3.347 \(\int \frac{a+b \sin ^{-1}(c x^2)}{x^5} \, dx\)

Optimal. Leaf size=41 \[ -\frac{a+b \sin ^{-1}\left (c x^2\right )}{4 x^4}-\frac{b c \sqrt{1-c^2 x^4}}{4 x^2} \]

[Out]

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

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Rubi [A]  time = 0.026372, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4842, 12, 264} \[ -\frac{a+b \sin ^{-1}\left (c x^2\right )}{4 x^4}-\frac{b c \sqrt{1-c^2 x^4}}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x^2])/x^5,x]

[Out]

-(b*c*Sqrt[1 - c^2*x^4])/(4*x^2) - (a + b*ArcSin[c*x^2])/(4*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 264

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0163505, size = 46, normalized size = 1.12 \[ -\frac{a}{4 x^4}-\frac{b c \sqrt{1-c^2 x^4}}{4 x^2}-\frac{b \sin ^{-1}\left (c x^2\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x^2])/x^5,x]

[Out]

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

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Maple [A]  time = 0.013, size = 54, normalized size = 1.3 \begin{align*} -{\frac{a}{4\,{x}^{4}}}+b \left ( -{\frac{\arcsin \left ( c{x}^{2} \right ) }{4\,{x}^{4}}}+{\frac{c \left ( c{x}^{2}-1 \right ) \left ( c{x}^{2}+1 \right ) }{4\,{x}^{2}}{\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x^2))/x^5,x)

[Out]

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

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Maxima [A]  time = 1.41391, size = 51, normalized size = 1.24 \begin{align*} -\frac{1}{4} \, b{\left (\frac{\sqrt{-c^{2} x^{4} + 1} c}{x^{2}} + \frac{\arcsin \left (c x^{2}\right )}{x^{4}}\right )} - \frac{a}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.50562, size = 92, normalized size = 2.24 \begin{align*} \frac{a x^{4} - \sqrt{-c^{2} x^{4} + 1} b c x^{2} - b \arcsin \left (c x^{2}\right ) - a}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.38465, size = 70, normalized size = 1.71 \begin{align*} - \frac{a}{4 x^{4}} + \frac{b c \left (\begin{cases} - \frac{i \sqrt{c^{2} x^{4} - 1}}{2 x^{2}} & \text{for}\: \left |{c^{2} x^{4}}\right | > 1 \\- \frac{\sqrt{- c^{2} x^{4} + 1}}{2 x^{2}} & \text{otherwise} \end{cases}\right )}{2} - \frac{b \operatorname{asin}{\left (c x^{2} \right )}}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x**2))/x**5,x)

[Out]

-a/(4*x**4) + b*c*Piecewise((-I*sqrt(c**2*x**4 - 1)/(2*x**2), Abs(c**2*x**4) > 1), (-sqrt(-c**2*x**4 + 1)/(2*x
**2), True))/2 - b*asin(c*x**2)/(4*x**4)

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Giac [B]  time = 1.23014, size = 238, normalized size = 5.8 \begin{align*} -\frac{\frac{b c^{5} x^{4} \arcsin \left (c x^{2}\right )}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac{a c^{5} x^{4}}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2}} - \frac{2 \, b c^{4} x^{2}}{\sqrt{-c^{2} x^{4} + 1} + 1} + 2 \, b c^{3} \arcsin \left (c x^{2}\right ) + 2 \, a c^{3} + \frac{2 \, b c^{2}{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}}{x^{2}} + \frac{b c{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2} \arcsin \left (c x^{2}\right )}{x^{4}} + \frac{a c{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2}}{x^{4}}}{16 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^5,x, algorithm="giac")

[Out]

-1/16*(b*c^5*x^4*arcsin(c*x^2)/(sqrt(-c^2*x^4 + 1) + 1)^2 + a*c^5*x^4/(sqrt(-c^2*x^4 + 1) + 1)^2 - 2*b*c^4*x^2
/(sqrt(-c^2*x^4 + 1) + 1) + 2*b*c^3*arcsin(c*x^2) + 2*a*c^3 + 2*b*c^2*(sqrt(-c^2*x^4 + 1) + 1)/x^2 + b*c*(sqrt
(-c^2*x^4 + 1) + 1)^2*arcsin(c*x^2)/x^4 + a*c*(sqrt(-c^2*x^4 + 1) + 1)^2/x^4)/c