3.541 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-\frac {c}{a x})^{3/2}} \, dx\)

Optimal. Leaf size=159 \[ -\frac {(1-a x)^{3/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{a^{5/2} x^{3/2} \left (c-\frac {c}{a x}\right )^{3/2}}+\frac {\sqrt {2} (1-a x)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {a x+1}}\right )}{a^{5/2} x^{3/2} \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {\sqrt {a x+1} (1-a x)^{3/2}}{a^2 x \left (c-\frac {c}{a x}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6134, 6129, 102, 21, 105, 54, 215, 93, 206} \[ -\frac {(1-a x)^{3/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{a^{5/2} x^{3/2} \left (c-\frac {c}{a x}\right )^{3/2}}+\frac {\sqrt {2} (1-a x)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {a x+1}}\right )}{a^{5/2} x^{3/2} \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {\sqrt {a x+1} (1-a x)^{3/2}}{a^2 x \left (c-\frac {c}{a x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - c/(a*x))^(3/2)),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 54

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 6129

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

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

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Mathematica [A]  time = 0.07, size = 105, normalized size = 0.66 \[ \frac {\sqrt {1-a x} \left (\sqrt {a} \sqrt {x} \sqrt {a x+1}+\sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )-\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {a x+1}}\right )\right )}{a^{3/2} c \sqrt {x} \sqrt {c-\frac {c}{a x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - c/(a*x))^(3/2)),x]

[Out]

(Sqrt[1 - a*x]*(Sqrt[a]*Sqrt[x]*Sqrt[1 + a*x] + ArcSinh[Sqrt[a]*Sqrt[x]] - Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sq
rt[x])/Sqrt[1 + a*x]]))/(a^(3/2)*c*Sqrt[c - c/(a*x)]*Sqrt[x])

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fricas [A]  time = 0.56, size = 453, normalized size = 2.85 \[ \left [\frac {4 \, \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}} + \sqrt {2} {\left (a c x - c\right )} \sqrt {-\frac {1}{c}} \log \left (-\frac {17 \, a^{3} x^{3} - 3 \, a^{2} x^{2} + 4 \, \sqrt {2} {\left (3 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {1}{c}} \sqrt {\frac {a c x - c}{a x}} - 13 \, a x - 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) - {\left (a x - 1\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x + 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right )}{4 \, {\left (a^{2} c^{2} x - a c^{2}\right )}}, \frac {2 \, \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}} + {\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - \frac {\sqrt {2} {\left (a c x - c\right )} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}}}{{\left (3 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt {c}}\right )}{\sqrt {c}}}{2 \, {\left (a^{2} c^{2} x - a c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*sqrt(-a^2*x^2 + 1)*a*x*sqrt((a*c*x - c)/(a*x)) + sqrt(2)*(a*c*x - c)*sqrt(-1/c)*log(-(17*a^3*x^3 - 3*a
^2*x^2 + 4*sqrt(2)*(3*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-1/c)*sqrt((a*c*x - c)/(a*x)) - 13*a*x - 1)/(a^3*
x^3 - 3*a^2*x^2 + 3*a*x - 1)) - (a*x - 1)*sqrt(-c)*log(-(8*a^3*c*x^3 - 7*a*c*x + 4*(2*a^2*x^2 + a*x)*sqrt(-a^2
*x^2 + 1)*sqrt(-c)*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)))/(a^2*c^2*x - a*c^2), 1/2*(2*sqrt(-a^2*x^2 + 1)*a*x
*sqrt((a*c*x - c)/(a*x)) + (a*x - 1)*sqrt(c)*arctan(2*sqrt(-a^2*x^2 + 1)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x))/(
2*a^2*c*x^2 - a*c*x - c)) - sqrt(2)*(a*c*x - c)*arctan(2*sqrt(2)*sqrt(-a^2*x^2 + 1)*a*x*sqrt((a*c*x - c)/(a*x)
)/((3*a^2*x^2 - 2*a*x - 1)*sqrt(c)))/sqrt(c))/(a^2*c^2*x - a*c^2)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.06, size = 168, normalized size = 1.06 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \sqrt {-a^{2} x^{2}+1}\, \left (2 \sqrt {-\left (a x +1\right ) x}\, a^{\frac {3}{2}} \sqrt {2}\, \sqrt {-\frac {1}{a}}-\arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) a \sqrt {2}\, \sqrt {-\frac {1}{a}}+2 \ln \left (\frac {2 \sqrt {2}\, \sqrt {-\frac {1}{a}}\, \sqrt {-\left (a x +1\right ) x}\, a -3 a x -1}{a x -1}\right ) \sqrt {a}\right ) \sqrt {2}}{4 a^{\frac {3}{2}} c^{2} \left (a x -1\right ) \sqrt {-\left (a x +1\right ) x}\, \sqrt {-\frac {1}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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