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

Optimal. Leaf size=90 \[ -\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{2 x^2}-\frac {b d \tanh ^{-1}\left (\frac {-c^2-c d x^2+1}{\sqrt {1-c^2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}\right )}{2 \sqrt {1-c^2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {4842, 12, 1114, 724, 206} \[ -\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{2 x^2}-\frac {b d \tanh ^{-1}\left (\frac {-c^2-c d x^2+1}{\sqrt {1-c^2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}\right )}{2 \sqrt {1-c^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x^2])/x^3,x]

[Out]

-(a + b*ArcSin[c + d*x^2])/(2*x^2) - (b*d*ArcTanh[(1 - c^2 - c*d*x^2)/(Sqrt[1 - c^2]*Sqrt[1 - c^2 - 2*c*d*x^2
- d^2*x^4])])/(2*Sqrt[1 - c^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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 81, normalized size = 0.90 \[ \frac {1}{2} \left (-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{x^2}-\frac {b d \tanh ^{-1}\left (\frac {-c^2-c d x^2+1}{\sqrt {1-c^2} \sqrt {1-\left (c+d x^2\right )^2}}\right )}{\sqrt {1-c^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 280, normalized size = 3.11 \[ \left [-\frac {\sqrt {-c^{2} + 1} b d x^{2} \log \left (\frac {{\left (2 \, c^{2} - 1\right )} d^{2} x^{4} + 2 \, c^{4} + 4 \, {\left (c^{3} - c\right )} d x^{2} + 2 \, \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (c d x^{2} + c^{2} - 1\right )} \sqrt {-c^{2} + 1} - 4 \, c^{2} + 2}{x^{4}}\right ) + 2 \, a c^{2} + 2 \, {\left (b c^{2} - b\right )} \arcsin \left (d x^{2} + c\right ) - 2 \, a}{4 \, {\left (c^{2} - 1\right )} x^{2}}, \frac {\sqrt {c^{2} - 1} b d x^{2} \arctan \left (\frac {\sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (c d x^{2} + c^{2} - 1\right )} \sqrt {c^{2} - 1}}{{\left (c^{2} - 1\right )} d^{2} x^{4} + c^{4} + 2 \, {\left (c^{3} - c\right )} d x^{2} - 2 \, c^{2} + 1}\right ) - a c^{2} - {\left (b c^{2} - b\right )} \arcsin \left (d x^{2} + c\right ) + a}{2 \, {\left (c^{2} - 1\right )} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(sqrt(-c^2 + 1)*b*d*x^2*log(((2*c^2 - 1)*d^2*x^4 + 2*c^4 + 4*(c^3 - c)*d*x^2 + 2*sqrt(-d^2*x^4 - 2*c*d*x
^2 - c^2 + 1)*(c*d*x^2 + c^2 - 1)*sqrt(-c^2 + 1) - 4*c^2 + 2)/x^4) + 2*a*c^2 + 2*(b*c^2 - b)*arcsin(d*x^2 + c)
- 2*a)/((c^2 - 1)*x^2), 1/2*(sqrt(c^2 - 1)*b*d*x^2*arctan(sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1)*(c*d*x^2 + c^2
- 1)*sqrt(c^2 - 1)/((c^2 - 1)*d^2*x^4 + c^4 + 2*(c^3 - c)*d*x^2 - 2*c^2 + 1)) - a*c^2 - (b*c^2 - b)*arcsin(d*
x^2 + c) + a)/((c^2 - 1)*x^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 89, normalized size = 0.99 \[ -\frac {a}{2 x^{2}}-\frac {b \arcsin \left (d \,x^{2}+c \right )}{2 x^{2}}-\frac {b d \ln \left (\frac {-2 c^{2}+2-2 c d \,x^{2}+2 \sqrt {-c^{2}+1}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{x^{2}}\right )}{2 \sqrt {-c^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c-1>0)', see `assume?` for mor
e details)Is c-1 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (d\,x^2+c\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x^2))/x^3,x)

[Out]

int((a + b*asin(c + d*x^2))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c + d x^{2} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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