Integrand size = 31, antiderivative size = 639 \[ \int \frac {1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx=-\frac {d (c+d x) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}{8 a \left (c^3+4 a d^2\right ) \left (\sqrt {c} \sqrt {c^3+4 a d^2}+(c+d x)^2\right )}-\frac {(c+d x) \left (c^3-4 a d^2-c (c+d x)^2\right )}{8 a c d \left (c^3+4 a d^2\right ) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac {\sqrt [4]{c} \sqrt {\frac {d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (c^3+4 a d^2\right ) \left (\sqrt {c}+\frac {(c+d x)^2}{\sqrt {c^3+4 a d^2}}\right )^2}} \left (\sqrt {c}+\frac {(c+d x)^2}{\sqrt {c^3+4 a d^2}}\right ) E\left (2 \arctan \left (\frac {c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac {1}{2} \left (1+\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}\right )\right )}{8 a d \sqrt [4]{c^3+4 a d^2} \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac {\left (c^3+4 a d^2-c^{3/2} \sqrt {c^3+4 a d^2}\right ) \sqrt {\frac {d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (c^3+4 a d^2\right ) \left (\sqrt {c}+\frac {(c+d x)^2}{\sqrt {c^3+4 a d^2}}\right )^2}} \left (\sqrt {c}+\frac {(c+d x)^2}{\sqrt {c^3+4 a d^2}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right ),\frac {1}{2} \left (1+\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}\right )\right )}{16 a c^{5/4} d \left (c^3+4 a d^2\right )^{3/4} \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \] Output:
-1/8*d*(d*x+c)*(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2)/a/(4*a*d^2+c^3)/( c^(1/2)*(4*a*d^2+c^3)^(1/2)+(d*x+c)^2)-1/8*(d*x+c)*(c^3-4*a*d^2-c*(d*x+c)^ 2)/a/c/d/(4*a*d^2+c^3)/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2)+1/8*c^(1/ 4)*(d^2*(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)/(4*a*d^2+c^3)/(c^(1/2)+(d*x+c) ^2/(4*a*d^2+c^3)^(1/2))^2)^(1/2)*(c^(1/2)+(d*x+c)^2/(4*a*d^2+c^3)^(1/2))*E llipticE(sin(2*arctan((d*x+c)/c^(1/4)/(4*a*d^2+c^3)^(1/4))),1/2*(2+2*c^(3/ 2)/(4*a*d^2+c^3)^(1/2))^(1/2))/a/d/(4*a*d^2+c^3)^(1/4)/(d^2*x^4+4*c*d*x^3+ 4*c^2*x^2+4*a*c)^(1/2)+1/16*(c^3+4*a*d^2-c^(3/2)*(4*a*d^2+c^3)^(1/2))*(d^2 *(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)/(4*a*d^2+c^3)/(c^(1/2)+(d*x+c)^2/(4*a *d^2+c^3)^(1/2))^2)^(1/2)*(c^(1/2)+(d*x+c)^2/(4*a*d^2+c^3)^(1/2))*InverseJ acobiAM(2*arctan((d*x+c)/c^(1/4)/(4*a*d^2+c^3)^(1/4)),1/2*(2+2*c^(3/2)/(4* a*d^2+c^3)^(1/2))^(1/2))/a/c^(5/4)/d/(4*a*d^2+c^3)^(3/4)/(d^2*x^4+4*c*d*x^ 3+4*c^2*x^2+4*a*c)^(1/2)
Result contains complex when optimal does not.
Time = 16.16 (sec) , antiderivative size = 5276, normalized size of antiderivative = 8.26 \[ \int \frac {1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx=\text {Result too large to show} \] Input:
Integrate[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^(-3/2),x]
Output:
Result too large to show
Time = 0.74 (sec) , antiderivative size = 777, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2458, 1405, 27, 1511, 27, 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.
