3.6.70 \(\int \frac {e^{\tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\) [570]

Optimal. Leaf size=84 \[ -\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{5 x^2 \sqrt {1-a x}}+\frac {4 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{15 x \sqrt {1-a x}} \]

[Out]

-2/5*(a*x+1)^(3/2)*(c-c/a/x)^(1/2)/x^2/(-a*x+1)^(1/2)+4/15*a*(a*x+1)^(3/2)*(c-c/a/x)^(1/2)/x/(-a*x+1)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6269, 6263, 862, 47, 37} \begin {gather*} \frac {4 a (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{15 x \sqrt {1-a x}}-\frac {2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{5 x^2 \sqrt {1-a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2))/(5*x^2*Sqrt[1 - a*x]) + (4*a*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2))/(15*x*S
qrt[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 47

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 862

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/e)*x)^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 6263

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 6269

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 {\sqrt {1-a^2 x^2}}{x^{7/2} \sqrt {1-a x}} \, dx}{\sqrt {1-a x}}\\ &=\frac {\left (\sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1+a x}}{x^{7/2}} \, dx}{\sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{5 x^2 \sqrt {1-a x}}-\frac {\left (2 a \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1+a x}}{x^{5/2}} \, dx}{5 \sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{5 x^2 \sqrt {1-a x}}+\frac {4 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{15 x \sqrt {1-a x}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 47, normalized size = 0.56 \begin {gather*} \frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2} (-3+2 a x)}{15 x^2 \sqrt {1-a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2)*(-3 + 2*a*x))/(15*x^2*Sqrt[1 - a*x])

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Maple [A]
time = 0.38, size = 51, normalized size = 0.61

method	result	size
gosper	\(\frac {2 \left (a x +1\right )^{2} \left (2 a x -3\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{15 x^{2} \sqrt {-a^{2} x^{2}+1}}\)	\(46\)
default	\(-\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) \left (2 a x -3\right )}{15 x^{2} \left (a x -1\right )}\)	\(51\)
risch	\(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {a c x \left (-a^{2} x^{2}+1\right )}{a x -1}}\, \left (2 a^{3} x^{3}+a^{2} x^{2}-4 a x -3\right )}{15 \sqrt {-a^{2} x^{2}+1}\, x^{2} \sqrt {-a c x \left (a x +1\right )}}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(c*(a*x-1)/a/x)^(1/2)/x^2*(-a^2*x^2+1)^(1/2)*(a*x+1)*(2*a*x-3)/(a*x-1)

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

[Out]

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

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Fricas [A]
time = 0.37, size = 58, normalized size = 0.69 \begin {gather*} -\frac {2 \, {\left (2 \, a^{2} x^{2} - a x - 3\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{15 \, {\left (a x^{3} - x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(2*a^2*x^2 - a*x - 3)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x))/(a*x^3 - x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.05, size = 52, normalized size = 0.62 \begin {gather*} -\frac {\sqrt {c-\frac {c}{a\,x}}\,\left (-\frac {4\,a^3\,x^3}{15}-\frac {2\,a^2\,x^2}{15}+\frac {8\,a\,x}{15}+\frac {2}{5}\right )}{x^2\,\sqrt {1-a^2\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c - c/(a*x))^(1/2)*((8*a*x)/15 - (2*a^2*x^2)/15 - (4*a^3*x^3)/15 + 2/5))/(x^2*(1 - a^2*x^2)^(1/2))

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