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]

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

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Rubi [A]  time = 0.09, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, 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 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 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 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]

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*}

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Mathematica [C]  time = 0.04, 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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fricas [A]  time = 0.46, size = 82, normalized size = 1.82 \[ \frac {4 \, a x + 2 \, {\left (a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (a x - 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - 4 \, \sqrt {-a^{2} x^{2} + 1} - 4}{a x - 1} \]

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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giac [B]  time = 0.54, size = 87, normalized size = 1.93 \[ -\frac {a \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} - \frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} + \frac {8 \, a}{{\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \]

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))

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maple [A]  time = 0.01, size = 75, normalized size = 1.67 \[ \frac {4 a x}{\sqrt {-a^{2} x^{2}+1}}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {4}{\sqrt {-a^{2} x^{2}+1}} \]

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 = 2.08, size = 65, normalized size = 1.44 \[ \frac {4 \, a x}{\sqrt {-a^{2} x^{2} + 1}} + \frac {4}{\sqrt {-a^{2} x^{2} + 1}} - \arcsin \left (a x\right ) - \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

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

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mupad [B]  time = 3.49, size = 82, normalized size = 1.82 \[ \frac {4\,a\,\sqrt {1-a^2\,x^2}}{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {a\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}-\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a*(1 - a^2*x^2)^(1/2))/((x*(-a^2)^(1/2) - (-a^2)^(1/2)/a)*(-a^2)^(1/2)) - (a*asinh(x*(-a^2)^(1/2)))/(-a^2)^
(1/2) - atanh((1 - a^2*x^2)^(1/2))

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

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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