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

Optimal. Leaf size=45 \[ \frac{4 (a x+1)}{\sqrt{1-a^2 x^2}}-\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\sin ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.0910878, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1805, 844, 216, 266, 63, 208} \[ \frac{4 (a x+1)}{\sqrt{1-a^2 x^2}}-\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.0376443, size = 59, normalized size = 1.31 \[ \frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};1-a^2 x^2\right )-\sqrt{1-a^2 x^2} \sin ^{-1}(a x)+4 a x+3}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 + 4*a*x - Sqrt[1 - a^2*x^2]*ArcSin[a*x] + Hypergeometric2F1[-1/2, 1, 1/2, 1 - a^2*x^2])/Sqrt[1 - a^2*x^2]

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Maple [A]  time = 0.012, size = 75, normalized size = 1.7 \begin{align*} 4\,{\frac{ax}{\sqrt{-{a}^{2}{x}^{2}+1}}}-{a\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+4\,{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}-{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.70773, size = 105, normalized size = 2.33 \begin{align*} \frac{4 \, a x}{\sqrt{-a^{2} x^{2} + 1}} - \frac{a \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}} + \frac{4}{\sqrt{-a^{2} x^{2} + 1}} - \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.53943, size = 193, normalized size = 4.29 \begin{align*} \frac{4 \, a x + 2 \,{\left (a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (a x - 1\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - 4 \, \sqrt{-a^{2} x^{2} + 1} - 4}{a x - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a*x + 2*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (a*x - 1)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - 4*sq
rt(-a^2*x^2 + 1) - 4)/(a*x - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a x + 1\right )^{3}}{x \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.18501, size = 117, normalized size = 2.6 \begin{align*} -\frac{a \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} - \frac{a \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 |}} + \frac{8 \, a}{{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*arcsin(a*x)*sgn(a)/abs(a) - a*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + 8*a/((
(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a))