Integrand size = 14, antiderivative size = 82 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^5} \, dx=-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{16 c x^3}-\frac {3 b \sqrt {1-\frac {c^2}{x^2}}}{32 c^3 x}+\frac {3 b \csc ^{-1}\left (\frac {x}{c}\right )}{32 c^4}-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{4 x^4} \]
3/32*b*arccsc(x/c)/c^4+1/4*(-a-b*arcsin(c/x))/x^4-1/16*b*(1-c^2/x^2)^(1/2) /c/x^3-3/32*b*(1-c^2/x^2)^(1/2)/c^3/x
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^5} \, dx=-\frac {a}{4 x^4}+b \left (-\frac {1}{16 c x^3}-\frac {3}{32 c^3 x}\right ) \sqrt {\frac {-c^2+x^2}{x^2}}+\frac {3 b \arcsin \left (\frac {c}{x}\right )}{32 c^4}-\frac {b \arcsin \left (\frac {c}{x}\right )}{4 x^4} \]
-1/4*a/x^4 + b*(-1/16*1/(c*x^3) - 3/(32*c^3*x))*Sqrt[(-c^2 + x^2)/x^2] + ( 3*b*ArcSin[c/x])/(32*c^4) - (b*ArcSin[c/x])/(4*x^4)
Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5341, 25, 27, 858, 262, 262, 223}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 5341 |
\(\displaystyle \frac {1}{4} b \int -\frac {c}{\sqrt {1-\frac {c^2}{x^2}} x^6}dx-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} b \int \frac {c}{\sqrt {1-\frac {c^2}{x^2}} x^6}dx-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{4} b c \int \frac {1}{\sqrt {1-\frac {c^2}{x^2}} x^6}dx-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {1}{4} b c \int \frac {1}{\sqrt {1-\frac {c^2}{x^2}} x^4}d\frac {1}{x}-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{4} b c \left (\frac {3 \int \frac {1}{\sqrt {1-\frac {c^2}{x^2}} x^2}d\frac {1}{x}}{4 c^2}-\frac {\sqrt {1-\frac {c^2}{x^2}}}{4 c^2 x^3}\right )-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{4} b c \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {1-\frac {c^2}{x^2}}}d\frac {1}{x}}{2 c^2}-\frac {\sqrt {1-\frac {c^2}{x^2}}}{2 c^2 x}\right )}{4 c^2}-\frac {\sqrt {1-\frac {c^2}{x^2}}}{4 c^2 x^3}\right )-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{4 x^4}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{4} b c \left (\frac {3 \left (\frac {\arcsin \left (\frac {c}{x}\right )}{2 c^3}-\frac {\sqrt {1-\frac {c^2}{x^2}}}{2 c^2 x}\right )}{4 c^2}-\frac {\sqrt {1-\frac {c^2}{x^2}}}{4 c^2 x^3}\right )-\frac {a+b \arcsin \left (\frac {c}{x}\right )}{4 x^4}\) |
-1/4*(a + b*ArcSin[c/x])/x^4 + (b*c*(-1/4*Sqrt[1 - c^2/x^2]/(c^2*x^3) + (3 *(-1/2*Sqrt[1 - c^2/x^2]/(c^2*x) + ArcSin[c/x]/(2*c^3)))/(4*c^2)))/4
3.4.78.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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] && Int egerQ[m]
Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Sim p[(c + d*x)^(m + 1)*((a + b*ArcSin[u])/(d*(m + 1))), x] - Simp[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]
Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91
method | result | size |
parts | \(-\frac {a}{4 x^{4}}-\frac {b \left (\frac {c^{4} \arcsin \left (\frac {c}{x}\right )}{4 x^{4}}+\frac {c^{3} \sqrt {1-\frac {c^{2}}{x^{2}}}}{16 x^{3}}+\frac {3 c \sqrt {1-\frac {c^{2}}{x^{2}}}}{32 x}-\frac {3 \arcsin \left (\frac {c}{x}\right )}{32}\right )}{c^{4}}\) | \(75\) |
derivativedivides | \(-\frac {\frac {a \,c^{4}}{4 x^{4}}+b \left (\frac {c^{4} \arcsin \left (\frac {c}{x}\right )}{4 x^{4}}+\frac {c^{3} \sqrt {1-\frac {c^{2}}{x^{2}}}}{16 x^{3}}+\frac {3 c \sqrt {1-\frac {c^{2}}{x^{2}}}}{32 x}-\frac {3 \arcsin \left (\frac {c}{x}\right )}{32}\right )}{c^{4}}\) | \(79\) |
default | \(-\frac {\frac {a \,c^{4}}{4 x^{4}}+b \left (\frac {c^{4} \arcsin \left (\frac {c}{x}\right )}{4 x^{4}}+\frac {c^{3} \sqrt {1-\frac {c^{2}}{x^{2}}}}{16 x^{3}}+\frac {3 c \sqrt {1-\frac {c^{2}}{x^{2}}}}{32 x}-\frac {3 \arcsin \left (\frac {c}{x}\right )}{32}\right )}{c^{4}}\) | \(79\) |
