3.764 \(\int (8 x-8 x^2+4 x^3-x^4)^{3/2} \, dx\)

Optimal. Leaf size=102 \[ \frac {1}{7} (x-1) \left (-(x-1)^4-2 (x-1)^2+3\right )^{3/2}+\frac {2}{35} \left (13-3 (x-1)^2\right ) (x-1) \sqrt {-(x-1)^4-2 (x-1)^2+3}-\frac {176}{35} \sqrt {3} F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )+\frac {16}{5} \sqrt {3} E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right ) \]

[Out]

1/7*(3-2*(-1+x)^2-(-1+x)^4)^(3/2)*(-1+x)-16/5*EllipticE(-1+x,1/3*I*3^(1/2))*3^(1/2)+176/35*EllipticF(-1+x,1/3*
I*3^(1/2))*3^(1/2)+2/35*(13-3*(-1+x)^2)*(-1+x)*(3-2*(-1+x)^2-(-1+x)^4)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {1106, 1091, 1176, 1180, 524, 424, 419} \[ \frac {1}{7} (x-1) \left (-(x-1)^4-2 (x-1)^2+3\right )^{3/2}+\frac {2}{35} \left (13-3 (x-1)^2\right ) (x-1) \sqrt {-(x-1)^4-2 (x-1)^2+3}-\frac {176}{35} \sqrt {3} F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )+\frac {16}{5} \sqrt {3} E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(13 - 3*(-1 + x)^2)*Sqrt[3 - 2*(-1 + x)^2 - (-1 + x)^4]*(-1 + x))/35 + ((3 - 2*(-1 + x)^2 - (-1 + x)^4)^(3/
2)*(-1 + x))/7 + (16*Sqrt[3]*EllipticE[ArcSin[1 - x], -1/3])/5 - (176*Sqrt[3]*EllipticF[ArcSin[1 - x], -1/3])/
35

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

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 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 1180

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[2*Sqrt[-c], Int[(d + e*x^2)/(Sqrt[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a, b, c,
d, e}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[c, 0]

Rubi steps

\begin {align*} \int \left (8 x-8 x^2+4 x^3-x^4\right )^{3/2} \, dx &=\operatorname {Subst}\left (\int \left (3-2 x^2-x^4\right )^{3/2} \, dx,x,-1+x\right )\\ &=\frac {1}{7} \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2} (-1+x)+\frac {3}{7} \operatorname {Subst}\left (\int \left (6-2 x^2\right ) \sqrt {3-2 x^2-x^4} \, dx,x,-1+x\right )\\ &=-\frac {2}{35} \left (13-3 (1-x)^2\right ) \sqrt {3-2 (1-x)^2-(1-x)^4} (1-x)+\frac {1}{7} \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2} (-1+x)-\frac {1}{35} \operatorname {Subst}\left (\int \frac {-192+112 x^2}{\sqrt {3-2 x^2-x^4}} \, dx,x,-1+x\right )\\ &=-\frac {2}{35} \left (13-3 (1-x)^2\right ) \sqrt {3-2 (1-x)^2-(1-x)^4} (1-x)+\frac {1}{7} \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2} (-1+x)-\frac {2}{35} \operatorname {Subst}\left (\int \frac {-192+112 x^2}{\sqrt {2-2 x^2} \sqrt {6+2 x^2}} \, dx,x,-1+x\right )\\ &=-\frac {2}{35} \left (13-3 (1-x)^2\right ) \sqrt {3-2 (1-x)^2-(1-x)^4} (1-x)+\frac {1}{7} \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2} (-1+x)-\frac {16}{5} \operatorname {Subst}\left (\int \frac {\sqrt {6+2 x^2}}{\sqrt {2-2 x^2}} \, dx,x,-1+x\right )+\frac {1056}{35} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-2 x^2} \sqrt {6+2 x^2}} \, dx,x,-1+x\right )\\ &=-\frac {2}{35} \left (13-3 (1-x)^2\right ) \sqrt {3-2 (1-x)^2-(1-x)^4} (1-x)+\frac {1}{7} \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2} (-1+x)+\frac {16}{5} \sqrt {3} E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )-\frac {176}{35} \sqrt {3} F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )\\ \end {align*}

