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

Optimal. Leaf size=272 \[ -\frac{\sqrt{x-1} (1-11 x)}{32 \left (x^2+1\right )}+\frac{\sqrt{x-1} x}{4 \left (x^2+1\right )^2}-\frac{1}{128} \sqrt{\frac{1}{2} \left (527+373 \sqrt{2}\right )} \log \left (-x-\sqrt{2 \left (\sqrt{2}-1\right )} \sqrt{x-1}-\sqrt{2}+1\right )+\frac{1}{128} \sqrt{\frac{1}{2} \left (527+373 \sqrt{2}\right )} \log \left (-x+\sqrt{2 \left (\sqrt{2}-1\right )} \sqrt{x-1}-\sqrt{2}+1\right )-\frac{1}{64} \sqrt{\frac{1}{2} \left (373 \sqrt{2}-527\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{2}-1\right )}-2 \sqrt{x-1}}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right )+\frac{1}{64} \sqrt{\frac{1}{2} \left (373 \sqrt{2}-527\right )} \tan ^{-1}\left (\frac{2 \sqrt{x-1}+\sqrt{2 \left (\sqrt{2}-1\right )}}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right ) \]

[Out]

(Sqrt[-1 + x]*x)/(4*(1 + x^2)^2) - ((1 - 11*x)*Sqrt[-1 + x])/(32*(1 + x^2)) - (Sqrt[(-527 + 373*Sqrt[2])/2]*Ar
cTan[(Sqrt[2*(-1 + Sqrt[2])] - 2*Sqrt[-1 + x])/Sqrt[2*(1 + Sqrt[2])]])/64 + (Sqrt[(-527 + 373*Sqrt[2])/2]*ArcT
an[(Sqrt[2*(-1 + Sqrt[2])] + 2*Sqrt[-1 + x])/Sqrt[2*(1 + Sqrt[2])]])/64 - (Sqrt[(527 + 373*Sqrt[2])/2]*Log[1 -
 Sqrt[2] - Sqrt[2*(-1 + Sqrt[2])]*Sqrt[-1 + x] - x])/128 + (Sqrt[(527 + 373*Sqrt[2])/2]*Log[1 - Sqrt[2] + Sqrt
[2*(-1 + Sqrt[2])]*Sqrt[-1 + x] - x])/128

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Rubi [A]  time = 0.364043, antiderivative size = 272, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {737, 823, 827, 1169, 634, 618, 204, 628} \[ -\frac{\sqrt{x-1} (1-11 x)}{32 \left (x^2+1\right )}+\frac{\sqrt{x-1} x}{4 \left (x^2+1\right )^2}-\frac{1}{128} \sqrt{\frac{1}{2} \left (527+373 \sqrt{2}\right )} \log \left (-x-\sqrt{2 \left (\sqrt{2}-1\right )} \sqrt{x-1}-\sqrt{2}+1\right )+\frac{1}{128} \sqrt{\frac{1}{2} \left (527+373 \sqrt{2}\right )} \log \left (-x+\sqrt{2 \left (\sqrt{2}-1\right )} \sqrt{x-1}-\sqrt{2}+1\right )-\frac{1}{64} \sqrt{\frac{1}{2} \left (373 \sqrt{2}-527\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{2}-1\right )}-2 \sqrt{x-1}}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right )+\frac{1}{64} \sqrt{\frac{1}{2} \left (373 \sqrt{2}-527\right )} \tan ^{-1}\left (\frac{2 \sqrt{x-1}+\sqrt{2 \left (\sqrt{2}-1\right )}}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + x]/(1 + x^2)^3,x]

[Out]

