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

Optimal. Leaf size=159 \[ -\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.0416802, antiderivative size = 159, 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 = {4816} \[ -\frac{x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac{x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[1 + d*x^2])^(-1),x]

[Out]

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

Rule 4816

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

Rubi steps

\begin{align*} \int \frac{1}{a+b \sin ^{-1}\left (1+d x^2\right )} \, dx &=-\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 )}{2 b \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 )}{2 b \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 = 0.680278, size = 120, normalized size = 0.75 \[ -\frac{x \left (\left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{1}{2} \left (\frac{a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )+\left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{1}{2} \left (\frac{a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )\right )}{2 b \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(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)])*Si
nIntegral[(a/b + ArcSin[1 + d*x^2])/2]))/(2*b*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))

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Maple [F]  time = 0.057, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arcsin \left ( d{x}^{2}+1 \right ) \right ) ^{-1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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