Integrand size = 25, antiderivative size = 351 \[ \int \frac {a+b x}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {2 x (7 a+5 b x)}{27 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {2 x (a+b x)}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}-\frac {10 b \left (1+x^3\right )}{27 \sqrt {1+x} \left (1+\sqrt {3}+x\right ) \sqrt {1-x+x^2}}+\frac {5 \sqrt {2-\sqrt {3}} b \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{9\ 3^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}}+\frac {2 \sqrt {2+\sqrt {3}} \left (7 a+5 \left (1-\sqrt {3}\right ) b\right ) \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \]
2/27*x*(5*b*x+7*a)/(1+x)^(1/2)/(x^2-x+1)^(1/2)+2/9*x*(b*x+a)/(x^3+1)/(1+x) ^(1/2)/(x^2-x+1)^(1/2)-10/27*b*(x^3+1)/(1+x+3^(1/2))/(1+x)^(1/2)/(x^2-x+1) ^(1/2)+5/27*3^(1/4)*b*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)+2/81*EllipticF((1+x-3^(1/2))/ (1+x+3^(1/2)),I*3^(1/2)+2*I)*(7*a+5*b*(1-3^(1/2)))*(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)
Result contains complex when optimal does not.
Time = 32.18 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.24 \[ \int \frac {a+b x}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {2 x \left (b x \left (8+5 x^3\right )+a \left (10+7 x^3\right )\right )}{27 (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}+\frac {(1+x)^{3/2} \left (-\frac {60 \sqrt {-\frac {i}{3 i+\sqrt {3}}} b \left (1-x+x^2\right )}{(1+x)^2}+\frac {15 i \sqrt {2} \left (i+\sqrt {3}\right ) b \sqrt {\frac {3 i+\sqrt {3}-\frac {6 i}{1+x}}{3 i+\sqrt {3}}} \sqrt {\frac {-3 i+\sqrt {3}+\frac {6 i}{1+x}}{-3 i+\sqrt {3}}} E\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {1+x}}+\frac {\sqrt {2} \left (14 i \sqrt {3} a+5 \left (3-i \sqrt {3}\right ) b\right ) \sqrt {\frac {3 i+\sqrt {3}-\frac {6 i}{1+x}}{3 i+\sqrt {3}}} \sqrt {\frac {-3 i+\sqrt {3}+\frac {6 i}{1+x}}{-3 i+\sqrt {3}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {1+x}}\right )}{162 \sqrt {-\frac {i}{3 i+\sqrt {3}}} \sqrt {1-x+x^2}} \]
(2*x*(b*x*(8 + 5*x^3) + a*(10 + 7*x^3)))/(27*(1 + x)^(3/2)*(1 - x + x^2)^( 3/2)) + ((1 + x)^(3/2)*((-60*Sqrt[(-I)/(3*I + Sqrt[3])]*b*(1 - x + x^2))/( 1 + x)^2 + ((15*I)*Sqrt[2]*(I + Sqrt[3])*b*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[1 + x] + (Sqrt[2]*((14*I)*Sqrt[3]*a + 5*( 3 - I*Sqrt[3])*b)*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])]*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])
Time = 0.52 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {1210, 2394, 27, 2394, 27, 2417, 759, 2416}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {a+b x}{(x+1)^{5/2} \left (x^2-x+1\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1210 |
| \(\displaystyle \frac {\sqrt {x^3+1} \int \frac {a+b x}{\left (x^3+1\right )^{5/2}}dx}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {2 x (a+b x)}{9 \left (x^3+1\right )^{3/2}}-\frac {2}{9} \int -\frac {7 a+5 b x}{2 \left (x^3+1\right )^{3/2}}dx\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {1}{9} \int \frac {7 a+5 b x}{\left (x^3+1\right )^{3/2}}dx+\frac {2 x (a+b x)}{9 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {1}{9} \left (\frac {2 x (7 a+5 b x)}{3 \sqrt {x^3+1}}-\frac {2}{3} \int -\frac {7 a-5 b x}{2 \sqrt {x^3+1}}dx\right )+\frac {2 x (a+b x)}{9 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {1}{9} \left (\frac {1}{3} \int \frac {7 a-5 b x}{\sqrt {x^3+1}}dx+\frac {2 x (7 a+5 b x)}{3 \sqrt {x^3+1}}\right )+\frac {2 x (a+b x)}{9 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 2417 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {1}{9} \left (\frac {1}{3} \left (\left (7 a+5 \left (1-\sqrt {3}\right ) b\right ) \int \frac {1}{\sqrt {x^3+1}}dx-5 b \int \frac {x-\sqrt {3}+1}{\sqrt {x^3+1}}dx\right )+\frac {2 x (7 a+5 b x)}{3 \sqrt {x^3+1}}\right )+\frac {2 x (a+b x)}{9 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 759 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {1}{9} \left (\frac {1}{3} \left (\frac {2 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \left (7 a+5 \left (1-\sqrt {3}\right ) b\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-5 b \int \frac {x-\sqrt {3}+1}{\sqrt {x^3+1}}dx\right )+\frac {2 x (7 a+5 b x)}{3 \sqrt {x^3+1}}\right )+\frac {2 x (a+b x)}{9 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 2416 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {1}{9} \left (\frac {1}{3} \left (\frac {2 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \left (7 a+5 \left (1-\sqrt {3}\right ) b\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-5 b \left (\frac {2 \sqrt {x^3+1}}{x+\sqrt {3}+1}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}\right )\right )+\frac {2 x (7 a+5 b x)}{3 \sqrt {x^3+1}}\right )+\frac {2 x (a+b x)}{9 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