(Sqrt[-1 + x]*x)/(4*(1 + x^2)^2) - ((1 - 11*x)*Sqrt[-1 + x])/(32*(1 + x^2)) - (Sqrt[(-527 + 373*Sqrt[2])/2]*Ar
cTan[(Sqrt[2*(-1 + Sqrt[2])] - 2*Sqrt[-1 + x])/Sqrt[2*(1 + Sqrt[2])]])/64 + (Sqrt[(-527 + 373*Sqrt[2])/2]*ArcT
an[(Sqrt[2*(-1 + Sqrt[2])] + 2*Sqrt[-1 + x])/Sqrt[2*(1 + Sqrt[2])]])/64 - (Sqrt[(527 + 373*Sqrt[2])/2]*Log[1 -
 Sqrt[2] - Sqrt[2*(-1 + Sqrt[2])]*Sqrt[-1 + x] - x])/128 + (Sqrt[(527 + 373*Sqrt[2])/2]*Log[1 - Sqrt[2] + Sqrt
[2*(-1 + Sqrt[2])]*Sqrt[-1 + x] - x])/128

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(d*(2*p + 3) + e*(m + 2*p + 3)*x)*(a + c*x^2
)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (LtQ[m, 1]
|| (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
 && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
 - d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
 && NeQ[c*d^2 + a*e^2, 0]

Rule 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
 b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{-1+x}}{\left (1+x^2\right )^3} \, dx &=\frac{\sqrt{-1+x} x}{4 \left (1+x^2\right )^2}-\frac{1}{4} \int \frac{3-\frac{5 x}{2}}{\sqrt{-1+x} \left (1+x^2\right )^2} \, dx\\ &=\frac{\sqrt{-1+x} x}{4 \left (1+x^2\right )^2}-\frac{(1-11 x) \sqrt{-1+x}}{32 \left (1+x^2\right )}+\frac{1}{16} \int \frac{-\frac{25}{4}+\frac{11 x}{4}}{\sqrt{-1+x} \left (1+x^2\right )} \, dx\\ &=\frac{\sqrt{-1+x} x}{4 \left (1+x^2\right )^2}-\frac{(1-11 x) \sqrt{-1+x}}{32 \left (1+x^2\right )}+\frac{1}{8} \operatorname{Subst}\left (\int \frac{-\frac{7}{2}+\frac{11 x^2}{4}}{2+2 x^2+x^4} \, dx,x,\sqrt{-1+x}\right )\\ &=\frac{\sqrt{-1+x} x}{4 \left (1+x^2\right )^2}-\frac{(1-11 x) \sqrt{-1+x}}{32 \left (1+x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-7 \sqrt{\frac{1}{2} \left (-1+\sqrt{2}\right )}-\left (-\frac{7}{2}-\frac{11}{2 \sqrt{2}}\right ) x}{\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{-1+x}\right )}{32 \sqrt{-1+\sqrt{2}}}+\frac{\operatorname{Subst}\left (\int \frac{-7 \sqrt{\frac{1}{2} \left (-1+\sqrt{2}\right )}+\left (-\frac{7}{2}-\frac{11}{2 \sqrt{2}}\right ) x}{\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{-1+x}\right )}{32 \sqrt{-1+\sqrt{2}}}\\ &=\frac{\sqrt{-1+x} x}{4 \left (1+x^2\right )^2}-\frac{(1-11 x) \sqrt{-1+x}}{32 \left (1+x^2\right )}+\frac{1}{128} \sqrt{219-154 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{-1+x}\right )+\frac{1}{128} \sqrt{219-154 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{-1+x}\right )+\frac{\left (14+11 \sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{-\sqrt{2 \left (-1+\sqrt{2}\right )}+2 x}{\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{-1+x}\right )}{256 \sqrt{-1+\sqrt{2}}}-\frac{\left (14+11 \sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2 \left (-1+\sqrt{2}\right )}+2 x}{\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{-1+x}\right )}{256 \sqrt{-1+\sqrt{2}}}\\ &=\frac{\sqrt{-1+x} x}{4 \left (1+x^2\right )^2}-\frac{(1-11 x) \sqrt{-1+x}}{32 \left (1+x^2\right )}-\frac{1}{256} \sqrt{1054+746 \sqrt{2}} \log \left (1-\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} \sqrt{-1+x}-x\right )+\frac{1}{256} \sqrt{1054+746 \sqrt{2}} \log \left (1-\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} \sqrt{-1+x}-x\right )-\frac{1}{64} \sqrt{219-154 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{-2 \left (1+\sqrt{2}\right )-x^2} \, dx,x,-\sqrt{2 \left (-1+\sqrt{2}\right )}+2 \sqrt{-1+x}\right )-\frac{1}{64} \sqrt{219-154 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{-2 \left (1+\sqrt{2}\right )-x^2} \, dx,x,\sqrt{2 \left (-1+\sqrt{2}\right )}+2 \sqrt{-1+x}\right )\\ &=\frac{\sqrt{-1+x} x}{4 \left (1+x^2\right )^2}-\frac{(1-11 x) \sqrt{-1+x}}{32 \left (1+x^2\right )}-\frac{1}{64} \sqrt{\frac{1}{2} \left (-527+373 \sqrt{2}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (-1+\sqrt{2}\right )}-2 \sqrt{-1+x}}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right )+\frac{1}{64} \sqrt{\frac{1}{2} \left (-527+373 \sqrt{2}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (-1+\sqrt{2}\right )}+2 \sqrt{-1+x}}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right )-\frac{1}{256} \sqrt{1054+746 \sqrt{2}} \log \left (1-\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} \sqrt{-1+x}-x\right )+\frac{1}{256} \sqrt{1054+746 \sqrt{2}} \log \left (1-\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} \sqrt{-1+x}-x\right )\\ \end{align*}

