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

Optimal. Leaf size=149 \[ \frac {x \left (5-11 x^2\right )}{2 \sqrt {x^4+3 x^2+2}}+\frac {261 x \left (x^2+2\right )}{2 \sqrt {x^4+3 x^2+2}}+\frac {169 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {261 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}} \]

[Out]

1/2*x*(-11*x^2+5)/(x^4+3*x^2+2)^(1/2)+261/2*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-261/2*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2
)*EllipticE(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+169/2*(x^2+1)^(3/
2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2
)

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Rubi [A]  time = 0.05, antiderivative size = 149, 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 = {1205, 1189, 1099, 1135} \[ \frac {x \left (5-11 x^2\right )}{2 \sqrt {x^4+3 x^2+2}}+\frac {261 x \left (x^2+2\right )}{2 \sqrt {x^4+3 x^2+2}}+\frac {169 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {261 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(5 - 11*x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (261*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) - (261*(1 + x^2)*Sqrt
[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (169*(1 + x^2)*Sqrt[(2 + x^
2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2 + x^4])

Rule 1099

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

Rule 1135

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b +
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)])/(2*c*Sqrt[a + b*x^2
+ c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1189

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

Rule 1205

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{f = Coeff[Polynom
ialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x
^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2))/(
2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToS
um[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c
*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*
a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\left (7+5 x^2\right )^3}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx &=\frac {x \left (5-11 x^2\right )}{2 \sqrt {2+3 x^2+x^4}}-\frac {1}{2} \int \frac {-338-261 x^2}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {x \left (5-11 x^2\right )}{2 \sqrt {2+3 x^2+x^4}}+\frac {261}{2} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+169 \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {x \left (5-11 x^2\right )}{2 \sqrt {2+3 x^2+x^4}}+\frac {261 x \left (2+x^2\right )}{2 \sqrt {2+3 x^2+x^4}}-\frac {261 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {169 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+3 x^2+x^4}}\\ \end {align*}

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Mathematica [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (125 \, x^{6} + 525 \, x^{4} + 735 \, x^{2} + 343\right )} \sqrt {x^{4} + 3 \, x^{2} + 2}}{x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((125*x^6 + 525*x^4 + 735*x^2 + 343)*sqrt(x^4 + 3*x^2 + 2)/(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.01, size = 196, normalized size = 1.32 \[ -\frac {169 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}-\frac {250 \left (\frac {5}{2} x^{3}+3 x \right )}{\sqrt {x^{4}+3 x^{2}+2}}+\frac {261 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )+\EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{4 \sqrt {x^{4}+3 x^{2}+2}}-\frac {1050 \left (-\frac {3}{2} x^{3}-2 x \right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {1470 \left (x^{3}+\frac {3}{2} x \right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {686 \left (-\frac {3}{4} x^{3}-\frac {5}{4} x \right )}{\sqrt {x^{4}+3 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-250*(5/2*x^3+3*x)/(x^4+3*x^2+2)^(1/2)-169/2*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*Ellip
ticF(1/2*I*2^(1/2)*x,2^(1/2))+261/4*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))-EllipticE(1/2*I*2^(1/2)*x,2^(1/2)))-1050*(-3/2*x^3-2*x)/(x^4+3*x^2+2)^(1/2)-1470*(x^3+3/
2*x)/(x^4+3*x^2+2)^(1/2)-686*(-3/4*x^3-5/4*x)/(x^4+3*x^2+2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (5 x^{2} + 7\right )^{3}}{\left (\left (x^{2} + 1\right ) \left (x^{2} + 2\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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