3.314 \(\int \frac{\left (7+5 x^2\right )^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx\)

Optimal. Leaf size=149 \[ -\frac{17 x \left (x^2+2\right )}{2 \sqrt{x^4+3 x^2+2}}+\frac{x \left (17 x^2+25\right )}{2 \sqrt{x^4+3 x^2+2}}+\frac{6 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{x^4+3 x^2+2}}+\frac{17 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4+3 x^2+2}} \]

[Out]

(-17*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (x*(25 + 17*x^2))/(2*Sqrt[2 + 3*x^
2 + x^4]) + (17*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(
Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (6*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*
EllipticF[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4]

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Rubi [A]  time = 0.129607, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167 \[ -\frac{17 x \left (x^2+2\right )}{2 \sqrt{x^4+3 x^2+2}}+\frac{x \left (17 x^2+25\right )}{2 \sqrt{x^4+3 x^2+2}}+\frac{6 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{x^4+3 x^2+2}}+\frac{17 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]  Int[(7 + 5*x^2)^2/(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

(-17*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (x*(25 + 17*x^2))/(2*Sqrt[2 + 3*x^
2 + x^4]) + (17*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(
Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (6*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*
EllipticF[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4]

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Rubi in Sympy [A]  time = 11.2299, size = 139, normalized size = 0.93 \[ - \frac{17 x \left (2 x^{2} + 4\right )}{4 \sqrt{x^{4} + 3 x^{2} + 2}} + \frac{x \left (153 x^{2} + 225\right )}{18 \sqrt{x^{4} + 3 x^{2} + 2}} + \frac{17 \sqrt{\frac{2 x^{2} + 4}{x^{2} + 1}} \left (4 x^{2} + 4\right ) E\left (\operatorname{atan}{\left (x \right )}\middle | \frac{1}{2}\right )}{8 \sqrt{x^{4} + 3 x^{2} + 2}} + \frac{3 \sqrt{\frac{2 x^{2} + 4}{x^{2} + 1}} \left (4 x^{2} + 4\right ) F\left (\operatorname{atan}{\left (x \right )}\middle | \frac{1}{2}\right )}{2 \sqrt{x^{4} + 3 x^{2} + 2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((5*x**2+7)**2/(x**4+3*x**2+2)**(3/2),x)

[Out]

-17*x*(2*x**2 + 4)/(4*sqrt(x**4 + 3*x**2 + 2)) + x*(153*x**2 + 225)/(18*sqrt(x**
4 + 3*x**2 + 2)) + 17*sqrt((2*x**2 + 4)/(x**2 + 1))*(4*x**2 + 4)*elliptic_e(atan
(x), 1/2)/(8*sqrt(x**4 + 3*x**2 + 2)) + 3*sqrt((2*x**2 + 4)/(x**2 + 1))*(4*x**2
+ 4)*elliptic_f(atan(x), 1/2)/(2*sqrt(x**4 + 3*x**2 + 2))

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Mathematica [C]  time = 0.081323, size = 99, normalized size = 0.66 \[ \frac{17 x^3-41 i \sqrt{x^2+1} \sqrt{x^2+2} F\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+17 i \sqrt{x^2+1} \sqrt{x^2+2} E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+25 x}{2 \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(7 + 5*x^2)^2/(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

(25*x + 17*x^3 + (17*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2
]], 2] - (41*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2])/
(2*Sqrt[2 + 3*x^2 + x^4])

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Maple [C]  time = 0.01, size = 173, normalized size = 1.2 \[ -98\,{\frac{-3/4\,{x}^{3}-5/4\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-{6\,i\sqrt{2}{\it EllipticF} \left ({\frac{i}{2}}\sqrt{2}x,\sqrt{2} \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}-{{\frac{17\,i}{4}}\sqrt{2} \left ({\it EllipticF} \left ({\frac{i}{2}}\sqrt{2}x,\sqrt{2} \right ) -{\it EllipticE} \left ({\frac{i}{2}}\sqrt{2}x,\sqrt{2} \right ) \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}-140\,{\frac{{x}^{3}+3/2\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-50\,{\frac{-3/2\,{x}^{3}-2\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((5*x^2+7)^2/(x^4+3*x^2+2)^(3/2),x)

[Out]

-98*(-3/4*x^3-5/4*x)/(x^4+3*x^2+2)^(1/2)-6*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/
2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2*I*2^(1/2)*x,2^(1/2))-17/4*I*2^(1/2)*(2*x^2+
4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*(EllipticF(1/2*I*2^(1/2)*x,2^(1/2))-E
llipticE(1/2*I*2^(1/2)*x,2^(1/2)))-140*(x^3+3/2*x)/(x^4+3*x^2+2)^(1/2)-50*(-3/2*
x^3-2*x)/(x^4+3*x^2+2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (5 \, x^{2} + 7\right )}^{2}}{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((5*x^2 + 7)^2/(x^4 + 3*x^2 + 2)^(3/2),x, algorithm="maxima")

[Out]

integrate((5*x^2 + 7)^2/(x^4 + 3*x^2 + 2)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{25 \, x^{4} + 70 \, x^{2} + 49}{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac{3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((5*x^2 + 7)^2/(x^4 + 3*x^2 + 2)^(3/2),x, algorithm="fricas")

[Out]

integral((25*x^4 + 70*x^2 + 49)/(x^4 + 3*x^2 + 2)^(3/2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (5 x^{2} + 7\right )^{2}}{\left (\left (x^{2} + 1\right ) \left (x^{2} + 2\right )\right )^{\frac{3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((5*x**2+7)**2/(x**4+3*x**2+2)**(3/2),x)

[Out]

Integral((5*x**2 + 7)**2/((x**2 + 1)*(x**2 + 2))**(3/2), x)

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (5 \, x^{2} + 7\right )}^{2}}{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((5*x^2 + 7)^2/(x^4 + 3*x^2 + 2)^(3/2),x, algorithm="giac")

[Out]

integrate((5*x^2 + 7)^2/(x^4 + 3*x^2 + 2)^(3/2), x)