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

Optimal. Leaf size=131 \[ -\frac {3 (1-a x)^{3/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{a^{5/2} x^{3/2} \left (c-\frac {c}{a x}\right )^{3/2}}+\frac {3 \sqrt {a x+1} (1-a x)^{3/2}}{a^2 x \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {2 (1-a x)^{3/2}}{a \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{3/2}} \]

[Out]

-3*(-a*x+1)^(3/2)*arcsinh(a^(1/2)*x^(1/2))/a^(5/2)/(c-c/a/x)^(3/2)/x^(3/2)-2*(-a*x+1)^(3/2)/a/(c-c/a/x)^(3/2)/
(a*x+1)^(1/2)+3*(-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 = 0.19, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {6134, 6128, 848, 47, 50, 54, 215} \[ -\frac {3 (1-a x)^{3/2} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{a^{5/2} x^{3/2} \left (c-\frac {c}{a x}\right )^{3/2}}+\frac {3 \sqrt {a x+1} (1-a x)^{3/2}}{a^2 x \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {2 (1-a x)^{3/2}}{a \sqrt {a x+1} \left (c-\frac {c}{a x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 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 848

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

Rule 6128

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

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

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Mathematica [C]  time = 0.04, size = 44, normalized size = 0.34 \[ \frac {2 x (1-a x)^{3/2} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};-a x\right )}{5 \left (c-\frac {c}{a x}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x*(1 - a*x)^(3/2)*Hypergeometric2F1[3/2, 5/2, 7/2, -(a*x)])/(5*(c - c/(a*x))^(3/2))

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fricas [A]  time = 0.45, size = 294, normalized size = 2.24 \[ \left [-\frac {3 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x + 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (a^{2} x^{2} + 3 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{4 \, {\left (a^{3} c^{2} x^{2} - a c^{2}\right )}}, \frac {3 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 2 \, {\left (a^{2} x^{2} + 3 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{3} c^{2} x^{2} - a c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(3*(a^2*x^2 - 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)) + 4*(a^2*x^2 + 3*a*x)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^3
*c^2*x^2 - a*c^2), 1/2*(3*(a^2*x^2 - 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)) - 2*(a^2*x^2 + 3*a*x)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^3*c^2*x^2 -
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)^3*(-a^2*x^2+1)^(3/2)/(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 = 0.05, size = 143, normalized size = 1.09 \[ -\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (2 a^{\frac {3}{2}} x \sqrt {-\left (a x +1\right ) x}+3 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) x a +6 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}+3 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right )\right ) \sqrt {-a^{2} x^{2}+1}}{2 \sqrt {a}\, c^{2} \left (a x +1\right ) \sqrt {-\left (a x +1\right ) x}\, \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*(a*x-1)/a/x)^(1/2)*x/a^(1/2)/c^2*(2*a^(3/2)*x*(-(a*x+1)*x)^(1/2)+3*arctan(1/2/a^(1/2)*(2*a*x+1)/(-(a*x
+1)*x)^(1/2))*x*a+6*a^(1/2)*(-(a*x+1)*x)^(1/2)+3*arctan(1/2/a^(1/2)*(2*a*x+1)/(-(a*x+1)*x)^(1/2)))*(-a^2*x^2+1
)^(1/2)/(a*x+1)/(-(a*x+1)*x)^(1/2)/(a*x-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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