3.778 \(\int \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx\)

Optimal. Leaf size=663 \[ \frac {1}{3} \left (\frac {d}{4 e}+x\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}-\frac {2 d^2 \left (\frac {d}{4 e}+x\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{\sqrt {256 a e^3+5 d^4} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )}+\frac {\sqrt [4]{256 a e^3+5 d^4} \left (-3 d^2 \sqrt {256 a e^3+5 d^4}+256 a e^3+5 d^4\right ) \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) F\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{48 \sqrt {2} e^2 \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}+\frac {d^2 \left (256 a e^3+5 d^4\right )^{3/4} \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) E\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{8 \sqrt {2} e^2 \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]

[Out]

1/3*(1/4*d/e+x)*(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2)-2*d^2*(1/4*d/e+x)*(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a
*e^2)^(1/2)/(1+16*e^2*(1/4*d/e+x)^2/(256*a*e^3+5*d^4)^(1/2))/(256*a*e^3+5*d^4)^(1/2)+1/16*d^2*(256*a*e^3+5*d^4
)^(3/4)*(cos(2*arctan((4*e*x+d)/(256*a*e^3+5*d^4)^(1/4)))^2)^(1/2)/cos(2*arctan((4*e*x+d)/(256*a*e^3+5*d^4)^(1
/4)))*EllipticE(sin(2*arctan((4*e*x+d)/(256*a*e^3+5*d^4)^(1/4))),1/2*(2+6*d^2/(256*a*e^3+5*d^4)^(1/2))^(1/2))*
(1+16*e^2*(1/4*d/e+x)^2/(256*a*e^3+5*d^4)^(1/2))*(e*(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)/(256*a*e^3+5*d^4)/(1
+16*e^2*(1/4*d/e+x)^2/(256*a*e^3+5*d^4)^(1/2))^2)^(1/2)/e^2*2^(1/2)/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2
)+1/96*(256*a*e^3+5*d^4)^(1/4)*(cos(2*arctan((4*e*x+d)/(256*a*e^3+5*d^4)^(1/4)))^2)^(1/2)/cos(2*arctan((4*e*x+
d)/(256*a*e^3+5*d^4)^(1/4)))*EllipticF(sin(2*arctan((4*e*x+d)/(256*a*e^3+5*d^4)^(1/4))),1/2*(2+6*d^2/(256*a*e^
3+5*d^4)^(1/2))^(1/2))*(1+16*e^2*(1/4*d/e+x)^2/(256*a*e^3+5*d^4)^(1/2))*(5*d^4+256*a*e^3-3*d^2*(256*a*e^3+5*d^
4)^(1/2))*(e*(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)/(256*a*e^3+5*d^4)/(1+16*e^2*(1/4*d/e+x)^2/(256*a*e^3+5*d^4)
^(1/2))^2)^(1/2)/e^2*2^(1/2)/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2)

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Rubi [A]  time = 0.81, antiderivative size = 663, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {1106, 1091, 1197, 1103, 1195} \[ -\frac {2 d^2 \left (\frac {d}{4 e}+x\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{\sqrt {256 a e^3+5 d^4} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )}+\frac {1}{3} \left (\frac {d}{4 e}+x\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}+\frac {\sqrt [4]{256 a e^3+5 d^4} \left (-3 d^2 \sqrt {256 a e^3+5 d^4}+256 a e^3+5 d^4\right ) \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) F\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{48 \sqrt {2} e^2 \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}+\frac {d^2 \left (256 a e^3+5 d^4\right )^{3/4} \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) E\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{8 \sqrt {2} e^2 \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4],x]

[Out]

((d/(4*e) + x)*Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4])/3 - (2*d^2*(d/(4*e) + x)*Sqrt[8*a*e^2 - d^3*x
+ 8*d*e^2*x^3 + 8*e^3*x^4])/(Sqrt[5*d^4 + 256*a*e^3]*(1 + (16*e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])) +
(d^2*(5*d^4 + 256*a*e^3)^(3/4)*Sqrt[(e*(8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4))/((5*d^4 + 256*a*e^3)*(1 +
(16*e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])^2)]*(1 + (16*e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])*
EllipticE[2*ArcTan[(d + 4*e*x)/(5*d^4 + 256*a*e^3)^(1/4)], (1 + (3*d^2)/Sqrt[5*d^4 + 256*a*e^3])/2])/(8*Sqrt[2
]*e^2*Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4]) + ((5*d^4 + 256*a*e^3)^(1/4)*(5*d^4 + 256*a*e^3 - 3*d^2
*Sqrt[5*d^4 + 256*a*e^3])*Sqrt[(e*(8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4))/((5*d^4 + 256*a*e^3)*(1 + (16*e
^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])^2)]*(1 + (16*e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])*Ellipt
icF[2*ArcTan[(d + 4*e*x)/(5*d^4 + 256*a*e^3)^(1/4)], (1 + (3*d^2)/Sqrt[5*d^4 + 256*a*e^3])/2])/(48*Sqrt[2]*e^2
*Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*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 1106

