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]

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

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Rubi [A]  time = 0.06, antiderivative size = 121, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.47, size = 122, normalized size = 1.01 \[ \frac {\frac {2 i \sqrt {2} b \Pi \left (\frac {b}{a};i \sinh ^{-1}(x)|\frac {1}{2}\right )}{a}-i \sqrt {2} \Pi \left (\frac {b}{a};i \sinh ^{-1}(x)|\frac {1}{2}\right )+\frac {2 \sqrt {x^2+2} x}{\sqrt {x^2+1}}-i \sqrt {2} F\left (i \sinh ^{-1}(x)|\frac {1}{2}\right )+2 i \sqrt {2} E\left (i \sinh ^{-1}(x)|\frac {1}{2}\right )}{2 a-2 b} \]

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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fricas [F]  time = 16.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{2} + 2} \sqrt {x^{2} + 1}}{b x^{6} + {\left (a + 2 \, b\right )} x^{4} + {\left (2 \, a + b\right )} x^{2} + a}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} + 2}}{{\left (b x^{2} + a\right )} {\left (x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.04, size = 147, normalized size = 1.21 \[ \frac {\left (a \,x^{3}+2 a x +i \sqrt {x^{2}+2}\, \sqrt {x^{2}+1}\, a \EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-i \sqrt {x^{2}+2}\, \sqrt {x^{2}+1}\, a \EllipticPi \left (\frac {i \sqrt {2}\, x}{2}, \frac {2 b}{a}, \sqrt {2}\right )+2 i \sqrt {x^{2}+2}\, \sqrt {x^{2}+1}\, b \EllipticPi \left (\frac {i \sqrt {2}\, x}{2}, \frac {2 b}{a}, \sqrt {2}\right )\right ) \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}{\left (x^{4}+3 x^{2}+2\right ) \left (a -b \right ) a} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} + 2}}{{\left (b x^{2} + a\right )} {\left (x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {x^2+2}}{{\left (x^2+1\right )}^{3/2}\,\left (b\,x^2+a\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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