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

Optimal. Leaf size=205 \[ -\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac{\sqrt{-d^2 x^4-2 d x^2}}{2 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )} \]

[Out]

-Sqrt[-2*d*x^2 - d^2*x^4]/(2*b*d*x*(a + b*ArcSin[1 + d*x^2])) - (x*CosIntegral[(a + b*ArcSin[1 + d*x^2])/(2*b)
]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(4*b^2*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2])) + (x*(Cos[a/(2*
b)] - Sin[a/(2*b)])*SinIntegral[(a + b*ArcSin[1 + d*x^2])/(2*b)])/(4*b^2*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSi
n[1 + d*x^2]/2]))

________________________________________________________________________________________

Rubi [A]  time = 0.0257812, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {4825} \[ -\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac{\sqrt{-d^2 x^4-2 d x^2}}{2 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[1 + d*x^2])^(-2),x]

[Out]

-Sqrt[-2*d*x^2 - d^2*x^4]/(2*b*d*x*(a + b*ArcSin[1 + d*x^2])) - (x*CosIntegral[(a + b*ArcSin[1 + d*x^2])/(2*b)
]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(4*b^2*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2])) + (x*(Cos[a/(2*
b)] - Sin[a/(2*b)])*SinIntegral[(a + b*ArcSin[1 + d*x^2])/(2*b)])/(4*b^2*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSi
n[1 + d*x^2]/2]))

Rule 4825

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-2), x_Symbol] :> -Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]/(2*b*d*x*(
a + b*ArcSin[c + d*x^2])), x] + (-Simp[(x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*CosIntegral[(c/(2*b))*(a + b*ArcSin[
c + d*x^2])])/(4*b^2*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2])), x] + Simp[(x*(Cos[a/(2*b)] - c*
Sin[a/(2*b)])*SinIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])])/(4*b^2*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSi
n[c + d*x^2]/2])), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^2} \, dx &=-\frac{\sqrt{-2 d x^2-d^2 x^4}}{2 b d x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )}-\frac{x \text{Ci}\left (\frac{a+b \sin ^{-1}\left (1+d x^2\right )}{2 b}\right ) \left (\cos \left (\frac{a}{2 b}\right )+\sin \left (\frac{a}{2 b}\right )\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+d x^2\right )\right )\right )}+\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (1+d x^2\right )}{2 b}\right )}{4 b^2 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1+d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1+d x^2\right )\right )\right )}\\ \end{align*}

Mathematica [A]  time = 1.33235, size = 164, normalized size = 0.8 \[ -\frac{\frac{x^2 \left (\left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{1}{2} \left (\frac{a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )+\left (\sin \left (\frac{a}{2 b}\right )-\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{1}{2} \left (\frac{a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )}+\frac{2 b \sqrt{-d x^2 \left (d x^2+2\right )}}{d \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )}}{4 b^2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[1 + d*x^2])^(-2),x]

[Out]

-((2*b*Sqrt[-(d*x^2*(2 + d*x^2))])/(d*(a + b*ArcSin[1 + d*x^2])) + (x^2*(CosIntegral[(a/b + ArcSin[1 + d*x^2])
/2]*(Cos[a/(2*b)] + Sin[a/(2*b)]) + (-Cos[a/(2*b)] + Sin[a/(2*b)])*SinIntegral[(a/b + ArcSin[1 + d*x^2])/2]))/
(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))/(4*b^2*x)

________________________________________________________________________________________

Maple [F]  time = 0.058, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arcsin \left ( d{x}^{2}+1 \right ) \right ) ^{-2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (b^{2} d \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right ) + a b d\right )} \sqrt{d} \int \frac{\sqrt{-d x^{2} - 2} x}{a b d x^{2} + 2 \, a b +{\left (b^{2} d x^{2} + 2 \, b^{2}\right )} \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right )}\,{d x} - \sqrt{-d x^{2} - 2} \sqrt{d}}{2 \,{\left (b^{2} d \arctan \left (d x^{2} + 1, \sqrt{-d x^{2} - 2} \sqrt{d} x\right ) + a b d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(d*x**2 + 1))**(-2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(d*x^2 + 1) + a)^(-2), x)