3.396 \(\int \frac{a+b \sin ^{-1}(c+d x^2)}{x^2} \, dx\)

Optimal. Leaf size=126 \[ \frac{2 b \sqrt{1-c} \sqrt{d} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right ),-\frac{1-c}{c+1}\right )}{\sqrt{-c^2-2 c d x^2-d^2 x^4+1}}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x} \]

[Out]

-((a + b*ArcSin[c + d*x^2])/x) + (2*b*Sqrt[1 - c]*Sqrt[d]*Sqrt[1 - (d*x^2)/(1 - c)]*Sqrt[1 + (d*x^2)/(1 + c)]*
EllipticF[ArcSin[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 - c)/(1 + c))])/Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4]

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Rubi [A]  time = 0.0764169, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4842, 12, 1104, 419} \[ \frac{2 b \sqrt{1-c} \sqrt{d} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} F\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{c+1}\right )}{\sqrt{-c^2-2 c d x^2-d^2 x^4+1}}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x^2])/x^2,x]

[Out]

-((a + b*ArcSin[c + d*x^2])/x) + (2*b*Sqrt[1 - c]*Sqrt[d]*Sqrt[1 - (d*x^2)/(1 - c)]*Sqrt[1 + (d*x^2)/(1 + c)]*
EllipticF[ArcSin[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 - c)/(1 + c))])/Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4]

Rule 4842

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 - u^2], x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

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

Rule 1104

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(Sqrt[1 + (2*
c*x^2)/(b - q)]*Sqrt[1 + (2*c*x^2)/(b + q)])/Sqrt[a + b*x^2 + c*x^4], Int[1/(Sqrt[1 + (2*c*x^2)/(b - q)]*Sqrt[
1 + (2*c*x^2)/(b + q)]), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[c/a]

Rule 419

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

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x^2} \, dx &=-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x}+b \int \frac{2 d}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x}+(2 b d) \int \frac{1}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x}+\frac{\left (2 b d \sqrt{1-\frac{2 d^2 x^2}{-2 d-2 c d}} \sqrt{1-\frac{2 d^2 x^2}{2 d-2 c d}}\right ) \int \frac{1}{\sqrt{1-\frac{2 d^2 x^2}{-2 d-2 c d}} \sqrt{1-\frac{2 d^2 x^2}{2 d-2 c d}}} \, dx}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}}\\ &=-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x}+\frac{2 b \sqrt{1-c} \sqrt{d} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{1+\frac{d x^2}{1+c}} F\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{1+c}\right )}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.230253, size = 140, normalized size = 1.11 \[ -\frac{2 i b d \sqrt{1-\frac{d x^2}{-c-1}} \sqrt{1-\frac{d x^2}{1-c}} \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{-\frac{d}{-c-1}}\right ),\frac{-c-1}{1-c}\right )}{\sqrt{-\frac{d}{-c-1}} \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}-\frac{a}{x}-\frac{b \sin ^{-1}\left (c+d x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x^2])/x^2,x]

[Out]

-(a/x) - (b*ArcSin[c + d*x^2])/x - ((2*I)*b*d*Sqrt[1 - (d*x^2)/(-1 - c)]*Sqrt[1 - (d*x^2)/(1 - c)]*EllipticF[I
*ArcSinh[Sqrt[-(d/(-1 - c))]*x], (-1 - c)/(1 - c)])/(Sqrt[-(d/(-1 - c))]*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])

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Maple [A]  time = 0.01, size = 114, normalized size = 0.9 \begin{align*} -{\frac{a}{x}}+b \left ( -{\frac{\arcsin \left ( d{x}^{2}+c \right ) }{x}}+2\,{\frac{d}{\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}\sqrt{1+{\frac{d{x}^{2}}{-1+c}}}\sqrt{1+{\frac{d{x}^{2}}{1+c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{d}{-1+c}}},\sqrt{-1+2\,{\frac{c}{1+c}}} \right ){\frac{1}{\sqrt{-{\frac{d}{-1+c}}}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x^2+c))/x^2,x)

[Out]

-a/x+b*(-1/x*arcsin(d*x^2+c)+2*d/(-d/(-1+c))^(1/2)*(1+d/(-1+c)*x^2)^(1/2)*(1+d*x^2/(1+c))^(1/2)/(-d^2*x^4-2*c*
d*x^2-c^2+1)^(1/2)*EllipticF(x*(-d/(-1+c))^(1/2),(-1+2*c/(1+c))^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+c))/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arcsin \left (d x^{2} + c\right ) + a}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+c))/x^2,x, algorithm="fricas")

[Out]

integral((b*arcsin(d*x^2 + c) + a)/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asin}{\left (c + d x^{2} \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x**2+c))/x**2,x)

[Out]

Integral((a + b*asin(c + d*x**2))/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (d x^{2} + c\right ) + a}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+c))/x^2,x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 + c) + a)/x^2, x)