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

Optimal. Leaf size=305 \[ \frac{4 \sqrt{x^4+3 x^2+4} x}{175 \left (x^2+2\right )}+\frac{22 \sqrt{x^4+3 x^2+4} x}{175 \left (5 x^2+7\right )}+\frac{1}{75} \sqrt{x^4+3 x^2+4} x+\frac{13}{350} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{x^4+3 x^2+4}}\right )+\frac{4 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{175 \sqrt{x^4+3 x^2+4}}-\frac{4 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{175 \sqrt{x^4+3 x^2+4}}+\frac{2431 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{36750 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

[Out]

(x*Sqrt[4 + 3*x^2 + x^4])/75 + (4*x*Sqrt[4 + 3*x^2 + x^4])/(175*(2 + x^2)) + (22
*x*Sqrt[4 + 3*x^2 + x^4])/(175*(7 + 5*x^2)) + (13*Sqrt[11/35]*ArcTan[(2*Sqrt[11/
35]*x)/Sqrt[4 + 3*x^2 + x^4]])/350 - (4*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)
/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(175*Sqrt[4 + 3*x^2 + x^4]) +
(4*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/S
qrt[2]], 1/8])/(175*Sqrt[4 + 3*x^2 + x^4]) + (2431*(2 + x^2)*Sqrt[(4 + 3*x^2 + x
^4)/(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(36750*Sqrt[2]*Sq
rt[4 + 3*x^2 + x^4])

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Rubi [A]  time = 0.711411, antiderivative size = 372, normalized size of antiderivative = 1.22, number of steps used = 18, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333 \[ \frac{4 \sqrt{x^4+3 x^2+4} x}{175 \left (x^2+2\right )}+\frac{22 \sqrt{x^4+3 x^2+4} x}{175 \left (5 x^2+7\right )}+\frac{1}{75} \sqrt{x^4+3 x^2+4} x+\frac{13}{350} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{x^4+3 x^2+4}}\right )+\frac{4 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{175 \sqrt{x^4+3 x^2+4}}-\frac{4 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{175 \sqrt{x^4+3 x^2+4}}+\frac{187 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{13125 \sqrt{x^4+3 x^2+4}}+\frac{6919 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{183750 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

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

[Out]

(x*Sqrt[4 + 3*x^2 + x^4])/75 + (4*x*Sqrt[4 + 3*x^2 + x^4])/(175*(2 + x^2)) + (22
*x*Sqrt[4 + 3*x^2 + x^4])/(175*(7 + 5*x^2)) + (13*Sqrt[11/35]*ArcTan[(2*Sqrt[11/
35]*x)/Sqrt[4 + 3*x^2 + x^4]])/350 - (4*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)
/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(175*Sqrt[4 + 3*x^2 + x^4]) +
(4*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/S
qrt[2]], 1/8])/(175*Sqrt[4 + 3*x^2 + x^4]) + (6919*(2 + x^2)*Sqrt[(4 + 3*x^2 + x
^4)/(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(183750*Sqrt[2]*S
qrt[4 + 3*x^2 + x^4]) + (187*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^
2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(13125*Sqrt[4 + 3*x^2 + x^4])

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

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Mathematica [C]  time = 0.726759, size = 309, normalized size = 1.01 \[ \frac{\frac{175 x \left (7 x^2+23\right ) \left (x^4+3 x^2+4\right )}{5 x^2+7}-i \sqrt{6+2 i \sqrt{7}} \sqrt{1-\frac{2 i x^2}{\sqrt{7}-3 i}} \sqrt{1+\frac{2 i x^2}{\sqrt{7}+3 i}} \left (7 \left (158+15 i \sqrt{7}\right ) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )+105 \left (3-i \sqrt{7}\right ) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )+429 \Pi \left (\frac{5}{14} \left (3+i \sqrt{7}\right );i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )\right )}{18375 \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

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

[Out]

((175*x*(23 + 7*x^2)*(4 + 3*x^2 + x^4))/(7 + 5*x^2) - I*Sqrt[6 + (2*I)*Sqrt[7]]*
Sqrt[1 - ((2*I)*x^2)/(-3*I + Sqrt[7])]*Sqrt[1 + ((2*I)*x^2)/(3*I + Sqrt[7])]*(10
5*(3 - I*Sqrt[7])*EllipticE[I*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - S
qrt[7])/(3*I + Sqrt[7])] + 7*(158 + (15*I)*Sqrt[7])*EllipticF[I*ArcSinh[Sqrt[(-2
*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] + 429*EllipticPi[(5*(
3 + I*Sqrt[7]))/14, I*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/
(3*I + Sqrt[7])]))/(18375*Sqrt[4 + 3*x^2 + x^4])

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Maple [C]  time = 0.034, size = 425, normalized size = 1.4 \[{\frac{22\,x}{875\,{x}^{2}+1225}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{x}{75}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{232}{375\,\sqrt{-6+2\,i\sqrt{7}}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticF} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}-{\frac{128}{175\,\sqrt{-6+2\,i\sqrt{7}} \left ( i\sqrt{7}+3 \right ) }\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticF} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}+{\frac{128}{175\,\sqrt{-6+2\,i\sqrt{7}} \left ( i\sqrt{7}+3 \right ) }\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticE} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}+{\frac{286}{6125\,\sqrt{-3/8+i/8\sqrt{7}}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticPi} \left ( \sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}x,-{\frac{5}{-{\frac{21}{8}}+{\frac{7\,i}{8}}\sqrt{7}}},{\frac{\sqrt{-{\frac{3}{8}}-{\frac{i}{8}}\sqrt{7}}}{\sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

22/175*x*(x^4+3*x^2+4)^(1/2)/(5*x^2+7)+1/75*x*(x^4+3*x^2+4)^(1/2)+232/375/(-6+2*
I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2
))^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticF(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7
^(1/2))^(1/2))-128/175/(-6+2*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2
)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2)/(I*7^(1/2)+3)*Elliptic
F(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2))^(1/2))+128/175/(-6+2*I*7^(1/2
))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)
/(x^4+3*x^2+4)^(1/2)/(I*7^(1/2)+3)*EllipticE(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2
+6*I*7^(1/2))^(1/2))+286/6125/(-3/8+1/8*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^
(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticPi(
(-3/8+1/8*I*7^(1/2))^(1/2)*x,-5/7/(-3/8+1/8*I*7^(1/2)),(-3/8-1/8*I*7^(1/2))^(1/2
)/(-3/8+1/8*I*7^(1/2))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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