3 \int \frac {e^{4 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^3}{x^6} \, dx}{a^6}\\ &=-\frac {c^3 \int \frac {(1-a x) (1+a x)^5}{x^6} \, dx}{a^6}\\ &=-\frac {c^3 \int \left (-a^6+\frac {1}{x^6}+\frac {4 a}{x^5}+\frac {5 a^2}{x^4}-\frac {5 a^4}{x^2}-\frac {4 a^5}{x}\right ) \, dx}{a^6}\\ &=\frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+c^3 x+\frac {4 c^3 \log (x)}{a}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 63, normalized size = 1.00 \begin {gather*} \frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+c^3 x+\frac {4 c^3 \log (x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcTanh[a*x])*(c - c/(a^2*x^2))^3,x]

[Out]

c^3/(5*a^6*x^5) + c^3/(a^5*x^4) + (5*c^3)/(3*a^4*x^3) - (5*c^3)/(a^2*x) + c^3*x + (4*c^3*Log[x])/a

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Maple [A]
time = 1.17, size = 47, normalized size = 0.75

method	result	size
default	\(\frac {c^{3} \left (a^{6} x +\frac {5 a^{2}}{3 x^{3}}+\frac {a}{x^{4}}-\frac {5 a^{4}}{x}+\frac {1}{5 x^{5}}+4 a^{5} \ln \left (x \right )\right )}{a^{6}}\)	\(47\)
risch	\(x \,c^{3}+\frac {-5 a^{4} c^{3} x^{4}+\frac {5}{3} a^{2} c^{3} x^{2}+a \,c^{3} x +\frac {1}{5} c^{3}}{a^{6} x^{5}}+\frac {4 c^{3} \ln \left (x \right )}{a}\)	\(58\)
norman	\(\frac {a^{2} c^{3} x^{3}+a^{7} c^{3} x^{8}-\frac {c^{3}}{5 a}-x \,c^{3}-\frac {22 a \,c^{3} x^{2}}{15}+\frac {20 a^{3} c^{3} x^{4}}{3}-6 a^{5} c^{3} x^{6}}{a^{5} \left (a^{2} x^{2}-1\right ) x^{5}}+\frac {4 c^{3} \ln \left (x \right )}{a}\)	\(96\)
meijerg	\(\frac {c^{3} \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}-\frac {3 c^{3} \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}-\frac {7 c^{3} \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \arctanh \left (a x \right )}{a}\right )}{\sqrt {-a^{2}}}-\frac {7 c^{3} \left (-\frac {2 \left (-3 a^{2} x^{2}+2\right )}{x \sqrt {-a^{2}}\, \left (-2 a^{2} x^{2}+2\right )}+\frac {3 a \arctanh \left (a x \right )}{\sqrt {-a^{2}}}\right )}{\sqrt {-a^{2}}}+\frac {2 c^{3} \left (\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {4 a \,c^{3} x^{2}}{-a^{2} x^{2}+1}-\frac {4 c^{3} \left (-\frac {3 a^{2} x^{2}}{-3 a^{2} x^{2}+3}+2 \ln \left (-a^{2} x^{2}+1\right )-1-4 \ln \left (x \right )-2 \ln \left (-a^{2}\right )+\frac {1}{a^{2} x^{2}}\right )}{a}-\frac {3 c^{3} \left (-\frac {2 \left (-15 a^{4} x^{4}+10 a^{2} x^{2}+2\right )}{3 x^{3} \left (-a^{2}\right )^{\frac {3}{2}} \left (-2 a^{2} x^{2}+2\right )}+\frac {5 a^{3} \arctanh \left (a x \right )}{\left (-a^{2}\right )^{\frac {3}{2}}}\right )}{2 \sqrt {-a^{2}}}-\frac {2 c^{3} \left (\frac {4 a^{2} x^{2}}{-4 a^{2} x^{2}+4}-3 \ln \left (-a^{2} x^{2}+1\right )+1+6 \ln \left (x \right )+3 \ln \left (-a^{2}\right )-\frac {1}{2 a^{4} x^{4}}-\frac {2}{a^{2} x^{2}}\right )}{a}+\frac {c^{3} \left (-\frac {2 \left (-105 a^{6} x^{6}+70 a^{4} x^{4}+14 a^{2} x^{2}+6\right )}{5 x^{5} \left (-a^{2}\right )^{\frac {5}{2}} \left (-6 a^{2} x^{2}+6\right )}+\frac {7 \arctanh \left (a x \right ) a^{5}}{\left (-a^{2}\right )^{\frac {5}{2}}}\right )}{2 \sqrt {-a^{2}}}\)	\(573\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^3,x,method=_RETURNVERBOSE)

