3.579 \(\int \frac {e^{2 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\)

Optimal. Leaf size=69 \[ -\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}+\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{c}-4 a^2 \sqrt {c-\frac {c}{a x}} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 69, 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 = {6133, 25, 514, 446, 77} \[ -\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}+\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{c}-4 a^2 \sqrt {c-\frac {c}{a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 77

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 446

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 514

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 6133

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^3} \, dx &=\int \frac {\sqrt {c-\frac {c}{a x}} (1+a x)}{x^3 (1-a x)} \, dx\\ &=-\frac {c \int \frac {1+a x}{\sqrt {c-\frac {c}{a x}} x^4} \, dx}{a}\\ &=-\frac {c \int \frac {a+\frac {1}{x}}{\sqrt {c-\frac {c}{a x}} x^3} \, dx}{a}\\ &=\frac {c \operatorname {Subst}\left (\int \frac {x (a+x)}{\sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=\frac {c \operatorname {Subst}\left (\int \left (\frac {2 a^2}{\sqrt {c-\frac {c x}{a}}}-\frac {3 a^2 \sqrt {c-\frac {c x}{a}}}{c}+\frac {a^2 \left (c-\frac {c x}{a}\right )^{3/2}}{c^2}\right ) \, dx,x,\frac {1}{x}\right )}{a}\\ &=-4 a^2 \sqrt {c-\frac {c}{a x}}+\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{c}-\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 36, normalized size = 0.52 \[ -\frac {2 \left (6 a^2 x^2+3 a x+1\right ) \sqrt {c-\frac {c}{a x}}}{5 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - c/(a*x)]*(1 + 3*a*x + 6*a^2*x^2))/(5*x^2)

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fricas [A]  time = 0.44, size = 36, normalized size = 0.52 \[ -\frac {2 \, {\left (6 \, a^{2} x^{2} + 3 \, a x + 1\right )} \sqrt {\frac {a c x - c}{a x}}}{5 \, x^{2}} \]

