3.92 \(\int \frac{\sqrt{2+x^2}}{(1+x^2)^{3/2} (a+b x^2)} \, dx\)

Optimal. Leaf size=121 \[ \frac{\sqrt{2} \sqrt{x^2+2} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)}-\frac{2 b \sqrt{x^2+1} \Pi \left (1-\frac{2 b}{a};\left .\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-1\right )}{a \sqrt{\frac{x^2+1}{x^2+2}} \sqrt{x^2+2} (a-b)} \]

[Out]

(Sqrt[2]*Sqrt[2 + x^2]*EllipticE[ArcTan[x], 1/2])/((a - b)*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)]) - (2*b*Sqr
t[1 + x^2]*EllipticPi[1 - (2*b)/a, ArcTan[x/Sqrt[2]], -1])/(a*(a - b)*Sqrt[(1 + x^2)/(2 + x^2)]*Sqrt[2 + x^2])

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Rubi [A]  time = 0.0597335, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {541, 539, 411} \[ \frac{\sqrt{2} \sqrt{x^2+2} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)}-\frac{2 b \sqrt{x^2+1} \Pi \left (1-\frac{2 b}{a};\left .\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-1\right )}{a \sqrt{\frac{x^2+1}{x^2+2}} \sqrt{x^2+2} (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + x^2]/((1 + x^2)^(3/2)*(a + b*x^2)),x]

[Out]

(Sqrt[2]*Sqrt[2 + x^2]*EllipticE[ArcTan[x], 1/2])/((a - b)*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)]) - (2*b*Sqr
t[1 + x^2]*EllipticPi[1 - (2*b)/a, ArcTan[x/Sqrt[2]], -1])/(a*(a - b)*Sqrt[(1 + x^2)/(2 + x^2)]*Sqrt[2 + x^2])

Rule 541

Int[Sqrt[(e_) + (f_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[b/(b*c -
a*d), Int[Sqrt[e + f*x^2]/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] - Dist[d/(b*c - a*d), Int[Sqrt[e + f*x^2]/(c +
d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c] && PosQ[f/e]

Rule 539

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(c*Sqrt[e +
f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcTan[Rt[d/c, 2]*x], 1 - (c*f)/(d*e)])/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sq
rt[(c*(e + f*x^2))/(e*(c + d*x^2))]), x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{\sqrt{2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx &=-\frac{b \int \frac{\sqrt{2+x^2}}{\sqrt{1+x^2} \left (a+b x^2\right )} \, dx}{a-b}-\frac{\int \frac{\sqrt{2+x^2}}{\left (1+x^2\right )^{3/2}} \, dx}{-a+b}\\ &=\frac{\sqrt{2} \sqrt{2+x^2} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{(a-b) \sqrt{1+x^2} \sqrt{\frac{2+x^2}{1+x^2}}}-\frac{2 b \sqrt{1+x^2} \Pi \left (1-\frac{2 b}{a};\left .\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-1\right )}{a (a-b) \sqrt{\frac{1+x^2}{2+x^2}} \sqrt{2+x^2}}\\ \end{align*}

Mathematica [C]  time = 0.346171, size = 122, normalized size = 1.01 \[ \frac{-i \sqrt{2} \text{EllipticF}\left (i \sinh ^{-1}(x),\frac{1}{2}\right )+\frac{2 i \sqrt{2} b \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )}{a}-i \sqrt{2} \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )+\frac{2 \sqrt{x^2+2} x}{\sqrt{x^2+1}}+2 i \sqrt{2} E\left (i \sinh ^{-1}(x)|\frac{1}{2}\right )}{2 a-2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + x^2]/((1 + x^2)^(3/2)*(a + b*x^2)),x]

[Out]

((2*x*Sqrt[2 + x^2])/Sqrt[1 + x^2] + (2*I)*Sqrt[2]*EllipticE[I*ArcSinh[x], 1/2] - I*Sqrt[2]*EllipticF[I*ArcSin
h[x], 1/2] - I*Sqrt[2]*EllipticPi[b/a, I*ArcSinh[x], 1/2] + ((2*I)*Sqrt[2]*b*EllipticPi[b/a, I*ArcSinh[x], 1/2
])/a)/(2*a - 2*b)

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Maple [A]  time = 0.033, size = 147, normalized size = 1.2 \begin{align*}{\frac{1}{a \left ({x}^{4}+3\,{x}^{2}+2 \right ) \left ( a-b \right ) } \left ( i{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) a\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}-i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ) a\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}+2\,i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ) b\sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}+a{x}^{3}+2\,ax \right ) \sqrt{{x}^{2}+1}\sqrt{{x}^{2}+2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+2)^(1/2)/(x^2+1)^(3/2)/(b*x^2+a),x)

[Out]

(I*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*a*(x^2+1)^(1/2)*(x^2+2)^(1/2)-I*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2)
)*a*(x^2+1)^(1/2)*(x^2+2)^(1/2)+2*I*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*b*(x^2+1)^(1/2)*(x^2+2)^(1/2)+a*
x^3+2*a*x)*(x^2+1)^(1/2)*(x^2+2)^(1/2)/a/(x^4+3*x^2+2)/(a-b)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + 2}}{{\left (b x^{2} + a\right )}{\left (x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)^(1/2)/(x^2+1)^(3/2)/(b*x^2+a),x, algorithm="maxima")

[Out]

integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(3/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{2} + 2} \sqrt{x^{2} + 1}}{b x^{6} +{\left (a + 2 \, b\right )} x^{4} +{\left (2 \, a + b\right )} x^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)^(1/2)/(x^2+1)^(3/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

integral(sqrt(x^2 + 2)*sqrt(x^2 + 1)/(b*x^6 + (a + 2*b)*x^4 + (2*a + b)*x^2 + a), x)

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Sympy [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((x**2+2)**(1/2)/(x**2+1)**(3/2)/(b*x**2+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + 2}}{{\left (b x^{2} + a\right )}{\left (x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)^(1/2)/(x^2+1)^(3/2)/(b*x^2+a),x, algorithm="giac")

[Out]

integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(3/2)), x)