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

Optimal. Leaf size=113 \[ -\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}-\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}-4 a^2 \sqrt {c-\frac {c}{a x}}+4 \sqrt {2} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6133, 25, 514, 446, 80, 50, 63, 208} \[ -\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}-\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}-4 a^2 \sqrt {c-\frac {c}{a x}}+4 \sqrt {2} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

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

Rule 208

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

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

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Mathematica [A]  time = 0.08, size = 79, normalized size = 0.70 \[ 4 \sqrt {2} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )-\frac {2 \left (38 a^2 x^2-11 a x+3\right ) \sqrt {c-\frac {c}{a x}}}{15 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - c/(a*x)]*(3 - 11*a*x + 38*a^2*x^2))/(15*x^2) + 4*Sqrt[2]*a^2*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*x)]/(S
qrt[2]*Sqrt[c])]

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fricas [A]  time = 0.57, size = 185, normalized size = 1.64 \[ \left [\frac {2 \, {\left (15 \, \sqrt {2} a^{2} \sqrt {c} x^{2} \log \left (-\frac {2 \, \sqrt {2} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + 3 \, a c x - c}{a x + 1}\right ) - {\left (38 \, a^{2} x^{2} - 11 \, a x + 3\right )} \sqrt {\frac {a c x - c}{a x}}\right )}}{15 \, x^{2}}, -\frac {2 \, {\left (30 \, \sqrt {2} a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{2 \, c}\right ) + {\left (38 \, a^{2} x^{2} - 11 \, a x + 3\right )} \sqrt {\frac {a c x - c}{a x}}\right )}}{15 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/15*(15*sqrt(2)*a^2*sqrt(c)*x^2*log(-(2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + 3*a*c*x - c)/(a*x + 1)
) - (38*a^2*x^2 - 11*a*x + 3)*sqrt((a*c*x - c)/(a*x)))/x^2, -2/15*(30*sqrt(2)*a^2*sqrt(-c)*x^2*arctan(1/2*sqrt
(2)*sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) + (38*a^2*x^2 - 11*a*x + 3)*sqrt((a*c*x - c)/(a*x)))/x^2]

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giac [B]  time = 1.18, size = 278, normalized size = 2.46 \[ -\frac {4 \, \sqrt {2} a^{3} c \arctan \left (\frac {\sqrt {2} {\left ({\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} a + \sqrt {c} {\left | a \right |}\right )}}{2 \, a \sqrt {-c}}\right )}{\sqrt {-c} {\left | a \right |} \mathrm {sgn}\relax (x)} - \frac {2 \, {\left (60 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{4} a^{5} c - 45 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{3} a^{4} c^{\frac {3}{2}} {\left | a \right |} + 35 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{2} a^{5} c^{2} - 15 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} a^{4} c^{\frac {5}{2}} {\left | a \right |} + 3 \, a^{5} c^{3}\right )}}{15 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{5} a^{2} {\left | a \right |} \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*sqrt(2)*a^3*c*arctan(1/2*sqrt(2)*((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*a + sqrt(c)*abs(a))/(a*sqrt(-c)
))/(sqrt(-c)*abs(a)*sgn(x)) - 2/15*(60*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^4*a^5*c - 45*(sqrt(a^2*c)*x -
sqrt(a^2*c*x^2 - a*c*x))^3*a^4*c^(3/2)*abs(a) + 35*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^2*a^5*c^2 - 15*(
sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*a^4*c^(5/2)*abs(a) + 3*a^5*c^3)/((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*
c*x))^5*a^2*abs(a)*sgn(x))

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maple [B]  time = 0.05, size = 278, normalized size = 2.46 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (-90 \sqrt {a \,x^{2}-x}\, a^{\frac {7}{2}} \sqrt {\frac {1}{a}}\, x^{4}+30 a^{\frac {7}{2}} \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, x^{4}+60 a^{\frac {5}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{2} \sqrt {\frac {1}{a}}+45 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, x^{4} a^{3}-30 a^{\frac {5}{2}} \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) x^{4}-45 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, x^{4} a^{3}-16 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x \sqrt {\frac {1}{a}}+6 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\, \sqrt {\frac {1}{a}}\right )}{15 x^{3} \sqrt {\left (a x -1\right ) x}\, \sqrt {a}\, \sqrt {\frac {1}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(c*(a*x-1)/a/x)^(1/2)/x^3*(-90*(a*x^2-x)^(1/2)*a^(7/2)*(1/a)^(1/2)*x^4+30*a^(7/2)*(1/a)^(1/2)*((a*x-1)*x)
^(1/2)*x^4+60*a^(5/2)*(a*x^2-x)^(3/2)*x^2*(1/a)^(1/2)+45*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(
1/a)^(1/2)*x^4*a^3-30*a^(5/2)*2^(1/2)*ln((2*2^(1/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*x^4-45*l
n(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*x^4*a^3-16*a^(3/2)*(a*x^2-x)^(3/2)*x*(1/a)^(1
/2)+6*(a*x^2-x)^(3/2)*a^(1/2)*(1/a)^(1/2))/((a*x-1)*x)^(1/2)/a^(1/2)/(1/a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.74, size = 96, normalized size = 0.85 \[ -4\,a^2\,\sqrt {c-\frac {c}{a\,x}}-\frac {2\,a^2\,{\left (c-\frac {c}{a\,x}\right )}^{3/2}}{3\,c}-\frac {2\,a^2\,{\left (c-\frac {c}{a\,x}\right )}^{5/2}}{5\,c^2}-\sqrt {2}\,a^2\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-\frac {c}{a\,x}}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,4{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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