3.2.55 \(\int \frac {(2+3 x^2) \sqrt {3+5 x^2+x^4}}{x^4} \, dx\) [155]

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

[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 = 0.12, 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 = {1285, 1295, 1203, 1113, 1149} \begin {gather*} \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 (\text {ArcTan}\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 (\text {ArcTan}\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 {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 {\sqrt {x^4+5 x^2+3} \left (2-9 x^2\right )}{3 x^3} \end {gather*}

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 1113

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)]/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*Elli
pticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], 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 1149

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)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q
/(b + q))], 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 1203

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 1285

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 1295

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] Result contains complex when optimal does not.
time = 9.47, size = 237, normalized size = 0.78 \begin {gather*} \frac {-2 \left (18+141 x^2+191 x^4+37 x^6\right )+32 i \sqrt {2} \left (-5+\sqrt {13}\right ) x^3 \sqrt {\frac {-5+\sqrt {13}-2 x^2}{-5+\sqrt {13}}} \sqrt {5+\sqrt {13}+2 x^2} E\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )-i \sqrt {2} \left (-13+32 \sqrt {13}\right ) x^3 \sqrt {\frac {-5+\sqrt {13}-2 x^2}{-5+\sqrt {13}}} \sqrt {5+\sqrt {13}+2 x^2} F\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {13}}} x\right )|\frac {19}{6}+\frac {5 \sqrt {13}}{6}\right )}{18 x^3 \sqrt {3+5 x^2+x^4}} \end {gather*}

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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Maple [A]
time = 0.06, size = 228, normalized size = 0.75

method	result	size
default	\(-\frac {37 \sqrt {x^{4}+5 x^{2}+3}}{9 x}+\frac {98 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}-\frac {256 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )-\EllipticE \left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (5+\sqrt {13}\right )}-\frac {2 \sqrt {x^{4}+5 x^{2}+3}}{3 x^{3}}\)	\(228\)
risch	\(-\frac {37 x^{6}+191 x^{4}+141 x^{2}+18}{9 x^{3} \sqrt {x^{4}+5 x^{2}+3}}-\frac {256 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )-\EllipticE \left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (5+\sqrt {13}\right )}+\frac {98 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}\)	\(228\)
elliptic	\(-\frac {37 \sqrt {x^{4}+5 x^{2}+3}}{9 x}+\frac {98 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}}-\frac {256 \sqrt {1-\left (-\frac {5}{6}+\frac {\sqrt {13}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{6}-\frac {\sqrt {13}}{6}\right ) x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )-\EllipticE \left (\frac {x \sqrt {-30+6 \sqrt {13}}}{6}, \frac {5 \sqrt {3}}{6}+\frac {\sqrt {39}}{6}\right )\right )}{\sqrt {-30+6 \sqrt {13}}\, \sqrt {x^{4}+5 x^{2}+3}\, \left (5+\sqrt {13}\right )}-\frac {2 \sqrt {x^{4}+5 x^{2}+3}}{3 x^{3}}\)	\(228\)

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,method=_RETURNVERBOSE)

[Out]

-37/9*(x^4+5*x^2+3)^(1/2)/x+98/(-30+6*13^(1/2))^(1/2)*(1-(-5/6+1/6*13^(1/2))*x^2)^(1/2)*(1-(-5/6-1/6*13^(1/2))
*x^2)^(1/2)/(x^4+5*x^2+3)^(1/2)*EllipticF(1/6*x*(-30+6*13^(1/2))^(1/2),5/6*3^(1/2)+1/6*39^(1/2))-256/(-30+6*13
^(1/2))^(1/2)*(1-(-5/6+1/6*13^(1/2))*x^2)^(1/2)*(1-(-5/6-1/6*13^(1/2))*x^2)^(1/2)/(x^4+5*x^2+3)^(1/2)/(5+13^(1
/2))*(EllipticF(1/6*x*(-30+6*13^(1/2))^(1/2),5/6*3^(1/2)+1/6*39^(1/2))-EllipticE(1/6*x*(-30+6*13^(1/2))^(1/2),
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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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