3.6.17 \(\int \frac {x^3}{(1+x)^{5/2} (1-x+x^2)^{5/2}} \, dx\) [517]

Optimal. Leaf size=168 \[ \frac {4 x}{27 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {2 x}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}+\frac {4 \sqrt {2+\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \]

[Out]

4/27*x/(1+x)^(1/2)/(x^2-x+1)^(1/2)-2/9*x/(x^3+1)/(1+x)^(1/2)/(x^2-x+1)^(1/2)+4/81*EllipticF((1+x-3^(1/2))/(1+x
+3^(1/2)),I*3^(1/2)+2*I)*(1+x)^(1/2)*(1/2*6^(1/2)+1/2*2^(1/2))*((x^2-x+1)/(1+x+3^(1/2))^2)^(1/2)*3^(3/4)/(x^2-
x+1)^(1/2)/((1+x)/(1+x+3^(1/2))^2)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {929, 294, 205, 224} \begin {gather*} \frac {4 \sqrt {2+\sqrt {3}} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\text {ArcSin}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}}+\frac {4 x}{27 \sqrt {x+1} \sqrt {x^2-x+1}}-\frac {2 x}{9 \sqrt {x+1} \sqrt {x^2-x+1} \left (x^3+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/((1 + x)^(5/2)*(1 - x + x^2)^(5/2)),x]

[Out]

(4*x)/(27*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (2*x)/(9*Sqrt[1 + x]*Sqrt[1 - x + x^2]*(1 + x^3)) + (4*Sqrt[2 + Sqr
t[3]]*Sqrt[1 + x]*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)
], -7 - 4*Sqrt[3]])/(27*3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 - x + x^2])

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt
[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq
rt[s*((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)
], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 929

Int[((g_.)*(x_))^(n_)*((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d
+ e*x)^FracPart[p]*((a + b*x + c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p]), Int[(g*x)^n*(a*d + c*e*x^3)^p,
x], x] /; FreeQ[{a, b, c, d, e, g, m, n, p}, x] && EqQ[m - p, 0] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0]

Rubi steps

\begin {align*} \int \frac {x^3}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx &=\frac {\sqrt {1+x^3} \int \frac {x^3}{\left (1+x^3\right )^{5/2}} \, dx}{\sqrt {1+x} \sqrt {1-x+x^2}}\\ &=-\frac {2 x}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}+\frac {\left (2 \sqrt {1+x^3}\right ) \int \frac {1}{\left (1+x^3\right )^{3/2}} \, dx}{9 \sqrt {1+x} \sqrt {1-x+x^2}}\\ &=\frac {4 x}{27 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {2 x}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}+\frac {\left (2 \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{27 \sqrt {1+x} \sqrt {1-x+x^2}}\\ &=\frac {4 x}{27 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {2 x}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}+\frac {4 \sqrt {2+\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 30.31, size = 178, normalized size = 1.06 \begin {gather*} \frac {\frac {6 x \left (-1+2 x^3\right )}{(1+x)^{3/2} \left (1-x+x^2\right )}+\frac {2 i (1+x) \sqrt {1+\frac {6 i}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {6-\frac {36 i}{\left (3 i+\sqrt {3}\right ) (1+x)}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{81 \sqrt {1-x+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/((1 + x)^(5/2)*(1 - x + x^2)^(5/2)),x]

[Out]

