Integrand size = 34, antiderivative size = 680 \[ \int \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx=\frac {96 d^2 e (d+4 e x) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{\left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right ) \left (\sqrt {5 d^4+256 a e^3}+(d+4 e x)^2\right )}+\frac {(d+4 e x) \left (13 d^4-256 a e^3-3 d^2 (d+4 e x)^2\right )}{\left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right ) \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac {3 d^2 \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right ) \sqrt {\frac {5 d^4+256 a e^3-6 d^2 (d+4 e x)^2+(d+4 e x)^4}{\left (5 d^4+256 a e^3\right ) \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right )^2}} E\left (2 \arctan \left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac {1}{2} \left (1+\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}\right )\right )}{\left (d^4-64 a e^3\right ) \sqrt [4]{5 d^4+256 a e^3} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac {\left (5 d^4+256 a e^3-3 d^2 \sqrt {5 d^4+256 a e^3}\right ) \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right ) \sqrt {\frac {5 d^4+256 a e^3-6 d^2 (d+4 e x)^2+(d+4 e x)^4}{\left (5 d^4+256 a e^3\right ) \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right ),\frac {1}{2} \left (1+\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}\right )\right )}{2 \left (d^4-64 a e^3\right ) \left (5 d^4+256 a e^3\right )^{3/4} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \] Output:
96*d^2*e*(4*e*x+d)*(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2)/(-16384*a^2 *e^6-64*a*d^4*e^3+5*d^8)/((256*a*e^3+5*d^4)^(1/2)+(4*e*x+d)^2)+(4*e*x+d)*( 13*d^4-256*a*e^3-3*d^2*(4*e*x+d)^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)-3*d^2*(1+(4*e*x+d)^2/(256*a*e^3+ 5*d^4)^(1/2))*((5*d^4+256*a*e^3-6*d^2*(4*e*x+d)^2+(4*e*x+d)^4)/(256*a*e^3+ 5*d^4)/(1+(4*e*x+d)^2/(256*a*e^3+5*d^4)^(1/2))^2)^(1/2)*EllipticE(sin(2*ar ctan((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))/(-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)-1/2*(5*d^4+256*a*e^3-3*d^2*(256*a*e^3+5*d^4)^(1/2))* (1+(4*e*x+d)^2/(256*a*e^3+5*d^4)^(1/2))*((5*d^4+256*a*e^3-6*d^2*(4*e*x+d)^ 2+(4*e*x+d)^4)/(256*a*e^3+5*d^4)/(1+(4*e*x+d)^2/(256*a*e^3+5*d^4)^(1/2))^2 )^(1/2)*InverseJacobiAM(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))/(-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)
Leaf count is larger than twice the leaf count of optimal. \(7629\) vs. \(2(680)=1360\).
Time = 16.18 (sec) , antiderivative size = 7629, normalized size of antiderivative = 11.22 \[ \int \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx=\text {Result too large to show} \] Input:
Integrate[(8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4)^(-3/2),x]
Output:
Result too large to show
Time = 1.37 (sec) , antiderivative size = 882, normalized size of antiderivative = 1.30, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2458, 1405, 27, 1511, 1416, 1509}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2458 |
| \(\displaystyle \int \frac {1}{\left (\frac {1}{32} \left (256 a e^2+\frac {5 d^4}{e}\right )-3 d^2 e \left (\frac {d}{4 e}+x\right )^2+8 e^3 \left (\frac {d}{4 e}+x\right )^4\right )^{3/2}}d\left (\frac {d}{4 e}+x\right )\) |
| \(\Big \downarrow \) 1405 |
| \(\displaystyle \frac {16 \sqrt {2} e \left (\frac {d}{4 e}+x\right ) \left (-256 a e^3+13 d^4-48 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2\right )}{\left (-16384 a^2 e^6-64 a d^4 e^3+5 d^8\right ) \sqrt {256 a e^2+\frac {5 d^4}{e}-96 d^2 e \left (\frac {d}{4 e}+x\right )^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4}}-\frac {8 \int \frac {2 \sqrt {2} e^2 \left (5 d^4-48 e^2 \left (\frac {d}{4 e}+x\right )^2 d^2+256 a e^3\right )}{\sqrt {\frac {5 d^4}{e}-96 e \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4+256 a e^2}}d\left (\frac {d}{4 e}+x\right )}{e \left (-16384 a^2 e^6-64 a d^4 e^3+5 d^8\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {16 \sqrt {2} e \left (\frac {d}{4 e}+x\right ) \left (-256 a e^3+13 d^4-48 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2\right )}{\left (-16384 a^2 e^6-64 a d^4 e^3+5 d^8\right ) \sqrt {256 a e^2+\frac {5 d^4}{e}-96 d^2 e \left (\frac {d}{4 e}+x\right )^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4}}-\frac {16 \sqrt {2} e \int \frac {5 d^4-48 e^2 \left (\frac {d}{4 e}+x\right )^2 d^2+256 a e^3}{\sqrt {\frac {5 d^4}{e}-96 e \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4+256 a e^2}}d\left (\frac {d}{4 e}+x\right )}{-16384 a^2 e^6-64 a d^4 e^3+5 d^8}\) |
