3.365 \(\int \frac {(7+5 x^2)^2}{\sqrt {4+3 x^2+x^4}} \, dx\)

Optimal. Leaf size=170 \[ \frac {20 \sqrt {x^4+3 x^2+4} x}{x^2+2}+\frac {25}{3} \sqrt {x^4+3 x^2+4} x+\frac {167 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{6 \sqrt {2} \sqrt {x^4+3 x^2+4}}-\frac {20 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{\sqrt {x^4+3 x^2+4}} \]

[Out]

25/3*x*(x^4+3*x^2+4)^(1/2)+20*x*(x^4+3*x^2+4)^(1/2)/(x^2+2)+167/12*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2)))^2)^(1
/2)/cos(2*arctan(1/2*x*2^(1/2)))*EllipticF(sin(2*arctan(1/2*x*2^(1/2))),1/4*2^(1/2))*((x^4+3*x^2+4)/(x^2+2)^2)
^(1/2)*2^(1/2)/(x^4+3*x^2+4)^(1/2)-20*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*x*2^(1/2
)))*EllipticE(sin(2*arctan(1/2*x*2^(1/2))),1/4*2^(1/2))*2^(1/2)*((x^4+3*x^2+4)/(x^2+2)^2)^(1/2)/(x^4+3*x^2+4)^
(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1206, 1197, 1103, 1195} \[ \frac {20 \sqrt {x^4+3 x^2+4} x}{x^2+2}+\frac {25}{3} \sqrt {x^4+3 x^2+4} x+\frac {167 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{6 \sqrt {2} \sqrt {x^4+3 x^2+4}}-\frac {20 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{\sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(25*x*Sqrt[4 + 3*x^2 + x^4])/3 + (20*x*Sqrt[4 + 3*x^2 + x^4])/(2 + x^2) - (20*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^
2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/Sqrt[4 + 3*x^2 + x^4] + (167*(2 + x^2)*Sqrt[(4 + 3*
x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(6*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1206

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e^q*x^(2*q - 3)*(
a + b*x^2 + c*x^4)^(p + 1))/(c*(4*p + 2*q + 1)), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + b*x^2 + c*x^4)^p*Ex
pandToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2*p + 2*q - 1)*e^q*x^(2*q - 2) -
c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && IGtQ[q, 1]

Rubi steps

\begin {align*} \int \frac {\left (7+5 x^2\right )^2}{\sqrt {4+3 x^2+x^4}} \, dx &=\frac {25}{3} x \sqrt {4+3 x^2+x^4}+\frac {1}{3} \int \frac {47+60 x^2}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {25}{3} x \sqrt {4+3 x^2+x^4}-40 \int \frac {1-\frac {x^2}{2}}{\sqrt {4+3 x^2+x^4}} \, dx+\frac {167}{3} \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {25}{3} x \sqrt {4+3 x^2+x^4}+\frac {20 x \sqrt {4+3 x^2+x^4}}{2+x^2}-\frac {20 \sqrt {2} \left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{\sqrt {4+3 x^2+x^4}}+\frac {167 \left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{6 \sqrt {2} \sqrt {4+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.43, size = 331, normalized size = 1.95 \[ \frac {\sqrt {2} \left (30 \sqrt {7}+43 i\right ) \sqrt {\frac {-2 i x^2+\sqrt {7}-3 i}{\sqrt {7}-3 i}} \sqrt {\frac {2 i x^2+\sqrt {7}+3 i}{\sqrt {7}+3 i}} F\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )-30 \sqrt {2} \left (\sqrt {7}+3 i\right ) \sqrt {\frac {-2 i x^2+\sqrt {7}-3 i}{\sqrt {7}-3 i}} \sqrt {\frac {2 i x^2+\sqrt {7}+3 i}{\sqrt {7}+3 i}} E\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )+50 \sqrt {-\frac {i}{\sqrt {7}-3 i}} x \left (x^4+3 x^2+4\right )}{6 \sqrt {-\frac {i}{\sqrt {7}-3 i}} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(50*Sqrt[(-I)/(-3*I + Sqrt[7])]*x*(4 + 3*x^2 + x^4) - 30*Sqrt[2]*(3*I + Sqrt[7])*Sqrt[(-3*I + Sqrt[7] - (2*I)*
x^2)/(-3*I + Sqrt[7])]*Sqrt[(3*I + Sqrt[7] + (2*I)*x^2)/(3*I + Sqrt[7])]*EllipticE[I*ArcSinh[Sqrt[(-2*I)/(-3*I
+ Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] + Sqrt[2]*(43*I + 30*Sqrt[7])*Sqrt[(-3*I + Sqrt[7] - (2*I)*x
^2)/(-3*I + Sqrt[7])]*Sqrt[(3*I + Sqrt[7] + (2*I)*x^2)/(3*I + Sqrt[7])]*EllipticF[I*ArcSinh[Sqrt[(-2*I)/(-3*I
+ Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])])/(6*Sqrt[(-I)/(-3*I + Sqrt[7])]*Sqrt[4 + 3*x^2 + x^4])

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {25 \, x^{4} + 70 \, x^{2} + 49}{\sqrt {x^{4} + 3 \, x^{2} + 4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+7)^2/(x^4+3*x^2+4)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x^{2} + 7\right )}^{2}}{\sqrt {x^{4} + 3 \, x^{2} + 4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+7)^2/(x^4+3*x^2+4)^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.01, size = 224, normalized size = 1.32 \[ \frac {25 \sqrt {x^{4}+3 x^{2}+4}\, x}{3}+\frac {188 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{3 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}}-\frac {640 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )+\EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )\right )}{\sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2+7)^2/(x^4+3*x^2+4)^(1/2),x)

[Out]

25/3*(x^4+3*x^2+4)^(1/2)*x+188/3/(-6+2*I*7^(1/2))^(1/2)*(-(-3/8+1/8*I*7^(1/2))*x^2+1)^(1/2)*(-(-3/8-1/8*I*7^(1
/2))*x^2+1)^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticF(1/4*(-6+2*I*7^(1/2))^(1/2)*x,1/4*(2+6*I*7^(1/2))^(1/2))-640/(-
6+2*I*7^(1/2))^(1/2)*(-(-3/8+1/8*I*7^(1/2))*x^2+1)^(1/2)*(-(-3/8-1/8*I*7^(1/2))*x^2+1)^(1/2)/(x^4+3*x^2+4)^(1/
2)/(I*7^(1/2)+3)*(EllipticF(1/4*(-6+2*I*7^(1/2))^(1/2)*x,1/4*(2+6*I*7^(1/2))^(1/2))-EllipticE(1/4*(-6+2*I*7^(1
/2))^(1/2)*x,1/4*(2+6*I*7^(1/2))^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x^{2} + 7\right )}^{2}}{\sqrt {x^{4} + 3 \, x^{2} + 4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+7)^2/(x^4+3*x^2+4)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (5\,x^2+7\right )}^2}{\sqrt {x^4+3\,x^2+4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2 + 7)^2/(3*x^2 + x^4 + 4)^(1/2),x)

[Out]

int((5*x^2 + 7)^2/(3*x^2 + x^4 + 4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (5 x^{2} + 7\right )^{2}}{\sqrt {\left (x^{2} - x + 2\right ) \left (x^{2} + x + 2\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+7)**2/(x**4+3*x**2+4)**(1/2),x)

[Out]

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

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