\(\int \frac {a+b \arcsin (\frac {c}{x})}{x^4} \, dx\) [377]

Optimal result

Integrand size = 14, antiderivative size = 62 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{3 c^3}+\frac {b \left (1-\frac {c^2}{x^2}\right )^{3/2}}{9 c^3}-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{3 x^3} \]

[Out]

1/9*b*(1-c^2/x^2)^(3/2)/c^3+1/3*(-a-b*arcsin(c/x))/x^3-1/3*b*(1-c^2/x^2)^(1/2)/c^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4926, 12, 272, 45} \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{3 x^3}+\frac {b \left (1-\frac {c^2}{x^2}\right )^{3/2}}{9 c^3}-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{3 c^3} \]

[In]

Int[(a + b*ArcSin[c/x])/x^4,x]

[Out]

-1/3*(b*Sqrt[1 - c^2/x^2])/c^3 + (b*(1 - c^2/x^2)^(3/2))/(9*c^3) - (a + b*ArcSin[c/x])/(3*x^3)

Rule 12

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4926

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*} \text {integral}& = -\frac {a+b \arcsin \left (\frac {c}{x}\right )}{3 x^3}-\frac {1}{3} b \int \frac {c}{\sqrt {1-\frac {c^2}{x^2}} x^5} \, dx \\ & = -\frac {a+b \arcsin \left (\frac {c}{x}\right )}{3 x^3}-\frac {1}{3} (b c) \int \frac {1}{\sqrt {1-\frac {c^2}{x^2}} x^5} \, dx \\ & = -\frac {a+b \arcsin \left (\frac {c}{x}\right )}{3 x^3}+\frac {1}{6} (b c) \text {Subst}\left (\int \frac {x}{\sqrt {1-c^2 x}} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {a+b \arcsin \left (\frac {c}{x}\right )}{3 x^3}+\frac {1}{6} (b c) \text {Subst}\left (\int \left (\frac {1}{c^2 \sqrt {1-c^2 x}}-\frac {\sqrt {1-c^2 x}}{c^2}\right ) \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {b \sqrt {1-\frac {c^2}{x^2}}}{3 c^3}+\frac {b \left (1-\frac {c^2}{x^2}\right )^{3/2}}{9 c^3}-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.97 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {a}{3 x^3}+b \left (-\frac {2}{9 c^3}-\frac {1}{9 c x^2}\right ) \sqrt {\frac {-c^2+x^2}{x^2}}-\frac {b \arcsin \left (\frac {c}{x}\right )}{3 x^3} \]

[In]

Integrate[(a + b*ArcSin[c/x])/x^4,x]

[Out]

-1/3*a/x^3 + b*(-2/(9*c^3) - 1/(9*c*x^2))*Sqrt[(-c^2 + x^2)/x^2] - (b*ArcSin[c/x])/(3*x^3)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02

method	result	size
parts	\(-\frac {a}{3 x^{3}}-\frac {b \left (\frac {c^{3} \arcsin \left (\frac {c}{x}\right )}{3 x^{3}}+\frac {c^{2} \sqrt {1-\frac {c^{2}}{x^{2}}}}{9 x^{2}}+\frac {2 \sqrt {1-\frac {c^{2}}{x^{2}}}}{9}\right )}{c^{3}}\)	\(63\)
derivativedivides	\(-\frac {\frac {a \,c^{3}}{3 x^{3}}+b \left (\frac {c^{3} \arcsin \left (\frac {c}{x}\right )}{3 x^{3}}+\frac {c^{2} \sqrt {1-\frac {c^{2}}{x^{2}}}}{9 x^{2}}+\frac {2 \sqrt {1-\frac {c^{2}}{x^{2}}}}{9}\right )}{c^{3}}\)	\(67\)
default	\(-\frac {\frac {a \,c^{3}}{3 x^{3}}+b \left (\frac {c^{3} \arcsin \left (\frac {c}{x}\right )}{3 x^{3}}+\frac {c^{2} \sqrt {1-\frac {c^{2}}{x^{2}}}}{9 x^{2}}+\frac {2 \sqrt {1-\frac {c^{2}}{x^{2}}}}{9}\right )}{c^{3}}\)	\(67\)

[In]

int((a+b*arcsin(c/x))/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*a/x^3-b/c^3*(1/3*c^3/x^3*arcsin(c/x)+1/9*c^2/x^2*(1-c^2/x^2)^(1/2)+2/9*(1-c^2/x^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {3 \, b c^{3} \arcsin \left (\frac {c}{x}\right ) + 3 \, a c^{3} + {\left (b c^{2} x + 2 \, b x^{3}\right )} \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}}}{9 \, c^{3} x^{3}} \]

[In]

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

[Out]

-1/9*(3*b*c^3*arcsin(c/x) + 3*a*c^3 + (b*c^2*x + 2*b*x^3)*sqrt(-(c^2 - x^2)/x^2))/(c^3*x^3)

Sympy [A] (verification not implemented)

Time = 1.67 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.81 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^4} \, dx=- \frac {a}{3 x^{3}} - \frac {b c \left (\begin {cases} \frac {\sqrt {-1 + \frac {x^{2}}{c^{2}}}}{3 c x^{3}} + \frac {2 \sqrt {-1 + \frac {x^{2}}{c^{2}}}}{3 c^{3} x} & \text {for}\: \left |{\frac {x^{2}}{c^{2}}}\right | > 1 \\\frac {i \sqrt {1 - \frac {x^{2}}{c^{2}}}}{3 c x^{3}} + \frac {2 i \sqrt {1 - \frac {x^{2}}{c^{2}}}}{3 c^{3} x} & \text {otherwise} \end {cases}\right )}{3} - \frac {b \operatorname {asin}{\left (\frac {c}{x} \right )}}{3 x^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^4} \, dx=\frac {1}{9} \, {\left (c {\left (\frac {{\left (-\frac {c^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}}}{c^{4}} - \frac {3 \, \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c^{4}}\right )} - \frac {3 \, \arcsin \left (\frac {c}{x}\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \]

[In]

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

[Out]

1/9*(c*((-c^2/x^2 + 1)^(3/2)/c^4 - 3*sqrt(-c^2/x^2 + 1)/c^4) - 3*arcsin(c/x)/x^3)*b - 1/3*a/x^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.42 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^4} \, dx=-\frac {\frac {3 \, b {\left (\frac {c^{2}}{x^{2}} - 1\right )} \arcsin \left (\frac {c}{x}\right )}{c x} - \frac {b {\left (-\frac {c^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}}}{c^{2}} + \frac {3 \, b \arcsin \left (\frac {c}{x}\right )}{c x} + \frac {3 \, b \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c^{2}} + \frac {3 \, a c}{x^{3}}}{9 \, c} \]

[In]

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

[Out]

-1/9*(3*b*(c^2/x^2 - 1)*arcsin(c/x)/(c*x) - b*(-c^2/x^2 + 1)^(3/2)/c^2 + 3*b*arcsin(c/x)/(c*x) + 3*b*sqrt(-c^2
/x^2 + 1)/c^2 + 3*a*c/x^3)/c

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^4} \, dx=\int \frac {a+b\,\mathrm {asin}\left (\frac {c}{x}\right )}{x^4} \,d x \]

[In]

int((a + b*asin(c/x))/x^4,x)

[Out]

int((a + b*asin(c/x))/x^4, x)