3.774 \(\int (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4)^{3/2} \, dx\)

Optimal. Leaf size=730 \[ \frac {1}{7} \left (\frac {c}{d}+x\right ) \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}-\frac {16 c^3 \left (8 a d^2+c^3\right ) \left (\frac {c}{d}+x\right ) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}{35 d^2 \sqrt {4 a d^2+c^3} \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )}+\frac {2 c \left (\frac {c}{d}+x\right ) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \left (20 a d^2+7 c^3-3 c d^2 \left (\frac {c}{d}+x\right )^2\right )}{35 d^2}+\frac {8 c^{7/4} \left (4 a d^2+c^3\right )^{3/4} \left (\sqrt {4 a d^2+c^3} \left (5 a d^2+c^3\right )-c^{3/2} \left (8 a d^2+c^3\right )\right ) \sqrt {\frac {d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )^2}} \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac {1}{2} \left (\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}+1\right )\right )}{35 d^5 \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac {16 c^{13/4} \left (4 a d^2+c^3\right )^{3/4} \left (8 a d^2+c^3\right ) \sqrt {\frac {d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )^2}} \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right ) E\left (2 \tan ^{-1}\left (\frac {c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac {1}{2} \left (\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}+1\right )\right )}{35 d^5 \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \]

[Out]

1/7*(c/d+x)*(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(3/2)+2/35*c*(c/d+x)*(7*c^3+20*a*d^2-3*c*d^2*(c/d+x)^2)*(d^2*x
^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2)/d^2-16/35*c^3*(8*a*d^2+c^3)*(c/d+x)*(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1
/2)/d^2/(c^(1/2)+d^2*(c/d+x)^2/(4*a*d^2+c^3)^(1/2))/(4*a*d^2+c^3)^(1/2)+16/35*c^(13/4)*(4*a*d^2+c^3)^(3/4)*(8*
a*d^2+c^3)*(cos(2*arctan((d*x+c)/c^(1/4)/(4*a*d^2+c^3)^(1/4)))^2)^(1/2)/cos(2*arctan((d*x+c)/c^(1/4)/(4*a*d^2+
c^3)^(1/4)))*EllipticE(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))*(c^(1/2)+d^2*(c/d+x)^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^2*(c/d+x)^2/(4*a*d^2+c^3)^(1/2))^2)^(1/2)/d^5/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2)+8/35*c^(7/4)*
(4*a*d^2+c^3)^(3/4)*(cos(2*arctan((d*x+c)/c^(1/4)/(4*a*d^2+c^3)^(1/4)))^2)^(1/2)/cos(2*arctan((d*x+c)/c^(1/4)/
(4*a*d^2+c^3)^(1/4)))*EllipticF(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))*(c^(1/2)+d^2*(c/d+x)^2/(4*a*d^2+c^3)^(1/2))*(-c^(3/2)*(8*a*d^2+c^3)+(5*a*d^2+c^3)*(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^2*(c/d+x)^2/(4*a*d^2+c^3)^(1/2))^
2)^(1/2)/d^5/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2)

