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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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 12

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

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 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.80 \[ a x-\frac {2}{3} b c x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^4\right )+b x \sin ^{-1}\left (c x^2\right ) \]

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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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b \arcsin \left (c x^{2}\right ) + a, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int b \arcsin \left (c x^{2}\right ) + a\,{d x} \]

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)

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maple [A]  time = 0.01, size = 71, normalized size = 1.45 \[ a x +b \left (x \arcsin \left (c \,x^{2}\right )+\frac {2 \sqrt {-c \,x^{2}+1}\, \sqrt {c \,x^{2}+1}\, \left (\EllipticF \left (x \sqrt {c}, i\right )-\EllipticE \left (x \sqrt {c}, i\right )\right )}{\sqrt {c}\, \sqrt {-c^{2} x^{4}+1}}\right ) \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (x \arctan \left (c x^{2}, \sqrt {c x^{2} + 1} \sqrt {-c x^{2} + 1}\right ) + 2 \, c \int \frac {x^{2} e^{\left (\frac {1}{2} \, \log \left (c x^{2} + 1\right ) + \frac {1}{2} \, \log \left (-c x^{2} + 1\right )\right )}}{c^{4} x^{8} - c^{2} x^{4} + {\left (c^{2} x^{4} - 1\right )} e^{\left (\log \left (c x^{2} + 1\right ) + \log \left (-c x^{2} + 1\right )\right )}}\,{d x}\right )} b + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*arctan2(c*x^2, sqrt(c*x^2 + 1)*sqrt(-c*x^2 + 1)) + 2*c*integrate(x^2*e^(1/2*log(c*x^2 + 1) + 1/2*log(-c*x^2
+ 1))/(c^4*x^8 - c^2*x^4 + (c^2*x^4 - 1)*e^(log(c*x^2 + 1) + log(-c*x^2 + 1))), x))*b + a*x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int a+b\,\mathrm {asin}\left (c\,x^2\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*asin(c*x^2),x)

[Out]

int(a + b*asin(c*x^2), x)

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sympy [A]  time = 1.00, size = 49, normalized size = 1.00 \[ a x + b \left (- \frac {c x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {c^{2} x^{4} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} + x \operatorname {asin}{\left (c x^{2} \right )}\right ) \]

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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