3.93 \(\int \frac{\sqrt{2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx\)

Optimal. Leaf size=215 \[ \frac{2 b^2 \sqrt{x^2+1} \Pi \left (1-\frac{2 b}{a};\left .\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-1\right )}{a \sqrt{\frac{x^2+1}{x^2+2}} \sqrt{x^2+2} (a-b)^2}+\frac{x \sqrt{x^2+2}}{3 \left (x^2+1\right )^{3/2} (a-b)}-\frac{\sqrt{2} \sqrt{x^2+2} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)}+\frac{\sqrt{2} \sqrt{x^2+2} (a-2 b) E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)^2} \]

[Out]

(x*Sqrt[2 + x^2])/(3*(a - b)*(1 + x^2)^(3/2)) + (Sqrt[2]*(a - 2*b)*Sqrt[2 + x^2]
*EllipticE[ArcTan[x], 1/2])/((a - b)^2*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)])
- (Sqrt[2]*Sqrt[2 + x^2]*EllipticF[ArcTan[x], 1/2])/(3*(a - b)*Sqrt[1 + x^2]*Sqr
t[(2 + x^2)/(1 + x^2)]) + (2*b^2*Sqrt[1 + x^2]*EllipticPi[1 - (2*b)/a, ArcTan[x/
Sqrt[2]], -1])/(a*(a - b)^2*Sqrt[(1 + x^2)/(2 + x^2)]*Sqrt[2 + x^2])

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Rubi [A]  time = 0.425159, antiderivative size = 215, 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 \[ \frac{2 b^2 \sqrt{x^2+1} \Pi \left (1-\frac{2 b}{a};\left .\tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-1\right )}{a \sqrt{\frac{x^2+1}{x^2+2}} \sqrt{x^2+2} (a-b)^2}+\frac{x \sqrt{x^2+2}}{3 \left (x^2+1\right )^{3/2} (a-b)}-\frac{\sqrt{2} \sqrt{x^2+2} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 \sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)}+\frac{\sqrt{2} \sqrt{x^2+2} (a-2 b) E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{x^2+1} \sqrt{\frac{x^2+2}{x^2+1}} (a-b)^2} \]

Antiderivative was successfully verified.

[In]  Int[Sqrt[2 + x^2]/((1 + x^2)^(5/2)*(a + b*x^2)),x]

[Out]

(x*Sqrt[2 + x^2])/(3*(a - b)*(1 + x^2)^(3/2)) + (Sqrt[2]*(a - 2*b)*Sqrt[2 + x^2]
*EllipticE[ArcTan[x], 1/2])/((a - b)^2*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)])
- (Sqrt[2]*Sqrt[2 + x^2]*EllipticF[ArcTan[x], 1/2])/(3*(a - b)*Sqrt[1 + x^2]*Sqr
t[(2 + x^2)/(1 + x^2)]) + (2*b^2*Sqrt[1 + x^2]*EllipticPi[1 - (2*b)/a, ArcTan[x/
Sqrt[2]], -1])/(a*(a - b)^2*Sqrt[(1 + x^2)/(2 + x^2)]*Sqrt[2 + x^2])

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Rubi in Sympy [A]  time = 55.113, size = 190, normalized size = 0.88 \[ \frac{x \sqrt{x^{2} + 2}}{3 \left (a - b\right ) \left (x^{2} + 1\right )^{\frac{3}{2}}} + \frac{\sqrt{2} \left (a - 2 b\right ) \sqrt{x^{2} + 2} E\left (\operatorname{atan}{\left (x \right )}\middle | \frac{1}{2}\right )}{\sqrt{\frac{x^{2} + 2}{x^{2} + 1}} \left (a - b\right )^{2} \sqrt{x^{2} + 1}} - \frac{\sqrt{2} \sqrt{x^{2} + 2} F\left (\operatorname{atan}{\left (x \right )}\middle | \frac{1}{2}\right )}{3 \sqrt{\frac{x^{2} + 2}{x^{2} + 1}} \left (a - b\right ) \sqrt{x^{2} + 1}} + \frac{2 \sqrt{2} b^{2} \sqrt{x^{2} + 1} \Pi \left (1 - \frac{2 b}{a}; \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}\middle | -1\right )}{a \sqrt{\frac{2 x^{2} + 2}{x^{2} + 2}} \left (a - b\right )^{2} \sqrt{x^{2} + 2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((x**2+2)**(1/2)/(x**2+1)**(5/2)/(b*x**2+a),x)

[Out]

x*sqrt(x**2 + 2)/(3*(a - b)*(x**2 + 1)**(3/2)) + sqrt(2)*(a - 2*b)*sqrt(x**2 + 2
)*elliptic_e(atan(x), 1/2)/(sqrt((x**2 + 2)/(x**2 + 1))*(a - b)**2*sqrt(x**2 + 1
)) - sqrt(2)*sqrt(x**2 + 2)*elliptic_f(atan(x), 1/2)/(3*sqrt((x**2 + 2)/(x**2 +
1))*(a - b)*sqrt(x**2 + 1)) + 2*sqrt(2)*b**2*sqrt(x**2 + 1)*elliptic_pi(1 - 2*b/
a, atan(sqrt(2)*x/2), -1)/(a*sqrt((2*x**2 + 2)/(x**2 + 2))*(a - b)**2*sqrt(x**2
+ 2))