Int[(P4_)^(p_), x_Symbol] :> With[{a = Coeff[P4, x, 0], b = Coeff[P4, x, 1], c = Coeff[P4, x, 2], d = Coeff[P4
, x, 3], e = Coeff[P4, x, 4]}, Subst[Int[SimplifyIntegrand[(a + d^4/(256*e^3) - (b*d)/(8*e) + (c - (3*d^2)/(8*
e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0] && NeQ[d, 0]] /; FreeQ[p, x] &&
PolyQ[P4, x, 4] && NeQ[p, 2] && NeQ[p, 3]

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 \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx &=\operatorname {Subst}\left (\int \sqrt {\frac {1}{32} \left (\frac {5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2+8 e^3 x^4} \, dx,x,\frac {d}{4 e}+x\right )\\ &=\frac {1}{3} \left (\frac {d}{4 e}+x\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {\frac {1}{16} \left (\frac {5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2}{\sqrt {\frac {1}{32} \left (\frac {5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2+8 e^3 x^4}} \, dx,x,\frac {d}{4 e}+x\right )\\ &=\frac {1}{3} \left (\frac {d}{4 e}+x\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}+\frac {\left (d^2 \sqrt {5 d^4+256 a e^3}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {16 e^2 x^2}{\sqrt {5 d^4+256 a e^3}}}{\sqrt {\frac {1}{32} \left (\frac {5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2+8 e^3 x^4}} \, dx,x,\frac {d}{4 e}+x\right )}{16 e}+\frac {\left (5 d^4+256 a e^3-3 d^2 \sqrt {5 d^4+256 a e^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {\frac {1}{32} \left (\frac {5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2+8 e^3 x^4}} \, dx,x,\frac {d}{4 e}+x\right )}{48 e}\\ &=\frac {1}{3} \left (\frac {d}{4 e}+x\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}-\frac {d^2 (d+4 e x) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{2 e \sqrt {5 d^4+256 a e^3} \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right )}+\frac {d^2 \left (5 d^4+256 a e^3\right )^{3/4} \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (5 d^4+256 a e^3\right ) \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right )^2}} \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right ) E\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (1+\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}\right )\right )}{8 \sqrt {2} e^2 \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}+\frac {\sqrt [4]{5 d^4+256 a e^3} \left (5 d^4+256 a e^3-3 d^2 \sqrt {5 d^4+256 a e^3}\right ) \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (5 d^4+256 a e^3\right ) \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right )^2}} \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right ) F\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (1+\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}\right )\right )}{48 \sqrt {2} e^2 \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}\\ \end {align*}

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Mathematica [B]  time = 6.12, size = 7543, normalized size = 11.38 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4],x]

[Out]

Result too large to show

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(8*e^3*x^4 + 8*d*e^2*x^3 - d^3*x + 8*a*e^2), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.27, size = 7887, normalized size = 11.90 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(8*e^3*x^4 + 8*d*e^2*x^3 - d^3*x + 8*a*e^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {-d^3\,x+8\,d\,e^2\,x^3+8\,e^3\,x^4+8\,a\,e^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*a*e^2 - d^3*x + 8*e^3*x^4 + 8*d*e^2*x^3)^(1/2),x)

[Out]

int((8*a*e^2 - d^3*x + 8*e^3*x^4 + 8*d*e^2*x^3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {8 a e^{2} - d^{3} x + 8 d e^{2} x^{3} + 8 e^{3} x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e**3*x**4+8*d*e**2*x**3-d**3*x+8*a*e**2)**(1/2),x)

[Out]

Integral(sqrt(8*a*e**2 - d**3*x + 8*d*e**2*x**3 + 8*e**3*x**4), x)

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