3.146 \(\int \frac {a+b \sin ^{-1}(c x)}{x^3} \, dx\)

Optimal. Leaf size=39 \[ -\frac {a+b \sin ^{-1}(c x)}{2 x^2}-\frac {b c \sqrt {1-c^2 x^2}}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4627, 264} \[ -\frac {a+b \sin ^{-1}(c x)}{2 x^2}-\frac {b c \sqrt {1-c^2 x^2}}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/x^3,x]

[Out]

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

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]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 44, normalized size = 1.13 \[ -\frac {a}{2 x^2}-\frac {b c \sqrt {1-c^2 x^2}}{2 x}-\frac {b \sin ^{-1}(c x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/x^3,x]

[Out]

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

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fricas [A]  time = 0.54, size = 35, normalized size = 0.90 \[ -\frac {\sqrt {-c^{2} x^{2} + 1} b c x - a x^{2} + b \arcsin \left (c x\right ) + a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.29, size = 163, normalized size = 4.18 \[ -\frac {b c^{4} x^{2} \arcsin \left (c x\right )}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}} - \frac {a c^{4} x^{2}}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}} + \frac {b c^{3} x}{4 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} - \frac {1}{4} \, b c^{2} \arcsin \left (c x\right ) - \frac {1}{4} \, a c^{2} - \frac {b c {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}}{4 \, x} - \frac {b {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2} \arcsin \left (c x\right )}{8 \, x^{2}} - \frac {a {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}}{8 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^3,x, algorithm="giac")

[Out]

-1/8*b*c^4*x^2*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1)^2 - 1/8*a*c^4*x^2/(sqrt(-c^2*x^2 + 1) + 1)^2 + 1/4*b*c^3*x
/(sqrt(-c^2*x^2 + 1) + 1) - 1/4*b*c^2*arcsin(c*x) - 1/4*a*c^2 - 1/4*b*c*(sqrt(-c^2*x^2 + 1) + 1)/x - 1/8*b*(sq
rt(-c^2*x^2 + 1) + 1)^2*arcsin(c*x)/x^2 - 1/8*a*(sqrt(-c^2*x^2 + 1) + 1)^2/x^2

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maple [A]  time = 0.01, size = 50, normalized size = 1.28 \[ c^{2} \left (-\frac {a}{2 c^{2} x^{2}}+b \left (-\frac {\arcsin \left (c x \right )}{2 c^{2} x^{2}}-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c x}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))/x^3,x)

[Out]

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

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maxima [A]  time = 0.47, size = 36, normalized size = 0.92 \[ -\frac {1}{2} \, b {\left (\frac {\sqrt {-c^{2} x^{2} + 1} c}{x} + \frac {\arcsin \left (c x\right )}{x^{2}}\right )} - \frac {a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/x^3,x)

[Out]

int((a + b*asin(c*x))/x^3, x)

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sympy [A]  time = 1.45, size = 61, normalized size = 1.56 \[ - \frac {a}{2 x^{2}} + \frac {b c \left (\begin {cases} - \frac {i \sqrt {c^{2} x^{2} - 1}}{x} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- c^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right )}{2} - \frac {b \operatorname {asin}{\left (c x \right )}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))/x**3,x)

[Out]

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

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