3.4.95 \(\int (a+b \text {ArcSin}(c+d x^2)) \, dx\) [395]

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

[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 = 0.18, 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 = {4924, 12, 1154, 507, 435, 430} \begin {gather*} 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 (\text {ArcSin}\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 (\text {ArcSin}\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 \text {ArcSin}\left (c+d x^2\right ) \end {gather*}

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 430

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

Rule 435

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

Rule 507

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 1154

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 4924

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] Result contains complex when optimal does not.
time = 10.09, size = 155, normalized size = 0.65 \begin {gather*} a x+b x \text {ArcSin}\left (c+d x^2\right )+\frac {2 i b (-1+c) \sqrt {\frac {-1+c+d x^2}{-1+c}} \sqrt {\frac {1+c+d x^2}{1+c}} \left (E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{1+c}} x\right )|\frac {1+c}{-1+c}\right )-F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{1+c}} x\right )|\frac {1+c}{-1+c}\right )\right )}{\sqrt {\frac {d}{1+c}} \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \end {gather*}

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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Maple [A]
time = 0.01, size = 153, normalized size = 0.65

method	result	size
default	\(a x +b \left (x \arcsin \left (d \,x^{2}+c \right )+\frac {4 d \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{\sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 d c +2 d \right )}\right )\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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

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Fricas [A]
time = 0.55, size = 55, normalized size = 0.23 \begin {gather*} \frac {b d x^{2} \arcsin \left (d x^{2} + c\right ) + a d x^{2} + 2 \, \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} b}{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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