3.5.85 \(\int \frac {e^{-3 \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a x}}} \, dx\) [485]

Optimal. Leaf size=118 \[ \frac {5 \sqrt {c-\frac {c}{a x}}}{a c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {c-\frac {c}{a x}} x}{c \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a \sqrt {c}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6312, 893, 883, 889, 214} \begin {gather*} \frac {x \sqrt {c-\frac {c}{a x}}}{c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {5 \sqrt {c-\frac {c}{a x}}}{a c \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*Sqrt[c - c/(a*x)]),x]

[Out]

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

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 883

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

Rule 889

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

Rule 893

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 6312

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.03, size = 69, normalized size = 0.58 \begin {gather*} \frac {\sqrt {1-\frac {1}{a x}} \left (a x+5 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};1+\frac {1}{a x}\right )\right )}{a \sqrt {1+\frac {1}{a x}} \sqrt {c-\frac {c}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*Sqrt[c - c/(a*x)]),x]

[Out]

(Sqrt[1 - 1/(a*x)]*(a*x + 5*Hypergeometric2F1[-1/2, 1, 1/2, 1 + 1/(a*x)]))/(a*Sqrt[1 + 1/(a*x)]*Sqrt[c - c/(a*
x)])

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Maple [A]
time = 0.10, size = 149, normalized size = 1.26

method	result	size
default	\(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (2 a^{\frac {3}{2}} x \sqrt {x \left (a x +1\right )}-5 \ln \left (\frac {2 \sqrt {x \left (a x +1\right )}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right ) a x +10 \sqrt {x \left (a x +1\right )}\, \sqrt {a}-5 \ln \left (\frac {2 \sqrt {x \left (a x +1\right )}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right )\right )}{2 \left (a x -1\right )^{2} \sqrt {a}\, c \sqrt {x \left (a x +1\right )}}\)	\(149\)
risch	\(\frac {\left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}{a \sqrt {\frac {c \left (a x -1\right )}{a x}}}+\frac {\left (-\frac {5 \ln \left (\frac {\frac {1}{2} a c +c \,a^{2} x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}+a c x}\right )}{2 a \sqrt {a^{2} c}}+\frac {4 \sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-\left (x +\frac {1}{a}\right ) a c}}{a^{3} c \left (x +\frac {1}{a}\right )}\right ) \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {c a x \left (a x +1\right )}}{\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x}\)	\(174\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x-1)/(a*x+1))^(3/2)/(a*x-1)^2*(a*x+1)*(c*(a*x-1)/a/x)^(1/2)*x*(2*a^(3/2)*x*(x*(a*x+1))^(1/2)-5*ln(1/2*
(2*(x*(a*x+1))^(1/2)*a^(1/2)+2*a*x+1)/a^(1/2))*a*x+10*(x*(a*x+1))^(1/2)*a^(1/2)-5*ln(1/2*(2*(x*(a*x+1))^(1/2)*
a^(1/2)+2*a*x+1)/a^(1/2)))/a^(1/2)/c/(x*(a*x+1))^(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(((a*x-1)/(a*x+1))^(3/2)/(c-c/a/x)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 303, normalized size = 2.57 \begin {gather*} \left [\frac {5 \, {\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{3} x^{3} + 3 \, 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 x - 1}\right ) + 4 \, {\left (a^{2} x^{2} + 5 \, a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{4 \, {\left (a^{2} c x - a c\right )}}, \frac {5 \, {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {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}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, {\left (a^{2} x^{2} + 5 \, a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{2} c x - a c\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(5*(a*x - 1)*sqrt(c)*log(-(8*a^3*c*x^3 - 7*a*c*x - 4*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt(c)*sqrt((a*x - 1)
/(a*x + 1))*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) + 4*(a^2*x^2 + 5*a*x)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*
x - c)/(a*x)))/(a^2*c*x - a*c), 1/2*(5*(a*x - 1)*sqrt(-c)*arctan(2*(a^2*x^2 + a*x)*sqrt(-c)*sqrt((a*x - 1)/(a*
x + 1))*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) + 2*(a^2*x^2 + 5*a*x)*sqrt((a*x - 1)/(a*x + 1))*sqr
t((a*c*x - c)/(a*x)))/(a^2*c*x - a*c)]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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