((6*x*(-1 + 2*x^3))/((1 + x)^(3/2)*(1 - x + x^2)) + ((2*I)*(1 + x)*Sqrt[1 + (6*I)/((-3*I + Sqrt[3])*(1 + x))]*
Sqrt[6 - (36*I)/((3*I + Sqrt[3])*(1 + x))]*EllipticF[I*ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I
+ Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[(-I)/(3*I + Sqrt[3])])/(81*Sqrt[1 - x + x^2])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 466 vs. \(2 (135 ) = 270\).
time = 0.10, size = 467, normalized size = 2.78

method	result	size
elliptic	\(\frac {\sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (-\frac {2 x}{9 \left (x^{3}+1\right )^{\frac {3}{2}}}+\frac {4 x}{27 \sqrt {x^{3}+1}}+\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{27 \sqrt {x^{3}+1}}\right )}{\sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(167\)
default	\(-\frac {2 \left (i \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) \sqrt {3}\, x^{3} \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}-3 \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) x^{3} \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}+i \sqrt {3}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-3 \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-2 x^{4}+x \right )}{27 \left (x^{2}-x +1\right )^{\frac {3}{2}} \left (1+x \right )^{\frac {3}{2}}}\)	\(467\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(1+x)^(5/2)/(x^2-x+1)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/27*(I*EllipticF((-2*(1+x)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(I*3^(1/2)+3))^(1/2))*3^(1/2)*x^3*((I*3^(1
/2)-2*x+1)/(I*3^(1/2)+3))^(1/2)*((I*3^(1/2)+2*x-1)/(-3+I*3^(1/2)))^(1/2)*(-2*(1+x)/(-3+I*3^(1/2)))^(1/2)-3*Ell
ipticF((-2*(1+x)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(I*3^(1/2)+3))^(1/2))*x^3*((I*3^(1/2)-2*x+1)/(I*3^(1/2
)+3))^(1/2)*((I*3^(1/2)+2*x-1)/(-3+I*3^(1/2)))^(1/2)*(-2*(1+x)/(-3+I*3^(1/2)))^(1/2)+I*3^(1/2)*(-2*(1+x)/(-3+I
*3^(1/2)))^(1/2)*((I*3^(1/2)-2*x+1)/(I*3^(1/2)+3))^(1/2)*((I*3^(1/2)+2*x-1)/(-3+I*3^(1/2)))^(1/2)*EllipticF((-
2*(1+x)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(I*3^(1/2)+3))^(1/2))-3*(-2*(1+x)/(-3+I*3^(1/2)))^(1/2)*((I*3^(
1/2)-2*x+1)/(I*3^(1/2)+3))^(1/2)*((I*3^(1/2)+2*x-1)/(-3+I*3^(1/2)))^(1/2)*EllipticF((-2*(1+x)/(-3+I*3^(1/2)))^
(1/2),(-(-3+I*3^(1/2))/(I*3^(1/2)+3))^(1/2))-2*x^4+x)/(x^2-x+1)^(3/2)/(1+x)^(3/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3/((x^2 - x + 1)^(5/2)*(x + 1)^(5/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.23, size = 56, normalized size = 0.33 \begin {gather*} \frac {2 \, {\left ({\left (2 \, x^{4} - x\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} + 2 \, {\left (x^{6} + 2 \, x^{3} + 1\right )} {\rm weierstrassPInverse}\left (0, -4, x\right )\right )}}{27 \, {\left (x^{6} + 2 \, x^{3} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="fricas")

[Out]

2/27*((2*x^4 - x)*sqrt(x^2 - x + 1)*sqrt(x + 1) + 2*(x^6 + 2*x^3 + 1)*weierstrassPInverse(0, -4, x))/(x^6 + 2*
x^3 + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (x + 1\right )^{\frac {5}{2}} \left (x^{2} - x + 1\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(1+x)**(5/2)/(x**2-x+1)**(5/2),x)

[Out]

Integral(x**3/((x + 1)**(5/2)*(x**2 - x + 1)**(5/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="giac")

[Out]

integrate(x^3/((x^2 - x + 1)^(5/2)*(x + 1)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{{\left (x+1\right )}^{5/2}\,{\left (x^2-x+1\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((x + 1)^(5/2)*(x^2 - x + 1)^(5/2)),x)

[Out]

int(x^3/((x + 1)^(5/2)*(x^2 - x + 1)^(5/2)), x)

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