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

Optimal. Leaf size=49 \[ \frac{2 b \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{c} x\right ),-1\right )}{\sqrt{c}}+a x+b x \sin ^{-1}\left (c x^2\right )-\frac{2 b E\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{\sqrt{c}} \]

[Out]

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

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Rubi [A]  time = 0.0426012, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4840, 12, 307, 221, 1199, 424} \[ a x+b x \sin ^{-1}\left (c x^2\right )+\frac{2 b F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{\sqrt{c}}-\frac{2 b E\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{\sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcSin[c*x^2],x]

[Out]

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

Rule 4840

Int[ArcSin[u_], x_Symbol] :> Simp[x*ArcSin[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/Sqrt[1 - u^2], x], x] /;
InverseFunctionFreeQ[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 307

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 424

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

Rubi steps

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

Mathematica [C]  time = 0.0051358, size = 39, normalized size = 0.8 \[ -\frac{2}{3} b c x^3 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},c^2 x^4\right )+a x+b x \sin ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcSin[c*x^2],x]

[Out]

a*x + b*x*ArcSin[c*x^2] - (2*b*c*x^3*Hypergeometric2F1[1/2, 3/4, 7/4, c^2*x^4])/3

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Maple [A]  time = 0.006, size = 71, normalized size = 1.5 \begin{align*} ax+b \left ( x\arcsin \left ( c{x}^{2} \right ) +2\,{\frac{\sqrt{-c{x}^{2}+1}\sqrt{c{x}^{2}+1} \left ({\it EllipticF} \left ( x\sqrt{c},i \right ) -{\it EllipticE} \left ( x\sqrt{c},i \right ) \right ) }{\sqrt{c}\sqrt{-{c}^{2}{x}^{4}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(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 (b \arcsin \left (c x^{2}\right ) + a, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b*arcsin(c*x^2) + a, x)

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Sympy [A]  time = 1.13843, size = 49, normalized size = 1. \begin{align*} a x + b \left (- \frac{c x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{c^{2} x^{4} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac{7}{4}\right )} + x \operatorname{asin}{\left (c x^{2} \right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*(-c*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c**2*x**4*exp_polar(2*I*pi))/(2*gamma(7/4)) + x*asin(c*x
**2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(b*arcsin(c*x^2) + a, x)