| \(\Big \downarrow \) 1511 |
| \(\displaystyle \frac {16 \sqrt {2} e \left (\frac {d}{4 e}+x\right ) \left (-256 a e^3+13 d^4-48 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2\right )}{\left (-16384 a^2 e^6-64 a d^4 e^3+5 d^8\right ) \sqrt {256 a e^2+\frac {5 d^4}{e}-96 d^2 e \left (\frac {d}{4 e}+x\right )^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4}}-\frac {16 \sqrt {2} e \left (3 d^2 \sqrt {256 a e^3+5 d^4} \int \frac {1-\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {5 d^4+256 a e^3}}}{\sqrt {\frac {5 d^4}{e}-96 e \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4+256 a e^2}}d\left (\frac {d}{4 e}+x\right )-\sqrt {256 a e^3+5 d^4} \left (3 d^2-\sqrt {256 a e^3+5 d^4}\right ) \int \frac {1}{\sqrt {\frac {5 d^4}{e}-96 e \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4+256 a e^2}}d\left (\frac {d}{4 e}+x\right )\right )}{-16384 a^2 e^6-64 a d^4 e^3+5 d^8}\) |
| \(\Big \downarrow \) 1416 |
| \(\displaystyle \frac {16 \sqrt {2} e \left (\frac {d}{4 e}+x\right ) \left (-256 a e^3+13 d^4-48 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2\right )}{\left (-16384 a^2 e^6-64 a d^4 e^3+5 d^8\right ) \sqrt {256 a e^2+\frac {5 d^4}{e}-96 d^2 e \left (\frac {d}{4 e}+x\right )^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4}}-\frac {16 \sqrt {2} e \left (3 d^2 \sqrt {256 a e^3+5 d^4} \int \frac {1-\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {5 d^4+256 a e^3}}}{\sqrt {\frac {5 d^4}{e}-96 e \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4+256 a e^2}}d\left (\frac {d}{4 e}+x\right )-\frac {\left (256 a e^3+5 d^4\right )^{3/4} \left (3 d^2-\sqrt {256 a e^3+5 d^4}\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) \sqrt {\frac {256 a e^3+5 d^4-96 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2+256 e^4 \left (\frac {d}{4 e}+x\right )^4}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {4 e \left (\frac {d}{4 e}+x\right )}{\sqrt [4]{5 d^4+256 a e^3}}\right ),\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{8 e \sqrt {256 a e^2+\frac {5 d^4}{e}-96 d^2 e \left (\frac {d}{4 e}+x\right )^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4}}\right )}{-16384 a^2 e^6-64 a d^4 e^3+5 d^8}\) |
| \(\Big \downarrow \) 1509 |
| \(\displaystyle \frac {16 \sqrt {2} e \left (\frac {d}{4 e}+x\right ) \left (13 d^4-48 e^2 \left (\frac {d}{4 e}+x\right )^2 d^2-256 a e^3\right )}{\left (5 d^8-64 a e^3 d^4-16384 a^2 e^6\right ) \sqrt {\frac {5 d^4}{e}-96 e \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4+256 a e^2}}-\frac {16 \sqrt {2} e \left (3 d^2 \sqrt {5 d^4+256 a e^3} \left (\frac {\sqrt [4]{5 d^4+256 a e^3} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {5 d^4+256 a e^3}}+1\right ) \sqrt {\frac {5 d^4-96 e^2 \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^4 \left (\frac {d}{4 e}+x\right )^4+256 a e^3}{\left (5 d^4+256 a e^3\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {5 d^4+256 a e^3}}+1\right )^2}} E\left (2 \arctan \left (\frac {4 e \left (\frac {d}{4 e}+x\right )}{\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 )}{4 e \sqrt {\frac {5 d^4}{e}-96 e \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4+256 a e^2}}-\frac {e \left (\frac {d}{4 e}+x\right ) \sqrt {\frac {5 d^4}{e}-96 e \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4+256 a e^2}}{\left (5 d^4+256 a e^3\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {5 d^4+256 a e^3}}+1\right )}\right )-\frac {\left (5 d^4+256 a e^3\right )^{3/4} \left (3 d^2-\sqrt {5 d^4+256 a e^3}\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {5 d^4+256 a e^3}}+1\right ) \sqrt {\frac {5 d^4-96 e^2 \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^4 \left (\frac {d}{4 e}+x\right )^4+256 a e^3}{\left (5 d^4+256 a e^3\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {5 d^4+256 a e^3}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {4 e \left (\frac {d}{4 e}+x\right )}{\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 e \sqrt {\frac {5 d^4}{e}-96 e \left (\frac {d}{4 e}+x\right )^2 d^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4+256 a e^2}}\right )}{5 d^8-64 a e^3 d^4-16384 a^2 e^6}\) |
