3.6.80 \(\int \frac {e^{2 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx\) [580]

Optimal. Leaf size=96 \[ -4 a^3 \sqrt {c-\frac {c}{a x}}+\frac {10 a^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}-\frac {8 a^3 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}+\frac {2 a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{7 c^3} \]

[Out]

10/3*a^3*(c-c/a/x)^(3/2)/c-8/5*a^3*(c-c/a/x)^(5/2)/c^2+2/7*a^3*(c-c/a/x)^(7/2)/c^3-4*a^3*(c-c/a/x)^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6268, 25, 528, 457, 78} \begin {gather*} \frac {2 a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{7 c^3}-\frac {8 a^3 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}+\frac {10 a^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}-4 a^3 \sqrt {c-\frac {c}{a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*a^3*Sqrt[c - c/(a*x)] + (10*a^3*(c - c/(a*x))^(3/2))/(3*c) - (8*a^3*(c - c/(a*x))^(5/2))/(5*c^2) + (2*a^3*(
c - c/(a*x))^(7/2))/(7*c^3)

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[u*((
a + b*x^n)^(m + p)/x^(n*p)), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 528

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 6268

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

Rubi steps

\begin {align*} \int \frac {e^{2 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx &=\int \frac {\sqrt {c-\frac {c}{a x}} (1+a x)}{x^4 (1-a x)} \, dx\\ &=-\frac {c \int \frac {1+a x}{\sqrt {c-\frac {c}{a x}} x^5} \, dx}{a}\\ &=-\frac {c \int \frac {a+\frac {1}{x}}{\sqrt {c-\frac {c}{a x}} x^4} \, dx}{a}\\ &=\frac {c \text {Subst}\left (\int \frac {x^2 (a+x)}{\sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=\frac {c \text {Subst}\left (\int \left (\frac {2 a^3}{\sqrt {c-\frac {c x}{a}}}-\frac {5 a^3 \sqrt {c-\frac {c x}{a}}}{c}+\frac {4 a^3 \left (c-\frac {c x}{a}\right )^{3/2}}{c^2}-\frac {a^3 \left (c-\frac {c x}{a}\right )^{5/2}}{c^3}\right ) \, dx,x,\frac {1}{x}\right )}{a}\\ &=-4 a^3 \sqrt {c-\frac {c}{a x}}+\frac {10 a^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}-\frac {8 a^3 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}+\frac {2 a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{7 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 44, normalized size = 0.46 \begin {gather*} -\frac {2 \sqrt {c-\frac {c}{a x}} \left (15+39 a x+52 a^2 x^2+104 a^3 x^3\right )}{105 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - c/(a*x)]*(15 + 39*a*x + 52*a^2*x^2 + 104*a^3*x^3))/(105*x^3)

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Maple [C] Result contains higher order function than in optimal. Order 3 vs. order 2.
time = 0.81, size = 211, normalized size = 2.20

method	result	size
gosper	\(-\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (104 a^{3} x^{3}+52 a^{2} x^{2}+39 a x +15\right )}{105 x^{3}}\)	\(43\)
trager	\(-\frac {2 \left (104 a^{3} x^{3}+52 a^{2} x^{2}+39 a x +15\right ) \sqrt {-\frac {-c x a +c}{a x}}}{105 x^{3}}\)	\(45\)
risch	\(-\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (104 a^{4} x^{4}-52 a^{3} x^{3}-13 a^{2} x^{2}-24 a x -15\right )}{105 \left (a x -1\right ) x^{3}}\)	\(58\)
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (210 \sqrt {\left (a x -1\right ) x}\, a^{\frac {9}{2}} x^{5}+210 a^{\frac {9}{2}} \sqrt {a \,x^{2}-x}\, x^{5}-420 a^{\frac {7}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{3}-105 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{4} x^{5}+105 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{4} x^{5}-212 a^{\frac {5}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{2}-108 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x -30 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\right )}{105 x^{4} \sqrt {\left (a x -1\right ) x}\, \sqrt {a}}\)	\(211\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(c*(a*x-1)/a/x)^(1/2)/x^4*(210*((a*x-1)*x)^(1/2)*a^(9/2)*x^5+210*a^(9/2)*(a*x^2-x)^(1/2)*x^5-420*a^(7/2
)*(a*x^2-x)^(3/2)*x^3-105*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*a^4*x^5+105*ln(1/2*(2*((a*x-1)*x
)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*a^4*x^5-212*a^(5/2)*(a*x^2-x)^(3/2)*x^2-108*a^(3/2)*(a*x^2-x)^(3/2)*x-30*(a*
x^2-x)^(3/2)*a^(1/2))/((a*x-1)*x)^(1/2)/a^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 44, normalized size = 0.46 \begin {gather*} -\frac {2 \, {\left (104 \, a^{3} x^{3} + 52 \, a^{2} x^{2} + 39 \, a x + 15\right )} \sqrt {\frac {a c x - c}{a x}}}{105 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(104*a^3*x^3 + 52*a^2*x^2 + 39*a*x + 15)*sqrt((a*c*x - c)/(a*x))/x^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\sqrt {c - \frac {c}{a x}}}{a x^{5} - x^{4}}\, dx - \int \frac {a x \sqrt {c - \frac {c}{a x}}}{a x^{5} - x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(sqrt(c - c/(a*x))/(a*x**5 - x**4), x) - Integral(a*x*sqrt(c - c/(a*x))/(a*x**5 - x**4), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 0.91, size = 77, normalized size = 0.80 \begin {gather*} -\frac {208\,a^3\,\sqrt {c-\frac {c}{a\,x}}}{105}-\frac {2\,\sqrt {c-\frac {c}{a\,x}}}{7\,x^3}-\frac {26\,a\,\sqrt {c-\frac {c}{a\,x}}}{35\,x^2}-\frac {104\,a^2\,\sqrt {c-\frac {c}{a\,x}}}{105\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (208*a^3*(c - c/(a*x))^(1/2))/105 - (2*(c - c/(a*x))^(1/2))/(7*x^3) - (26*a*(c - c/(a*x))^(1/2))/(35*x^2) -
(104*a^2*(c - c/(a*x))^(1/2))/(105*x)

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