3.346 \(\int \frac{a+b \sin ^{-1}(c x^2)}{x^3} \, dx\)

Optimal. Leaf size=39 \[ -\frac{a+b \sin ^{-1}\left (c x^2\right )}{2 x^2}-\frac{1}{2} b c \tanh ^{-1}\left (\sqrt{1-c^2 x^4}\right ) \]

[Out]

-(a + b*ArcSin[c*x^2])/(2*x^2) - (b*c*ArcTanh[Sqrt[1 - c^2*x^4]])/2

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Rubi [A]  time = 0.0331595, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4842, 12, 266, 63, 208} \[ -\frac{a+b \sin ^{-1}\left (c x^2\right )}{2 x^2}-\frac{1}{2} b c \tanh ^{-1}\left (\sqrt{1-c^2 x^4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x^2])/x^3,x]

[Out]

-(a + b*ArcSin[c*x^2])/(2*x^2) - (b*c*ArcTanh[Sqrt[1 - c^2*x^4]])/2

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]

Rule 12

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

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

Mathematica [A]  time = 0.0063874, size = 44, normalized size = 1.13 \[ -\frac{a}{2 x^2}-\frac{1}{2} b c \tanh ^{-1}\left (\sqrt{1-c^2 x^4}\right )-\frac{b \sin ^{-1}\left (c x^2\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x^2])/x^3,x]

[Out]

-a/(2*x^2) - (b*ArcSin[c*x^2])/(2*x^2) - (b*c*ArcTanh[Sqrt[1 - c^2*x^4]])/2

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Maple [A]  time = 0.01, size = 38, normalized size = 1. \begin{align*} -{\frac{a}{2\,{x}^{2}}}+b \left ( -{\frac{\arcsin \left ( c{x}^{2} \right ) }{2\,{x}^{2}}}-{\frac{c}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x^2))/x^3,x)

[Out]

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

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Maxima [A]  time = 1.43334, size = 77, normalized size = 1.97 \begin{align*} -\frac{1}{4} \,{\left (c{\left (\log \left (\sqrt{-c^{2} x^{4} + 1} + 1\right ) - \log \left (\sqrt{-c^{2} x^{4} + 1} - 1\right )\right )} + \frac{2 \, \arcsin \left (c x^{2}\right )}{x^{2}}\right )} b - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.47718, size = 151, normalized size = 3.87 \begin{align*} -\frac{b c x^{2} \log \left (\sqrt{-c^{2} x^{4} + 1} + 1\right ) - b c x^{2} \log \left (\sqrt{-c^{2} x^{4} + 1} - 1\right ) + 2 \, b \arcsin \left (c x^{2}\right ) + 2 \, a}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.01823, size = 54, normalized size = 1.38 \begin{align*} - \frac{a}{2 x^{2}} + b c \left (\begin{cases} - \frac{\operatorname{acosh}{\left (\frac{1}{c x^{2}} \right )}}{2} & \text{for}\: \frac{1}{\left |{c^{2} x^{4}}\right |} > 1 \\\frac{i \operatorname{asin}{\left (\frac{1}{c x^{2}} \right )}}{2} & \text{otherwise} \end{cases}\right ) - \frac{b \operatorname{asin}{\left (c x^{2} \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x**2))/x**3,x)

[Out]

-a/(2*x**2) + b*c*Piecewise((-acosh(1/(c*x**2))/2, 1/Abs(c**2*x**4) > 1), (I*asin(1/(c*x**2))/2, True)) - b*as
in(c*x**2)/(2*x**2)

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Giac [B]  time = 1.37022, size = 478, normalized size = 12.26 \begin{align*} -\frac{\frac{\sqrt{-c^{2} x^{4} + 1} b c^{3} x^{2} \arcsin \left (c x^{2}\right )}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac{b c^{3} x^{2} \arcsin \left (c x^{2}\right )}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac{\sqrt{-c^{2} x^{4} + 1} a c^{3} x^{2}}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac{a c^{3} x^{2}}{{\left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}^{2}} - \frac{2 \, \sqrt{-c^{2} x^{4} + 1} b c^{2} \log \left (x^{2}{\left | c \right |}\right )}{\sqrt{-c^{2} x^{4} + 1} + 1} + \frac{2 \, \sqrt{-c^{2} x^{4} + 1} b c^{2} \log \left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}{\sqrt{-c^{2} x^{4} + 1} + 1} - \frac{2 \, b c^{2} \log \left (x^{2}{\left | c \right |}\right )}{\sqrt{-c^{2} x^{4} + 1} + 1} + \frac{2 \, b c^{2} \log \left (\sqrt{-c^{2} x^{4} + 1} + 1\right )}{\sqrt{-c^{2} x^{4} + 1} + 1} + \frac{\sqrt{-c^{2} x^{4} + 1} b c \arcsin \left (c x^{2}\right )}{x^{2}} + \frac{b c \arcsin \left (c x^{2}\right )}{x^{2}} + \frac{\sqrt{-c^{2} x^{4} + 1} a c}{x^{2}} + \frac{a c}{x^{2}}}{4 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^3,x, algorithm="giac")

[Out]

-1/4*(sqrt(-c^2*x^4 + 1)*b*c^3*x^2*arcsin(c*x^2)/(sqrt(-c^2*x^4 + 1) + 1)^2 + b*c^3*x^2*arcsin(c*x^2)/(sqrt(-c
^2*x^4 + 1) + 1)^2 + sqrt(-c^2*x^4 + 1)*a*c^3*x^2/(sqrt(-c^2*x^4 + 1) + 1)^2 + a*c^3*x^2/(sqrt(-c^2*x^4 + 1) +
1)^2 - 2*sqrt(-c^2*x^4 + 1)*b*c^2*log(x^2*abs(c))/(sqrt(-c^2*x^4 + 1) + 1) + 2*sqrt(-c^2*x^4 + 1)*b*c^2*log(s
qrt(-c^2*x^4 + 1) + 1)/(sqrt(-c^2*x^4 + 1) + 1) - 2*b*c^2*log(x^2*abs(c))/(sqrt(-c^2*x^4 + 1) + 1) + 2*b*c^2*l
og(sqrt(-c^2*x^4 + 1) + 1)/(sqrt(-c^2*x^4 + 1) + 1) + sqrt(-c^2*x^4 + 1)*b*c*arcsin(c*x^2)/x^2 + b*c*arcsin(c*
x^2)/x^2 + sqrt(-c^2*x^4 + 1)*a*c/x^2 + a*c/x^2)/c