3.8.26 \(\int \frac {\sqrt {1+x} (1+x^3)}{1+x^2} \, dx\) [726]

Optimal. Leaf size=80 \[ -2 \sqrt {1+x}-\frac {2}{3} (1+x)^{3/2}+\frac {2}{5} (1+x)^{5/2}+(1-i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {1-i}}\right )+(1+i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {1+i}}\right ) \]

[Out]

-2/3*(1+x)^(3/2)+2/5*(1+x)^(5/2)+(1-I)^(3/2)*arctanh((1+x)^(1/2)/(1-I)^(1/2))+(1+I)^(3/2)*arctanh((1+x)^(1/2)/
(1+I)^(1/2))-2*(1+x)^(1/2)

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(224\) vs. \(2(80)=160\).
time = 0.21, antiderivative size = 224, normalized size of antiderivative = 2.80, number of steps used = 16, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1639, 1643, 839, 12, 722, 1108, 648, 632, 210, 642} \begin {gather*} -\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {x+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {2 \sqrt {x+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\frac {2}{5} (x+1)^{5/2}-\frac {2}{3} (x+1)^{3/2}-2 \sqrt {x+1}-\frac {\log \left (x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{2 \sqrt {1+\sqrt {2}}}+\frac {\log \left (x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {x+1}+\sqrt {2}+1\right )}{2 \sqrt {1+\sqrt {2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*Sqrt[1 + x] - (2*(1 + x)^(3/2))/3 + (2*(1 + x)^(5/2))/5 - Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] -
2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]] + Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + x])/Sqr
t[2*(-1 + Sqrt[2])]] - Log[1 + Sqrt[2] + x - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]]/(2*Sqrt[1 + Sqrt[2]]) + Log[1
+ Sqrt[2] + x + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]]/(2*Sqrt[1 + Sqrt[2]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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

Rule 632

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 642

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]

Rule 648

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 722

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(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}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 839

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(
c*m)), x] + Dist[1/c, Int[(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d + c*e*f)*x, x]/(a + c*x^2)), x], x] /
; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[m] && GtQ[m, 0]

Rule 1108

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

Rule 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + 1)*Polynomial
Quotient[Pq, d + e*x, x]*(a + c*x^2)^p, x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[PolynomialRe
mainder[Pq, d + e*x, x], 0]

