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

Optimal. Leaf size=128 \[ -\frac {2 \sqrt {1-a^2 x^2} \sqrt {c-\frac {c}{a x}}}{5 x^2 (1-a x)}-\frac {12 a^2 \sqrt {a x+1} \sqrt {c-\frac {c}{a x}}}{5 \sqrt {1-a x}}+\frac {6 a \sqrt {a x+1} \sqrt {c-\frac {c}{a x}}}{5 x \sqrt {1-a x}} \]

[Out]

-12/5*a^2*(c-c/a/x)^(1/2)*(a*x+1)^(1/2)/(-a*x+1)^(1/2)+6/5*a*(c-c/a/x)^(1/2)*(a*x+1)^(1/2)/x/(-a*x+1)^(1/2)-2/
5*(c-c/a/x)^(1/2)*(-a^2*x^2+1)^(1/2)/x^2/(-a*x+1)

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Rubi [A]  time = 0.27, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6134, 6128, 879, 848, 45, 37} \[ -\frac {2 \sqrt {1-a^2 x^2} \sqrt {c-\frac {c}{a x}}}{5 x^2 (1-a x)}-\frac {12 a^2 \sqrt {a x+1} \sqrt {c-\frac {c}{a x}}}{5 \sqrt {1-a x}}+\frac {6 a \sqrt {a x+1} \sqrt {c-\frac {c}{a x}}}{5 x \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*a^2*Sqrt[c - c/(a*x)]*Sqrt[1 + a*x])/(5*Sqrt[1 - a*x]) + (6*a*Sqrt[c - c/(a*x)]*Sqrt[1 + a*x])/(5*x*Sqrt[
1 - a*x]) - (2*Sqrt[c - c/(a*x)]*Sqrt[1 - a^2*x^2])/(5*x^2*(1 - a*x))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

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

Rule 848

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

Rule 879

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(e*f
- d*g)*(d + e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + 1)*(e*f + d*g)), x] - Dist[(e*(e*f*
(p + 1) - d*g*(2*n + p + 3)))/(g*(n + 1)*(e*f + d*g)), Int[(d + e*x)^(m - 1)*(f + g*x)^(n + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&
EqQ[m + p - 1, 0] && LtQ[n, -1] && IntegerQ[2*p]

Rule 6128

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

Rule 6134

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

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx &=\frac {\left (\sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {e^{-\tanh ^{-1}(a x)} \sqrt {1-a x}}{x^{7/2}} \, dx}{\sqrt {1-a x}}\\ &=\frac {\left (\sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {(1-a x)^{3/2}}{x^{7/2} \sqrt {1-a^2 x^2}} \, dx}{\sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} \sqrt {1-a^2 x^2}}{5 x^2 (1-a x)}-\frac {\left (9 a \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1-a x}}{x^{5/2} \sqrt {1-a^2 x^2}} \, dx}{5 \sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} \sqrt {1-a^2 x^2}}{5 x^2 (1-a x)}-\frac {\left (9 a \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {1}{x^{5/2} \sqrt {1+a x}} \, dx}{5 \sqrt {1-a x}}\\ &=\frac {6 a \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{5 x \sqrt {1-a x}}-\frac {2 \sqrt {c-\frac {c}{a x}} \sqrt {1-a^2 x^2}}{5 x^2 (1-a x)}+\frac {\left (6 a^2 \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {1}{x^{3/2} \sqrt {1+a x}} \, dx}{5 \sqrt {1-a x}}\\ &=-\frac {12 a^2 \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{5 \sqrt {1-a x}}+\frac {6 a \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{5 x \sqrt {1-a x}}-\frac {2 \sqrt {c-\frac {c}{a x}} \sqrt {1-a^2 x^2}}{5 x^2 (1-a x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 55, normalized size = 0.43 \[ -\frac {2 \sqrt {a x+1} \left (6 a^2 x^2-3 a x+1\right ) \sqrt {c-\frac {c}{a x}}}{5 x^2 \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 58, normalized size = 0.45 \[ \frac {2 \, {\left (6 \, a^{2} x^{2} - 3 \, a x + 1\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{5 \, {\left (a x^{3} - x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 54, normalized size = 0.42 \[ \frac {2 \left (6 a^{2} x^{2}-3 a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}}{5 x^{2} \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.05, size = 79, normalized size = 0.62 \[ \frac {\sqrt {c-\frac {c}{a\,x}}\,\left (\frac {2\,\sqrt {1-a^2\,x^2}}{5\,a}-\frac {6\,x\,\sqrt {1-a^2\,x^2}}{5}+\frac {12\,a\,x^2\,\sqrt {1-a^2\,x^2}}{5}\right )}{x^3-\frac {x^2}{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c - c/(a*x))^(1/2)*((2*(1 - a^2*x^2)^(1/2))/(5*a) - (6*x*(1 - a^2*x^2)^(1/2))/5 + (12*a*x^2*(1 - a^2*x^2)^(1
/2))/5))/(x^3 - x^2/a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-c*(-1 + 1/(a*x)))*sqrt(-(a*x - 1)*(a*x + 1))/(x**3*(a*x + 1)), x)

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