3.780 \(\int \frac {1}{(8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=748 \[ \frac {4 e \left (\frac {d}{4 e}+x\right ) \left (-256 a e^3+13 d^4-48 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2\right )}{\left (-16384 a^2 e^6-64 a d^4 e^3+5 d^8\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}+\frac {384 d^2 e^2 \left (\frac {d}{4 e}+x\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{\left (d^4-64 a e^3\right ) \left (256 a e^3+5 d^4\right )^{3/2} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )}-\frac {2 \sqrt {2} \left (-3 d^2 \sqrt {256 a e^3+5 d^4}+256 a e^3+5 d^4\right ) \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) F\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{\left (d^4-64 a e^3\right ) \left (256 a e^3+5 d^4\right )^{3/4} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac {12 \sqrt {2} d^2 \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) E\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{\left (d^4-64 a e^3\right ) \sqrt [4]{256 a e^3+5 d^4} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]

[Out]

4*e*(1/4*d/e+x)*(13*d^4-256*a*e^3-48*d^2*e^2*(1/4*d/e+x)^2)/(-16384*a^2*e^6-64*a*d^4*e^3+5*d^8)/(8*e^3*x^4+8*d
*e^2*x^3-d^3*x+8*a*e^2)^(1/2)+384*d^2*e^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)/(-64*a*e^3+d
^4)/(256*a*e^3+5*d^4)^(3/2)/(1+16*e^2*(1/4*d/e+x)^2/(256*a*e^3+5*d^4)^(1/2))-12*d^2*(cos(2*arctan((4*e*x+d)/(2
56*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))*2^(1/2)*(1+16*e^2*(1/4*d/e+x)^2/(2
56*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)/(-64*a*e^3+d^4)/(256*a*e^3+5*d^4)^(1/4)/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1
/2)-2*(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))*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)/
(-64*a*e^3+d^4)/(256*a*e^3+5*d^4)^(3/4)/(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 = 0.79, antiderivative size = 748, 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, 1092, 1197, 1103, 1195} \[ \frac {4 e \left (\frac {d}{4 e}+x\right ) \left (-256 a e^3-48 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2+13 d^4\right )}{\left (-16384 a^2 e^6-64 a d^4 e^3+5 d^8\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}+\frac {384 d^2 e^2 \left (\frac {d}{4 e}+x\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{\left (d^4-64 a e^3\right ) \left (256 a e^3+5 d^4\right )^{3/2} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )}-\frac {2 \sqrt {2} \left (-3 d^2 \sqrt {256 a e^3+5 d^4}+256 a e^3+5 d^4\right ) \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) F\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{\left (d^4-64 a e^3\right ) \left (256 a e^3+5 d^4\right )^{3/4} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac {12 \sqrt {2} d^2 \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) E\left (2 \tan ^{-1}\left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{\left (d^4-64 a e^3\right ) \sqrt [4]{256 a e^3+5 d^4} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*e*(d/(4*e) + x)*(13*d^4 - 256*a*e^3 - 48*d^2*e^2*(d/(4*e) + x)^2))/((5*d^8 - 64*a*d^4*e^3 - 16384*a^2*e^6)*
Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4]) + (384*d^2*e^2*(d/(4*e) + x)*Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x
^3 + 8*e^3*x^4])/((d^4 - 64*a*e^3)*(5*d^4 + 256*a*e^3)^(3/2)*(1 + (16*e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*
e^3])) - (12*Sqrt[2]*d^2*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])*Ellipti
cE[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])/((d^4 - 64*a*e^3
)*(5*d^4 + 256*a*e^3)^(1/4)*Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4]) - (2*Sqrt[2]*(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])*
EllipticF[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])/((d^4 - 6
4*a*e^3)*(5*d^4 + 256*a*e^3)^(3/4)*Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4])

Rule 1092

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(b^2 - 2*a*c + b*c*x^2)*(a + b*x^2 + c*x^
4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(b^2 - 2*a*c + 2*(p + 1)
*(b^2 - 4*a*c) + b*c*(4*p + 7)*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] && LtQ[p, -1] && 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 \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (\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\right )^{3/2}} \, dx,x,\frac {d}{4 e}+x\right )\\ &=\frac {4 e \left (\frac {d}{4 e}+x\right ) \left (13 d^4-256 a e^3-48 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2\right )}{\left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac {8 \operatorname {Subst}\left (\int \frac {\frac {1}{2} e^3 \left (\frac {5 d^4}{e}+256 a e^2\right )-24 d^2 e^4 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 )}{e \left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right )}\\ &=\frac {4 e \left (\frac {d}{4 e}+x\right ) \left (13 d^4-256 a e^3-48 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2\right )}{\left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac {\left (12 d^2 e\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 )}{\left (d^4-64 a e^3\right ) \sqrt {5 d^4+256 a e^3}}-\frac {\left (4 e \left (5 d^4+256 a e^3-3 d^2 \sqrt {5 d^4+256 a e^3}\right )\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 )}{5 d^8-64 a d^4 e^3-16384 a^2 e^6}\\ &=\frac {4 e \left (\frac {d}{4 e}+x\right ) \left (13 d^4-256 a e^3-48 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2\right )}{\left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}+\frac {96 d^2 e (d+4 e x) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{\left (d^4-64 a e^3\right ) \left (5 d^4+256 a e^3\right )^{3/2} \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right )}-\frac {12 \sqrt {2} d^2 \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 )}{\left (d^4-64 a e^3\right ) \sqrt [4]{5 d^4+256 a e^3} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac {2 \sqrt {2} \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 )}{\left (d^4-64 a e^3\right ) \left (5 d^4+256 a e^3\right )^{3/4} \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 = 6.15, size = 7629, normalized size = 10.20 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}}{64 \, e^{6} x^{8} + 128 \, d e^{5} x^{7} + 64 \, d^{2} e^{4} x^{6} - 16 \, d^{3} e^{3} x^{5} + 128 \, a d e^{4} x^{3} + d^{6} x^{2} - 16 \, a d^{3} e^{2} x + 64 \, a^{2} e^{4} - 16 \, {\left (d^{4} e^{2} - 8 \, a e^{5}\right )} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(8*e^3*x^4 + 8*d*e^2*x^3 - d^3*x + 8*a*e^2)/(64*e^6*x^8 + 128*d*e^5*x^7 + 64*d^2*e^4*x^6 - 16*d^3
*e^3*x^5 + 128*a*d*e^4*x^3 + d^6*x^2 - 16*a*d^3*e^2*x + 64*a^2*e^4 - 16*(d^4*e^2 - 8*a*e^5)*x^4), 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(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.08, size = 8103, normalized size = 10.83 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (-d^3\,x+8\,d\,e^2\,x^3+8\,e^3\,x^4+8\,a\,e^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (8 a e^{2} - d^{3} x + 8 d e^{2} x^{3} + 8 e^{3} x^{4}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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