3.4.96 \(\int \frac {a+b \text {ArcSin}(c+d x^2)}{x^2} \, dx\) [396]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4926, 12, 1118, 430} \begin {gather*} \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 (\text {ArcSin}\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 \text {ArcSin}\left (c+d x^2\right )}{x} \end {gather*}

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 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 1118

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 4926

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]

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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.16, size = 140, normalized size = 1.11 \begin {gather*} -\frac {a}{x}-\frac {b \text {ArcSin}\left (c+d x^2\right )}{x}-\frac {2 i b d \sqrt {1-\frac {d x^2}{-1-c}} \sqrt {1-\frac {d x^2}{1-c}} F\left (i \sinh ^{-1}\left (\sqrt {-\frac {d}{-1-c}} x\right )|\frac {-1-c}{1-c}\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^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 = 0.90

method	result	size
default	\(-\frac {a}{x}+b \left (-\frac {\arcsin \left (d \,x^{2}+c \right )}{x}+\frac {2 d \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )}{\sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}\right )\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, 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^2,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 = 0.71, size = 17, normalized size = 0.13 \begin {gather*} -\frac {b \arcsin \left (d x^{2} + c\right ) + a}{x} \end {gather*}

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]

-(b*arcsin(d*x^2 + c) + a)/x

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

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 = 0.00, 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^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {asin}\left (d\,x^2+c\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x^2))/x^2,x)

[Out]

int((a + b*asin(c + d*x^2))/x^2, x)

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