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

Optimal. Leaf size=82 \[ -\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{4 x^4}+\frac {3 b \csc ^{-1}\left (\frac {x}{c}\right )}{32 c^4}-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{16 c x^3}-\frac {3 b \sqrt {1-\frac {c^2}{x^2}}}{32 c^3 x} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{4 x^4}-\frac {3 b \sqrt {1-\frac {c^2}{x^2}}}{32 c^3 x}-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{16 c x^3}+\frac {3 b \csc ^{-1}\left (\frac {x}{c}\right )}{32 c^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c/x])/x^5,x]

[Out]

-(b*Sqrt[1 - c^2/x^2])/(16*c*x^3) - (3*b*Sqrt[1 - c^2/x^2])/(32*c^3*x) + (3*b*ArcCsc[x/c])/(32*c^4) - (a + b*A
rcSin[c/x])/(4*x^4)

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^5} \, dx &=-\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{4 x^4}-\frac {1}{4} b \int \frac {c}{\sqrt {1-\frac {c^2}{x^2}} x^6} \, dx\\ &=-\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{4 x^4}-\frac {1}{4} (b c) \int \frac {1}{\sqrt {1-\frac {c^2}{x^2}} x^6} \, dx\\ &=-\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{4 x^4}+\frac {1}{4} (b c) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1-c^2 x^2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {b \sqrt {1-\frac {c^2}{x^2}}}{16 c x^3}-\frac {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{4 x^4}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx,x,\frac {1}{x}\right )}{16 c}\\ &=-\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 {a+b \sin ^{-1}\left (\frac {c}{x}\right )}{4 x^4}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-c^2 x^2}} \, dx,x,\frac {1}{x}\right )}{32 c^3}\\ &=-\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 \sin ^{-1}\left (\frac {c}{x}\right )}{4 x^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 77, normalized size = 0.94 \[ -\frac {a}{4 x^4}+\frac {3 b \sin ^{-1}\left (\frac {c}{x}\right )}{32 c^4}+b \left (-\frac {3}{32 c^3 x}-\frac {1}{16 c x^3}\right ) \sqrt {\frac {x^2-c^2}{x^2}}-\frac {b \sin ^{-1}\left (\frac {c}{x}\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c/x])/x^5,x]

[Out]

-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*ArcSi
n[c/x])/(4*x^4)

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fricas [A]  time = 0.65, size = 67, normalized size = 0.82 \[ -\frac {8 \, a c^{4} + {\left (8 \, b c^{4} - 3 \, b x^{4}\right )} \arcsin \left (\frac {c}{x}\right ) + {\left (2 \, b c^{3} x + 3 \, b c x^{3}\right )} \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}}}{32 \, c^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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giac [A]  time = 0.24, size = 111, normalized size = 1.35 \[ -\frac {\frac {8 \, b {\left (\frac {c^{2}}{x^{2}} - 1\right )}^{2} \arcsin \left (\frac {c}{x}\right )}{c^{3}} + \frac {16 \, b {\left (\frac {c^{2}}{x^{2}} - 1\right )} \arcsin \left (\frac {c}{x}\right )}{c^{3}} - \frac {2 \, b {\left (-\frac {c^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}}}{c^{2} x} + \frac {5 \, b \arcsin \left (\frac {c}{x}\right )}{c^{3}} + \frac {5 \, b \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c^{2} x} + \frac {8 \, a c}{x^{4}}}{32 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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maple [A]  time = 0.01, size = 79, normalized size = 0.96 \[ -\frac {\frac {c^{4} a}{4 x^{4}}+b \left (\frac {c^{4} \arcsin \left (\frac {c}{x}\right )}{4 x^{4}}+\frac {c^{3} \sqrt {1-\frac {c^{2}}{x^{2}}}}{16 x^{3}}+\frac {3 c \sqrt {1-\frac {c^{2}}{x^{2}}}}{32 x}-\frac {3 \arcsin \left (\frac {c}{x}\right )}{32}\right )}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.06, size = 126, normalized size = 1.54 \[ -\frac {1}{32} \, {\left (c {\left (\frac {3 \, x^{3} {\left (-\frac {c^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}} + 5 \, c^{2} x \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c^{4} x^{4} {\left (\frac {c^{2}}{x^{2}} - 1\right )}^{2} - 2 \, c^{6} x^{2} {\left (\frac {c^{2}}{x^{2}} - 1\right )} + c^{8}} + \frac {3 \, \arctan \left (\frac {x \sqrt {-\frac {c^{2}}{x^{2}} + 1}}{c}\right )}{c^{5}}\right )} + \frac {8 \, \arcsin \left (\frac {c}{x}\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (\frac {c}{x}\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c/x))/x^5,x)

[Out]

int((a + b*asin(c/x))/x^5, x)

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sympy [A]  time = 5.06, size = 180, normalized size = 2.20 \[ - \frac {a}{4 x^{4}} - \frac {b c \left (\begin {cases} \frac {i}{4 x^{5} \sqrt {\frac {c^{2}}{x^{2}} - 1}} + \frac {i}{8 c^{2} x^{3} \sqrt {\frac {c^{2}}{x^{2}} - 1}} - \frac {3 i}{8 c^{4} x \sqrt {\frac {c^{2}}{x^{2}} - 1}} + \frac {3 i \operatorname {acosh}{\left (\frac {c}{x} \right )}}{8 c^{5}} & \text {for}\: \left |{\frac {c^{2}}{x^{2}}}\right | > 1 \\- \frac {1}{4 x^{5} \sqrt {- \frac {c^{2}}{x^{2}} + 1}} - \frac {1}{8 c^{2} x^{3} \sqrt {- \frac {c^{2}}{x^{2}} + 1}} + \frac {3}{8 c^{4} x \sqrt {- \frac {c^{2}}{x^{2}} + 1}} - \frac {3 \operatorname {asin}{\left (\frac {c}{x} \right )}}{8 c^{5}} & \text {otherwise} \end {cases}\right )}{4} - \frac {b \operatorname {asin}{\left (\frac {c}{x} \right )}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*a
sin(c/x)/(4*x**4)

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