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

Optimal. Leaf size=61 \[ -\frac {a+b \sin ^{-1}\left (c x^2\right )}{5 x^5}+\frac {2}{15} b c^{5/2} F\left (\left .\sin ^{-1}\left (\sqrt {c} x\right )\right |-1\right )-\frac {2 b c \sqrt {1-c^2 x^4}}{15 x^3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4842, 12, 325, 221} \[ -\frac {a+b \sin ^{-1}\left (c x^2\right )}{5 x^5}-\frac {2 b c \sqrt {1-c^2 x^4}}{15 x^3}+\frac {2}{15} b c^{5/2} F\left (\left .\sin ^{-1}\left (\sqrt {c} x\right )\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x^2])/x^6,x]

[Out]

(-2*b*c*Sqrt[1 - c^2*x^4])/(15*x^3) - (a + b*ArcSin[c*x^2])/(5*x^5) + (2*b*c^(5/2)*EllipticF[ArcSin[Sqrt[c]*x]
, -1])/15

Rule 12

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

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

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Mathematica [C]  time = 0.15, size = 72, normalized size = 1.18 \[ -\frac {3 a+2 b c x^2 \sqrt {1-c^2 x^4}-2 i b (-c)^{5/2} x^5 F\left (\left .i \sinh ^{-1}\left (\sqrt {-c} x\right )\right |-1\right )+3 b \sin ^{-1}\left (c x^2\right )}{15 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(3*a + 2*b*c*x^2*Sqrt[1 - c^2*x^4] + 3*b*ArcSin[c*x^2] - (2*I)*b*(-c)^(5/2)*x^5*EllipticF[I*ArcSinh[Sqrt
[-c]*x], -1])/x^5

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (c x^{2}\right ) + a}{x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 87, normalized size = 1.43 \[ -\frac {a}{5 x^{5}}+b \left (-\frac {\arcsin \left (c \,x^{2}\right )}{5 x^{5}}+\frac {2 c \left (-\frac {\sqrt {-c^{2} x^{4}+1}}{3 x^{3}}+\frac {c^{\frac {3}{2}} \sqrt {-c \,x^{2}+1}\, \sqrt {c \,x^{2}+1}\, \EllipticF \left (x \sqrt {c}, i\right )}{3 \sqrt {-c^{2} x^{4}+1}}\right )}{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*a/x^5+b*(-1/5/x^5*arcsin(c*x^2)+2/5*c*(-1/3*(-c^2*x^4+1)^(1/2)/x^3+1/3*c^(3/2)*(-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)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (2 \, c x^{5} \int \frac {e^{\left (\frac {1}{2} \, \log \left (c x^{2} + 1\right ) + \frac {1}{2} \, \log \left (-c x^{2} + 1\right )\right )}}{c^{4} x^{12} - c^{2} x^{8} - {\left (c^{2} x^{8} - x^{4}\right )} {\left (c x^{2} + 1\right )} {\left (c x^{2} - 1\right )}}\,{d x} + \arctan \left (c x^{2}, \sqrt {c x^{2} + 1} \sqrt {-c x^{2} + 1}\right )\right )} b}{5 \, x^{5}} - \frac {a}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(10*c*x^5*integrate(1/5*e^(1/2*log(c*x^2 + 1) + 1/2*log(-c*x^2 + 1))/(c^4*x^12 - c^2*x^8 + (c^2*x^8 - x^4
)*e^(log(c*x^2 + 1) + log(-c*x^2 + 1))), x) + arctan2(c*x^2, sqrt(c*x^2 + 1)*sqrt(-c*x^2 + 1)))*b/x^5 - 1/5*a/
x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x^2\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x^2))/x^6,x)

[Out]

int((a + b*asin(c*x^2))/x^6, x)

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sympy [A]  time = 2.11, size = 61, normalized size = 1.00 \[ - \frac {a}{5 x^{5}} + \frac {b c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {c^{2} x^{4} e^{2 i \pi }} \right )}}{10 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {b \operatorname {asin}{\left (c x^{2} \right )}}{5 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(5*x**5) + b*c*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), c**2*x**4*exp_polar(2*I*pi))/(10*x**3*gamma(1/4)) - b
*asin(c*x**2)/(5*x**5)

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