3.360 \(\int (4+3 x^2+x^4)^{3/2} \, dx\)

Optimal. Leaf size=198 \[ \frac {1}{7} x \left (x^4+3 x^2+4\right )^{3/2}+\frac {1}{35} x \left (9 x^2+49\right ) \sqrt {x^4+3 x^2+4}+\frac {138 x \sqrt {x^4+3 x^2+4}}{35 \left (x^2+2\right )}+\frac {4 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{\sqrt {x^4+3 x^2+4}}-\frac {138 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{35 \sqrt {x^4+3 x^2+4}} \]

[Out]

1/7*x*(x^4+3*x^2+4)^(3/2)+138/35*x*(x^4+3*x^2+4)^(1/2)/(x^2+2)+1/35*x*(9*x^2+49)*(x^4+3*x^2+4)^(1/2)-138/35*(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)+4*(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)

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Rubi [A]  time = 0.07, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1091, 1176, 1197, 1103, 1195} \[ \frac {1}{7} x \left (x^4+3 x^2+4\right )^{3/2}+\frac {1}{35} x \left (9 x^2+49\right ) \sqrt {x^4+3 x^2+4}+\frac {138 x \sqrt {x^4+3 x^2+4}}{35 \left (x^2+2\right )}+\frac {4 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{\sqrt {x^4+3 x^2+4}}-\frac {138 \sqrt {2} \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{35 \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Int[(4 + 3*x^2 + x^4)^(3/2),x]

[Out]

(138*x*Sqrt[4 + 3*x^2 + x^4])/(35*(2 + x^2)) + (x*(49 + 9*x^2)*Sqrt[4 + 3*x^2 + x^4])/35 + (x*(4 + 3*x^2 + x^4
)^(3/2))/7 - (138*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(
35*Sqrt[4 + 3*x^2 + x^4]) + (4*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt
[2]], 1/8])/Sqrt[4 + 3*x^2 + x^4]

Rule 1091

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a + b*x^2 + c*x^4)^p)/(4*p + 1), x] + Dis
t[(2*p)/(4*p + 1), Int[(2*a + b*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4
*a*c, 0] && GtQ[p, 0] && IntegerQ[2*p]

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 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 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]

Rubi steps

\begin {align*} \int \left (4+3 x^2+x^4\right )^{3/2} \, dx &=\frac {1}{7} x \left (4+3 x^2+x^4\right )^{3/2}+\frac {3}{7} \int \left (8+3 x^2\right ) \sqrt {4+3 x^2+x^4} \, dx\\ &=\frac {1}{35} x \left (49+9 x^2\right ) \sqrt {4+3 x^2+x^4}+\frac {1}{7} x \left (4+3 x^2+x^4\right )^{3/2}+\frac {1}{35} \int \frac {284+138 x^2}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {1}{35} x \left (49+9 x^2\right ) \sqrt {4+3 x^2+x^4}+\frac {1}{7} x \left (4+3 x^2+x^4\right )^{3/2}-\frac {276}{35} \int \frac {1-\frac {x^2}{2}}{\sqrt {4+3 x^2+x^4}} \, dx+16 \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {138 x \sqrt {4+3 x^2+x^4}}{35 \left (2+x^2\right )}+\frac {1}{35} x \left (49+9 x^2\right ) \sqrt {4+3 x^2+x^4}+\frac {1}{7} x \left (4+3 x^2+x^4\right )^{3/2}-\frac {138 \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 )}{35 \sqrt {4+3 x^2+x^4}}+\frac {4 \sqrt {2} \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 )}{\sqrt {4+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.42, size = 343, normalized size = 1.73 \[ \frac {\sqrt {2} \left (69 \sqrt {7}-77 i\right ) \sqrt {\frac {-2 i x^2+\sqrt {7}-3 i}{\sqrt {7}-3 i}} \sqrt {\frac {2 i x^2+\sqrt {7}+3 i}{\sqrt {7}+3 i}} F\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )-69 \sqrt {2} \left (\sqrt {7}+3 i\right ) \sqrt {\frac {-2 i x^2+\sqrt {7}-3 i}{\sqrt {7}-3 i}} \sqrt {\frac {2 i x^2+\sqrt {7}+3 i}{\sqrt {7}+3 i}} E\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )+2 \sqrt {-\frac {i}{\sqrt {7}-3 i}} x \left (5 x^8+39 x^6+161 x^4+303 x^2+276\right )}{70 \sqrt {-\frac {i}{\sqrt {7}-3 i}} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(4 + 3*x^2 + x^4)^(3/2),x]

[Out]

(2*Sqrt[(-I)/(-3*I + Sqrt[7])]*x*(276 + 303*x^2 + 161*x^4 + 39*x^6 + 5*x^8) - 69*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*A
rcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] + Sqrt[2]*(-77*I + 69*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*A
rcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])])/(70*Sqrt[(-I)/(-3*I + Sqrt[7])]*Sqr
t[4 + 3*x^2 + x^4])

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (x^{4} + 3 \, x^{2} + 4\right )}^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((x^4 + 3*x^2 + 4)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + 3*x^2 + 4)^(3/2), x)

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maple [C]  time = 0.00, size = 258, normalized size = 1.30 \[ \frac {\sqrt {x^{4}+3 x^{2}+4}\, x^{5}}{7}+\frac {24 \sqrt {x^{4}+3 x^{2}+4}\, x^{3}}{35}+\frac {69 \sqrt {x^{4}+3 x^{2}+4}\, x}{35}+\frac {1136 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{35 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}}-\frac {4416 \sqrt {-\left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {3}{8}-\frac {i \sqrt {7}}{8}\right ) x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )+\EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )\right )}{35 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(x^4+3*x^2+4)^(1/2)*x^5+24/35*(x^4+3*x^2+4)^(1/2)*x^3+69/35*(x^4+3*x^2+4)^(1/2)*x+1136/35/(-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))-4416/35/(-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 \[ \int {\left (x^{4} + 3 \, x^{2} + 4\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + 3*x^2 + 4)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2 + x^4 + 4)^(3/2),x)

[Out]

int((3*x^2 + x^4 + 4)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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