3.502 \(\int \frac {(1+x)^{3/2} (1-x+x^2)^{3/2}}{x^3} \, dx\)

Optimal. Leaf size=175 \[ \frac {9}{10} x \sqrt {x^2-x+1} \sqrt {x+1}-\frac {\sqrt {x^2-x+1} \left (x^3+1\right ) \sqrt {x+1}}{2 x^2}+\frac {9\ 3^{3/4} \sqrt {2+\sqrt {3}} \sqrt {x^2-x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} (x+1)^{3/2} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{10 \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \left (x^3+1\right )} \]

[Out]

9/10*x*(1+x)^(1/2)*(x^2-x+1)^(1/2)-1/2*(x^3+1)*(1+x)^(1/2)*(x^2-x+1)^(1/2)/x^2+9/10*3^(3/4)*(1+x)^(3/2)*Ellipt
icF((1+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*(x^2-x+1)^(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^3+1)/((1+x)/(1+x+3^(1/2))^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 175, 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 = {915, 277, 195, 218} \[ -\frac {\sqrt {x^2-x+1} \left (x^3+1\right ) \sqrt {x+1}}{2 x^2}+\frac {9}{10} x \sqrt {x^2-x+1} \sqrt {x+1}+\frac {9\ 3^{3/4} \sqrt {2+\sqrt {3}} \sqrt {x^2-x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} (x+1)^{3/2} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{10 \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \left (x^3+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*x*Sqrt[1 + x]*Sqrt[1 - x + x^2])/10 - (Sqrt[1 + x]*Sqrt[1 - x + x^2]*(1 + x^3))/(2*x^2) + (9*3^(3/4)*Sqrt[2
+ Sqrt[3]]*(1 + x)^(3/2)*Sqrt[1 - x + x^2]*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt
[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(10*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*(1 + x^3))

Rule 195

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

Rule 218

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

Rule 277

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

Rule 915

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 {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x^3} \, dx &=\frac {\left (\sqrt {1+x} \sqrt {1-x+x^2}\right ) \int \frac {\left (1+x^3\right )^{3/2}}{x^3} \, dx}{\sqrt {1+x^3}}\\ &=-\frac {\sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}{2 x^2}+\frac {\left (9 \sqrt {1+x} \sqrt {1-x+x^2}\right ) \int \sqrt {1+x^3} \, dx}{4 \sqrt {1+x^3}}\\ &=\frac {9}{10} x \sqrt {1+x} \sqrt {1-x+x^2}-\frac {\sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}{2 x^2}+\frac {\left (27 \sqrt {1+x} \sqrt {1-x+x^2}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{20 \sqrt {1+x^3}}\\ &=\frac {9}{10} x \sqrt {1+x} \sqrt {1-x+x^2}-\frac {\sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}{2 x^2}+\frac {9\ 3^{3/4} \sqrt {2+\sqrt {3}} (1+x)^{3/2} \sqrt {1-x+x^2} \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 )}{10 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \left (1+x^3\right )}\\ \end {align*}

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Mathematica [C]  time = 0.35, size = 192, normalized size = 1.10 \[ \frac {\sqrt {x+1} \left (\frac {2 \left (x^2-x+1\right ) \left (4 x^3-5\right )}{x^2}-\frac {27 i \sqrt {2} \sqrt {\frac {-2 i x+\sqrt {3}+i}{\sqrt {3}+3 i}} \sqrt {\frac {2 i x+\sqrt {3}-i}{\sqrt {3}-3 i}} F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {i (x+1)}{3 i+\sqrt {3}}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i (x+1)}{\sqrt {3}+3 i}}}\right )}{20 \sqrt {x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + x]*((2*(1 - x + x^2)*(-5 + 4*x^3))/x^2 - ((27*I)*Sqrt[2]*Sqrt[(I + Sqrt[3] - (2*I)*x)/(3*I + Sqrt[3]
)]*Sqrt[(-I + Sqrt[3] + (2*I)*x)/(-3*I + Sqrt[3])]*EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[((-I)*(1 + x))/(3*I + Sqrt
[3])]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[((-I)*(1 + x))/(3*I + Sqrt[3])]))/(20*Sqrt[1 - x + x^2])

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fricas [F]  time = 1.16, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (x^{3} + 1\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((x^3 + 1)*sqrt(x^2 - x + 1)*sqrt(x + 1)/x^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 264, normalized size = 1.51 \[ -\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (-8 x^{6}+2 x^{3}+27 i \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, \sqrt {3}\, x^{2} \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-81 \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, x^{2} \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )+10\right )}{20 \left (x^{3}+1\right ) x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(x+1)^(1/2)*(x^2-x+1)^(1/2)*(27*I*(-2*(x+1)/(-3+I*3^(1/2)))^(1/2)*((-2*x+I*3^(1/2)+1)/(I*3^(1/2)+3))^(1/
2)*((2*x+I*3^(1/2)-1)/(-3+I*3^(1/2)))^(1/2)*EllipticF((-2*(x+1)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(I*3^(1
/2)+3))^(1/2))*3^(1/2)*x^2-81*(-2*(x+1)/(-3+I*3^(1/2)))^(1/2)*((-2*x+I*3^(1/2)+1)/(I*3^(1/2)+3))^(1/2)*((2*x+I
*3^(1/2)-1)/(-3+I*3^(1/2)))^(1/2)*EllipticF((-2*(x+1)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(I*3^(1/2)+3))^(1
/2))*x^2-8*x^6+2*x^3+10)/(x^3+1)/x^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x + 1\right )^{\frac {3}{2}} \left (x^{2} - x + 1\right )^{\frac {3}{2}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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