3.2304 \(\int \frac{a+b x}{(1+x)^{3/2} (1-x+x^2)^{3/2}} \, dx\)

Optimal. Leaf size=304 \[ \frac{2 \sqrt{2+\sqrt{3}} \sqrt{x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} \left (a-\sqrt{3} b+b\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^2-x+1}}+\frac{2 x (a+b x)}{3 \sqrt{x+1} \sqrt{x^2-x+1}}-\frac{2 b \left (x^3+1\right )}{3 \sqrt{x+1} \left (x+\sqrt{3}+1\right ) \sqrt{x^2-x+1}}+\frac{\sqrt{2-\sqrt{3}} b \sqrt{x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{3^{3/4} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^2-x+1}} \]

[Out]

(2*x*(a + b*x))/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (2*b*(1 + x^3))/(3*Sqrt[1 + x]*(1 + Sqrt[3] + x)*Sqrt[1 -
x + x^2]) + (Sqrt[2 - Sqrt[3]]*b*Sqrt[1 + x]*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticE[ArcSin[(1 - Sqr
t[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3^(3/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 - x + x^2]) +
(2*Sqrt[2 + Sqrt[3]]*(a + b - Sqrt[3]*b)*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]])/(3*3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 -
x + x^2])

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Rubi [A]  time = 0.160442, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {809, 1855, 1878, 218, 1877} \[ \frac{2 x (a+b x)}{3 \sqrt{x+1} \sqrt{x^2-x+1}}+\frac{2 \sqrt{2+\sqrt{3}} \sqrt{x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} \left (a-\sqrt{3} b+b\right ) F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^2-x+1}}-\frac{2 b \left (x^3+1\right )}{3 \sqrt{x+1} \left (x+\sqrt{3}+1\right ) \sqrt{x^2-x+1}}+\frac{\sqrt{2-\sqrt{3}} b \sqrt{x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{3^{3/4} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(a + b*x))/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (2*b*(1 + x^3))/(3*Sqrt[1 + x]*(1 + Sqrt[3] + x)*Sqrt[1 -
x + x^2]) + (Sqrt[2 - Sqrt[3]]*b*Sqrt[1 + x]*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticE[ArcSin[(1 - Sqr
t[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3^(3/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 - x + x^2]) +
(2*Sqrt[2 + Sqrt[3]]*(a + b - Sqrt[3]*b)*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]])/(3*3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 -
x + x^2])

Rule 809

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 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]

Rule 1878

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

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 1877

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]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), 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{a+b x}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx &=\frac{\sqrt{1+x^3} \int \frac{a+b x}{\left (1+x^3\right )^{3/2}} \, dx}{\sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2 x (a+b x)}{3 \sqrt{1+x} \sqrt{1-x+x^2}}-\frac{\left (2 \sqrt{1+x^3}\right ) \int \frac{-\frac{a}{2}+\frac{b x}{2}}{\sqrt{1+x^3}} \, dx}{3 \sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2 x (a+b x)}{3 \sqrt{1+x} \sqrt{1-x+x^2}}-\frac{\left (b \sqrt{1+x^3}\right ) \int \frac{1-\sqrt{3}+x}{\sqrt{1+x^3}} \, dx}{3 \sqrt{1+x} \sqrt{1-x+x^2}}-\frac{\left (2 \left (-\frac{a}{2}-\frac{1}{2} \left (1-\sqrt{3}\right ) b\right ) \sqrt{1+x^3}\right ) \int \frac{1}{\sqrt{1+x^3}} \, dx}{3 \sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2 x (a+b x)}{3 \sqrt{1+x} \sqrt{1-x+x^2}}-\frac{2 b \left (1+x^3\right )}{3 \sqrt{1+x} \left (1+\sqrt{3}+x\right ) \sqrt{1-x+x^2}}+\frac{\sqrt{2-\sqrt{3}} b \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 )}{3^{3/4} \sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \sqrt{1-x+x^2}}+\frac{2 \sqrt{2+\sqrt{3}} \left (a+\left (1-\sqrt{3}\right ) b\right ) \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 )}{3 \sqrt [4]{3} \sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \sqrt{1-x+x^2}}\\ \end{align*}