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Mathematica [C]  time = 0.47772, size = 357, normalized size = 1.66 \[ \frac{8 a^2 \sqrt{x^2+1} \sqrt{x^2+2} x+6 a^2 \sqrt{x^2+1} \sqrt{x^2+2} x^3-6 i \sqrt{2} b^2 x^4 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )-12 i \sqrt{2} b^2 x^2 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )-6 i \sqrt{2} b^2 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )+3 i \sqrt{2} a b x^4 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )-14 a b \sqrt{x^2+1} \sqrt{x^2+2} x-i \sqrt{2} a \left (x^2+1\right )^2 (4 a-7 b) F\left (i \sinh ^{-1}(x)|\frac{1}{2}\right )+6 i \sqrt{2} a \left (x^2+1\right )^2 (a-2 b) E\left (i \sinh ^{-1}(x)|\frac{1}{2}\right )+6 i \sqrt{2} a b x^2 \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )-12 a b \sqrt{x^2+1} \sqrt{x^2+2} x^3+3 i \sqrt{2} a b \Pi \left (\frac{b}{a};i \sinh ^{-1}(x)|\frac{1}{2}\right )}{6 a \left (x^2+1\right )^2 (a-b)^2} \]

Antiderivative was successfully verified.

[In]  Integrate[Sqrt[2 + x^2]/((1 + x^2)^(5/2)*(a + b*x^2)),x]

[Out]

(8*a^2*x*Sqrt[1 + x^2]*Sqrt[2 + x^2] - 14*a*b*x*Sqrt[1 + x^2]*Sqrt[2 + x^2] + 6*
a^2*x^3*Sqrt[1 + x^2]*Sqrt[2 + x^2] - 12*a*b*x^3*Sqrt[1 + x^2]*Sqrt[2 + x^2] + (
6*I)*Sqrt[2]*a*(a - 2*b)*(1 + x^2)^2*EllipticE[I*ArcSinh[x], 1/2] - I*Sqrt[2]*a*
(4*a - 7*b)*(1 + x^2)^2*EllipticF[I*ArcSinh[x], 1/2] + (3*I)*Sqrt[2]*a*b*Ellipti
cPi[b/a, I*ArcSinh[x], 1/2] - (6*I)*Sqrt[2]*b^2*EllipticPi[b/a, I*ArcSinh[x], 1/
2] + (6*I)*Sqrt[2]*a*b*x^2*EllipticPi[b/a, I*ArcSinh[x], 1/2] - (12*I)*Sqrt[2]*b
^2*x^2*EllipticPi[b/a, I*ArcSinh[x], 1/2] + (3*I)*Sqrt[2]*a*b*x^4*EllipticPi[b/a
, I*ArcSinh[x], 1/2] - (6*I)*Sqrt[2]*b^2*x^4*EllipticPi[b/a, I*ArcSinh[x], 1/2])
/(6*a*(a - b)^2*(1 + x^2)^2)

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Maple [B]  time = 0.051, size = 477, normalized size = 2.2 \[ -{\frac{1}{3\, \left ( a-b \right ) ^{2}a} \left ( -3\,i{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){a}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}+6\,i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){x}^{2}{b}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}+6\,i{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){x}^{2}ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}+6\,i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){b}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-3\,i{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){x}^{2}{a}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-i{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-3\,{x}^{5}{a}^{2}+6\,{x}^{5}ab+i{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){a}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}+6\,i{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-3\,i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ){x}^{2}ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-3\,i{\it EllipticPi} \left ({\frac{i}{2}}x\sqrt{2},2\,{\frac{b}{a}},\sqrt{2} \right ) ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-i{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){x}^{2}ab\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}+i{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ){x}^{2}{a}^{2}\sqrt{{x}^{2}+2}\sqrt{{x}^{2}+1}-10\,{x}^{3}{a}^{2}+19\,ab{x}^{3}-8\,x{a}^{2}+14\,abx \right ){\frac{1}{\sqrt{{x}^{2}+2}}} \left ({x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x)

[Out]

-1/3*(-3*I*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*a^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)+6*
I*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*x^2*b^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)+
6*I*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*x^2*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)+6*I*E
llipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*b^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)-3*I*Ell
ipticE(1/2*I*x*2^(1/2),2^(1/2))*x^2*a^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)-I*EllipticF(
1/2*I*x*2^(1/2),2^(1/2))*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)-3*x^5*a^2+6*x^5*a*b+I*E
llipticF(1/2*I*x*2^(1/2),2^(1/2))*a^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)+6*I*EllipticE(
1/2*I*x*2^(1/2),2^(1/2))*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)-3*I*EllipticPi(1/2*I*x*
2^(1/2),2*b/a,2^(1/2))*x^2*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)-3*I*EllipticPi(1/2*I*
x*2^(1/2),2*b/a,2^(1/2))*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)-I*EllipticF(1/2*I*x*2^(
1/2),2^(1/2))*x^2*a*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)+I*EllipticF(1/2*I*x*2^(1/2),2^
(1/2))*x^2*a^2*(x^2+2)^(1/2)*(x^2+1)^(1/2)-10*x^3*a^2+19*a*b*x^3-8*x*a^2+14*a*b*
x)/(x^2+2)^(1/2)/(a-b)^2/a/(x^2+1)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\sqrt{x^{2} + 2}}{{\left (b x^{2} + a\right )}{\left (x^{2} + 1\right )}^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(5/2)),x, algorithm="maxima")

[Out]

integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(5/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(5/2)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((x**2+2)**(1/2)/(x**2+1)**(5/2)/(b*x**2+a),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\sqrt{x^{2} + 2}}{{\left (b x^{2} + a\right )}{\left (x^{2} + 1\right )}^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(5/2)),x, algorithm="giac")

[Out]

integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(5/2)), x)