3.398 \(\int \frac{a+b \sin ^{-1}(c+d x^2)}{x^6} \, dx\)

Optimal. Leaf size=355 \[ \frac{2 b (3 c+1) d^{5/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right ),-\frac{1-c}{c+1}\right )}{15 \sqrt{1-c} \left (1-c^2\right ) \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{5 x^5}-\frac{8 b c d^2 \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{15 \left (1-c^2\right )^2 x}-\frac{2 b d \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{15 \left (1-c^2\right ) x^3}-\frac{8 b c d^{5/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{c+1}\right )}{15 \sqrt{1-c} \left (1-c^2\right ) \sqrt{-c^2-2 c d x^2-d^2 x^4+1}} \]

[Out]

(-2*b*d*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(15*(1 - c^2)*x^3) - (8*b*c*d^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^
4])/(15*(1 - c^2)^2*x) - (a + b*ArcSin[c + d*x^2])/(5*x^5) - (8*b*c*d^(5/2)*Sqrt[1 - (d*x^2)/(1 - c)]*Sqrt[1 +
(d*x^2)/(1 + c)]*EllipticE[ArcSin[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 - c)/(1 + c))])/(15*Sqrt[1 - c]*(1 - c^2)*Sq
rt[1 - c^2 - 2*c*d*x^2 - d^2*x^4]) + (2*b*(1 + 3*c)*d^(5/2)*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))])/(15*Sqrt[1 - c]*(1 - c^2)*Sqrt[1 - c^2 - 2*c
*d*x^2 - d^2*x^4])

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Rubi [A]  time = 0.356087, antiderivative size = 355, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4842, 12, 1123, 1281, 1202, 524, 424, 419} \[ -\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{5 x^5}-\frac{8 b c d^2 \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{15 \left (1-c^2\right )^2 x}-\frac{2 b d \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{15 \left (1-c^2\right ) x^3}+\frac{2 b (3 c+1) d^{5/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} F\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{c+1}\right )}{15 \sqrt{1-c} \left (1-c^2\right ) \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}-\frac{8 b c d^{5/2} \sqrt{1-\frac{d x^2}{1-c}} \sqrt{\frac{d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{1-c}}\right )|-\frac{1-c}{c+1}\right )}{15 \sqrt{1-c} \left (1-c^2\right ) \sqrt{-c^2-2 c d x^2-d^2 x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x^2])/x^6,x]

[Out]

(-2*b*d*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(15*(1 - c^2)*x^3) - (8*b*c*d^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^
4])/(15*(1 - c^2)^2*x) - (a + b*ArcSin[c + d*x^2])/(5*x^5) - (8*b*c*d^(5/2)*Sqrt[1 - (d*x^2)/(1 - c)]*Sqrt[1 +
(d*x^2)/(1 + c)]*EllipticE[ArcSin[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 - c)/(1 + c))])/(15*Sqrt[1 - c]*(1 - c^2)*Sq
rt[1 - c^2 - 2*c*d*x^2 - d^2*x^4]) + (2*b*(1 + 3*c)*d^(5/2)*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))])/(15*Sqrt[1 - c]*(1 - c^2)*Sqrt[1 - c^2 - 2*c
*d*x^2 - d^2*x^4])

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 1123

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*x^2 +
c*x^4)^(p + 1))/(a*d*(m + 1)), x] - Dist[1/(a*d^2*(m + 1)), Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p +
5)*x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && In
tegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(
f*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + b
*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1202

Int[((d_) + (e_.)*(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[(d + e*x^2)/(Sqr
t[1 + (2*c*x^2)/(b - q)]*Sqrt[1 + (2*c*x^2)/(b + q)]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c
, 0] && NegQ[c/a]

