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

Optimal. Leaf size=284 \[ -\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}-\frac {2 b d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x}+\frac {2 b d^{3/2} \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} F\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{3 \sqrt {1-c} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {2 b d^{3/2} \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{3 \sqrt {1-c} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}} \]

[Out]

1/3*(-a-b*arcsin(d*x^2+c))/x^3-2/3*b*d^(3/2)*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)/(1-c)^(1/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)+2/3*b*d^(3/2)*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)/(1-c)^(1/2)/(-d^2*x^4-2*c*d*x^
2-c^2+1)^(1/2)-2/3*b*d*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/(-c^2+1)/x

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4842, 12, 1123, 1140, 493, 424, 419} \[ -\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{3 x^3}-\frac {2 b d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (1-c^2\right ) x}+\frac {2 b d^{3/2} \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} F\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{3 \sqrt {1-c} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {2 b d^{3/2} \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{3 \sqrt {1-c} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x^2])/x^4,x]

[Out]

(-2*b*d*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(3*(1 - c^2)*x) - (a + b*ArcSin[c + d*x^2])/(3*x^3) - (2*b*d^(3/2
)*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))])/(3*Sqrt[1 - c]*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4]) + (2*b*d^(3/2)*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))])/(3*Sqrt[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 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 493

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 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 1140

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

________________________________________________________________________________________

Mathematica [C]  time = 0.42, size = 243, normalized size = 0.86 \[ -\frac {a}{3 x^3}+\frac {2 b d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{3 \left (c^2-1\right ) x}+\frac {2 i b (1-c) d^2 \sqrt {1-\frac {d x^2}{-c-1}} \sqrt {1-\frac {d x^2}{1-c}} \left (E\left (i \sinh ^{-1}\left (\sqrt {-\frac {d}{-c-1}} x\right )|\frac {-c-1}{1-c}\right )-F\left (i \sinh ^{-1}\left (\sqrt {-\frac {d}{-c-1}} x\right )|\frac {-c-1}{1-c}\right )\right )}{3 (c-1) (c+1) \sqrt {-\frac {d}{-c-1}} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {b \sin ^{-1}\left (c+d x^2\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 1.38, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 207, normalized size = 0.73 \[ -\frac {a}{3 x^{3}}+b \left (-\frac {\arcsin \left (d \,x^{2}+c \right )}{3 x^{3}}+\frac {2 d \left (\frac {\sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{\left (c^{2}-1\right ) x}-\frac {2 d^{2} \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{\left (c^{2}-1\right ) \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 d c +2 d \right )}\right )}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a/x^3+b*(-1/3/x^3*arcsin(d*x^2+c)+2/3*d*(1/(c^2-1)*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/x-2*d^2/(c^2-1)*(-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)))))

________________________________________________________________________________________

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^4,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?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asin}\left (d\,x^2+c\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x^2))/x^4,x)

[Out]

int((a + b*asin(c + d*x^2))/x^4, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c + d x^{2} \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________