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]

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

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 12

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

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Mathematica [A]  time = 0.01, 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]

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

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fricas [A]  time = 0.47, size = 61, normalized size = 1.56 \[ -\frac {b c x^{2} \log \left (\sqrt {-c^{2} x^{4} + 1} + 1\right ) - b c x^{2} \log \left (\sqrt {-c^{2} x^{4} + 1} - 1\right ) + 2 \, b \arcsin \left (c x^{2}\right ) + 2 \, a}{4 \, x^{2}} \]

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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giac [B]  time = 0.35, size = 354, normalized size = 9.08 \[ -\frac {\frac {\sqrt {-c^{2} x^{4} + 1} b c^{3} x^{2} \arcsin \left (c x^{2}\right )}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac {b c^{3} x^{2} \arcsin \left (c x^{2}\right )}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac {\sqrt {-c^{2} x^{4} + 1} a c^{3} x^{2}}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac {a c^{3} x^{2}}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2}} - \frac {2 \, \sqrt {-c^{2} x^{4} + 1} b c^{2} \log \left (x^{2} {\left | c \right |}\right )}{\sqrt {-c^{2} x^{4} + 1} + 1} + \frac {2 \, \sqrt {-c^{2} x^{4} + 1} b c^{2} \log \left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}{\sqrt {-c^{2} x^{4} + 1} + 1} - \frac {2 \, b c^{2} \log \left (x^{2} {\left | c \right |}\right )}{\sqrt {-c^{2} x^{4} + 1} + 1} + \frac {2 \, b c^{2} \log \left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}{\sqrt {-c^{2} x^{4} + 1} + 1} + \frac {\sqrt {-c^{2} x^{4} + 1} b c \arcsin \left (c x^{2}\right )}{x^{2}} + \frac {b c \arcsin \left (c x^{2}\right )}{x^{2}} + \frac {\sqrt {-c^{2} x^{4} + 1} a c}{x^{2}} + \frac {a c}{x^{2}}}{4 \, c} \]

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

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maple [A]  time = 0.02, size = 38, normalized size = 0.97 \[ -\frac {a}{2 x^{2}}+b \left (-\frac {\arcsin \left (c \,x^{2}\right )}{2 x^{2}}-\frac {c \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{4}+1}}\right )}{2}\right ) \]

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 = 0.42, size = 57, normalized size = 1.46 \[ -\frac {1}{4} \, {\left (c {\left (\log \left (\sqrt {-c^{2} x^{4} + 1} + 1\right ) - \log \left (\sqrt {-c^{2} x^{4} + 1} - 1\right )\right )} + \frac {2 \, \arcsin \left (c x^{2}\right )}{x^{2}}\right )} b - \frac {a}{2 \, x^{2}} \]

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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mupad [B]  time = 0.32, size = 36, normalized size = 0.92 \[ -\frac {a}{2\,x^2}-\frac {b\,c\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-c^2\,x^4}}\right )}{2}-\frac {b\,\mathrm {asin}\left (c\,x^2\right )}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x^2))/x^3,x)

[Out]

- a/(2*x^2) - (b*c*atanh(1/(1 - c^2*x^4)^(1/2)))/2 - (b*asin(c*x^2))/(2*x^2)

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sympy [A]  time = 1.74, size = 54, normalized size = 1.38 \[ - \frac {a}{2 x^{2}} + b c \left (\begin {cases} - \frac {\operatorname {acosh}{\left (\frac {1}{c x^{2}} \right )}}{2} & \text {for}\: \frac {1}{\left |{c^{2} x^{4}}\right |} > 1 \\\frac {i \operatorname {asin}{\left (\frac {1}{c x^{2}} \right )}}{2} & \text {otherwise} \end {cases}\right ) - \frac {b \operatorname {asin}{\left (c x^{2} \right )}}{2 x^{2}} \]

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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