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

Optimal. Leaf size=305 \[ -\frac {64 \sqrt {x^4+5 x^2+3}}{9 x}+\frac {32 x \left (2 x^2+\sqrt {13}+5\right )}{9 \sqrt {x^4+5 x^2+3}}+\frac {49 \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{3 \sqrt {6 \left (5+\sqrt {13}\right )} \sqrt {x^4+5 x^2+3}}-\frac {16 \sqrt {\frac {2}{3} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{9 \sqrt {x^4+5 x^2+3}}-\frac {\sqrt {x^4+5 x^2+3} \left (2-9 x^2\right )}{3 x^3} \]

[Out]

32/9*x*(5+2*x^2+13^(1/2))/(x^4+5*x^2+3)^(1/2)-64/9*(x^4+5*x^2+3)^(1/2)/x-1/3*(-9*x^2+2)*(x^4+5*x^2+3)^(1/2)/x^
3-16/27*(1/(36+x^2*(30+6*13^(1/2))))^(1/2)*(36+x^2*(30+6*13^(1/2)))^(1/2)*EllipticE(x*(30+6*13^(1/2))^(1/2)/(3
6+x^2*(30+6*13^(1/2)))^(1/2),1/6*(-78+30*13^(1/2))^(1/2))*(6+x^2*(5+13^(1/2)))*(30+6*13^(1/2))^(1/2)*((6+x^2*(
5-13^(1/2)))/(6+x^2*(5+13^(1/2))))^(1/2)/(x^4+5*x^2+3)^(1/2)+49/3*(1/(36+x^2*(30+6*13^(1/2))))^(1/2)*(36+x^2*(
30+6*13^(1/2)))^(1/2)*EllipticF(x*(30+6*13^(1/2))^(1/2)/(36+x^2*(30+6*13^(1/2)))^(1/2),1/6*(-78+30*13^(1/2))^(
1/2))*(6+x^2*(5+13^(1/2)))*((6+x^2*(5-13^(1/2)))/(6+x^2*(5+13^(1/2))))^(1/2)/(x^4+5*x^2+3)^(1/2)/(30+6*13^(1/2
))^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1271, 1281, 1189, 1099, 1135} \[ -\frac {\sqrt {x^4+5 x^2+3} \left (2-9 x^2\right )}{3 x^3}-\frac {64 \sqrt {x^4+5 x^2+3}}{9 x}+\frac {32 x \left (2 x^2+\sqrt {13}+5\right )}{9 \sqrt {x^4+5 x^2+3}}+\frac {49 \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{3 \sqrt {6 \left (5+\sqrt {13}\right )} \sqrt {x^4+5 x^2+3}}-\frac {16 \sqrt {\frac {2}{3} \left (5+\sqrt {13}\right )} \sqrt {\frac {\left (5-\sqrt {13}\right ) x^2+6}{\left (5+\sqrt {13}\right ) x^2+6}} \left (\left (5+\sqrt {13}\right ) x^2+6\right ) E\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{9 \sqrt {x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*x*(5 + Sqrt[13] + 2*x^2))/(9*Sqrt[3 + 5*x^2 + x^4]) - (64*Sqrt[3 + 5*x^2 + x^4])/(9*x) - ((2 - 9*x^2)*Sqrt
[3 + 5*x^2 + x^4])/(3*x^3) - (16*Sqrt[(2*(5 + Sqrt[13]))/3]*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[13])*
x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticE[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/(9*Sqrt[3 +
5*x^2 + x^4]) + (49*Sqrt[(6 + (5 - Sqrt[13])*x^2)/(6 + (5 + Sqrt[13])*x^2)]*(6 + (5 + Sqrt[13])*x^2)*EllipticF
[ArcTan[Sqrt[(5 + Sqrt[13])/6]*x], (-13 + 5*Sqrt[13])/6])/(3*Sqrt[6*(5 + Sqrt[13])]*Sqrt[3 + 5*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 1271