Mathematica [C]  time = 0.0941321, size = 90, normalized size = 0.33 \[ \frac{1}{64} \left (\frac{2 \sqrt{x-1} \left (11 x^3-x^2+19 x-1\right )}{\left (x^2+1\right )^2}-(7-18 i) \sqrt{1-i} \tan ^{-1}\left (\frac{\sqrt{x-1}}{\sqrt{1-i}}\right )-(7+18 i) \sqrt{1+i} \tan ^{-1}\left (\frac{\sqrt{x-1}}{\sqrt{1+i}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + x]/(1 + x^2)^3,x]

[Out]

((2*Sqrt[-1 + x]*(-1 + 19*x - x^2 + 11*x^3))/(1 + x^2)^2 - (7 - 18*I)*Sqrt[1 - I]*ArcTan[Sqrt[-1 + x]/Sqrt[1 -
 I]] - (7 + 18*I)*Sqrt[1 + I]*ArcTan[Sqrt[-1 + x]/Sqrt[1 + I]])/64

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Maple [B]  time = 0.365, size = 639, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(-4/23*(-759-506*2^(1/2))/(-6-4*2^(1/2))*(-1+x)^(3/2)-1/23/(-6-4*2^(1/2))*(-5336-3588*2^(1/2))*(-2+2*2^
(1/2))^(1/2)*(-1+x)-2/23*(-2392*2^(1/2)-3036)/(-6-4*2^(1/2))*(-1+x)^(1/2)-1/46*(-3312*2^(1/2)-4416)*(-2+2*2^(1
/2))^(1/2)/(-6-4*2^(1/2)))/(-1+x+(-1+x)^(1/2)*(-2+2*2^(1/2))^(1/2)+2^(1/2))^2-13/32/(3+2*2^(1/2))*ln(-1+x+(-1+
x)^(1/2)*(-2+2*2^(1/2))^(1/2)+2^(1/2))*2^(1/2)*(-2+2*2^(1/2))^(1/2)-147/256/(3+2*2^(1/2))*ln(-1+x+(-1+x)^(1/2)
*(-2+2*2^(1/2))^(1/2)+2^(1/2))*(-2+2*2^(1/2))^(1/2)+1/64/(3+2*2^(1/2))/(2+2*2^(1/2))^(1/2)*arctan((2*(-1+x)^(1
/2)+(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))*2^(1/2)+5/64/(3+2*2^(1/2))/(2+2*2^(1/2))^(1/2)*arctan((2*(-1+x)
^(1/2)+(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))+1/128*(4/23*(-759-506*2^(1/2))/(-6-4*2^(1/2))*(-1+x)^(3/2)-1
/23/(-6-4*2^(1/2))*(-5336-3588*2^(1/2))*(-2+2*2^(1/2))^(1/2)*(-1+x)+2/23*(-2392*2^(1/2)-3036)/(-6-4*2^(1/2))*(
-1+x)^(1/2)-1/46*(-3312*2^(1/2)-4416)*(-2+2*2^(1/2))^(1/2)/(-6-4*2^(1/2)))/(-1+x-(-1+x)^(1/2)*(-2+2*2^(1/2))^(
1/2)+2^(1/2))^2+13/32/(3+2*2^(1/2))*ln(-1+x-(-1+x)^(1/2)*(-2+2*2^(1/2))^(1/2)+2^(1/2))*2^(1/2)*(-2+2*2^(1/2))^
(1/2)+147/256/(3+2*2^(1/2))*ln(-1+x-(-1+x)^(1/2)*(-2+2*2^(1/2))^(1/2)+2^(1/2))*(-2+2*2^(1/2))^(1/2)+1/64/(3+2*
2^(1/2))/(2+2*2^(1/2))^(1/2)*arctan((2*(-1+x)^(1/2)-(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))*2^(1/2)+5/64/(3
+2*2^(1/2))/(2+2*2^(1/2))^(1/2)*arctan((2*(-1+x)^(1/2)-(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x - 1)/(x^2 + 1)^3, x)