[Out]

c^3/a^6*(a^6*x+5/3*a^2/x^3+a/x^4-5*a^4/x+1/5/x^5+4*a^5*ln(x))

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Maxima [A]
time = 0.26, size = 59, normalized size = 0.94 \begin {gather*} c^{3} x + \frac {4 \, c^{3} \log \left (x\right )}{a} - \frac {75 \, a^{4} c^{3} x^{4} - 25 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 3 \, c^{3}}{15 \, a^{6} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^3,x, algorithm="maxima")

[Out]

c^3*x + 4*c^3*log(x)/a - 1/15*(75*a^4*c^3*x^4 - 25*a^2*c^3*x^2 - 15*a*c^3*x - 3*c^3)/(a^6*x^5)

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Fricas [A]
time = 0.40, size = 67, normalized size = 1.06 \begin {gather*} \frac {15 \, a^{6} c^{3} x^{6} + 60 \, a^{5} c^{3} x^{5} \log \left (x\right ) - 75 \, a^{4} c^{3} x^{4} + 25 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x + 3 \, c^{3}}{15 \, a^{6} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^3,x, algorithm="fricas")

[Out]

1/15*(15*a^6*c^3*x^6 + 60*a^5*c^3*x^5*log(x) - 75*a^4*c^3*x^4 + 25*a^2*c^3*x^2 + 15*a*c^3*x + 3*c^3)/(a^6*x^5)

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Sympy [A]
time = 0.18, size = 65, normalized size = 1.03 \begin {gather*} \frac {a^{6} c^{3} x + 4 a^{5} c^{3} \log {\left (x \right )} + \frac {- 75 a^{4} c^{3} x^{4} + 25 a^{2} c^{3} x^{2} + 15 a c^{3} x + 3 c^{3}}{15 x^{5}}}{a^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2*(c-c/a**2/x**2)**3,x)

[Out]

(a**6*c**3*x + 4*a**5*c**3*log(x) + (-75*a**4*c**3*x**4 + 25*a**2*c**3*x**2 + 15*a*c**3*x + 3*c**3)/(15*x**5))
/a**6

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Giac [A]
time = 0.42, size = 60, normalized size = 0.95 \begin {gather*} c^{3} x + \frac {4 \, c^{3} \log \left ({\left | x \right |}\right )}{a} - \frac {75 \, a^{4} c^{3} x^{4} - 25 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 3 \, c^{3}}{15 \, a^{6} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^3,x, algorithm="giac")

[Out]

c^3*x + 4*c^3*log(abs(x))/a - 1/15*(75*a^4*c^3*x^4 - 25*a^2*c^3*x^2 - 15*a*c^3*x - 3*c^3)/(a^6*x^5)

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Mupad [B]
time = 0.06, size = 48, normalized size = 0.76 \begin {gather*} \frac {c^3\,\left (a\,x+\frac {5\,a^2\,x^2}{3}-5\,a^4\,x^4+a^6\,x^6+4\,a^5\,x^5\,\ln \left (x\right )+\frac {1}{5}\right )}{a^6\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - c/(a^2*x^2))^3*(a*x + 1)^4)/(a^2*x^2 - 1)^2,x)

[Out]

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

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