(Sqrt[1 + x^3]*((2*x*(a + b*x))/(9*(1 + x^3)^(3/2)) + ((2*x*(7*a + 5*b*x)) /(3*Sqrt[1 + x^3]) + (-5*b*((2*Sqrt[1 + x^3])/(1 + Sqrt[3] + x) - (3^(1/4) *Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*Ellipti cE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])) + (2*Sqrt[2 + Sqrt[3]]*(7*a + 5* (1 - Sqrt[3])*b)*(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^(1/4)*Sq rt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]))/3)/9))/(Sqrt[1 + x]*Sqrt[1 - x + x^2])
3.24.5.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]*Sqrt[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]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(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)^n*(a*d + c* e*x^3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && EqQ[m, p]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-x)*Pq*((a + b *x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[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]
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N umer[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] - S imp[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] && Eq Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N umer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(c*r - (1 - Sqrt[3])*d*s)/r Int[1/Sqrt[a + b*x^3], x], x] + Simp[d/r Int[((1 - Sqrt[3])*s + r*x)/Sq rt[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]
Time = 1.29 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.03
| method | result | size |
| elliptic | \(\frac {\sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (\frac {\frac {2}{9} b \,x^{2}+\frac {2}{9} a x}{\left (x^{3}+1\right )^{\frac {3}{2}}}-\frac {2 \left (-\frac {5}{27} b \,x^{2}-\frac {7}{27} a x \right )}{\sqrt {x^{3}+1}}+\frac {14 a \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{27 \sqrt {x^{3}+1}}-\frac {10 b \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) E\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{27 \sqrt {x^{3}+1}}\right )}{\sqrt {1+x}\, \sqrt {x^{2}-x +1}}\) | \(360\) |
| default | \(\text {Expression too large to display}\) | \(1152\) |
((1+x)*(x^2-x+1))^(1/2)/(1+x)^(1/2)/(x^2-x+1)^(1/2)*((2/9*b*x^2+2/9*a*x)/( x^3+1)^(3/2)-2*(-5/27*b*x^2-7/27*a*x)/(x^3+1)^(1/2)+14/27*a*(3/2-1/2*I*3^( 1/2))*((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2-1/2*I*3^(1/2))/(-3/2-1/2*I *3^(1/2)))^(1/2)*((x-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/(x^3+1 )^(1/2)*EllipticF(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1/2))/ (-3/2-1/2*I*3^(1/2)))^(1/2))-10/27*b*(3/2-1/2*I*3^(1/2))*((1+x)/(3/2-1/2*I *3^(1/2)))^(1/2)*((x-1/2-1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/ 2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/(x^3+1)^(1/2)*((-3/2-1/2*I*3^ (1/2))*EllipticE(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1/2))/( -3/2-1/2*I*3^(1/2)))^(1/2))+(1/2+1/2*I*3^(1/2))*EllipticF(((1+x)/(3/2-1/2* I*3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.27 \[ \int \frac {a+b x}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left (5 \, b x^{5} + 7 \, a x^{4} + 8 \, b x^{2} + 10 \, a x\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} + 7 \, {\left (a x^{6} + 2 \, a x^{3} + a\right )} {\rm weierstrassPInverse}\left (0, -4, x\right ) + 5 \, {\left (b x^{6} + 2 \, b x^{3} + b\right )} {\rm weierstrassZeta}\left (0, -4, {\rm weierstrassPInverse}\left (0, -4, x\right )\right )\right )}}{27 \, {\left (x^{6} + 2 \, x^{3} + 1\right )}} \]
2/27*((5*b*x^5 + 7*a*x^4 + 8*b*x^2 + 10*a*x)*sqrt(x^2 - x + 1)*sqrt(x + 1) + 7*(a*x^6 + 2*a*x^3 + a)*weierstrassPInverse(0, -4, x) + 5*(b*x^6 + 2*b* x^3 + b)*weierstrassZeta(0, -4, weierstrassPInverse(0, -4, x)))/(x^6 + 2*x ^3 + 1)
\[ \int \frac {a+b x}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {a + b x}{\left (x + 1\right )^{\frac {5}{2}} \left (x^{2} - x + 1\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {a+b x}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int { \frac {b x + a}{{\left (x^{2} - x + 1\right )}^{\frac {5}{2}} {\left (x + 1\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {a+b x}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int { \frac {b x + a}{{\left (x^{2} - x + 1\right )}^{\frac {5}{2}} {\left (x + 1\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b x}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {a+b\,x}{{\left (x+1\right )}^{5/2}\,{\left (x^2-x+1\right )}^{5/2}} \,d x \]