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

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

[Out]

1/5*(-a-b*arcsin(d*x^2+c))/x^5-8/15*b*c*d^(5/2)*EllipticE(x*d^(1/2)/(1-c)^(1/2),((-1+c)/(1+c))^(1/2))*(1-d*x^2
/(1-c))^(1/2)*(1+d*x^2/(1+c))^(1/2)/(-c^2+1)/(1-c)^(1/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)+2/15*b*(1+3*c)*d^(5/
2)*EllipticF(x*d^(1/2)/(1-c)^(1/2),((-1+c)/(1+c))^(1/2))*(1-d*x^2/(1-c))^(1/2)*(1+d*x^2/(1+c))^(1/2)/(-c^2+1)/
(1-c)^(1/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)-2/15*b*d*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/(-c^2+1)/x^3-8/15*b*c*d
^2*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/(-c^2+1)^2/x

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Rubi [A]  time = 0.36, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.85, size = 370, normalized size = 1.04 \[ \frac {\sqrt {\frac {d}{c+1}} \left (-3 a \left (c^2-1\right )^2 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}-3 b \left (c^2-1\right )^2 \sqrt {-c^2-2 c d x^2-d^2 x^4+1} \sin ^{-1}\left (c+d x^2\right )+2 b d x^2 \left (-c^4+2 c^3 d x^2+c^2 \left (7 d^2 x^4+2\right )+c \left (4 d^3 x^6-2 d x^2\right )+d^2 x^4-1\right )\right )-2 i b \left (3 c^2-4 c+1\right ) d^3 x^5 \sqrt {\frac {c+d x^2-1}{c-1}} \sqrt {\frac {c+d x^2+1}{c+1}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c+1}} x\right )|\frac {c+1}{c-1}\right )+8 i b (c-1) c d^3 x^5 \sqrt {\frac {c+d x^2-1}{c-1}} \sqrt {\frac {c+d x^2+1}{c+1}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c+1}} x\right )|\frac {c+1}{c-1}\right )}{15 \left (c^2-1\right )^2 x^5 \sqrt {\frac {d}{c+1}} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d/(1 + c)]*(-3*a*(-1 + c^2)^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4] + 2*b*d*x^2*(-1 - c^4 + 2*c^3*d*x^2 +
d^2*x^4 + c^2*(2 + 7*d^2*x^4) + c*(-2*d*x^2 + 4*d^3*x^6)) - 3*b*(-1 + c^2)^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^
4]*ArcSin[c + d*x^2]) + (8*I)*b*(-1 + c)*c*d^3*x^5*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)] - (2*I)*b*(1 - 4*c + 3*c^2)*d^3*x^5*Sqrt[(-1 + c
+ d*x^2)/(-1 + c)]*Sqrt[(1 + c + d*x^2)/(1 + c)]*EllipticF[I*ArcSinh[Sqrt[d/(1 + c)]*x], (1 + c)/(-1 + c)])/(1
5*(-1 + c^2)^2*Sqrt[d/(1 + c)]*x^5*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])

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fricas [F]  time = 1.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{6}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{6}}\,{d x} \]

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)

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maple [A]  time = 0.02, size = 346, normalized size = 0.97 \[ -\frac {a}{5 x^{5}}+b \left (-\frac {\arcsin \left (d \,x^{2}+c \right )}{5 x^{5}}+\frac {2 d \left (\frac {\sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{3 \left (c^{2}-1\right ) x^{3}}-\frac {4 c d \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{3 \left (c^{2}-1\right )^{2} x}-\frac {d^{2} \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )}{3 \left (c^{2}-1\right ) \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}+\frac {8 c \,d^{3} \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{3 \left (c^{2}-1\right )^{2} \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 d c +2 d \right )}\right )}{5}\right ) \]

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 = 0.00, 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^6,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 \[ \int \frac {a+b\,\mathrm {asin}\left (d\,x^2+c\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x^2))/x^6,x)

[Out]

int((a + b*asin(c + d*x^2))/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c + d x^{2} \right )}}{x^{6}}\, dx \]

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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