3.6.19 \(\int \frac {x}{(1+x)^{5/2} (1-x+x^2)^{5/2}} \, dx\) [519]

Optimal. Leaf size=318 \[ \frac {10 x^2}{27 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {2 x^2}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}-\frac {10 \left (1+x^3\right )}{27 \sqrt {1+x} \left (1+\sqrt {3}+x\right ) \sqrt {1-x+x^2}}+\frac {5 \sqrt {2-\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{9\ 3^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}}-\frac {10 \sqrt {2} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \]

[Out]

10/27*x^2/(1+x)^(1/2)/(x^2-x+1)^(1/2)+2/9*x^2/(x^3+1)/(1+x)^(1/2)/(x^2-x+1)^(1/2)-10/27*(x^3+1)/(1+x+3^(1/2))/
(1+x)^(1/2)/(x^2-x+1)^(1/2)-10/81*EllipticF((1+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*2^(1/2)*(1+x)^(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)+5/27*3^(1/4)*EllipticE((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)/(x^2-x+1)^(1/2)/((1+x)/(1+x+3^(1/2))^2)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {823, 296, 309, 224, 1891} \begin {gather*} -\frac {10 \sqrt {2} \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 {5 \sqrt {2-\sqrt {3}} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} E\left (\text {ArcSin}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{9\ 3^{3/4} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}}+\frac {10 x^2}{27 \sqrt {x+1} \sqrt {x^2-x+1}}+\frac {2 x^2}{9 \sqrt {x+1} \sqrt {x^2-x+1} \left (x^3+1\right )}-\frac {10 \left (x^3+1\right )}{27 \sqrt {x+1} \left (x+\sqrt {3}+1\right ) \sqrt {x^2-x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*x^2)/(27*Sqrt[1 + x]*Sqrt[1 - x + x^2]) + (2*x^2)/(9*Sqrt[1 + x]*Sqrt[1 - x + x^2]*(1 + x^3)) - (10*(1 + x
^3))/(27*Sqrt[1 + x]*(1 + Sqrt[3] + x)*Sqrt[1 - x + x^2]) + (5*Sqrt[2 - Sqrt[3]]*Sqrt[1 + x]*Sqrt[(1 - x + x^2
)/(1 + Sqrt[3] + x)^2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(9*3^(3/4)*Sqrt
[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 - x + x^2]) - (10*Sqrt[2]*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 + Sqr
t[3] + x)^2]*Sqrt[1 - x + x^2])

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 296

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

Rule 309

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((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[(f + g*x)*(a*d + c*e*x^
3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[m, p] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0]