| \(\displaystyle \int \frac {1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2458 |
| \(\displaystyle \int \frac {1}{\left (c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4\right )^{3/2}}d\left (\frac {c}{d}+x\right )\) |
| \(\Big \downarrow \) 1405 |
| \(\displaystyle \frac {\int \frac {2 c \left (c^3-d^2 \left (\frac {c}{d}+x\right )^2 c+4 a d^2\right )}{\sqrt {d^2 \left (\frac {c}{d}+x\right )^4-2 c^2 \left (\frac {c}{d}+x\right )^2+c \left (\frac {c^3}{d^2}+4 a\right )}}d\left (\frac {c}{d}+x\right )}{16 a c^2 \left (4 a d^2+c^3\right )}-\frac {\left (\frac {c}{d}+x\right ) \left (-4 a d^2+c^3-c d^2 \left (\frac {c}{d}+x\right )^2\right )}{8 a c \left (4 a d^2+c^3\right ) \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {c^3-d^2 \left (\frac {c}{d}+x\right )^2 c+4 a d^2}{\sqrt {d^2 \left (\frac {c}{d}+x\right )^4-2 c^2 \left (\frac {c}{d}+x\right )^2+c \left (\frac {c^3}{d^2}+4 a\right )}}d\left (\frac {c}{d}+x\right )}{8 a c \left (4 a d^2+c^3\right )}-\frac {\left (\frac {c}{d}+x\right ) \left (-4 a d^2+c^3-c d^2 \left (\frac {c}{d}+x\right )^2\right )}{8 a c \left (4 a d^2+c^3\right ) \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}\) |
| \(\Big \downarrow \) 1511 |
| \(\displaystyle \frac {c^{3/2} \sqrt {4 a d^2+c^3} \int \frac {\sqrt {c}-\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {c^3+4 a d^2}}}{\sqrt {c} \sqrt {d^2 \left (\frac {c}{d}+x\right )^4-2 c^2 \left (\frac {c}{d}+x\right )^2+c \left (\frac {c^3}{d^2}+4 a\right )}}d\left (\frac {c}{d}+x\right )-\sqrt {4 a d^2+c^3} \left (c^{3/2}-\sqrt {4 a d^2+c^3}\right ) \int \frac {1}{\sqrt {d^2 \left (\frac {c}{d}+x\right )^4-2 c^2 \left (\frac {c}{d}+x\right )^2+c \left (\frac {c^3}{d^2}+4 a\right )}}d\left (\frac {c}{d}+x\right )}{8 a c \left (4 a d^2+c^3\right )}-\frac {\left (\frac {c}{d}+x\right ) \left (-4 a d^2+c^3-c d^2 \left (\frac {c}{d}+x\right )^2\right )}{8 a c \left (4 a d^2+c^3\right ) \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c \sqrt {4 a d^2+c^3} \int \frac {\sqrt {c}-\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {c^3+4 a d^2}}}{\sqrt {d^2 \left (\frac {c}{d}+x\right )^4-2 c^2 \left (\frac {c}{d}+x\right )^2+c \left (\frac {c^3}{d^2}+4 a\right )}}d\left (\frac {c}{d}+x\right )-\sqrt {4 a d^2+c^3} \left (c^{3/2}-\sqrt {4 a d^2+c^3}\right ) \int \frac {1}{\sqrt {d^2 \left (\frac {c}{d}+x\right )^4-2 c^2 \left (\frac {c}{d}+x\right )^2+c \left (\frac {c^3}{d^2}+4 a\right )}}d\left (\frac {c}{d}+x\right )}{8 a c \left (4 a d^2+c^3\right )}-\frac {\left (\frac {c}{d}+x\right ) \left (-4 a d^2+c^3-c d^2 \left (\frac {c}{d}+x\right )^2\right )}{8 a c \left (4 a d^2+c^3\right ) \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}\) |
| \(\Big \downarrow \) 1416 |
| \(\displaystyle \frac {c \sqrt {4 a d^2+c^3} \int \frac {\sqrt {c}-\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {c^3+4 a d^2}}}{\sqrt {d^2 \left (\frac {c}{d}+x\right )^4-2 c^2 \left (\frac {c}{d}+x\right )^2+c \left (\frac {c^3}{d^2}+4 a\right )}}d\left (\frac {c}{d}+x\right )-\frac {\left (4 a d^2+c^3\right )^{3/4} \left (c^{3/2}-\sqrt {4 a d^2+c^3}\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right ) \sqrt {\frac {d^2 \left (c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {d \left (\frac {c}{d}+x\right )}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right ),\frac {1}{2} \left (\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}+1\right )\right )}{2 \sqrt [4]{c} d \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}}{8 a c \left (4 a d^2+c^3\right )}-\frac {\left (\frac {c}{d}+x\right ) \left (-4 a d^2+c^3-c d^2 \left (\frac {c}{d}+x\right )^2\right )}{8 a c \left (4 a d^2+c^3\right ) \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}\) |
| \(\Big \downarrow \) 1509 |
