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

Optimal. Leaf size=57 \[ -\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{2 x^2}-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{4 c x}+\frac {b \csc ^{-1}\left (\frac {x}{c}\right )}{4 c^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4842, 12, 335, 321, 216} \[ -\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{2 x^2}-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{4 c x}+\frac {b \csc ^{-1}\left (\frac {x}{c}\right )}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c/x])/x^3,x]

[Out]

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

Rule 12

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && 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 (\frac {c}{x}\right )}{x^3} \, dx &=-\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{2 x^2}-\frac {1}{2} b \int \frac {c}{\sqrt {1-\frac {c^2}{x^2}} x^4} \, dx\\ &=-\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{2 x^2}-\frac {1}{2} (b c) \int \frac {1}{\sqrt {1-\frac {c^2}{x^2}} x^4} \, dx\\ &=-\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{2 x^2}+\frac {1}{2} (b c) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{4 c x}-\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{2 x^2}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-c^2 x^2}} \, dx,x,\frac {1}{x}\right )}{4 c}\\ &=-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{4 c x}+\frac {b \csc ^{-1}\left (\frac {x}{c}\right )}{4 c^2}-\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 65, normalized size = 1.14 \[ -\frac {a}{2 x^2}-\frac {b \sqrt {\frac {x^2-c^2}{x^2}}}{4 c x}+\frac {b \sin ^{-1}\left (\frac {c}{x}\right )}{4 c^2}-\frac {b \sin ^{-1}\left (\frac {c}{x}\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c/x])/x^3,x]

[Out]

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

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fricas [A]  time = 0.90, size = 55, normalized size = 0.96 \[ -\frac {b c x \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} + 2 \, a c^{2} + {\left (2 \, b c^{2} - b x^{2}\right )} \arcsin \left (\frac {c}{x}\right )}{4 \, c^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 70, normalized size = 1.23 \[ -\frac {\frac {2 \, b {\left (\frac {c^{2}}{x^{2}} - 1\right )} \arcsin \left (\frac {c}{x}\right )}{c} + \frac {2 \, a {\left (\frac {c^{2}}{x^{2}} - 1\right )}}{c} + \frac {b \arcsin \left (\frac {c}{x}\right )}{c} + \frac {b \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{x}}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 59, normalized size = 1.04 \[ -\frac {\frac {c^{2} a}{2 x^{2}}+b \left (\frac {\arcsin \left (\frac {c}{x}\right ) c^{2}}{2 x^{2}}+\frac {c \sqrt {1-\frac {c^{2}}{x^{2}}}}{4 x}-\frac {\arcsin \left (\frac {c}{x}\right )}{4}\right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 86, normalized size = 1.51 \[ \frac {1}{4} \, {\left (c {\left (\frac {x \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c^{2} x^{2} {\left (\frac {c^{2}}{x^{2}} - 1\right )} - c^{4}} - \frac {\arctan \left (\frac {x \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c}\right )}{c^{3}}\right )} - \frac {2 \, \arcsin \left (\frac {c}{x}\right )}{x^{2}}\right )} b - \frac {a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.34, size = 50, normalized size = 0.88 \[ -\frac {a}{2\,x^2}-\frac {b\,\sqrt {1-\frac {c^2}{x^2}}}{4\,c\,x}-\frac {b\,\mathrm {asin}\left (\frac {c}{x}\right )\,\left (\frac {2\,c^2}{x^2}-1\right )}{4\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c/x))/x^3,x)

[Out]

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

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sympy [A]  time = 3.04, size = 112, normalized size = 1.96 \[ - \frac {a}{2 x^{2}} - \frac {b c \left (\begin {cases} \frac {i \sqrt {\frac {c^{2}}{x^{2}} - 1}}{2 c^{2} x} + \frac {i \operatorname {acosh}{\left (\frac {c}{x} \right )}}{2 c^{3}} & \text {for}\: \left |{\frac {c^{2}}{x^{2}}}\right | > 1 \\- \frac {1}{2 x^{3} \sqrt {- \frac {c^{2}}{x^{2}} + 1}} + \frac {1}{2 c^{2} x \sqrt {- \frac {c^{2}}{x^{2}} + 1}} - \frac {\operatorname {asin}{\left (\frac {c}{x} \right )}}{2 c^{3}} & \text {otherwise} \end {cases}\right )}{2} - \frac {b \operatorname {asin}{\left (\frac {c}{x} \right )}}{2 x^{2}} \]

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)/(2*c**2*x) + I*acosh(c/x)/(2*c**3), Abs(c**2/x**2) > 1), (-
1/(2*x**3*sqrt(-c**2/x**2 + 1)) + 1/(2*c**2*x*sqrt(-c**2/x**2 + 1)) - asin(c/x)/(2*c**3), True))/2 - b*asin(c/
x)/(2*x**2)

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