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^3,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Unable to divide, perhaps due to rounding error%%%{%%%{%%{[-5,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,5]%%%},[6]%%%
}+%%%{%%{[%%%{25,[0,4]%%%},0]:[1,0,%%%{%%%{-1,[1]%%%},[2,2]%%%}+%%%{%%%{1,[1]%%%},[1,1]%%%}]%%},[5]%%%}+%%%{%%
%{%%{[-50,0]:[1,0,%%%{-1,[1]%%%}]%%},[2,5]%%%}+%%%{%%{[50,0]:[1,0,%%%{-1,[1]%%%}]%%},[1,4]%%%},[4]%%%}+%%%{%%{
[%%%{50,[2,4]%%%}+%%%{-50,[1,3]%%%},0]:[1,0,%%%{%%%{-1,[1]%%%},[2,2]%%%}+%%%{%%%{1,[1]%%%},[1,1]%%%}]%%},[3]%%
%}+%%%{%%%{%%{[-25,0]:[1,0,%%%{-1,[1]%%%}]%%},[4,5]%%%}+%%%{%%{[50,0]:[1,0,%%%{-1,[1]%%%}]%%},[3,4]%%%}+%%%{%%
{[-25,0]:[1,0,%%%{-1,[1]%%%}]%%},[2,3]%%%},[2]%%%}+%%%{%%{[%%%{5,[4,4]%%%}+%%%{-10,[3,3]%%%}+%%%{5,[2,2]%%%},0
]:[1,0,%%%{%%%{-1,[1]%%%},[2,2]%%%}+%%%{%%%{1,[1]%%%},[1,1]%%%}]%%},[1]%%%} / %%%{%%%{%%{poly1[%%%{5,[6]%%%},0
]:[1,0,%%%{-1,[1]%%%}]%%},[12,12]%%%}+%%%{%%{poly1[%%%{-30,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[11,11]%%%}+%%%{
%%{poly1[%%%{75,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[10,10]%%%}+%%%{%%{poly1[%%%{-100,[6]%%%},0]:[1,0,%%%{-1,[1
]%%%}]%%},[9,9]%%%}+%%%{%%{poly1[%%%{75,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[8,8]%%%}+%%%{%%{poly1[%%%{-30,[6]%
%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[7,7]%%%}+%%%{%%{poly1[%%%{5,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[6,6]%%%},[6]%%
%}+%%%{%%{[%%%{%%%{-25,[6]%%%},[12,11]%%%}+%%%{%%%{150,[6]%%%},[11,10]%%%}+%%%{%%%{-375,[6]%%%},[10,9]%%%}+%%%
{%%%{500,[6]%%%},[9,8]%%%}+%%%{%%%{-375,[6]%%%},[8,7]%%%}+%%%{%%%{150,[6]%%%},[7,6]%%%}+%%%{%%%{-25,[6]%%%},[6
,5]%%%},0]:[1,0,%%%{%%%{-1,[1]%%%},[2,2]%%%}+%%%{%%%{1,[1]%%%},[1,1]%%%}]%%},[5]%%%}+%%%{%%%{%%{poly1[%%%{50,[
6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[14,12]%%%}+%%%{%%{poly1[%%%{-350,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[13,11]
%%%}+%%%{%%{poly1[%%%{1050,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[12,10]%%%}+%%%{%%{poly1[%%%{-1750,[6]%%%},0]:[1
,0,%%%{-1,[1]%%%}]%%},[11,9]%%%}+%%%{%%{poly1[%%%{1750,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[10,8]%%%}+%%%{%%{po
ly1[%%%{-1050,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[9,7]%%%}+%%%{%%{poly1[%%%{350,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}
]%%},[8,6]%%%}+%%%{%%{poly1[%%%{-50,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[7,5]%%%},[4]%%%}+%%%{%%{[%%%{%%%{-50,[
6]%%%},[14,11]%%%}+%%%{%%%{350,[6]%%%},[13,10]%%%}+%%%{%%%{-1050,[6]%%%},[12,9]%%%}+%%%{%%%{1750,[6]%%%},[11,8
]%%%}+%%%{%%%{-1750,[6]%%%},[10,7]%%%}+%%%{%%%{1050,[6]%%%},[9,6]%%%}+%%%{%%%{-350,[6]%%%},[8,5]%%%}+%%%{%%%{5
0,[6]%%%},[7,4]%%%},0]:[1,0,%%%{%%%{-1,[1]%%%},[2,2]%%%}+%%%{%%%{1,[1]%%%},[1,1]%%%}]%%},[3]%%%}+%%%{%%%{%%{po
ly1[%%%{25,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[16,12]%%%}+%%%{%%{poly1[%%%{-200,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}
]%%},[15,11]%%%}+%%%{%%{poly1[%%%{700,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[14,10]%%%}+%%%{%%{poly1[%%%{-1400,[6
]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[13,9]%%%}+%%%{%%{poly1[%%%{1750,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[12,8]%%%
}+%%%{%%{poly1[%%%{-1400,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[11,7]%%%}+%%%{%%{poly1[%%%{700,[6]%%%},0]:[1,0,%%
%{-1,[1]%%%}]%%},[10,6]%%%}+%%%{%%{poly1[%%%{-200,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[9,5]%%%}+%%%{%%{poly1[%%
%{25,[6]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[8,4]%%%},[2]%%%}+%%%{%%{[%%%{%%%{-5,[6]%%%},[16,11]%%%}+%%%{%%%{40,[6
]%%%},[15,10]%%%}+%%%{%%%{-140,[6]%%%},[14,9]%%%}+%%%{%%%{280,[6]%%%},[13,8]%%%}+%%%{%%%{-350,[6]%%%},[12,7]%%
%}+%%%{%%%{280,[6]%%%},[11,6]%%%}+%%%{%%%{-140,[6]%%%},[10,5]%%%}+%%%{%%%{40,[6]%%%},[9,4]%%%}+%%%{%%%{-5,[6]%
%%},[8,3]%%%},0]:[1,0,%%%{%%%{-1,[1]%%%},[2,2]%%%}+%%%{%%%{1,[1]%%%},[1,1]%%%}]%%},[1]%%%} Error: Bad Argument
Value

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maple [A]  time = 0.03, size = 35, normalized size = 0.51 \[ -\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (6 a^{2} x^{2}+3 a x +1\right )}{5 x^{2}} \]

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^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (a x + 1\right )}^{2} \sqrt {c - \frac {c}{a x}}}{{\left (a^{2} x^{2} - 1\right )} x^{3}}\,{d x} \]

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^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.91, size = 32, normalized size = 0.46 \[ -\frac {2\,\sqrt {c-\frac {c}{a\,x}}\,\left (6\,a^2\,x^2+3\,a\,x+1\right )}{5\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\sqrt {c - \frac {c}{a x}}}{a x^{4} - x^{3}}\, dx - \int \frac {a x \sqrt {c - \frac {c}{a x}}}{a x^{4} - x^{3}}\, dx \]

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**3,x)

[Out]

-Integral(sqrt(c - c/(a*x))/(a*x**4 - x**3), x) - Integral(a*x*sqrt(c - c/(a*x))/(a*x**4 - x**3), x)

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