3.147 \(\int \frac {a+b \sin ^{-1}(c x)}{x^4} \, dx\)

Optimal. Leaf size=62 \[ -\frac {a+b \sin ^{-1}(c x)}{3 x^3}-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2}-\frac {1}{6} b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right ) \]

[Out]

1/3*(-a-b*arcsin(c*x))/x^3-1/6*b*c^3*arctanh((-c^2*x^2+1)^(1/2))-1/6*b*c*(-c^2*x^2+1)^(1/2)/x^2

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Rubi [A]  time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4627, 266, 51, 63, 208} \[ -\frac {a+b \sin ^{-1}(c x)}{3 x^3}-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2}-\frac {1}{6} b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/x^4,x]

[Out]

-(b*c*Sqrt[1 - c^2*x^2])/(6*x^2) - (a + b*ArcSin[c*x])/(3*x^3) - (b*c^3*ArcTanh[Sqrt[1 - c^2*x^2]])/6

Rule 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

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 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^4} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{3 x^3}+\frac {1}{3} (b c) \int \frac {1}{x^3 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {a+b \sin ^{-1}(c x)}{3 x^3}+\frac {1}{6} (b c) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1-c^2 x}} \, dx,x,x^2\right )\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2}-\frac {a+b \sin ^{-1}(c x)}{3 x^3}+\frac {1}{12} \left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2}-\frac {a+b \sin ^{-1}(c x)}{3 x^3}-\frac {1}{6} (b c) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2}-\frac {a+b \sin ^{-1}(c x)}{3 x^3}-\frac {1}{6} b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 67, normalized size = 1.08 \[ -\frac {a}{3 x^3}-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2}-\frac {1}{6} b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )-\frac {b \sin ^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/x^4,x]

[Out]

-1/3*a/x^3 - (b*c*Sqrt[1 - c^2*x^2])/(6*x^2) - (b*ArcSin[c*x])/(3*x^3) - (b*c^3*ArcTanh[Sqrt[1 - c^2*x^2]])/6

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fricas [A]  time = 0.57, size = 80, normalized size = 1.29 \[ -\frac {b c^{3} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - b c^{3} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) + 2 \, \sqrt {-c^{2} x^{2} + 1} b c x + 4 \, b \arcsin \left (c x\right ) + 4 \, a}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(b*c^3*x^3*log(sqrt(-c^2*x^2 + 1) + 1) - b*c^3*x^3*log(sqrt(-c^2*x^2 + 1) - 1) + 2*sqrt(-c^2*x^2 + 1)*b*
c*x + 4*b*arcsin(c*x) + 4*a)/x^3

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giac [B]  time = 0.92, size = 284, normalized size = 4.58 \[ -\frac {b c^{6} x^{3} \arcsin \left (c x\right )}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}} - \frac {a c^{6} x^{3}}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}} + \frac {b c^{5} x^{2}}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}} - \frac {b c^{4} x \arcsin \left (c x\right )}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} - \frac {a c^{4} x}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} + \frac {1}{6} \, b c^{3} \log \left ({\left | c \right |} {\left | x \right |}\right ) - \frac {1}{6} \, b c^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - \frac {b c^{2} {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )} \arcsin \left (c x\right )}{8 \, x} - \frac {a c^{2} {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}}{8 \, x} - \frac {b c {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}}{24 \, x^{2}} - \frac {b {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3} \arcsin \left (c x\right )}{24 \, x^{3}} - \frac {a {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}}{24 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*b*c^6*x^3*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1)^3 - 1/24*a*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + 1/24*b*c^
5*x^2/(sqrt(-c^2*x^2 + 1) + 1)^2 - 1/8*b*c^4*x*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1) - 1/8*a*c^4*x/(sqrt(-c^2*x
^2 + 1) + 1) + 1/6*b*c^3*log(abs(c)*abs(x)) - 1/6*b*c^3*log(sqrt(-c^2*x^2 + 1) + 1) - 1/8*b*c^2*(sqrt(-c^2*x^2
+ 1) + 1)*arcsin(c*x)/x - 1/8*a*c^2*(sqrt(-c^2*x^2 + 1) + 1)/x - 1/24*b*c*(sqrt(-c^2*x^2 + 1) + 1)^2/x^2 - 1/
24*b*(sqrt(-c^2*x^2 + 1) + 1)^3*arcsin(c*x)/x^3 - 1/24*a*(sqrt(-c^2*x^2 + 1) + 1)^3/x^3

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maple [A]  time = 0.00, size = 65, normalized size = 1.05 \[ c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {\arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(-1/3*a/c^3/x^3+b*(-1/3*arcsin(c*x)/c^3/x^3-1/6/c^2/x^2*(-c^2*x^2+1)^(1/2)-1/6*arctanh(1/(-c^2*x^2+1)^(1/2
))))

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maxima [A]  time = 0.44, size = 69, normalized size = 1.11 \[ -\frac {1}{6} \, {\left ({\left (c^{2} \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{2}}\right )} c + \frac {2 \, \arcsin \left (c x\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((c^2*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + sqrt(-c^2*x^2 + 1)/x^2)*c + 2*arcsin(c*x)/x^3)*b - 1/
3*a/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/x^4,x)

[Out]

int((a + b*asin(c*x))/x^4, x)

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sympy [A]  time = 2.61, size = 119, normalized size = 1.92 \[ - \frac {a}{3 x^{3}} + \frac {b c \left (\begin {cases} - \frac {c^{2} \operatorname {acosh}{\left (\frac {1}{c x} \right )}}{2} - \frac {c \sqrt {-1 + \frac {1}{c^{2} x^{2}}}}{2 x} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac {i c^{2} \operatorname {asin}{\left (\frac {1}{c x} \right )}}{2} - \frac {i c}{2 x \sqrt {1 - \frac {1}{c^{2} x^{2}}}} + \frac {i}{2 c x^{3} \sqrt {1 - \frac {1}{c^{2} x^{2}}}} & \text {otherwise} \end {cases}\right )}{3} - \frac {b \operatorname {asin}{\left (c x \right )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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