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Fricas [B]  time = 2.42514, size = 1778, normalized size = 6.54 \begin{align*} -\frac{92 \cdot 278258^{\frac{1}{4}} \sqrt{2}{\left (x^{4} + 2 \, x^{2} + 1\right )} \sqrt{-393142 \, \sqrt{2} + 556516} \arctan \left (\frac{1}{109810067572} \cdot 278258^{\frac{3}{4}} \sqrt{46} \sqrt{373 \cdot 278258^{\frac{1}{4}} \sqrt{x - 1}{\left (11 \, \sqrt{2} + 14\right )} \sqrt{-393142 \, \sqrt{2} + 556516} + 6399934 \, x + 6399934 \, \sqrt{2} - 6399934}{\left (7 \, \sqrt{2} + 11\right )} \sqrt{-393142 \, \sqrt{2} + 556516} - \frac{1}{6399934} \cdot 278258^{\frac{3}{4}} \sqrt{x - 1}{\left (7 \, \sqrt{2} + 11\right )} \sqrt{-393142 \, \sqrt{2} + 556516} - \sqrt{2} + 1\right ) + 92 \cdot 278258^{\frac{1}{4}} \sqrt{2}{\left (x^{4} + 2 \, x^{2} + 1\right )} \sqrt{-393142 \, \sqrt{2} + 556516} \arctan \left (\frac{1}{109810067572} \cdot 278258^{\frac{3}{4}} \sqrt{46} \sqrt{-373 \cdot 278258^{\frac{1}{4}} \sqrt{x - 1}{\left (11 \, \sqrt{2} + 14\right )} \sqrt{-393142 \, \sqrt{2} + 556516} + 6399934 \, x + 6399934 \, \sqrt{2} - 6399934}{\left (7 \, \sqrt{2} + 11\right )} \sqrt{-393142 \, \sqrt{2} + 556516} - \frac{1}{6399934} \cdot 278258^{\frac{3}{4}} \sqrt{x - 1}{\left (7 \, \sqrt{2} + 11\right )} \sqrt{-393142 \, \sqrt{2} + 556516} + \sqrt{2} - 1\right ) + 278258^{\frac{1}{4}}{\left (746 \, x^{4} + 1492 \, x^{2} + 527 \, \sqrt{2}{\left (x^{4} + 2 \, x^{2} + 1\right )} + 746\right )} \sqrt{-393142 \, \sqrt{2} + 556516} \log \left (\frac{373}{46} \cdot 278258^{\frac{1}{4}} \sqrt{x - 1}{\left (11 \, \sqrt{2} + 14\right )} \sqrt{-393142 \, \sqrt{2} + 556516} + 139129 \, x + 139129 \, \sqrt{2} - 139129\right ) - 278258^{\frac{1}{4}}{\left (746 \, x^{4} + 1492 \, x^{2} + 527 \, \sqrt{2}{\left (x^{4} + 2 \, x^{2} + 1\right )} + 