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

Optimal. Leaf size=51 \[ \frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{a c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6127, 661, 208} \[ \frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{a c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

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

Rule 661

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 6127

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 44, normalized size = 0.86 \[ \frac {\sqrt {2-2 a x} \tanh ^{-1}\left (\frac {\sqrt {a x+1}}{\sqrt {2}}\right )}{a c \sqrt {c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 133, normalized size = 2.61 \[ \left [\frac {\sqrt {2} \log \left (-\frac {a^{2} x^{2} + 2 \, a x - \frac {2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{\sqrt {c}} - 3}{a^{2} x^{2} - 2 \, a x + 1}\right )}{2 \, a c^{\frac {3}{2}}}, \frac {\sqrt {2} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-\frac {1}{c}}}{a^{2} x^{2} - 1}\right )}{a c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 62, normalized size = 1.22 \[ -\frac {{\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c}} - \frac {\sqrt {2} \arctan \left (\frac {\sqrt {c}}{\sqrt {-c}}\right )}{a \sqrt {-c}}\right )} {\left | c \right |}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/(a*sqrt(-c)) - sqrt(2)*arctan(sqrt(c)/sqrt(-c))/(a*sqrt
(-c)))*abs(c)/c^2

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maple [A]  time = 0.04, size = 68, normalized size = 1.33 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{\left (-a x +1\right ) \sqrt {c \left (a x +1\right )}\, c^{\frac {3}{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-a\,c\,x\right )}^{3/2}\,\left (a\,x+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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