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

Optimal. Leaf size=179 \[ \frac {1}{63} x \left (35 x^2+108\right ) \left (x^4+3 x^2+2\right )^{3/2}+\frac {1}{105} x \left (149 x^2+519\right ) \sqrt {x^4+3 x^2+2}+\frac {116 x \left (x^2+2\right )}{15 \sqrt {x^4+3 x^2+2}}+\frac {197 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \sqrt {x^4+3 x^2+2}}-\frac {116 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{15 \sqrt {x^4+3 x^2+2}} \]

[Out]

1/63*x*(35*x^2+108)*(x^4+3*x^2+2)^(3/2)+116/15*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-116/15*(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)+197/35*(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)+1/105*x*(149*x^2+519)*(x^4+3*x^2+2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1176, 1189, 1099, 1135} \[ \frac {1}{63} x \left (35 x^2+108\right ) \left (x^4+3 x^2+2\right )^{3/2}+\frac {1}{105} x \left (149 x^2+519\right ) \sqrt {x^4+3 x^2+2}+\frac {116 x \left (x^2+2\right )}{15 \sqrt {x^4+3 x^2+2}}+\frac {197 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \sqrt {x^4+3 x^2+2}}-\frac {116 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{15 \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(116*x*(2 + x^2))/(15*Sqrt[2 + 3*x^2 + x^4]) + (x*(519 + 149*x^2)*Sqrt[2 + 3*x^2 + x^4])/105 + (x*(108 + 35*x^
2)*(2 + 3*x^2 + x^4)^(3/2))/63 - (116*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(
15*Sqrt[2 + 3*x^2 + x^4]) + (197*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(35*Sq
rt[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 1176

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(2*b*e*p + c*d*(4*p
+ 3) + c*e*(4*p + 1)*x^2)*(a + b*x^2 + c*x^4)^p)/(c*(4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/(c*(4*p + 1)*(4*p +
3)), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*x^2, x]*(a +
b*x^2 + c*x^4)^(p - 1), 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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]

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]

Rubi steps

\begin {align*} \int \left (7+5 x^2\right ) \left (2+3 x^2+x^4\right )^{3/2} \, dx &=\frac {1}{63} x \left (108+35 x^2\right ) \left (2+3 x^2+x^4\right )^{3/2}+\frac {1}{21} \int \left (222+149 x^2\right ) \sqrt {2+3 x^2+x^4} \, dx\\ &=\frac {1}{105} x \left (519+149 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {1}{63} x \left (108+35 x^2\right ) \left (2+3 x^2+x^4\right )^{3/2}+\frac {1}{315} \int \frac {3546+2436 x^2}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {1}{105} x \left (519+149 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {1}{63} x \left (108+35 x^2\right ) \left (2+3 x^2+x^4\right )^{3/2}+\frac {116}{15} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {394}{35} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {116 x \left (2+x^2\right )}{15 \sqrt {2+3 x^2+x^4}}+\frac {1}{105} x \left (519+149 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {1}{63} x \left (108+35 x^2\right ) \left (2+3 x^2+x^4\right )^{3/2}-\frac {116 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{15 \sqrt {2+3 x^2+x^4}}+\frac {197 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \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)*(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 \[ {\rm integral}\left ({\left (5 \, x^{6} + 22 \, x^{4} + 31 \, x^{2} + 14\right )} \sqrt {x^{4} + 3 \, x^{2} + 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((5*x^6 + 22*x^4 + 31*x^2 + 14)*sqrt(x^4 + 3*x^2 + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.00, size = 172, normalized size = 0.96 \[ \frac {5 \sqrt {x^{4}+3 x^{2}+2}\, x^{7}}{9}+\frac {71 \sqrt {x^{4}+3 x^{2}+2}\, x^{5}}{21}+\frac {2417 \sqrt {x^{4}+3 x^{2}+2}\, x^{3}}{315}+\frac {293 \sqrt {x^{4}+3 x^{2}+2}\, x}{35}-\frac {197 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{35 \sqrt {x^{4}+3 x^{2}+2}}+\frac {58 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )+\EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{15 \sqrt {x^{4}+3 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/9*(x^4+3*x^2+2)^(1/2)*x^7+71/21*(x^4+3*x^2+2)^(1/2)*x^5+2417/315*(x^4+3*x^2+2)^(1/2)*x^3+293/35*(x^4+3*x^2+2
)^(1/2)*x-197/35*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)
)+58/15*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))-Ellipt
icE(1/2*I*2^(1/2)*x,2^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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