3.118 \(\int \frac {(1-a^2 x^2)^{3/2}}{x^2 (1-a x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {850, 813, 844, 216, 266, 63, 208} \[ -\frac {\sqrt {1-a^2 x^2} (1-a x)}{x}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-a \sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 216

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

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

Rule 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 49, normalized size = 0.96 \[ \frac {\sqrt {1-a^2 x^2} (a x-1)}{x}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-a \sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 74, normalized size = 1.45 \[ \frac {2 \, a x \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + a x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + a x + \sqrt {-a^{2} x^{2} + 1} {\left (a x - 1\right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.19, size = 125, normalized size = 2.45 \[ \frac {a^{4} x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} - \frac {a^{2} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} - \frac {a^{2} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} + \sqrt {-a^{2} x^{2} + 1} a - \frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{2 \, x {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^4*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*abs(a)) - a^2*arcsin(a*x)*sgn(a)/abs(a) - a^2*log(1/2*abs(-2*sqrt(-
a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + sqrt(-a^2*x^2 + 1)*a - 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(
x*abs(a))

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maple [B]  time = 0.02, size = 238, normalized size = 4.67 \[ \frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}\, a^{2} x}{2}-\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} a^{2} x -\frac {3 \sqrt {-a^{2} x^{2}+1}\, a^{2} x}{2}+\frac {a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}\right )}{2 \sqrt {a^{2}}}-\frac {3 a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}-a \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a \right )^{\frac {3}{2}} a}{3}+\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} a}{3}+\sqrt {-a^{2} x^{2}+1}\, a -\frac {\left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*(-(x-1/a)^2*a^2-2*(x-1/a)*a)^(3/2)+1/2*a^2*(-(x-1/a)^2*a^2-2*(x-1/a)*a)^(1/2)*x+1/2*a^2/(a^2)^(1/2)*arc
tan((a^2)^(1/2)*x/(-(x-1/a)^2*a^2-2*(x-1/a)*a)^(1/2))-1/x*(-a^2*x^2+1)^(5/2)-a^2*x*(-a^2*x^2+1)^(3/2)-3/2*a^2*
x*(-a^2*x^2+1)^(1/2)-3/2*a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)/(-a^2*x^2+1)^(1/2)*x)+1/3*a*(-a^2*x^2+1)^(3/2)+a*(
-a^2*x^2+1)^(1/2)-a*arctanh(1/(-a^2*x^2+1)^(1/2))

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maxima [A]  time = 0.99, size = 68, normalized size = 1.33 \[ -a \arcsin \left (a x\right ) - a \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \sqrt {-a^{2} x^{2} + 1} a - \frac {\sqrt {-a^{2} x^{2} + 1}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*arcsin(a*x) - a*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) + sqrt(-a^2*x^2 + 1)*a - sqrt(-a^2*x^2 + 1)/x

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mupad [B]  time = 0.05, size = 74, normalized size = 1.45 \[ a\,\sqrt {1-a^2\,x^2}-\frac {\sqrt {1-a^2\,x^2}}{x}-\frac {a^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+a\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*atan((1 - a^2*x^2)^(1/2)*1i)*1i + a*(1 - a^2*x^2)^(1/2) - (1 - a^2*x^2)^(1/2)/x - (a^2*asinh(x*(-a^2)^(1/2))
)/(-a^2)^(1/2)

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sympy [C]  time = 6.53, size = 170, normalized size = 3.33 \[ a \left (\begin {cases} i \sqrt {a^{2} x^{2} - 1} - \log {\left (a x \right )} + \frac {\log {\left (a^{2} x^{2} \right )}}{2} + i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\sqrt {- a^{2} x^{2} + 1} + \frac {\log {\left (a^{2} x^{2} \right )}}{2} - \log {\left (\sqrt {- a^{2} x^{2} + 1} + 1 \right )} & \text {otherwise} \end {cases}\right ) + \begin {cases} - \frac {i a^{2} x}{\sqrt {a^{2} x^{2} - 1}} + i a \operatorname {acosh}{\left (a x \right )} + \frac {i}{x \sqrt {a^{2} x^{2} - 1}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {a^{2} x}{\sqrt {- a^{2} x^{2} + 1}} - a \operatorname {asin}{\left (a x \right )} - \frac {1}{x \sqrt {- a^{2} x^{2} + 1}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((I*sqrt(a**2*x**2 - 1) - log(a*x) + log(a**2*x**2)/2 + I*asin(1/(a*x)), Abs(a**2*x**2) > 1), (sqrt
(-a**2*x**2 + 1) + log(a**2*x**2)/2 - log(sqrt(-a**2*x**2 + 1) + 1), True)) + Piecewise((-I*a**2*x/sqrt(a**2*x
**2 - 1) + I*a*acosh(a*x) + I/(x*sqrt(a**2*x**2 - 1)), Abs(a**2*x**2) > 1), (a**2*x/sqrt(-a**2*x**2 + 1) - a*a
sin(a*x) - 1/(x*sqrt(-a**2*x**2 + 1)), True))

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