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((f
*x)^(m + 1)*(a + b*x^2 + c*x^4)^p*(d*(m + 4*p + 3) + e*(m + 1)*x^2))/(f*(m + 1)*(m + 4*p + 3)), x] + Dist[(2*p
)/(f^2*(m + 1)*(m + 4*p + 3)), Int[(f*x)^(m + 2)*(a + b*x^2 + c*x^4)^(p - 1)*Simp[2*a*e*(m + 1) - b*d*(m + 4*p
+ 3) + (b*e*(m + 1) - 2*c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c
, 0] && GtQ[p, 0] && LtQ[m, -1] && m + 4*p + 3 != 0 && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(
f*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + b
*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {\left (2+3 x^2\right ) \sqrt {3+5 x^2+x^4}}{x^4} \, dx &=-\frac {\left (2-9 x^2\right ) \sqrt {3+5 x^2+x^4}}{3 x^3}-\frac {1}{3} \int \frac {-64-49 x^2}{x^2 \sqrt {3+5 x^2+x^4}} \, dx\\ &=-\frac {64 \sqrt {3+5 x^2+x^4}}{9 x}-\frac {\left (2-9 x^2\right ) \sqrt {3+5 x^2+x^4}}{3 x^3}+\frac {1}{9} \int \frac {147+64 x^2}{\sqrt {3+5 x^2+x^4}} \, dx\\ &=-\frac {64 \sqrt {3+5 x^2+x^4}}{9 x}-\frac {\left (2-9 x^2\right ) \sqrt {3+5 x^2+x^4}}{3 x^3}+\frac {64}{9} \int \frac {x^2}{\sqrt {3+5 x^2+x^4}} \, dx+\frac {49}{3} \int \frac {1}{\sqrt {3+5 x^2+x^4}} \, dx\\ &=\frac {32 x \left (5+\sqrt {13}+2 x^2\right )}{9 \sqrt {3+5 x^2+x^4}}-\frac {64 \sqrt {3+5 x^2+x^4}}{9 x}-\frac {\left (2-9 x^2\right ) \sqrt {3+5 x^2+x^4}}{3 x^3}-\frac {16 \sqrt {\frac {2}{3} \left (5+\sqrt {13}\right )} \sqrt {\frac {6+\left (5-\sqrt {13}\right ) x^2}{6+\left (5+\sqrt {13}\right ) x^2}} \left (6+\left (5+\sqrt {13}\right ) x^2\right ) E\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{9 \sqrt {3+5 x^2+x^4}}+\frac {49 \sqrt {\frac {6+\left (5-\sqrt {13}\right ) x^2}{6+\left (5+\sqrt {13}\right ) x^2}} \left (6+\left (5+\sqrt {13}\right ) x^2\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (5+\sqrt {13}\right )} x\right )|\frac {1}{6} \left (-13+5 \sqrt {13}\right )\right )}{3 \sqrt {6 \left (5+\sqrt {13}\right )} \sqrt {3+5 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 237, normalized size = 0.78 \[ \frac {-i \sqrt {2} \left (32 \sqrt {13}-13\right ) \sqrt {\frac {-2 x^2+\sqrt {13}-5}{\sqrt {13}-5}} \sqrt {2 x^2+\sqrt {13}+5} x^3 F\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )+32 i \sqrt {2} \left (\sqrt {13}-5\right ) \sqrt {\frac {-2 x^2+\sqrt {13}-5}{\sqrt {13}-5}} \sqrt {2 x^2+\sqrt {13}+5} x^3 E\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )-2 \left (37 x^6+191 x^4+141 x^2+18\right )}{18 x^3 \sqrt {x^4+5 x^2+3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*(18 + 141*x^2 + 191*x^4 + 37*x^6) + (32*I)*Sqrt[2]*(-5 + Sqrt[13])*x^3*Sqrt[(-5 + Sqrt[13] - 2*x^2)/(-5 +
Sqrt[13])]*Sqrt[5 + Sqrt[13] + 2*x^2]*EllipticE[I*ArcSinh[Sqrt[2/(5 + Sqrt[13])]*x], 19/6 + (5*Sqrt[13])/6] -
I*Sqrt[2]*(-13 + 32*Sqrt[13])*x^3*Sqrt[(-5 + Sqrt[13] - 2*x^2)/(-5 + Sqrt[13])]*Sqrt[5 + Sqrt[13] + 2*x^2]*Ell
ipticF[I*ArcSinh[Sqrt[2/(5 + Sqrt[13])]*x], 19/6 + (5*Sqrt[13])/6])/(18*x^3*Sqrt[3 + 5*x^2 + x^4])

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 5 \, x^{2} + 3} {\left (3 \, x^{2} + 2\right )}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 228, normalized size = 0.75 \[ \frac {98 \sqrt {-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}-\frac {37 \sqrt {x^{4}+5 x^{2}+3}}{9 x}-\frac {2 \sqrt {x^{4}+5 x^{2}+3}}{3 x^{3}}-\frac {256 \sqrt {-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )+\EllipticF \left (\frac {\sqrt {-30+6 \sqrt {13}}\, x}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (\sqrt {13}+5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-37/9*(x^4+5*x^2+3)^(1/2)/x+98/(-30+6*13^(1/2))^(1/2)*(-(-5/6+1/6*13^(1/2))*x^2+1)^(1/2)*(-(-5/6-1/6*13^(1/2))
*x^2+1)^(1/2)/(x^4+5*x^2+3)^(1/2)*EllipticF(1/6*(-30+6*13^(1/2))^(1/2)*x,5/6*3^(1/2)+1/6*39^(1/2))-256/(-30+6*
13^(1/2))^(1/2)*(-(-5/6+1/6*13^(1/2))*x^2+1)^(1/2)*(-(-5/6-1/6*13^(1/2))*x^2+1)^(1/2)/(x^4+5*x^2+3)^(1/2)/(13^
(1/2)+5)*(EllipticF(1/6*(-30+6*13^(1/2))^(1/2)*x,5/6*3^(1/2)+1/6*39^(1/2))-EllipticE(1/6*(-30+6*13^(1/2))^(1/2
)*x,5/6*3^(1/2)+1/6*39^(1/2)))-2/3*(x^4+5*x^2+3)^(1/2)/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((3*x**2 + 2)*sqrt(x**4 + 5*x**2 + 3)/x**4, x)

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