Input:
Int[(8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4)^(-3/2),x]
Output:
(16*Sqrt[2]*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[(5*d^4)/e + 256*a*e^2 - 96*d^2*e*(d/(4*e) + x)^2 + 256*e^3*(d/(4*e) + x)^4]) - (16*Sqrt[2]*e*(3*d^ 2*Sqrt[5*d^4 + 256*a*e^3]*(-((e*(d/(4*e) + x)*Sqrt[(5*d^4)/e + 256*a*e^2 - 96*d^2*e*(d/(4*e) + x)^2 + 256*e^3*(d/(4*e) + 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]))) + ((5*d^4 + 256* a*e^3)^(1/4)*(1 + (16*e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])*Sqrt[( 5*d^4 + 256*a*e^3 - 96*d^2*e^2*(d/(4*e) + x)^2 + 256*e^4*(d/(4*e) + 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)]*EllipticE[2*ArcTan[(4*e*(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])/(4*e*Sqrt[(5*d^4)/e + 256*a*e^2 - 96*d^2*e*(d/(4*e) + x)^2 + 256*e^3*(d/(4*e) + x)^4])) - ((5*d^4 + 256*a*e ^3)^(3/4)*(3*d^2 - 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])*Sqrt[(5*d^4 + 256*a*e^3 - 96*d^2*e^2*(d/(4*e) + x )^2 + 256*e^4*(d/(4*e) + 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)]*EllipticF[2*ArcTan[(4*e*(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 *e*Sqrt[(5*d^4)/e + 256*a*e^2 - 96*d^2*e*(d/(4*e) + x)^2 + 256*e^3*(d/(4*e ) + x)^4])))/(5*d^8 - 64*a*d^4*e^3 - 16384*a^2*e^6)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp[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] /; Fr eeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p]
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)]/ (2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) ], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> 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)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 /(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + b*x^2 + c*x^ 4], x], x] - Simp[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] && Pos Q[c/a]
Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
Leaf count of result is larger than twice the leaf count of optimal. \(8102\) vs. \(2(643)=1286\).
Time = 0.79 (sec) , antiderivative size = 8103, normalized size of antiderivative = 11.92
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(8103\) |
| elliptic | \(\text {Expression too large to display}\) | \(8103\) |
Input:
int(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
result too large to display
\[ \int \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(3/2),x, algorithm="fric as")
Output:
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)
\[ \int \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (8 a e^{2} - d^{3} x + 8 d e^{2} x^{3} + 8 e^{3} x^{4}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(8*e**3*x**4+8*d*e**2*x**3-d**3*x+8*a*e**2)**(3/2),x)
Output:
Integral((8*a*e**2 - d**3*x + 8*d*e**2*x**3 + 8*e**3*x**4)**(-3/2), x)
\[ \int \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(3/2),x, algorithm="maxi ma")
Output:
integrate((8*e^3*x^4 + 8*d*e^2*x^3 - d^3*x + 8*a*e^2)^(-3/2), x)
\[ \int \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
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 ")
Output:
integrate((8*e^3*x^4 + 8*d*e^2*x^3 - d^3*x + 8*a*e^2)^(-3/2), x)
Timed out. \[ \int \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (-d^3\,x+8\,d\,e^2\,x^3+8\,e^3\,x^4+8\,a\,e^2\right )}^{3/2}} \,d x \] Input:
int(1/(8*a*e^2 - d^3*x + 8*e^3*x^4 + 8*d*e^2*x^3)^(3/2),x)
Output:
int(1/(8*a*e^2 - d^3*x + 8*e^3*x^4 + 8*d*e^2*x^3)^(3/2), x)
\[ \int \frac {1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {8 e^{3} x^{4}+8 d \,e^{2} x^{3}-d^{3} x +8 a \,e^{2}}}{64 e^{6} x^{8}+128 d \,e^{5} x^{7}+64 d^{2} e^{4} x^{6}-16 d^{3} e^{3} x^{5}+128 a \,e^{5} x^{4}-16 d^{4} e^{2} x^{4}+128 a d \,e^{4} x^{3}+d^{6} x^{2}-16 a \,d^{3} e^{2} x +64 a^{2} e^{4}}d x \] Input:
int(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(3/2),x)
Output:
int(sqrt(8*a*e**2 - d**3*x + 8*d*e**2*x**3 + 8*e**3*x**4)/(64*a**2*e**4 - 16*a*d**3*e**2*x + 128*a*d*e**4*x**3 + 128*a*e**5*x**4 + d**6*x**2 - 16*d* *4*e**2*x**4 - 16*d**3*e**3*x**5 + 64*d**2*e**4*x**6 + 128*d*e**5*x**7 + 6 4*e**6*x**8),x)