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

Optimal. Leaf size=64 \[ -\frac {b \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{\sqrt {1-a^2}}-\frac {\sin ^{-1}(a+b x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4805, 4743, 725, 206} \[ -\frac {b \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{\sqrt {1-a^2}}-\frac {\sin ^{-1}(a+b x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]/x^2,x]

[Out]

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

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 725

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

Rule 4743

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

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 66, normalized size = 1.03 \[ -\frac {b \tanh ^{-1}\left (\frac {-a^2-a b x+1}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{\sqrt {1-a^2}}-\frac {\sin ^{-1}(a+b x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]/x^2,x]

[Out]

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

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fricas [A]  time = 0.50, size = 233, normalized size = 3.64 \[ \left [-\frac {\sqrt {-a^{2} + 1} b x \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x + 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) + 2 \, {\left (a^{2} - 1\right )} \arcsin \left (b x + a\right )}{2 \, {\left (a^{2} - 1\right )} x}, \frac {\sqrt {a^{2} - 1} b x \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - {\left (a^{2} - 1\right )} \arcsin \left (b x + a\right )}{{\left (a^{2} - 1\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.34, size = 79, normalized size = 1.23 \[ \frac {2 \, b^{2} \arctan \left (\frac {\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{\sqrt {a^{2} - 1} {\left | b \right |}} - \frac {\arcsin \left (b x + a\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*arctan(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) - 1)/sqrt(a^2 - 1))/(sqrt(a^2 -
1)*abs(b)) - arcsin(b*x + a)/x

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maple [A]  time = 0.02, size = 78, normalized size = 1.22 \[ -\frac {\arcsin \left (b x +a \right )}{x}-\frac {b \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b x}\right )}{\sqrt {-a^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(arcsin(b*x+a)/x^2,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(a-1>0)', see `assume?` for mor
e details)Is a-1 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asin}\left (a+b\,x\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)/x^2,x)

[Out]

int(asin(a + b*x)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a + b*x)/x**2, x)

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