746\right )} \sqrt{-393142 \, \sqrt{2} + 556516} \log \left (-\frac{373}{46} \cdot 278258^{\frac{1}{4}} \sqrt{x - 1}{\left (11 \, \sqrt{2} + 14\right )} \sqrt{-393142 \, \sqrt{2} + 556516} + 139129 \, x + 139129 \, \sqrt{2} - 139129\right ) - 137264 \,{\left (11 \, x^{3} - x^{2} + 19 \, x - 1\right )} \sqrt{x - 1}}{4392448 \,{\left (x^{4} + 2 \, x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4392448*(92*278258^(1/4)*sqrt(2)*(x^4 + 2*x^2 + 1)*sqrt(-393142*sqrt(2) + 556516)*arctan(1/109810067572*278
258^(3/4)*sqrt(46)*sqrt(373*278258^(1/4)*sqrt(x - 1)*(11*sqrt(2) + 14)*sqrt(-393142*sqrt(2) + 556516) + 639993
4*x + 6399934*sqrt(2) - 6399934)*(7*sqrt(2) + 11)*sqrt(-393142*sqrt(2) + 556516) - 1/6399934*278258^(3/4)*sqrt
(x - 1)*(7*sqrt(2) + 11)*sqrt(-393142*sqrt(2) + 556516) - sqrt(2) + 1) + 92*278258^(1/4)*sqrt(2)*(x^4 + 2*x^2
+ 1)*sqrt(-393142*sqrt(2) + 556516)*arctan(1/109810067572*278258^(3/4)*sqrt(46)*sqrt(-373*278258^(1/4)*sqrt(x
- 1)*(11*sqrt(2) + 14)*sqrt(-393142*sqrt(2) + 556516) + 6399934*x + 6399934*sqrt(2) - 6399934)*(7*sqrt(2) + 11
)*sqrt(-393142*sqrt(2) + 556516) - 1/6399934*278258^(3/4)*sqrt(x - 1)*(7*sqrt(2) + 11)*sqrt(-393142*sqrt(2) +
556516) + sqrt(2) - 1) + 278258^(1/4)*(746*x^4 + 1492*x^2 + 527*sqrt(2)*(x^4 + 2*x^2 + 1) + 746)*sqrt(-393142*
sqrt(2) + 556516)*log(373/46*278258^(1/4)*sqrt(x - 1)*(11*sqrt(2) + 14)*sqrt(-393142*sqrt(2) + 556516) + 13912
9*x + 139129*sqrt(2) - 139129) - 278258^(1/4)*(746*x^4 + 1492*x^2 + 527*sqrt(2)*(x^4 + 2*x^2 + 1) + 746)*sqrt(
-393142*sqrt(2) + 556516)*log(-373/46*278258^(1/4)*sqrt(x - 1)*(11*sqrt(2) + 14)*sqrt(-393142*sqrt(2) + 556516
) + 139129*x + 139129*sqrt(2) - 139129) - 137264*(11*x^3 - x^2 + 19*x - 1)*sqrt(x - 1))/(x^4 + 2*x^2 + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x - 1)/(x^2 + 1)^3, x)