Rule 1891

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[(1 - Sqrt[3])*(d/c)]]
, s = Denom[Simplify[(1 - Sqrt[3])*(d/c)]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x
] - Simp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(r^2
*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1
+ Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin {align*} \int \frac {x}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx &=\frac {\sqrt {1+x^3} \int \frac {x}{\left (1+x^3\right )^{5/2}} \, dx}{\sqrt {1+x} \sqrt {1-x+x^2}}\\ &=\frac {2 x^2}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}+\frac {\left (5 \sqrt {1+x^3}\right ) \int \frac {x}{\left (1+x^3\right )^{3/2}} \, dx}{9 \sqrt {1+x} \sqrt {1-x+x^2}}\\ &=\frac {10 x^2}{27 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {2 x^2}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}-\frac {\left (5 \sqrt {1+x^3}\right ) \int \frac {x}{\sqrt {1+x^3}} \, dx}{27 \sqrt {1+x} \sqrt {1-x+x^2}}\\ &=\frac {10 x^2}{27 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {2 x^2}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}-\frac {\left (5 \sqrt {1+x^3}\right ) \int \frac {1-\sqrt {3}+x}{\sqrt {1+x^3}} \, dx}{27 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {\left (5 \sqrt {2 \left (2-\sqrt {3}\right )} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{27 \sqrt {1+x} \sqrt {1-x+x^2}}\\ &=\frac {10 x^2}{27 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {2 x^2}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}-\frac {10 \left (1+x^3\right )}{27 \sqrt {1+x} \left (1+\sqrt {3}+x\right ) \sqrt {1-x+x^2}}+\frac {5 \sqrt {2-\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{9\ 3^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}}-\frac {10 \sqrt {2} \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 = 10.49, size = 409, normalized size = 1.29 \begin {gather*} \frac {2 x^2 \left (8+5 x^3\right )}{27 (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}-\frac {5 (1+x)^{3/2} \left (\frac {12 \sqrt {-\frac {i}{3 i+\sqrt {3}}} \left (1-x+x^2\right )}{(1+x)^2}+\frac {3 \sqrt {2} \left (1-i \sqrt {3}\right ) \sqrt {\frac {3 i+\sqrt {3}-\frac {6 i}{1+x}}{3 i+\sqrt {3}}} \sqrt {\frac {-3 i+\sqrt {3}+\frac {6 i}{1+x}}{-3 i+\sqrt {3}}} E\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 {1+x}}+\frac {i \sqrt {2} \left (3 i+\sqrt {3}\right ) \sqrt {\frac {3 i+\sqrt {3}-\frac {6 i}{1+x}}{3 i+\sqrt {3}}} \sqrt {\frac {-3 i+\sqrt {3}+\frac {6 i}{1+x}}{-3 i+\sqrt {3}}} 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 {1+x}}\right )}{162 \sqrt {-\frac {i}{3 i+\sqrt {3}}} \sqrt {1-x+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^2*(8 + 5*x^3))/(27*(1 + x)^(3/2)*(1 - x + x^2)^(3/2)) - (5*(1 + x)^(3/2)*((12*Sqrt[(-I)/(3*I + Sqrt[3])]*
(1 - x + x^2))/(1 + x)^2 + (3*Sqrt[2]*(1 - I*Sqrt[3])*Sqrt[(3*I + Sqrt[3] - (6*I)/(1 + x))/(3*I + Sqrt[3])]*Sq
rt[(-3*I + Sqrt[3] + (6*I)/(1 + x))/(-3*I + Sqrt[3])]*EllipticE[I*ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1
+ x]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[1 + x] + (I*Sqrt[2]*(3*I + Sqrt[3])*Sqrt[(3*I + Sqrt[3] - (6*I)/
(1 + x))/(3*I + Sqrt[3])]*Sqrt[(-3*I + Sqrt[3] + (6*I)/(1 + x))/(-3*I + Sqrt[3])]*EllipticF[I*ArcSinh[Sqrt[(-6
*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[1 + x]))/(162*Sqrt[(-I)/(3*I + Sqrt[
3])]*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. 687 vs. \(2 (252 ) = 504\).
time = 0.11, size = 688, normalized size = 2.16

method	result	size
elliptic	\(\frac {\sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (\frac {2 x^{2}}{9 \left (x^{3}+1\right )^{\frac {3}{2}}}+\frac {10 x^{2}}{27 \sqrt {x^{3}+1}}-\frac {10 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \EllipticE \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \EllipticF \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{27 \sqrt {x^{3}+1}}\right )}{\sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(228\)
default	\(-\frac {5 i \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) \sqrt {3}\, x^{3} \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}+15 \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) x^{3} \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}-30 \EllipticE \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) x^{3} \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}+5 i \sqrt {3}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-10 x^{5}+15 \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-30 \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \EllipticE \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-16 x^{2}}{27 \left (x^{2}-x +1\right )^{\frac {3}{2}} \left (1+x \right )^{\frac {3}{2}}}\)	\(688\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*(5*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)/(-3+I*3^(1/2)))^(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)+15*
EllipticF((-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)-30*EllipticE((-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)/(-3+I*3^(1/2)))^(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)+5*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)*((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*x+1)/(I*3^(1/2)+3))^(1/2)-10*x^5+15*(-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))-30*(-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)*EllipticE((-2*(1+x)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(I*3^(
1/2)+3))^(1/2))-16*x^2)/(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/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="maxima")

[Out]

integrate(x/((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 = 0.19, size = 61, normalized size = 0.19 \begin {gather*} \frac {2 \, {\left ({\left (5 \, x^{5} + 8 \, x^{2}\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} + 5 \, {\left (x^{6} + 2 \, x^{3} + 1\right )} {\rm weierstrassZeta}\left (0, -4, {\rm weierstrassPInverse}\left (0, -4, x\right )\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/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="fricas")

[Out]

2/27*((5*x^5 + 8*x^2)*sqrt(x^2 - x + 1)*sqrt(x + 1) + 5*(x^6 + 2*x^3 + 1)*weierstrassZeta(0, -4, weierstrassPI
nverse(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}{\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/(1+x)**(5/2)/(x**2-x+1)**(5/2),x)

[Out]

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

[Out]

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

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

[Out]

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

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