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Mathematica [C]  time = 0.77, size = 278, normalized size = 2.73 \[ \frac {5 x^9-45 x^8+206 x^7-602 x^6+1152 x^5-1420 x^4+848 x^3+352 x^2-304 i \sqrt {2} \sqrt {-\frac {i (x-2)}{\left (\sqrt {3}-i\right ) x}} \sqrt {\frac {x^2-2 x+4}{x^2}} x^2 F\left (\sin ^{-1}\left (\frac {\sqrt {\sqrt {3}+i-\frac {4 i}{x}}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{-i+\sqrt {3}}\right )+\frac {112 i \sqrt {2} (x-2) \sqrt {\frac {x^2-2 x+4}{x^2}} x E\left (\sin ^{-1}\left (\frac {\sqrt {\sqrt {3}+i-\frac {4 i}{x}}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{-i+\sqrt {3}}\right )}{\sqrt {-\frac {i (x-2)}{\left (\sqrt {3}-i\right ) x}}}-1056 x+896}{35 \sqrt {-x \left (x^3-4 x^2+8 x-8\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(896 - 1056*x + 352*x^2 + 848*x^3 - 1420*x^4 + 1152*x^5 - 602*x^6 + 206*x^7 - 45*x^8 + 5*x^9 + ((112*I)*Sqrt[2
]*(-2 + x)*x*Sqrt[(4 - 2*x + x^2)/x^2]*EllipticE[ArcSin[Sqrt[I + Sqrt[3] - (4*I)/x]/(Sqrt[2]*3^(1/4))], (2*Sqr
t[3])/(-I + Sqrt[3])])/Sqrt[((-I)*(-2 + x))/((-I + Sqrt[3])*x)] - (304*I)*Sqrt[2]*Sqrt[((-I)*(-2 + x))/((-I +
Sqrt[3])*x)]*x^2*Sqrt[(4 - 2*x + x^2)/x^2]*EllipticF[ArcSin[Sqrt[I + Sqrt[3] - (4*I)/x]/(Sqrt[2]*3^(1/4))], (2
*Sqrt[3])/(-I + Sqrt[3])])/(35*Sqrt[-(x*(-8 + 8*x - 4*x^2 + x^3))])

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-x^{4} + 4 \, x^{3} - 8 \, x^{2} + 8 \, x\right )}^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-x^4 + 4*x^3 - 8*x^2 + 8*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.15, size = 1050, normalized size = 10.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^4+4*x^3-8*x^2+8*x)^(3/2),x)

[Out]

-1/7*x^5*(-x^4+4*x^3-8*x^2+8*x)^(1/2)+5/7*x^4*(-x^4+4*x^3-8*x^2+8*x)^(1/2)-66/35*x^3*(-x^4+4*x^3-8*x^2+8*x)^(1
/2)+14/5*x^2*(-x^4+4*x^3-8*x^2+8*x)^(1/2)-32/35*x*(-x^4+4*x^3-8*x^2+8*x)^(1/2)-4/7*(-x^4+4*x^3-8*x^2+8*x)^(1/2
)+32/7*(-1-I*3^(1/2))*((I*3^(1/2)-1)*x/(1+I*3^(1/2))/(x-2))^(1/2)*(x-2)^2*((x-1+I*3^(1/2))/(1-I*3^(1/2))/(x-2)
)^(1/2)*((x-1-I*3^(1/2))/(1+I*3^(1/2))/(x-2))^(1/2)/(I*3^(1/2)-1)/(-x*(x-2)*(x-1+I*3^(1/2))*(x-1-I*3^(1/2)))^(
1/2)*EllipticF(((I*3^(1/2)-1)*x/(1+I*3^(1/2))/(x-2))^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^
(1/2)))^(1/2))+64/5*(-1-I*3^(1/2))*((I*3^(1/2)-1)*x/(1+I*3^(1/2))/(x-2))^(1/2)*(x-2)^2*((x-1+I*3^(1/2))/(1-I*3
^(1/2))/(x-2))^(1/2)*((x-1-I*3^(1/2))/(1+I*3^(1/2))/(x-2))^(1/2)/(I*3^(1/2)-1)/(-x*(x-2)*(x-1+I*3^(1/2))*(x-1-
I*3^(1/2)))^(1/2)*(2*EllipticF(((I*3^(1/2)-1)*x/(1+I*3^(1/2))/(x-2))^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^
(1/2)-1)/(1-I*3^(1/2)))^(1/2))-2*EllipticPi(((I*3^(1/2)-1)*x/(1+I*3^(1/2))/(x-2))^(1/2),(1+I*3^(1/2))/(I*3^(1/
2)-1),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^(1/2)))^(1/2)))-16/5*(x*(x-1+I*3^(1/2))*(x-1-I*3^(1/2
))+2*(-1-I*3^(1/2))*((I*3^(1/2)-1)*x/(1+I*3^(1/2))/(x-2))^(1/2)*(x-2)^2*((x-1+I*3^(1/2))/(1-I*3^(1/2))/(x-2))^
(1/2)*((x-1-I*3^(1/2))/(1+I*3^(1/2))/(x-2))^(1/2)*(1/2*(6+2*I*3^(1/2))/(I*3^(1/2)-1)*EllipticF(((I*3^(1/2)-1)*
x/(1+I*3^(1/2))/(x-2))^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^(1/2)))^(1/2))+1/2*(I*3^(1/2)-
1)*EllipticE(((I*3^(1/2)-1)*x/(1+I*3^(1/2))/(x-2))^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^(1
/2)))^(1/2))-4/(I*3^(1/2)-1)*EllipticPi(((I*3^(1/2)-1)*x/(1+I*3^(1/2))/(x-2))^(1/2),(-1-I*3^(1/2))/(1-I*3^(1/2
)),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^(1/2)))^(1/2))))/(-x*(x-2)*(x-1+I*3^(1/2))*(x-1-I*3^(1/2
)))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (-x^4+4\,x^3-8\,x^2+8\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - 8*x^2 + 4*x^3 - x^4)^(3/2),x)

[Out]

int((8*x - 8*x^2 + 4*x^3 - x^4)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**4+4*x**3-8*x**2+8*x)**(3/2),x)

[Out]

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

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