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Rubi [A]  time = 0.90, antiderivative size = 730, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {1106, 1091, 1176, 1197, 1103, 1195} \[ -\frac {16 c^3 \left (8 a d^2+c^3\right ) \left (\frac {c}{d}+x\right ) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}{35 d^2 \sqrt {4 a d^2+c^3} \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )}+\frac {2 c \left (\frac {c}{d}+x\right ) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \left (20 a d^2+7 c^3-3 c d^2 \left (\frac {c}{d}+x\right )^2\right )}{35 d^2}+\frac {1}{7} \left (\frac {c}{d}+x\right ) \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}+\frac {8 c^{7/4} \left (4 a d^2+c^3\right )^{3/4} \left (\sqrt {4 a d^2+c^3} \left (5 a d^2+c^3\right )-c^{3/2} \left (8 a d^2+c^3\right )\right ) \sqrt {\frac {d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )^2}} \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac {1}{2} \left (\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}+1\right )\right )}{35 d^5 \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac {16 c^{13/4} \left (4 a d^2+c^3\right )^{3/4} \left (8 a d^2+c^3\right ) \sqrt {\frac {d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right )^2}} \left (\frac {d^2 \left (\frac {c}{d}+x\right )^2}{\sqrt {4 a d^2+c^3}}+\sqrt {c}\right ) E\left (2 \tan ^{-1}\left (\frac {c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac {1}{2} \left (\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}+1\right )\right )}{35 d^5 \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c/d + x)*(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^(3/2))/7 + (2*c*(c/d + x)*Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*
x^3 + d^2*x^4]*(7*c^3 + 20*a*d^2 - 3*c*d^2*(c/d + x)^2))/(35*d^2) - (16*c^3*(c^3 + 8*a*d^2)*(c/d + x)*Sqrt[4*a
*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4])/(35*d^2*Sqrt[c^3 + 4*a*d^2]*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a
*d^2])) + (16*c^(13/4)*(c^3 + 4*a*d^2)^(3/4)*(c^3 + 8*a*d^2)*Sqrt[(d^2*(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^
4))/((c^3 + 4*a*d^2)*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])^2)]*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c
^3 + 4*a*d^2])*EllipticE[2*ArcTan[(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])/(35*d^5*Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4]) + (8*c^(7/4)*(c^3 + 4*a*d^2)^(3/4)*(Sqrt[c^3 + 4*
a*d^2]*(c^3 + 5*a*d^2) - c^(3/2)*(c^3 + 8*a*d^2))*Sqrt[(d^2*(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4))/((c^3 +
4*a*d^2)*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^2])^2)]*(Sqrt[c] + (d^2*(c/d + x)^2)/Sqrt[c^3 + 4*a*d^
2])*EllipticF[2*ArcTan[(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])/(35*d
^5*Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^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 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 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 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2} \, dx &=\operatorname {Subst}\left (\int \left (c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4\right )^{3/2} \, dx,x,\frac {c}{d}+x\right )\\ &=\frac {1}{7} \left (\frac {c}{d}+x\right ) \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}+\frac {3}{7} \operatorname {Subst}\left (\int \left (2 c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 x^2\right ) \sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4} \, dx,x,\frac {c}{d}+x\right )\\ &=\frac {1}{7} \left (\frac {c}{d}+x\right ) \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}+\frac {2 c (c+d x) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \left (7 c^3+20 a d^2-3 c (c+d x)^2\right )}{35 d^3}+\frac {\operatorname {Subst}\left (\int \frac {\frac {16 c^2 \left (c^3+4 a d^2\right ) \left (c^3+5 a d^2\right )}{d^2}-16 c^3 \left (c^3+8 a d^2\right ) x^2}{\sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4}} \, dx,x,\frac {c}{d}+x\right )}{35 d^2}\\ &=\frac {1}{7} \left (\frac {c}{d}+x\right ) \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}+\frac {2 c (c+d x) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \left (7 c^3+20 a d^2-3 c (c+d x)^2\right )}{35 d^3}+\frac {\left (16 c^{7/2} \sqrt {c^3+4 a d^2} \left (c^3+8 a d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1-\frac {d^2 x^2}{\sqrt {c} \sqrt {c^3+4 a d^2}}}{\sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4}} \, dx,x,\frac {c}{d}+x\right )}{35 d^4}+\frac {\left (16 c^2 \sqrt {c^3+4 a d^2} \left (\sqrt {c^3+4 a d^2} \left (c^3+5 a d^2\right )-c^{3/2} \left (c^3+8 a d^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4}} \, dx,x,\frac {c}{d}+x\right )}{35 d^4}\\ &=\frac {1}{7} \left (\frac {c}{d}+x\right ) \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^{3/2}+\frac {2 c (c+d x) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \left (7 c^3+20 a d^2-3 c (c+d x)^2\right )}{35 d^3}-\frac {16 c^3 \left (c^3+8 a d^2\right ) (c+d x) \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}{35 d^3 \sqrt {c^3+4 a d^2} \left (\sqrt {c}+\frac {(c+d x)^2}{\sqrt {c^3+4 a d^2}}\right )}+\frac {16 c^{13/4} \left (c^3+4 a d^2\right )^{3/4} \left (c^3+8 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 ) E\left (2 \tan ^{-1}\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 )}{35 d^5 \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}+\frac {8 c^{7/4} \left (c^3+4 a d^2\right )^{3/4} \left (\sqrt {c^3+4 a d^2} \left (c^3+5 a d^2\right )-c^{3/2} \left (c^3+8 a d^2\right )\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 ) F\left (2 \tan ^{-1}\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 )}{35 d^5 \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 6.20, size = 10468, normalized size = 14.34 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c)^(3/2), x)

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maple [B]  time = 0.20, size = 5229, normalized size = 7.16 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d**2*x**4+4*c*d*x**3+4*c**2*x**2+4*a*c)**(3/2),x)

[Out]

Integral((4*a*c + 4*c**2*x**2 + 4*c*d*x**3 + d**2*x**4)**(3/2), x)

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