Rule 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {\sqrt {1+x} \left (1+x^3\right )}{1+x^2} \, dx &=\int \frac {(1+x)^{3/2} \left (1-x+x^2\right )}{1+x^2} \, dx\\ &=\int \left ((1+x)^{3/2}-\frac {x (1+x)^{3/2}}{1+x^2}\right ) \, dx\\ &=\frac {2}{5} (1+x)^{5/2}-\int \frac {x (1+x)^{3/2}}{1+x^2} \, dx\\ &=-\frac {2}{3} (1+x)^{3/2}+\frac {2}{5} (1+x)^{5/2}-\int \frac {(-1+x) \sqrt {1+x}}{1+x^2} \, dx\\ &=-2 \sqrt {1+x}-\frac {2}{3} (1+x)^{3/2}+\frac {2}{5} (1+x)^{5/2}-\int -\frac {2}{\sqrt {1+x} \left (1+x^2\right )} \, dx\\ &=-2 \sqrt {1+x}-\frac {2}{3} (1+x)^{3/2}+\frac {2}{5} (1+x)^{5/2}+2 \int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx\\ &=-2 \sqrt {1+x}-\frac {2}{3} (1+x)^{3/2}+\frac {2}{5} (1+x)^{5/2}+4 \text {Subst}\left (\int \frac {1}{2-2 x^2+x^4} \, dx,x,\sqrt {1+x}\right )\\ &=-2 \sqrt {1+x}-\frac {2}{3} (1+x)^{3/2}+\frac {2}{5} (1+x)^{5/2}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-x}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+x}\right )}{\sqrt {1+\sqrt {2}}}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}+x}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+x}\right )}{\sqrt {1+\sqrt {2}}}\\ &=-2 \sqrt {1+x}-\frac {2}{3} (1+x)^{3/2}+\frac {2}{5} (1+x)^{5/2}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+x}\right )}{\sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+x}\right )}{\sqrt {2}}-\frac {\text {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 )}{2 \sqrt {1+\sqrt {2}}}+\frac {\text {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 )}{2 \sqrt {1+\sqrt {2}}}\\ &=-2 \sqrt {1+x}-\frac {2}{3} (1+x)^{3/2}+\frac {2}{5} (1+x)^{5/2}-\frac {\log \left (1+\sqrt {2}+x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )}{2 \sqrt {1+\sqrt {2}}}+\frac {\log \left (1+\sqrt {2}+x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )}{2 \sqrt {1+\sqrt {2}}}-\sqrt {2} \text {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 )-\sqrt {2} \text {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 )\\ &=-2 \sqrt {1+x}-\frac {2}{3} (1+x)^{3/2}+\frac {2}{5} (1+x)^{5/2}-\frac {\tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+x}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{\sqrt {-1+\sqrt {2}}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+x}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{\sqrt {-1+\sqrt {2}}}-\frac {\log \left (1+\sqrt {2}+x-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )}{2 \sqrt {1+\sqrt {2}}}+\frac {\log \left (1+\sqrt {2}+x+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+x}\right )}{2 \sqrt {1+\sqrt {2}}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 76, normalized size = 0.95 \begin {gather*} \frac {2}{15} \sqrt {1+x} \left (-17+x+3 x^2\right )+\sqrt {2+2 i} \tan ^{-1}\left (\sqrt {-\frac {1}{2}-\frac {i}{2}} \sqrt {1+x}\right )+\sqrt {2-2 i} \tan ^{-1}\left (\sqrt {-\frac {1}{2}+\frac {i}{2}} \sqrt {1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 + x]*(-17 + x + 3*x^2))/15 + Sqrt[2 + 2*I]*ArcTan[Sqrt[-1/2 - I/2]*Sqrt[1 + x]] + Sqrt[2 - 2*I]*ArcT
an[Sqrt[-1/2 + I/2]*Sqrt[1 + x]]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(288\) vs. \(2(58)=116\).
time = 0.64, size = 289, normalized size = 3.61

method	result	size
derivativedivides	\(\frac {2 \left (1+x \right )^{\frac {5}{2}}}{5}-\frac {2 \left (1+x \right )^{\frac {3}{2}}}{3}-2 \sqrt {1+x}+\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \ln \left (1+x +\sqrt {2}+\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right )}{4}+\frac {\left (2 \sqrt {2}-\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \ln \left (1+x +\sqrt {2}-\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right )}{4}-\frac {\left (-2 \sqrt {2}+\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\)	\(289\)
default	\(\frac {2 \left (1+x \right )^{\frac {5}{2}}}{5}-\frac {2 \left (1+x \right )^{\frac {3}{2}}}{3}-2 \sqrt {1+x}+\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \ln \left (1+x +\sqrt {2}+\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right )}{4}+\frac {\left (2 \sqrt {2}-\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \ln \left (1+x +\sqrt {2}-\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right )}{4}-\frac {\left (-2 \sqrt {2}+\frac {\left (-\sqrt {2+2 \sqrt {2}}\, \sqrt {2}+2 \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\)	\(289\)
trager	\(\left (\frac {2}{5} x^{2}+\frac {2}{15} x -\frac {34}{15}\right ) \sqrt {1+x}+\RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2}+1\right ) \ln \left (\frac {-1024 \RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2}+1\right ) \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{4} x -192 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2} \RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2}+1\right ) x +192 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2} \sqrt {1+x}-160 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2} \RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2}+1\right )-9 \RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2}+1\right ) x -2 \sqrt {1+x}-15 \RootOf \left (\textit {\_Z}^{2}+16 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2}+1\right )}{32 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2} x +x -1}\right )+4 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2048 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{5} x -128 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{3} x +96 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2} \sqrt {1+x}-320 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{3}+2 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right ) x +7 \sqrt {1+x}+10 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )}{32 \RootOf \left (512 \textit {\_Z}^{4}+32 \textit {\_Z}^{2}+1\right )^{2} x +x +1}\right )\)	\(425\)
risch	\(\frac {2 \left (3 x^{2}+x -17\right ) \sqrt {1+x}}{15}+\frac {\ln \left (1+x +\sqrt {2}-\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}\, \sqrt {2}}{4}-\frac {\ln \left (1+x +\sqrt {2}-\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}+\frac {\arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \left (2+2 \sqrt {2}\right ) \sqrt {2}}{2 \sqrt {-2+2 \sqrt {2}}}-\frac {\arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \left (2+2 \sqrt {2}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {2 \arctan \left (\frac {2 \sqrt {1+x}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{\sqrt {-2+2 \sqrt {2}}}-\frac {\ln \left (1+x +\sqrt {2}+\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}\, \sqrt {2}}{4}+\frac {\ln \left (1+x +\sqrt {2}+\sqrt {1+x}\, \sqrt {2+2 \sqrt {2}}\right ) \sqrt {2+2 \sqrt {2}}}{2}+\frac {\arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \left (2+2 \sqrt {2}\right ) \sqrt {2}}{2 \sqrt {-2+2 \sqrt {2}}}-\frac {\arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \left (2+2 \sqrt {2}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {2 \arctan \left (\frac {2 \sqrt {1+x}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{\sqrt {-2+2 \sqrt {2}}}\)	\(437\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+1)*(1+x)^(1/2)/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