-1/4*a/x^4-b/c^4*(1/4*c^4/x^4*arcsin(c/x)+1/16*c^3/x^3*(1-c^2/x^2)^(1/2)+3 /32*c/x*(1-c^2/x^2)^(1/2)-3/32*arcsin(c/x))
Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^5} \, dx=-\frac {8 \, a c^{4} + {\left (8 \, b c^{4} - 3 \, b x^{4}\right )} \arcsin \left (\frac {c}{x}\right ) + {\left (2 \, b c^{3} x + 3 \, b c x^{3}\right )} \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}}}{32 \, c^{4} x^{4}} \]
-1/32*(8*a*c^4 + (8*b*c^4 - 3*b*x^4)*arcsin(c/x) + (2*b*c^3*x + 3*b*c*x^3) *sqrt(-(c^2 - x^2)/x^2))/(c^4*x^4)
Time = 3.78 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.20 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^5} \, dx=- \frac {a}{4 x^{4}} - \frac {b c \left (\begin {cases} \frac {i}{4 x^{5} \sqrt {\frac {c^{2}}{x^{2}} - 1}} + \frac {i}{8 c^{2} x^{3} \sqrt {\frac {c^{2}}{x^{2}} - 1}} - \frac {3 i}{8 c^{4} x \sqrt {\frac {c^{2}}{x^{2}} - 1}} + \frac {3 i \operatorname {acosh}{\left (\frac {c}{x} \right )}}{8 c^{5}} & \text {for}\: \left |{\frac {c^{2}}{x^{2}}}\right | > 1 \\- \frac {1}{4 x^{5} \sqrt {- \frac {c^{2}}{x^{2}} + 1}} - \frac {1}{8 c^{2} x^{3} \sqrt {- \frac {c^{2}}{x^{2}} + 1}} + \frac {3}{8 c^{4} x \sqrt {- \frac {c^{2}}{x^{2}} + 1}} - \frac {3 \operatorname {asin}{\left (\frac {c}{x} \right )}}{8 c^{5}} & \text {otherwise} \end {cases}\right )}{4} - \frac {b \operatorname {asin}{\left (\frac {c}{x} \right )}}{4 x^{4}} \]
-a/(4*x**4) - b*c*Piecewise((I/(4*x**5*sqrt(c**2/x**2 - 1)) + I/(8*c**2*x* *3*sqrt(c**2/x**2 - 1)) - 3*I/(8*c**4*x*sqrt(c**2/x**2 - 1)) + 3*I*acosh(c /x)/(8*c**5), Abs(c**2/x**2) > 1), (-1/(4*x**5*sqrt(-c**2/x**2 + 1)) - 1/( 8*c**2*x**3*sqrt(-c**2/x**2 + 1)) + 3/(8*c**4*x*sqrt(-c**2/x**2 + 1)) - 3* asin(c/x)/(8*c**5), True))/4 - b*asin(c/x)/(4*x**4)
Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.54 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^5} \, dx=-\frac {1}{32} \, {\left (c {\left (\frac {3 \, x^{3} {\left (-\frac {c^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}} + 5 \, c^{2} x \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c^{4} x^{4} {\left (\frac {c^{2}}{x^{2}} - 1\right )}^{2} - 2 \, c^{6} x^{2} {\left (\frac {c^{2}}{x^{2}} - 1\right )} + c^{8}} + \frac {3 \, \arctan \left (\frac {x \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c}\right )}{c^{5}}\right )} + \frac {8 \, \arcsin \left (\frac {c}{x}\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \]
-1/32*(c*((3*x^3*(-c^2/x^2 + 1)^(3/2) + 5*c^2*x*sqrt(-c^2/x^2 + 1))/(c^4*x ^4*(c^2/x^2 - 1)^2 - 2*c^6*x^2*(c^2/x^2 - 1) + c^8) + 3*arctan(x*sqrt(-c^2 /x^2 + 1)/c)/c^5) + 8*arcsin(c/x)/x^4)*b - 1/4*a/x^4
Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.35 \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^5} \, dx=-\frac {\frac {8 \, b {\left (\frac {c^{2}}{x^{2}} - 1\right )}^{2} \arcsin \left (\frac {c}{x}\right )}{c^{3}} + \frac {16 \, b {\left (\frac {c^{2}}{x^{2}} - 1\right )} \arcsin \left (\frac {c}{x}\right )}{c^{3}} - \frac {2 \, b {\left (-\frac {c^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}}}{c^{2} x} + \frac {5 \, b \arcsin \left (\frac {c}{x}\right )}{c^{3}} + \frac {5 \, b \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c^{2} x} + \frac {8 \, a c}{x^{4}}}{32 \, c} \]
-1/32*(8*b*(c^2/x^2 - 1)^2*arcsin(c/x)/c^3 + 16*b*(c^2/x^2 - 1)*arcsin(c/x )/c^3 - 2*b*(-c^2/x^2 + 1)^(3/2)/(c^2*x) + 5*b*arcsin(c/x)/c^3 + 5*b*sqrt( -c^2/x^2 + 1)/(c^2*x) + 8*a*c/x^4)/c
Timed out. \[ \int \frac {a+b \arcsin \left (\frac {c}{x}\right )}{x^5} \, dx=\int \frac {a+b\,\mathrm {asin}\left (\frac {c}{x}\right )}{x^5} \,d x \]