| \(\displaystyle \frac {c \sqrt {4 a d^2+c^3} \left (\frac {\sqrt [4]{c} \sqrt [4]{4 a d^2+c^3} \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right ) \sqrt {\frac {d^2 \left (c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )^2}} E\left (2 \arctan \left (\frac {d \left (\frac {c}{d}+x\right )}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac {1}{2} \left (\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}+1\right )\right )}{d \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}-\frac {\left (\frac {c}{d}+x\right ) \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}{\left (4 a+\frac {c^3}{d^2}\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )}\right )-\frac {\left (4 a d^2+c^3\right )^{3/4} \left (c^{3/2}-\sqrt {4 a d^2+c^3}\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right ) \sqrt {\frac {d^2 \left (c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {d \left (\frac {c}{d}+x\right )}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right ),\frac {1}{2} \left (\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}+1\right )\right )}{2 \sqrt [4]{c} d \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}}{8 a c \left (4 a d^2+c^3\right )}-\frac {\left (\frac {c}{d}+x\right ) \left (-4 a d^2+c^3-c d^2 \left (\frac {c}{d}+x\right )^2\right )}{8 a c \left (4 a d^2+c^3\right ) \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}\) |
Input:
Int[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^(-3/2),x]
Output:
-1/8*((c/d + x)*(c^3 - 4*a*d^2 - c*d^2*(c/d + x)^2))/(a*c*(c^3 + 4*a*d^2)* Sqrt[c*(4*a + c^3/d^2) - 2*c^2*(c/d + x)^2 + d^2*(c/d + x)^4]) + (c*Sqrt[c ^3 + 4*a*d^2]*(-(((c/d + x)*Sqrt[c*(4*a + c^3/d^2) - 2*c^2*(c/d + x)^2 + d ^2*(c/d + x)^4])/((4*a + c^3/d^2)*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2]))) + (c^(1/4)*(c^3 + 4*a*d^2)^(1/4)*(Sqrt[c] + (d^2*(c/d + x)^2)/ Sqrt[c^3 + 4*a*d^2])*Sqrt[(d^2*(c*(4*a + c^3/d^2) - 2*c^2*(c/d + x)^2 + d^ 2*(c/d + x)^4))/((c^3 + 4*a*d^2)*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4 *a*d^2])^2)]*EllipticE[2*ArcTan[(d*(c/d + x))/(c^(1/4)*(c^3 + 4*a*d^2)^(1/ 4))], (1 + c^(3/2)/Sqrt[c^3 + 4*a*d^2])/2])/(d*Sqrt[c*(4*a + c^3/d^2) - 2* c^2*(c/d + x)^2 + d^2*(c/d + x)^4])) - ((c^3 + 4*a*d^2)^(3/4)*(c^(3/2) - S qrt[c^3 + 4*a*d^2])*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])*Sqrt [(d^2*(c*(4*a + c^3/d^2) - 2*c^2*(c/d + x)^2 + d^2*(c/d + x)^4))/((c^3 + 4 *a*d^2)*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])^2)]*EllipticF[2* ArcTan[(d*(c/d + x))/(c^(1/4)*(c^3 + 4*a*d^2)^(1/4))], (1 + c^(3/2)/Sqrt[c ^3 + 4*a*d^2])/2])/(2*c^(1/4)*d*Sqrt[c*(4*a + c^3/d^2) - 2*c^2*(c/d + x)^2 + d^2*(c/d + x)^4]))/(8*a*c*(c^3 + 4*a*d^2))
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. \(5023\) vs. \(2(574)=1148\).
Time = 1.06 (sec) , antiderivative size = 5024, normalized size of antiderivative = 7.86
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(5024\) |
| elliptic | \(\text {Expression too large to display}\) | \(5024\) |
Input:
int(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(3/2),x,method=_RETURNVERBOSE)
Output:
result too large to display
\[ \int \frac {1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(3/2),x, algorithm="fricas ")
Output:
integral(sqrt(d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c)/(d^4*x^8 + 8*c*d^3* x^7 + 24*c^2*d^2*x^6 + 32*c^3*d*x^5 + 32*a*c^2*d*x^3 + 32*a*c^3*x^2 + 8*(2 *c^4 + a*c*d^2)*x^4 + 16*a^2*c^2), x)
\[ \int \frac {1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (4 a c + 4 c^{2} x^{2} + 4 c d x^{3} + d^{2} x^{4}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(d**2*x**4+4*c*d*x**3+4*c**2*x**2+4*a*c)**(3/2),x)
Output:
Integral((4*a*c + 4*c**2*x**2 + 4*c*d*x**3 + d**2*x**4)**(-3/2), x)
\[ \int \frac {1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(3/2),x, algorithm="maxima ")
Output:
integrate((d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c)^(-3/2), x)
\[ \int \frac {1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(3/2),x, algorithm="giac")
Output:
integrate((d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c)^(-3/2), x)
Timed out. \[ \int \frac {1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (4\,c^2\,x^2+4\,c\,d\,x^3+4\,a\,c+d^2\,x^4\right )}^{3/2}} \,d x \] Input:
int(1/(4*a*c + 4*c^2*x^2 + d^2*x^4 + 4*c*d*x^3)^(3/2),x)
Output:
int(1/(4*a*c + 4*c^2*x^2 + d^2*x^4 + 4*c*d*x^3)^(3/2), x)
\[ \int \frac {1}{\left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {d^{2} x^{4}+4 c d \,x^{3}+4 c^{2} x^{2}+4 a c}}{d^{4} x^{8}+8 c \,d^{3} x^{7}+24 c^{2} d^{2} x^{6}+32 c^{3} d \,x^{5}+8 a c \,d^{2} x^{4}+16 c^{4} x^{4}+32 a \,c^{2} d \,x^{3}+32 a \,c^{3} x^{2}+16 a^{2} c^{2}}d x \] Input:
int(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(3/2),x)
Output:
int(sqrt(4*a*c + 4*c**2*x**2 + 4*c*d*x**3 + d**2*x**4)/(16*a**2*c**2 + 32* a*c**3*x**2 + 32*a*c**2*d*x**3 + 8*a*c*d**2*x**4 + 16*c**4*x**4 + 32*c**3* d*x**5 + 24*c**2*d**2*x**6 + 8*c*d**3*x**7 + d**4*x**8),x)