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

Optimal. Leaf size=192 \[ \frac{\sqrt{x^2+2} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{\sqrt{2} b \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}}}-\frac{\sqrt{x^2+2} (a-2 b) \Pi \left (1-\frac{b}{a};\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} a b \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}}}+\frac{\sqrt{x^2+2} x}{b \sqrt{x^2+1}}-\frac{\sqrt{2} \sqrt{x^2+2} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{b \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}}} \]

[Out]

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

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Rubi [A]  time = 0.0918102, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {534, 422, 418, 492, 411, 539} \[ -\frac{\sqrt{x^2+2} (a-2 b) \Pi \left (1-\frac{b}{a};\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} a b \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}}}+\frac{\sqrt{x^2+2} x}{b \sqrt{x^2+1}}+\frac{\sqrt{x^2+2} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} b \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}}}-\frac{\sqrt{2} \sqrt{x^2+2} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{b \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 534

Int[(Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2])/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d/b, Int[Sq
rt[e + f*x^2]/Sqrt[c + d*x^2], x], x] + Dist[(b*c - a*d)/b, Int[Sqrt[e + f*x^2]/((a + b*x^2)*Sqrt[c + d*x^2]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[d/c, 0] && GtQ[f/e, 0] &&  !SimplerSqrtQ[d/c, f/e]

Rule 422

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

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*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[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, 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]

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]

Rubi steps

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

Mathematica [C]  time = 0.191546, size = 71, normalized size = 0.37 \[ \frac{i \left ((a-b) \left (a \text{EllipticF}\left (i \sinh ^{-1}(x),\frac{1}{2}\right )-(a-2 b) \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )\right )-2 a b E\left (i \sinh ^{-1}(x)|\frac{1}{2}\right )\right )}{\sqrt{2} a b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.008, size = 120, normalized size = 0.6 \begin{align*}{\frac{-i}{a{b}^{2}} \left ({\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) ba-{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){a}^{2}+2\,{\it EllipticF} \left ( i/2x\sqrt{2},\sqrt{2} \right ) ba+{a}^{2}{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ) -3\,{\it EllipticPi} \left ( i/2x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ) ba+2\,{\it EllipticPi} \left ( i/2x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){b}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^2 + 2)*sqrt(x^2 + 1)/(b*x^2 + a), 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^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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