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

Optimal. Leaf size=126 \[ -\frac {5 \sqrt {a} x^{3/2} \left (c-\frac {c}{a x}\right )^{3/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{3/2}}+\frac {a x^2 \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{3/2}}{(1-a x)^{3/2}}-\frac {2 x \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{3/2}}{(1-a x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6134, 6129, 89, 80, 54, 215} \[ \frac {a x^2 \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{3/2}}{(1-a x)^{3/2}}-\frac {5 \sqrt {a} x^{3/2} \left (c-\frac {c}{a x}\right )^{3/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{3/2}}-\frac {2 x \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{3/2}}{(1-a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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

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Mathematica [A]  time = 0.05, size = 71, normalized size = 0.56 \[ -\frac {c \sqrt {c-\frac {c}{a x}} \left ((a x-2) \sqrt {a x+1}-5 \sqrt {a} \sqrt {x} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )\right )}{a \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 268, normalized size = 2.13 \[ \left [\frac {5 \, {\left (a c x - c\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x + 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, \sqrt {-a^{2} x^{2} + 1} {\left (a c x - 2 \, c\right )} \sqrt {\frac {a c x - c}{a x}}}{4 \, {\left (a^{2} x - a\right )}}, -\frac {5 \, {\left (a c x - c\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} {\left (a c x - 2 \, c\right )} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{2} x - a\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]sym2poly/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 = 109, normalized size = 0.87 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c \sqrt {-a^{2} x^{2}+1}\, \left (2 a^{\frac {3}{2}} x \sqrt {-\left (a x +1\right ) x}+5 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) x a -4 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}\right )}{2 a^{\frac {3}{2}} \left (a x -1\right ) \sqrt {-\left (a x +1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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