2/5*(1+x)^(5/2)-2/3*(1+x)^(3/2)-2*(1+x)^(1/2)+1/4*(-(2+2*2^(1/2))^(1/2)*2^(1/2)+2*(2+2*2^(1/2))^(1/2))*ln(1+x+
2^(1/2)+(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))+(2*2^(1/2)-1/2*(-(2+2*2^(1/2))^(1/2)*2^(1/2)+2*(2+2*2^(1/2))^(1/2))*(
2+2*2^(1/2))^(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))-1/4*
(-(2+2*2^(1/2))^(1/2)*2^(1/2)+2*(2+2*2^(1/2))^(1/2))*ln(1+x+2^(1/2)-(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))-(-2*2^(1/
2)+1/2*(-(2+2*2^(1/2))^(1/2)*2^(1/2)+2*(2+2*2^(1/2))^(1/2))*(2+2*2^(1/2))^(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 = 0.00, 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^3+1)*(1+x)^(1/2)/(x^2+1),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (50) = 100\).
time = 0.36, size = 302, normalized size = 3.78 \begin {gather*} -\frac {1}{8} \cdot 8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 4} {\left (\sqrt {2} - 2\right )} \log \left (2 \cdot 8^{\frac {1}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 4 \, x + 4 \, \sqrt {2} + 4\right ) + \frac {1}{8} \cdot 8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 4} {\left (\sqrt {2} - 2\right )} \log \left (-2 \cdot 8^{\frac {1}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 4 \, x + 4 \, \sqrt {2} + 4\right ) - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, \sqrt {2} + 4} \arctan \left (\frac {1}{16} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {2 \cdot 8^{\frac {1}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 4 \, x + 4 \, \sqrt {2} + 4} \sqrt {2 \, \sqrt {2} + 4} - \frac {1}{8} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} - \sqrt {2} - 1\right ) - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, \sqrt {2} + 4} \arctan \left (\frac {1}{16} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {-2 \cdot 8^{\frac {1}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 4 \, x + 4 \, \sqrt {2} + 4} \sqrt {2 \, \sqrt {2} + 4} - \frac {1}{8} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + \sqrt {2} + 1\right ) + \frac {2}{15} \, {\left (3 \, x^{2} + x - 17\right )} \sqrt {x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*8^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 2)*log(2*8^(1/4)*sqrt(x + 1)*sqrt(2*sqrt(2) + 4) + 4*x + 4*sqrt(2)
+ 4) + 1/8*8^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 2)*log(-2*8^(1/4)*sqrt(x + 1)*sqrt(2*sqrt(2) + 4) + 4*x + 4
*sqrt(2) + 4) - 1/2*8^(1/4)*sqrt(2)*sqrt(2*sqrt(2) + 4)*arctan(1/16*8^(3/4)*sqrt(2)*sqrt(2*8^(1/4)*sqrt(x + 1)
*sqrt(2*sqrt(2) + 4) + 4*x + 4*sqrt(2) + 4)*sqrt(2*sqrt(2) + 4) - 1/8*8^(3/4)*sqrt(2)*sqrt(x + 1)*sqrt(2*sqrt(
2) + 4) - sqrt(2) - 1) - 1/2*8^(1/4)*sqrt(2)*sqrt(2*sqrt(2) + 4)*arctan(1/16*8^(3/4)*sqrt(2)*sqrt(-2*8^(1/4)*s
qrt(x + 1)*sqrt(2*sqrt(2) + 4) + 4*x + 4*sqrt(2) + 4)*sqrt(2*sqrt(2) + 4) - 1/8*8^(3/4)*sqrt(2)*sqrt(x + 1)*sq
rt(2*sqrt(2) + 4) + sqrt(2) + 1) + 2/15*(3*x^2 + x - 17)*sqrt(x + 1)

