3.357 \(\int \frac{a+b \sin ^{-1}(c x^2)}{x^4} \, dx\)

Optimal. Leaf size=81 \[ \frac{2}{3} b c^{3/2} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{c} x\right ),-1\right )-\frac{a+b \sin ^{-1}\left (c x^2\right )}{3 x^3}-\frac{2 b c \sqrt{1-c^2 x^4}}{3 x}-\frac{2}{3} b c^{3/2} E\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right ) \]

[Out]

(-2*b*c*Sqrt[1 - c^2*x^4])/(3*x) - (a + b*ArcSin[c*x^2])/(3*x^3) - (2*b*c^(3/2)*EllipticE[ArcSin[Sqrt[c]*x], -
1])/3 + (2*b*c^(3/2)*EllipticF[ArcSin[Sqrt[c]*x], -1])/3

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Rubi [A]  time = 0.0575011, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4842, 12, 325, 307, 221, 1199, 424} \[ -\frac{a+b \sin ^{-1}\left (c x^2\right )}{3 x^3}-\frac{2 b c \sqrt{1-c^2 x^4}}{3 x}+\frac{2}{3} b c^{3/2} F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )-\frac{2}{3} b c^{3/2} E\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x^2])/x^4,x]

[Out]

(-2*b*c*Sqrt[1 - c^2*x^4])/(3*x) - (a + b*ArcSin[c*x^2])/(3*x^3) - (2*b*c^(3/2)*EllipticE[ArcSin[Sqrt[c]*x], -
1])/3 + (2*b*c^(3/2)*EllipticF[ArcSin[Sqrt[c]*x], -1])/3

Rule 4842

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]

Rule 12

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, 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 \frac{a+b \sin ^{-1}\left (c x^2\right )}{x^4} \, dx &=-\frac{a+b \sin ^{-1}\left (c x^2\right )}{3 x^3}+\frac{1}{3} b \int \frac{2 c}{x^2 \sqrt{1-c^2 x^4}} \, dx\\ &=-\frac{a+b \sin ^{-1}\left (c x^2\right )}{3 x^3}+\frac{1}{3} (2 b c) \int \frac{1}{x^2 \sqrt{1-c^2 x^4}} \, dx\\ &=-\frac{2 b c \sqrt{1-c^2 x^4}}{3 x}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{3 x^3}-\frac{1}{3} \left (2 b c^3\right ) \int \frac{x^2}{\sqrt{1-c^2 x^4}} \, dx\\ &=-\frac{2 b c \sqrt{1-c^2 x^4}}{3 x}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{3 x^3}+\frac{1}{3} \left (2 b c^2\right ) \int \frac{1}{\sqrt{1-c^2 x^4}} \, dx-\frac{1}{3} \left (2 b c^2\right ) \int \frac{1+c x^2}{\sqrt{1-c^2 x^4}} \, dx\\ &=-\frac{2 b c \sqrt{1-c^2 x^4}}{3 x}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{3 x^3}+\frac{2}{3} b c^{3/2} F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )-\frac{1}{3} \left (2 b c^2\right ) \int \frac{\sqrt{1+c x^2}}{\sqrt{1-c x^2}} \, dx\\ &=-\frac{2 b c \sqrt{1-c^2 x^4}}{3 x}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{3 x^3}-\frac{2}{3} b c^{3/2} E\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )+\frac{2}{3} b c^{3/2} F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )\\ \end{align*}

Mathematica [C]  time = 0.180652, size = 89, normalized size = 1.1 \[ -\frac{2 i b \sqrt{-c} c x^3 \left (E\left (\left .i \sinh ^{-1}\left (\sqrt{-c} x\right )\right |-1\right )-\text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-c} x\right ),-1\right )\right )+a+2 b c x^2 \sqrt{1-c^2 x^4}+b \sin ^{-1}\left (c x^2\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x^2])/x^4,x]

[Out]

-(a + 2*b*c*x^2*Sqrt[1 - c^2*x^4] + b*ArcSin[c*x^2] + (2*I)*b*Sqrt[-c]*c*x^3*(EllipticE[I*ArcSinh[Sqrt[-c]*x],
-1] - EllipticF[I*ArcSinh[Sqrt[-c]*x], -1]))/(3*x^3)

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Maple [A]  time = 0.012, size = 97, normalized size = 1.2 \begin{align*} -{\frac{a}{3\,{x}^{3}}}+b \left ( -{\frac{\arcsin \left ( c{x}^{2} \right ) }{3\,{x}^{3}}}+{\frac{2\,c}{3} \left ( -{\frac{1}{x}\sqrt{-{c}^{2}{x}^{4}+1}}+{\sqrt{c}\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 ){\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x^2))/x^4,x)

[Out]

-1/3*a/x^3+b*(-1/3/x^3*arcsin(c*x^2)+2/3*c*(-(-c^2*x^4+1)^(1/2)/x+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)-EllipticE(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^4,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 (\frac{b \arcsin \left (c x^{2}\right ) + a}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^4,x, algorithm="fricas")

[Out]

integral((b*arcsin(c*x^2) + a)/x^4, x)

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Sympy [A]  time = 2.01972, size = 60, normalized size = 0.74 \begin{align*} - \frac{a}{3 x^{3}} + \frac{b c \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \frac{3}{4} \end{matrix}\middle |{c^{2} x^{4} e^{2 i \pi }} \right )}}{6 x \Gamma \left (\frac{3}{4}\right )} - \frac{b \operatorname{asin}{\left (c x^{2} \right )}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x**2))/x**4,x)

[Out]

-a/(3*x**3) + b*c*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), c**2*x**4*exp_polar(2*I*pi))/(6*x*gamma(3/4)) - b*asi
n(c*x**2)/(3*x**3)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (c x^{2}\right ) + a}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^4,x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x^2) + a)/x^4, x)