3.54 \(\int \frac{\cos ^2(\frac{\sqrt{1-a x}}{\sqrt{1+a x}})}{1-a^2 x^2} \, dx\)

Optimal. Leaf size=58 \[ -\frac{\text{CosIntegral}\left (\frac{2 \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a}-\frac{\log \left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a} \]

[Out]

-CosIntegral[(2*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/(2*a) - Log[Sqrt[1 - a*x]/Sqrt[1 + a*x]]/(2*a)

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Rubi [A]  time = 0.0799495, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {6681, 3312, 3302} \[ -\frac{\text{CosIntegral}\left (\frac{2 \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a}-\frac{\log \left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Cos[Sqrt[1 - a*x]/Sqrt[1 + a*x]]^2/(1 - a^2*x^2),x]

[Out]

-CosIntegral[(2*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/(2*a) - Log[Sqrt[1 - a*x]/Sqrt[1 + a*x]]/(2*a)

Rule 6681

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[(2*e*g)/(C*(e*f - d*g)), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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

Mathematica [A]  time = 0.0381715, size = 51, normalized size = 0.88 \[ -\frac{\text{CosIntegral}\left (\frac{2 \sqrt{1-a x}}{\sqrt{a x+1}}\right )+\log \left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[Sqrt[1 - a*x]/Sqrt[1 + a*x]]^2/(1 - a^2*x^2),x]

[Out]

-(CosIntegral[(2*Sqrt[1 - a*x])/Sqrt[1 + a*x]] + Log[Sqrt[1 - a*x]/Sqrt[1 + a*x]])/(2*a)

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Maple [F]  time = 0.088, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{-{a}^{2}{x}^{2}+1} \left ( \cos \left ({\sqrt{-ax+1}{\frac{1}{\sqrt{ax+1}}}} \right ) \right ) ^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos((-a*x+1)^(1/2)/(a*x+1)^(1/2))^2/(-a^2*x^2+1),x)

[Out]

int(cos((-a*x+1)^(1/2)/(a*x+1)^(1/2))^2/(-a^2*x^2+1),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a \int \frac{\cos \left (\frac{2 \, \sqrt{-a x + 1}}{\sqrt{a x + 1}}\right )}{a^{2} x^{2} - 1}\,{d x} + a \int \frac{\cos \left (\frac{2 \, \sqrt{-a x + 1}}{\sqrt{a x + 1}}\right )}{{\left (a^{2} x^{2} - 1\right )} \cos \left (\frac{2 \, \sqrt{-a x + 1}}{\sqrt{a x + 1}}\right )^{2} +{\left (a^{2} x^{2} - 1\right )} \sin \left (\frac{2 \, \sqrt{-a x + 1}}{\sqrt{a x + 1}}\right )^{2}}\,{d x} - \log \left (a x + 1\right ) + \log \left (a x - 1\right )}{4 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*a*integrate(1/4*cos(2*sqrt(-a*x + 1)/sqrt(a*x + 1))/(a^2*x^2 - 1), x) + 4*a*integrate(1/4*cos(2*sqrt(-
a*x + 1)/sqrt(a*x + 1))/((a^2*x^2 - 1)*cos(2*sqrt(-a*x + 1)/sqrt(a*x + 1))^2 + (a^2*x^2 - 1)*sin(2*sqrt(-a*x +
1)/sqrt(a*x + 1))^2), x) - log(a*x + 1) + log(a*x - 1))/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-cos(sqrt(-a*x + 1)/sqrt(a*x + 1))^2/(a^2*x^2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos((-a*x+1)**(1/2)/(a*x+1)**(1/2))**2/(-a**2*x**2+1),x)

[Out]

-Integral(cos(sqrt(-a*x + 1)/sqrt(a*x + 1))**2/(a**2*x**2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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