3.5.76 \(\int e^{4 \tanh ^{-1}(a x)} (c-\frac {c}{a x})^4 \, dx\) [476]

Optimal. Leaf size=30 \[ -\frac {c^4}{3 a^4 x^3}+\frac {2 c^4}{a^2 x}+c^4 x \]

[Out]

-1/3*c^4/a^4/x^3+2*c^4/a^2/x+c^4*x

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Rubi [A]
time = 0.06, antiderivative size = 30, 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 = {6266, 6264, 74, 276} \begin {gather*} -\frac {c^4}{3 a^4 x^3}+\frac {2 c^4}{a^2 x}+c^4 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 74

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m] && (NeQ[m, -1] || (EqQ[e, 0] && (EqQ[p, 1] ||  !IntegerQ[p])))

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 6264

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

Rule 6266

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

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 30, normalized size = 1.00 \begin {gather*} -\frac {c^4}{3 a^4 x^3}+\frac {2 c^4}{a^2 x}+c^4 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.77, size = 27, normalized size = 0.90

method	result	size
default	\(\frac {c^{4} \left (a^{4} x -\frac {1}{3 x^{3}}+\frac {2 a^{2}}{x}\right )}{a^{4}}\)	\(27\)
gosper	\(\frac {\left (3 a^{4} x^{4}+6 a^{2} x^{2}-1\right ) c^{4}}{3 x^{3} a^{4}}\)	\(30\)
risch	\(c^{4} x +\frac {2 a^{2} c^{4} x^{2}-\frac {1}{3} c^{4}}{a^{4} x^{3}}\)	\(31\)
norman	\(\frac {c^{4} a^{5} x^{6}+c^{4} x^{4} a^{3}+\frac {c^{4}}{3 a}-\frac {7 c^{4} a \,x^{2}}{3}}{\left (a^{2} x^{2}-1\right ) a^{3} x^{3}}\)	\(57\)
meijerg	\(\frac {c^{4} \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \arctanh \left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}+\frac {2 c^{4} \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \arctanh \left (a x \right )}{a^{3}}\right )}{\sqrt {-a^{2}}}+\frac {3 c^{4} \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \arctanh \left (a x \right )}{a}\right )}{\sqrt {-a^{2}}}+\frac {c^{4} \left (-\frac {2 \left (-15 a^{4} x^{4}+10 a^{2} x^{2}+2\right )}{3 x^{3} \left (-a^{2}\right )^{\frac {3}{2}} \left (-2 a^{2} x^{2}+2\right )}+\frac {5 a^{3} \arctanh \left (a x \right )}{\left (-a^{2}\right )^{\frac {3}{2}}}\right )}{2 \sqrt {-a^{2}}}+\frac {2 c^{4} \left (-\frac {2 \left (-3 a^{2} x^{2}+2\right )}{x \sqrt {-a^{2}}\, \left (-2 a^{2} x^{2}+2\right )}+\frac {3 a \arctanh \left (a x \right )}{\sqrt {-a^{2}}}\right )}{\sqrt {-a^{2}}}\)	\(301\)

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/x)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 31, normalized size = 1.03 \begin {gather*} c^{4} x + \frac {6 \, a^{2} c^{4} x^{2} - c^{4}}{3 \, a^{4} x^{3}} \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/x)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 36, normalized size = 1.20 \begin {gather*} \frac {3 \, a^{4} c^{4} x^{4} + 6 \, a^{2} c^{4} x^{2} - c^{4}}{3 \, a^{4} x^{3}} \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/x)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.08, size = 31, normalized size = 1.03 \begin {gather*} \frac {a^{4} c^{4} x + \frac {6 a^{2} c^{4} x^{2} - c^{4}}{3 x^{3}}}{a^{4}} \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/x)**4,x)

[Out]

(a**4*c**4*x + (6*a**2*c**4*x**2 - c**4)/(3*x**3))/a**4

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Giac [A]
time = 0.38, size = 31, normalized size = 1.03 \begin {gather*} c^{4} x + \frac {6 \, a^{2} c^{4} x^{2} - c^{4}}{3 \, a^{4} x^{3}} \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/x)^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.81, size = 27, normalized size = 0.90 \begin {gather*} \frac {c^4\,\left (a^4\,x^4+2\,a^2\,x^2-\frac {1}{3}\right )}{a^4\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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