3.6.13 \(\int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\) [513]

Optimal. Leaf size=170 \[ \frac {2 a^2 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{5 \left (c-\frac {c}{a x}\right )^{5/2}}+\frac {2 a^2 c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{3 \left (c-\frac {c}{a x}\right )^{3/2}}+\frac {4 a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}-4 \sqrt {2} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {2} \sqrt {c-\frac {c}{a x}}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 170, 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 = {6313, 809, 679, 675, 214} \begin {gather*} \frac {2 a^2 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{5 \left (c-\frac {c}{a x}\right )^{5/2}}+\frac {2 a^2 c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{3 \left (c-\frac {c}{a x}\right )^{3/2}}+\frac {4 a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}-4 \sqrt {2} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {2} \sqrt {c-\frac {c}{a x}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*c^3*(1 - 1/(a^2*x^2))^(5/2))/(5*(c - c/(a*x))^(5/2)) + (2*a^2*c^2*(1 - 1/(a^2*x^2))^(3/2))/(3*(c - c/(a
*x))^(3/2)) + (4*a^2*c*Sqrt[1 - 1/(a^2*x^2)])/Sqrt[c - c/(a*x)] - 4*Sqrt[2]*a^2*Sqrt[c]*ArcTanh[(Sqrt[c]*Sqrt[
1 - 1/(a^2*x^2)])/(Sqrt[2]*Sqrt[c - c/(a*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 675

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(2*c*d + e^2*x^2
), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0]

Rule 679

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

Rule 809

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

Rule 6313

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[-c^n, Subst[Int[(c +
d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps

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

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Mathematica [A]
time = 0.18, size = 162, normalized size = 0.95 \begin {gather*} \frac {2 a \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}} \left (3+11 a x+38 a^2 x^2\right )}{15 x (-1+a x)}+2 \sqrt {2} a^2 \sqrt {c} \log \left ((-1+a x)^2\right )-2 \sqrt {2} a^2 \sqrt {c} \log \left (2 \sqrt {2} a^2 \sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}} x^2+c \left (-1-2 a x+3 a^2 x^2\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*(3 + 11*a*x + 38*a^2*x^2))/(15*x*(-1 + a*x)) + 2*Sqrt[2]*a^2*Sqrt
[c]*Log[(-1 + a*x)^2] - 2*Sqrt[2]*a^2*Sqrt[c]*Log[2*Sqrt[2]*a^2*Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)
]*x^2 + c*(-1 - 2*a*x + 3*a^2*x^2)]

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Maple [A]
time = 0.14, size = 165, normalized size = 0.97

method	result	size
default	\(\frac {2 \left (a x -1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (-15 a^{2} \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {x \left (a x +1\right )}\, a +3 a x +1}{a x -1}\right ) x^{3}+38 x^{2} \sqrt {x \left (a x +1\right )}\, a^{2} \sqrt {\frac {1}{a}}+11 x \sqrt {x \left (a x +1\right )}\, a \sqrt {\frac {1}{a}}+3 \sqrt {x \left (a x +1\right )}\, \sqrt {\frac {1}{a}}\right )}{15 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) x^{2} \sqrt {x \left (a x +1\right )}\, \sqrt {\frac {1}{a}}}\)	\(165\)
risch	\(\frac {2 \left (38 a^{3} x^{3}+49 a^{2} x^{2}+14 a x +3\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{15 x^{2} \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}-\frac {2 a^{2} \sqrt {2}\, \ln \left (\frac {4 c +3 \left (x -\frac {1}{a}\right ) a c +2 \sqrt {2}\, \sqrt {c}\, \sqrt {a^{2} c \left (x -\frac {1}{a}\right )^{2}+3 \left (x -\frac {1}{a}\right ) a c +2 c}}{x -\frac {1}{a}}\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {c a x \left (a x +1\right )}}{\sqrt {c}\, \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\)	\(190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/((a*x-1)/(a*x+1))^(3/2)*(a*x-1)/(a*x+1)*(c*(a*x-1)/a/x)^(1/2)*(-15*a^2*2^(1/2)*ln((2*2^(1/2)*(1/a)^(1/2)*
(x*(a*x+1))^(1/2)*a+3*a*x+1)/(a*x-1))*x^3+38*x^2*(x*(a*x+1))^(1/2)*a^2*(1/a)^(1/2)+11*x*(x*(a*x+1))^(1/2)*a*(1
/a)^(1/2)+3*(x*(a*x+1))^(1/2)*(1/a)^(1/2))/x^2/(x*(a*x+1))^(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(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a/x)^(1/2)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 381, normalized size = 2.24 \begin {gather*} \left [\frac {15 \, \sqrt {2} {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \sqrt {c} \log \left (-\frac {17 \, a^{3} c x^{3} - 3 \, a^{2} c x^{2} - 13 \, a c x - 4 \, \sqrt {2} {\left (3 \, a^{3} x^{3} + 4 \, a^{2} x^{2} + a x\right )} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}} - c}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) + 2 \, {\left (38 \, a^{3} x^{3} + 49 \, a^{2} x^{2} + 14 \, a x + 3\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{15 \, {\left (a x^{3} - x^{2}\right )}}, \frac {2 \, {\left (15 \, \sqrt {2} {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} {\left (a^{2} x^{2} + a x\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{3 \, a^{2} c x^{2} - 2 \, a c x - c}\right ) + {\left (38 \, a^{3} x^{3} + 49 \, a^{2} x^{2} + 14 \, a x + 3\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}\right )}}{15 \, {\left (a x^{3} - x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*sqrt(2)*(a^3*x^3 - a^2*x^2)*sqrt(c)*log(-(17*a^3*c*x^3 - 3*a^2*c*x^2 - 13*a*c*x - 4*sqrt(2)*(3*a^3*x
^3 + 4*a^2*x^2 + a*x)*sqrt(c)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)) - c)/(a^3*x^3 - 3*a^2*x^2 + 3*
a*x - 1)) + 2*(38*a^3*x^3 + 49*a^2*x^2 + 14*a*x + 3)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)))/(a*x^3
- x^2), 2/15*(15*sqrt(2)*(a^3*x^3 - a^2*x^2)*sqrt(-c)*arctan(2*sqrt(2)*(a^2*x^2 + a*x)*sqrt(-c)*sqrt((a*x - 1
)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(3*a^2*c*x^2 - 2*a*c*x - c)) + (38*a^3*x^3 + 49*a^2*x^2 + 14*a*x + 3)*sqr
t((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)))/(a*x^3 - x^2)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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(1/((a*x-1)/(a*x+1))^(3/2)*(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,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 [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c-\frac {c}{a\,x}}}{x^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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