3.2.50 \(\int \frac {x^3}{(1+a x) \sqrt {1-a^2 x^2}} \, dx\)

Optimal. Leaf size=66 \[ \frac {3 \sin ^{-1}(a x)}{2 a^4}+\frac {x^2 (1-a x)}{a^2 \sqrt {1-a^2 x^2}}+\frac {(4-3 a x) \sqrt {1-a^2 x^2}}{2 a^4} \]

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Rubi [A]  time = 0.05, antiderivative size = 66, 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 = {850, 819, 780, 216} \begin {gather*} \frac {x^2 (1-a x)}{a^2 \sqrt {1-a^2 x^2}}+\frac {(4-3 a x) \sqrt {1-a^2 x^2}}{2 a^4}+\frac {3 \sin ^{-1}(a x)}{2 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(1 - a*x))/(a^2*Sqrt[1 - a^2*x^2]) + ((4 - 3*a*x)*Sqrt[1 - a^2*x^2])/(2*a^4) + (3*ArcSin[a*x])/(2*a^4)

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 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 {x^3}{(1+a x) \sqrt {1-a^2 x^2}} \, dx &=\int \frac {x^3 (1-a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=\frac {x^2 (1-a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {\int \frac {x (2-3 a x)}{\sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=\frac {x^2 (1-a x)}{a^2 \sqrt {1-a^2 x^2}}+\frac {(4-3 a x) \sqrt {1-a^2 x^2}}{2 a^4}+\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{2 a^3}\\ &=\frac {x^2 (1-a x)}{a^2 \sqrt {1-a^2 x^2}}+\frac {(4-3 a x) \sqrt {1-a^2 x^2}}{2 a^4}+\frac {3 \sin ^{-1}(a x)}{2 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 54, normalized size = 0.82 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (-a^2 x^2+a x+4\right )+3 (a x+1) \sin ^{-1}(a x)}{2 a^4 (a x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(4 + a*x - a^2*x^2) + 3*(1 + a*x)*ArcSin[a*x])/(2*a^4*(1 + a*x))

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IntegrateAlgebraic [A]  time = 0.43, size = 86, normalized size = 1.30 \begin {gather*} \frac {3 \sqrt {-a^2} \log \left (\sqrt {1-a^2 x^2}-\sqrt {-a^2} x\right )}{2 a^5}+\frac {\sqrt {1-a^2 x^2} \left (-a^2 x^2+a x+4\right )}{2 a^4 (a x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(4 + a*x - a^2*x^2))/(2*a^4*(1 + a*x)) + (3*Sqrt[-a^2]*Log[-(Sqrt[-a^2]*x) + Sqrt[1 - a^2*x
^2]])/(2*a^5)

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fricas [A]  time = 0.42, size = 75, normalized size = 1.14 \begin {gather*} \frac {4 \, a x - 6 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (a^{2} x^{2} - a x - 4\right )} \sqrt {-a^{2} x^{2} + 1} + 4}{2 \, {\left (a^{5} x + a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a*x - 6*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (a^2*x^2 - a*x - 4)*sqrt(-a^2*x^2 + 1) + 4)/
(a^5*x + a^4)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(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:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.01, size = 100, normalized size = 1.52 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2}+1}\, x}{2 a^{3}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}\, a^{3}}+\frac {\sqrt {-a^{2} x^{2}+1}}{a^{4}}+\frac {\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2}+2 \left (x +\frac {1}{a}\right ) a}}{\left (x +\frac {1}{a}\right ) a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 68, normalized size = 1.03 \begin {gather*} \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{5} x + a^{4}} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{3}} + \frac {3 \, \arcsin \left (a x\right )}{2 \, a^{4}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-a^2*x^2 + 1)/(a^5*x + a^4) - 1/2*sqrt(-a^2*x^2 + 1)*x/a^3 + 3/2*arcsin(a*x)/a^4 + sqrt(-a^2*x^2 + 1)/a^4

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mupad [B]  time = 0.07, size = 116, normalized size = 1.76 \begin {gather*} \frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a^3\,\sqrt {-a^2}}-\frac {\left (\frac {1}{a^2\,\sqrt {-a^2}}+\frac {x\,\sqrt {-a^2}}{2\,a^3}\right )\,\sqrt {1-a^2\,x^2}}{\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a^3\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\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**3/(a*x+1)/(-a**2*x**2+1)**(1/2),x)

[Out]

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

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