Rule 524

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

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x^6} \, dx &=-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{5 x^5}+\frac{1}{5} b \int \frac{2 d}{x^4 \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 )}{5 x^5}+\frac{1}{5} (2 b d) \int \frac{1}{x^4 \sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right ) x^3}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{5 x^5}+\frac{(2 b d) \int \frac{4 c d+d^2 x^2}{x^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx}{15 \left (1-c^2\right )}\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right ) x^3}-\frac{8 b c d^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right )^2 x}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{5 x^5}-\frac{(2 b d) \int \frac{-\left (1-c^2\right ) d^2+4 c d^3 x^2}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx}{15 \left (1-c^2\right )^2}\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right ) x^3}-\frac{8 b c d^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right )^2 x}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{5 x^5}-\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{-\left (1-c^2\right ) d^2+4 c d^3 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}{15 \left (1-c^2\right )^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right ) x^3}-\frac{8 b c d^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right )^2 x}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{5 x^5}-\frac{\left (8 b c (1+c) d^3 \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}{15 \left (1-c^2\right )^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}+\frac{\left (2 b (1+c) (1+3 c) d^3 \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}{15 \left (1-c^2\right )^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}\\ &=-\frac{2 b d \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right ) x^3}-\frac{8 b c d^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{15 \left (1-c^2\right )^2 x}-\frac{a+b \sin ^{-1}\left (c+d x^2\right )}{5 x^5}-\frac{8 b c d^{5/2} \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 )}{15 (1-c)^{3/2} (1+c) \sqrt{1-c^2-2 c d x^2-d^2 x^4}}+\frac{2 b (1+3 c) d^{5/2} \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 )}{15 (1-c)^{3/2} (1+c) \sqrt{1-c^2-2 c d x^2-d^2 x^4}}\\ \end{align*}

Mathematica [F]  time = 0.798054, size = 0, normalized size = 0. \[ \int \frac{a+b \sin ^{-1}\left (c+d x^2\right )}{x^6} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*ArcSin[c + d*x^2])/x^6,x]

[Out]

Integrate[(a + b*ArcSin[c + d*x^2])/x^6, x]

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Maple [A]  time = 0.016, size = 346, normalized size = 1. \begin{align*} -{\frac{a}{5\,{x}^{5}}}+b \left ( -{\frac{\arcsin \left ( d{x}^{2}+c \right ) }{5\,{x}^{5}}}+{\frac{2\,d}{5} \left ({\frac{1}{ \left ( 3\,{c}^{2}-3 \right ){x}^{3}}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}-{\frac{4\,dc}{3\, \left ({c}^{2}-1 \right ) ^{2}x}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}-{\frac{{d}^{2}}{3\,{c}^{2}-3}\sqrt{1+{\frac{d{x}^{2}}{-1+c}}}\sqrt{1+{\frac{d{x}^{2}}{1+c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{d}{-1+c}}},\sqrt{-1+2\,{\frac{c}{1+c}}} \right ){\frac{1}{\sqrt{-{\frac{d}{-1+c}}}}}{\frac{1}{\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}}}+{\frac{8\,c{d}^{3} \left ( -{c}^{2}+1 \right ) }{3\, \left ({c}^{2}-1 \right ) ^{2} \left ( -2\,dc+2\,d \right ) }\sqrt{1+{\frac{d{x}^{2}}{-1+c}}}\sqrt{1+{\frac{d{x}^{2}}{1+c}}} \left ({\it EllipticF} \left ( x\sqrt{-{\frac{d}{-1+c}}},\sqrt{-1+2\,{\frac{c}{1+c}}} \right ) -{\it EllipticE} \left ( x\sqrt{-{\frac{d}{-1+c}}},\sqrt{-1+2\,{\frac{c}{1+c}}} \right ) \right ){\frac{1}{\sqrt{-{\frac{d}{-1+c}}}}}{\frac{1}{\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x^2+c))/x^6,x)

[Out]

-1/5*a/x^5+b*(-1/5/x^5*arcsin(d*x^2+c)+2/5*d*(1/3/(c^2-1)*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/x^3-4/3*c*d/(c^2-1)
^2*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/x-1/3*d^2/(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)*EllipticF(x*(-d/(-1+c))^(1/2),(-1+2*c/(1+c))^(1/2))+8/3*c*d^3/(c^2-1)^
2*(-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., 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(d*x^2+c))/x^6,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 (d x^{2} + c\right ) + a}{x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+c))/x^6,x, algorithm="fricas")

[Out]

integral((b*arcsin(d*x^2 + c) + a)/x^6, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asin}{\left (c + d x^{2} \right )}}{x^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x**2+c))/x**6,x)

[Out]

Integral((a + b*asin(c + d*x**2))/x**6, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (d x^{2} + c\right ) + a}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+c))/x^6,x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 + c) + a)/x^6, x)