3.2.51 \(\int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx\) [151]

Optimal. Leaf size=55 \[ -\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\sqrt {1-a^2 x^2}}{a^3 (1+a x)}-\frac {\sin ^{-1}(a x)}{a^3} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1653, 12, 807, 222} \begin {gather*} -\frac {\text {ArcSin}(a x)}{a^3}-\frac {\sqrt {1-a^2 x^2}}{a^3 (a x+1)}-\frac {\sqrt {1-a^2 x^2}}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - a^2*x^2]/a^3) - Sqrt[1 - a^2*x^2]/(a^3*(1 + a*x)) - ArcSin[a*x]/a^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 222

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 807

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

Rule 1653

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx &=-\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\int \frac {a^3 x}{(1+a x) \sqrt {1-a^2 x^2}} \, dx}{a^4}\\ &=-\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\int \frac {x}{(1+a x) \sqrt {1-a^2 x^2}} \, dx}{a}\\ &=-\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\sqrt {1-a^2 x^2}}{a^3 (1+a x)}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\sqrt {1-a^2 x^2}}{a^3 (1+a x)}-\frac {\sin ^{-1}(a x)}{a^3}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 71, normalized size = 1.29 \begin {gather*} \frac {(-2-a x) \sqrt {1-a^2 x^2}}{a^3 (1+a x)}-\frac {\log \left (-\sqrt {-a^2} x+\sqrt {1-a^2 x^2}\right )}{\left (-a^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2 - a*x)*Sqrt[1 - a^2*x^2])/(a^3*(1 + a*x)) - Log[-(Sqrt[-a^2]*x) + Sqrt[1 - a^2*x^2]]/(-a^2)^(3/2)

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Maple [A]
time = 0.06, size = 84, normalized size = 1.53

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}}{a^{3}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}-\frac {\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2}+2 \left (x +\frac {1}{a}\right ) a}}{a^{4} \left (x +\frac {1}{a}\right )}\)	\(84\)
risch	\(\frac {a^{2} x^{2}-1}{a^{3} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}-\frac {\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2}+2 \left (x +\frac {1}{a}\right ) a}}{a^{4} \left (x +\frac {1}{a}\right )}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x+1)/(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 52, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2} + 1}}{a^{4} x + a^{3}} - \frac {\arcsin \left (a x\right )}{a^{3}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-a^2*x^2 + 1)/(a^4*x + a^3) - arcsin(a*x)/a^3 - sqrt(-a^2*x^2 + 1)/a^3

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Fricas [A]
time = 2.54, size = 66, normalized size = 1.20 \begin {gather*} -\frac {2 \, a x - 2 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x + 2\right )} + 2}{a^{4} x + a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (a x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.82, size = 70, normalized size = 1.27 \begin {gather*} -\frac {\arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{a^{2} {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{3}} + \frac {2}{a^{2} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(a*x)*sgn(a)/(a^2*abs(a)) - sqrt(-a^2*x^2 + 1)/a^3 + 2/(a^2*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) +
1)*abs(a))

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Mupad [B]
time = 0.07, size = 84, normalized size = 1.53 \begin {gather*} \frac {\sqrt {1-a^2\,x^2}}{\left (a\,\sqrt {-a^2}+a^2\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a^2\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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