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]

-(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

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Rubi [A]  time = 0.0382465, antiderivative size = 62, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [A]  time = 0.0183823, 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]

-a/(3*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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Maple [A]  time = 0.003, size = 65, normalized size = 1.1 \begin{align*}{c}^{3} \left ( -{\frac{a}{3\,{c}^{3}{x}^{3}}}+b \left ( -{\frac{\arcsin \left ( cx \right ) }{3\,{c}^{3}{x}^{3}}}-{\frac{1}{6\,{c}^{2}{x}^{2}}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{1}{6}{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) } \right ) \right ) \end{align*}

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/c^3/x^3*arcsin(c*x)-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 = 1.54952, size = 93, normalized size = 1.5 \begin{align*} -\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}} \end{align*}

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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Fricas [A]  time = 1.66349, size = 194, normalized size = 3.13 \begin{align*} -\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}} \end{align*}

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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Sympy [A]  time = 5.74433, size = 119, normalized size = 1.92 \begin{align*} - \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}} \end{align*}

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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Giac [B]  time = 2.03371, size = 383, normalized size = 6.18 \begin{align*} -\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}} \end{align*}

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