Mathematica [C]  time = 1.05844, size = 417, normalized size = 1.37 \[ \frac{\sqrt{2} \sqrt{\frac{-\frac{6 i}{x+1}+\sqrt{3}+3 i}{\sqrt{3}+3 i}} \sqrt{-\frac{\frac{6 i}{x+1}+\sqrt{3}-3 i}{-\sqrt{3}+3 i}} (x+1)^{3/2} \left (2 i \sqrt{3} a+\left (3-i \sqrt{3}\right ) b\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{6 i}{\sqrt{3}+3 i}}}{\sqrt{x+1}}\right ),\frac{\sqrt{3}+3 i}{-\sqrt{3}+3 i}\right )+12 \sqrt{-\frac{i}{\sqrt{3}+3 i}} x (a+b x)-12 \sqrt{-\frac{i}{\sqrt{3}+3 i}} b \left (x^2-x+1\right )+3 i \sqrt{2} \left (\sqrt{3}+i\right ) b \sqrt{\frac{-\frac{6 i}{x+1}+\sqrt{3}+3 i}{\sqrt{3}+3 i}} \sqrt{-\frac{\frac{6 i}{x+1}+\sqrt{3}-3 i}{-\sqrt{3}+3 i}} (x+1)^{3/2} E\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{6 i}{3 i+\sqrt{3}}}}{\sqrt{x+1}}\right )|\frac{3 i+\sqrt{3}}{3 i-\sqrt{3}}\right )}{18 \sqrt{-\frac{i}{\sqrt{3}+3 i}} \sqrt{x+1} \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*Sqrt[(-I)/(3*I + Sqrt[3])]*x*(a + b*x) - 12*Sqrt[(-I)/(3*I + Sqrt[3])]*b*(1 - x + x^2) + (3*I)*Sqrt[2]*(I
+ Sqrt[3])*b*(1 + x)^(3/2)*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]))]*EllipticE[I*ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3
*I - Sqrt[3])] + Sqrt[2]*((2*I)*Sqrt[3]*a + (3 - I*Sqrt[3])*b)*(1 + x)^(3/2)*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])])/(18*Sqrt[(-I)/(3*I + Sqrt[3])]*Sqrt[1 + x]*S
qrt[1 - x + x^2])

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Maple [B]  time = 0.06, size = 584, normalized size = 1.9 \begin{align*} -{\frac{1}{3\,{x}^{3}+3}\sqrt{1+x}\sqrt{{x}^{2}-x+1} \left ( i\sqrt{3}{\it EllipticF} \left ( \sqrt{-2\,{\frac{1+x}{-3+i\sqrt{3}}}},\sqrt{-{\frac{-3+i\sqrt{3}}{i\sqrt{3}+3}}} \right ) a\sqrt{-2\,{\frac{1+x}{-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}}}}+i\sqrt{3}{\it EllipticF} \left ( \sqrt{-2\,{\frac{1+x}{-3+i\sqrt{3}}}},\sqrt{-{\frac{-3+i\sqrt{3}}{i\sqrt{3}+3}}} \right ) b\sqrt{-2\,{\frac{1+x}{-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}}}}-3\,\sqrt{-2\,{\frac{1+x}{-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}}}}{\it EllipticF} \left ( \sqrt{-2\,{\frac{1+x}{-3+i\sqrt{3}}}},\sqrt{-{\frac{-3+i\sqrt{3}}{i\sqrt{3}+3}}} \right ) a+3\,\sqrt{-2\,{\frac{1+x}{-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}}}}{\it EllipticF} \left ( \sqrt{-2\,{\frac{1+x}{-3+i\sqrt{3}}}},\sqrt{-{\frac{-3+i\sqrt{3}}{i\sqrt{3}+3}}} \right ) b-6\,\sqrt{-2\,{\frac{1+x}{-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}}}}{\it EllipticE} \left ( \sqrt{-2\,{\frac{1+x}{-3+i\sqrt{3}}}},\sqrt{-{\frac{-3+i\sqrt{3}}{i\sqrt{3}+3}}} \right ) b-2\,b{x}^{2}-2\,ax \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(1+x)^(1/2)*(x^2-x+1)^(1/2)*(I*3^(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))*a*(-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)+I*3^(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))
*b*(-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)-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))*a+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))*b-6*(-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))*b-2*b*x^2-2*a*x)/(x^3+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b x + a}{{\left (x^{2} - x + 1\right )}^{\frac{3}{2}}{\left (x + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )} \sqrt{x^{2} - x + 1} \sqrt{x + 1}}{x^{6} + 2 \, x^{3} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x + a)*sqrt(x^2 - x + 1)*sqrt(x + 1)/(x^6 + 2*x^3 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b x + a}{{\left (x^{2} - x + 1\right )}^{\frac{3}{2}}{\left (x + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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