3.7.4 \(\int \frac {e^{-2 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx\) [604]

Optimal. Leaf size=82 \[ 4 a \sqrt {c-\frac {c}{a x}}+\frac {2 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}-4 \sqrt {2} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6268, 25, 528, 455, 52, 65, 214} \begin {gather*} \frac {2 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}+4 a \sqrt {c-\frac {c}{a x}}-4 \sqrt {2} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4*a*Sqrt[c - c/(a*x)] + (2*a*(c - c/(a*x))^(3/2))/(3*c) - 4*Sqrt[2]*a*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 52

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 65

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 214

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

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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^2} \, dx &=\int \frac {\sqrt {c-\frac {c}{a x}} (1-a x)}{x^2 (1+a x)} \, dx\\ &=-\frac {a \int \frac {\left (c-\frac {c}{a x}\right )^{3/2}}{x (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^2} \, dx}{c}\\ &=\frac {a \text {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 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}+(2 a) \text {Subst}\left (\int \frac {\sqrt {c-\frac {c x}{a}}}{a+x} \, dx,x,\frac {1}{x}\right )\\ &=4 a \sqrt {c-\frac {c}{a x}}+\frac {2 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}+(4 a c) \text {Subst}\left (\int \frac {1}{(a+x) \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )\\ &=4 a \sqrt {c-\frac {c}{a x}}+\frac {2 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}-\left (8 a^2\right ) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )\\ &=4 a \sqrt {c-\frac {c}{a x}}+\frac {2 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}-4 \sqrt {2} a \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 = 69, normalized size = 0.84 \begin {gather*} \frac {2 \sqrt {c-\frac {c}{a x}} (-1+7 a x)}{3 x}-4 \sqrt {2} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c - c/(a*x)]*(-1 + 7*a*x))/(3*x) - 4*Sqrt[2]*a*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*x)]/(Sqrt[2]*Sqrt[c])]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(253\) vs. \(2(67)=134\).
time = 0.42, size = 254, normalized size = 3.10

method	result	size
risch	\(\frac {2 \left (7 a^{2} x^{2}-8 a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{3 x \left (a x -1\right )}+\frac {2 a \sqrt {2}\, \ln \left (\frac {4 c -3 \left (x +\frac {1}{a}\right ) a c +2 \sqrt {2}\, \sqrt {c}\, \sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-3 \left (x +\frac {1}{a}\right ) a c +2 c}}{x +\frac {1}{a}}\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {a c x \left (a x -1\right )}}{\sqrt {c}\, \left (a x -1\right )}\)	\(142\)
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (6 a^{\frac {5}{2}} \sqrt {\left (a x -1\right ) x}\, \sqrt {\frac {1}{a}}\, x^{3}-18 a^{\frac {5}{2}} \sqrt {a \,x^{2}-x}\, \sqrt {\frac {1}{a}}\, x^{3}-6 a^{\frac {3}{2}} \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) x^{3}+12 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x \sqrt {\frac {1}{a}}+9 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, a^{2} x^{3}-9 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, a^{2} x^{3}-2 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\, \sqrt {\frac {1}{a}}\right )}{3 x^{2} \sqrt {\left (a x -1\right ) x}\, \sqrt {a}\, \sqrt {\frac {1}{a}}}\)	\(254\)

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^2,x,method=_RETURNVERBOSE)

[Out]

-1/3*(c*(a*x-1)/a/x)^(1/2)*(6*a^(5/2)*((a*x-1)*x)^(1/2)*(1/a)^(1/2)*x^3-18*a^(5/2)*(a*x^2-x)^(1/2)*(1/a)^(1/2)
*x^3-6*a^(3/2)*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^3+12*a^(3/2)*(a*x^2-x
)^(3/2)*x*(1/a)^(1/2)+9*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*a^2*x^3-9*ln(1/2*(2*((
a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*a^2*x^3-2*(a*x^2-x)^(3/2)*a^(1/2)*(1/a)^(1/2))/x^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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 157, normalized size = 1.91 \begin {gather*} \left [\frac {2 \, {\left (3 \, \sqrt {2} a \sqrt {c} x \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 (7 \, a x - 1\right )} \sqrt {\frac {a c x - c}{a x}}\right )}}{3 \, x}, \frac {2 \, {\left (6 \, \sqrt {2} a \sqrt {-c} x \arctan \left (\frac {\sqrt {2} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{2 \, c}\right ) + {\left (7 \, a x - 1\right )} \sqrt {\frac {a c x - c}{a x}}\right )}}{3 \, x}\right ] \end {gather*}

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^2,x, algorithm="fricas")

[Out]

[2/3*(3*sqrt(2)*a*sqrt(c)*x*log((2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) - 3*a*c*x + c)/(a*x + 1)) + (7*
a*x - 1)*sqrt((a*c*x - c)/(a*x)))/x, 2/3*(6*sqrt(2)*a*sqrt(-c)*x*arctan(1/2*sqrt(2)*sqrt(-c)*sqrt((a*c*x - c)/
(a*x))/c) + (7*a*x - 1)*sqrt((a*c*x - c)/(a*x)))/x]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \left (- \frac {\sqrt {c - \frac {c}{a x}}}{a x^{3} + x^{2}}\right )\, dx - \int \frac {a x \sqrt {c - \frac {c}{a x}}}{a x^{3} + x^{2}}\, dx \end {gather*}

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**2,x)

[Out]

-Integral(-sqrt(c - c/(a*x))/(a*x**3 + x**2), x) - Integral(a*x*sqrt(c - c/(a*x))/(a*x**3 + x**2), 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((c-c/a/x)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^2,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 = 1.42, size = 67, normalized size = 0.82 \begin {gather*} 4\,a\,\sqrt {c-\frac {c}{a\,x}}+\frac {2\,a\,{\left (c-\frac {c}{a\,x}\right )}^{3/2}}{3\,c}-4\,\sqrt {2}\,a\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-\frac {c}{a\,x}}}{2\,\sqrt {c}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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