3.2.49 \(\int \frac {1}{\text {ArcSin}(a+b x)^2} \, dx\) [149]

Optimal. Leaf size=41 \[ -\frac {\sqrt {1-(a+b x)^2}}{b \text {ArcSin}(a+b x)}-\frac {\text {Si}(\text {ArcSin}(a+b x))}{b} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4887, 4717, 4809, 3380} \begin {gather*} -\frac {\text {Si}(\text {ArcSin}(a+b x))}{b}-\frac {\sqrt {1-(a+b x)^2}}{b \text {ArcSin}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^(-2),x]

[Out]

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

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 4717

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Sqrt[1 - c^2*x^2]*((a + b*ArcSin[c*x])^(n + 1)/
(b*c*(n + 1))), x] + Dist[c/(b*(n + 1)), Int[x*((a + b*ArcSin[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x] /; Free
Q[{a, b, c}, x] && LtQ[n, -1]

Rule 4809

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

Rule 4887

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSin[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 37, normalized size = 0.90 \begin {gather*} -\frac {\frac {\sqrt {1-(a+b x)^2}}{\text {ArcSin}(a+b x)}+\text {Si}(\text {ArcSin}(a+b x))}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^(-2),x]

[Out]

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

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Maple [A]
time = 0.07, size = 38, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {1-\left (b x +a \right )^{2}}}{\arcsin \left (b x +a \right )}-\sinIntegral \left (\arcsin \left (b x +a \right )\right )}{b}\)	\(38\)
default	\(\frac {-\frac {\sqrt {1-\left (b x +a \right )^{2}}}{\arcsin \left (b x +a \right )}-\sinIntegral \left (\arcsin \left (b x +a \right )\right )}{b}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arcsin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsin(b*x+a)^2,x, algorithm="maxima")

[Out]

(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))*integrate(sqrt(b*x + a + 1)*(b*x + a)*sqrt(-b*x - a
+ 1)/((b^2*x^2 + 2*a*b*x + a^2 - 1)*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))), x) - sqrt(b*x + a
+ 1)*sqrt(-b*x - a + 1))/(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsin(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(arcsin(b*x + a)^(-2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\operatorname {asin}^{2}{\left (a + b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/asin(b*x+a)**2,x)

[Out]

Integral(asin(a + b*x)**(-2), x)

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Giac [A]
time = 0.39, size = 39, normalized size = 0.95 \begin {gather*} -\frac {\operatorname {Si}\left (\arcsin \left (b x + a\right )\right )}{b} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{b \arcsin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsin(b*x+a)^2,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/asin(a + b*x)^2,x)

[Out]

int(1/asin(a + b*x)^2, x)

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