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

Optimal. Leaf size=93 \[ \frac{\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^2 d}-\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{b^2 d}-\frac{\sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]

[Out]

-(Sqrt[1 - (c + d*x)^2]/(b*d*(a + b*ArcSin[c + d*x]))) + (CosIntegral[(a + b*ArcSin[c + d*x])/b]*Sin[a/b])/(b^
2*d) - (Cos[a/b]*SinIntegral[(a + b*ArcSin[c + d*x])/b])/(b^2*d)

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Rubi [A]  time = 0.171562, antiderivative size = 89, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4803, 4621, 4723, 3303, 3299, 3302} \[ \frac{\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac{\sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^(-2),x]

[Out]

-(Sqrt[1 - (c + d*x)^2]/(b*d*(a + b*ArcSin[c + d*x]))) + (CosIntegral[a/b + ArcSin[c + d*x]]*Sin[a/b])/(b^2*d)
- (Cos[a/b]*SinIntegral[a/b + ArcSin[c + d*x]])/(b^2*d)

Rule 4803

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]

Rule 4621

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 4723

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

Rule 3303

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

Rule 3299

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 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

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

Mathematica [A]  time = 0.286741, size = 182, normalized size = 1.96 \[ \frac{\sin \left (\frac{a}{b}\right ) \left (a+b \sin ^{-1}(c+d x)\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )-\cos \left (\frac{a}{b}\right ) \left (a+b \sin ^{-1}(c+d x)\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+b c \sin ^{-1}(c+d x) \log \left (a+b \sin ^{-1}(c+d x)\right )-b c \sin ^{-1}(c+d x) \log \left (d \left (a+b \sin ^{-1}(c+d x)\right )\right )+a c \log \left (a+b \sin ^{-1}(c+d x)\right )-a c \log \left (d \left (a+b \sin ^{-1}(c+d x)\right )\right )-b \sqrt{-c^2-2 c d x-d^2 x^2+1}}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(b*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2]) + a*c*Log[a + b*ArcSin[c + d*x]] + b*c*ArcSin[c + d*x]*Log[a + b*ArcSi
n[c + d*x]] - a*c*Log[d*(a + b*ArcSin[c + d*x])] - b*c*ArcSin[c + d*x]*Log[d*(a + b*ArcSin[c + d*x])] + (a + b
*ArcSin[c + d*x])*CosIntegral[a/b + ArcSin[c + d*x]]*Sin[a/b] - (a + b*ArcSin[c + d*x])*Cos[a/b]*SinIntegral[a
/b + ArcSin[c + d*x]])/(b^2*d*(a + b*ArcSin[c + d*x]))

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Maple [A]  time = 0.034, size = 83, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( -{\frac{1}{ \left ( a+b\arcsin \left ( dx+c \right ) \right ) b}\sqrt{1- \left ( dx+c \right ) ^{2}}}-{\frac{1}{{b}^{2}} \left ({\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) -{\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsin(d*x+c))^2,x)

[Out]

1/d*(-(1-(d*x+c)^2)^(1/2)/(a+b*arcsin(d*x+c))/b-(Si(arcsin(d*x+c)+a/b)*cos(a/b)-Ci(arcsin(d*x+c)+a/b)*sin(a/b)
)/b^2)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/(b^2*arcsin(d*x + c)^2 + 2*a*b*arcsin(d*x + c) + a^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.17332, size = 290, normalized size = 3.12 \begin{align*} \frac{b \arcsin \left (d x + c\right ) \operatorname{Ci}\left (\frac{a}{b} + \arcsin \left (d x + c\right )\right ) \sin \left (\frac{a}{b}\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac{b \arcsin \left (d x + c\right ) \cos \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac{a \operatorname{Ci}\left (\frac{a}{b} + \arcsin \left (d x + c\right )\right ) \sin \left (\frac{a}{b}\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac{a \cos \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac{\sqrt{-{\left (d x + c\right )}^{2} + 1} b}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*arcsin(d*x + c)*cos_integral(a/b + arcsin(d*x + c))*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) - b*arcsin(d*
x + c)*cos(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + a*cos_integral(a/b + a
rcsin(d*x + c))*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) - a*cos(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b
^3*d*arcsin(d*x + c) + a*b^2*d) - sqrt(-(d*x + c)^2 + 1)*b/(b^3*d*arcsin(d*x + c) + a*b^2*d)