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

Optimal. Leaf size=237 \[ a x+\frac {2 b \sqrt {1-c} (c+1) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} F\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{\sqrt {d} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {2 b \sqrt {1-c} (c+1) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{\sqrt {d} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}+b x \sin ^{-1}\left (c+d x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4840, 12, 1140, 493, 424, 419} \[ a x+\frac {2 b \sqrt {1-c} (c+1) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} F\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{\sqrt {d} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {2 b \sqrt {1-c} (c+1) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{\sqrt {d} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}+b x \sin ^{-1}\left (c+d x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcSin[c + d*x^2],x]

[Out]

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

Rule 12

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

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)])

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 493

Int[(x_)^(n_)/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/b, Int[Sqrt[a +
b*x^n]/Sqrt[c + d*x^n], x], x] - Dist[a/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b, c,
d}, x] && NeQ[b*c - a*d, 0] && (EqQ[n, 2] || EqQ[n, 4]) &&  !(EqQ[n, 2] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 1140

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

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Mathematica [C]  time = 0.15, size = 155, normalized size = 0.65 \[ a x+\frac {2 i b (c-1) \sqrt {\frac {c+d x^2-1}{c-1}} \sqrt {\frac {c+d x^2+1}{c+1}} \left (E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c+1}} x\right )|\frac {c+1}{c-1}\right )-F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c+1}} x\right )|\frac {c+1}{c-1}\right )\right )}{\sqrt {\frac {d}{c+1}} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}+b x \sin ^{-1}\left (c+d x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcSin[c + d*x^2],x]

[Out]

a*x + b*x*ArcSin[c + d*x^2] + ((2*I)*b*(-1 + c)*Sqrt[(-1 + c + d*x^2)/(-1 + c)]*Sqrt[(1 + c + d*x^2)/(1 + c)]*
(EllipticE[I*ArcSinh[Sqrt[d/(1 + c)]*x], (1 + c)/(-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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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b \arcsin \left (d x^{2} + c\right ) + a, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b*arcsin(d*x^2 + c) + a, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(b*arcsin(d*x^2 + c) + a, x)

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maple [A]  time = 0.01, size = 153, normalized size = 0.65 \[ a x +b \left (x \arcsin \left (d \,x^{2}+c \right )+\frac {4 d \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{\sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 d c +2 d \right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c-1>0)', see `assume?` for mor
e details)Is c-1 positive or negative?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*asin(c + d*x^2),x)

[Out]

int(a + b*asin(c + d*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(a + b*asin(c + d*x**2), x)

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