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Sympy [A]
time = 5.72, size = 56, normalized size = 0.70 \begin {gather*} \frac {2 \left (x + 1\right )^{\frac {5}{2}}}{5} - \frac {2 \left (x + 1\right )^{\frac {3}{2}}}{3} - 2 \sqrt {x + 1} + 4 \operatorname {RootSum} {\left (512 t^{4} + 32 t^{2} + 1, \left ( t \mapsto t \log {\left (- 128 t^{3} + \sqrt {x + 1} \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(x + 1)**(5/2)/5 - 2*(x + 1)**(3/2)/3 - 2*sqrt(x + 1) + 4*RootSum(512*_t**4 + 32*_t**2 + 1, Lambda(_t, _t*lo
g(-128*_t**3 + sqrt(x + 1))))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (50) = 100\).
time = 2.18, size = 171, normalized size = 2.14 \begin {gather*} \frac {2}{5} \, {\left (x + 1\right )}^{\frac {5}{2}} - \frac {2}{3} \, {\left (x + 1\right )}^{\frac {3}{2}} + \sqrt {\sqrt {2} + 1} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {x + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) + \sqrt {\sqrt {2} + 1} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {x + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{2} \, \sqrt {\sqrt {2} - 1} \log \left (2^{\frac {1}{4}} \sqrt {x + 1} \sqrt {\sqrt {2} + 2} + x + \sqrt {2} + 1\right ) - \frac {1}{2} \, \sqrt {\sqrt {2} - 1} \log \left (-2^{\frac {1}{4}} \sqrt {x + 1} \sqrt {\sqrt {2} + 2} + x + \sqrt {2} + 1\right ) - 2 \, \sqrt {x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x + 1)^(5/2) - 2/3*(x + 1)^(3/2) + sqrt(sqrt(2) + 1)*arctan(1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) + 2*sq
rt(x + 1))/sqrt(-sqrt(2) + 2)) + sqrt(sqrt(2) + 1)*arctan(-1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) - 2*sqrt(x +
1))/sqrt(-sqrt(2) + 2)) + 1/2*sqrt(sqrt(2) - 1)*log(2^(1/4)*sqrt(x + 1)*sqrt(sqrt(2) + 2) + x + sqrt(2) + 1)
- 1/2*sqrt(sqrt(2) - 1)*log(-2^(1/4)*sqrt(x + 1)*sqrt(sqrt(2) + 2) + x + sqrt(2) + 1) - 2*sqrt(x + 1)

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Mupad [B]
time = 0.10, size = 255, normalized size = 3.19 \begin {gather*} \frac {2\,{\left (x+1\right )}^{5/2}}{5}-\frac {2\,{\left (x+1\right )}^{3/2}}{3}-2\,\sqrt {x+1}-\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-\frac {\sqrt {2}}{4}-\frac {1}{4}}\,\sqrt {x+1}\,64{}\mathrm {i}}{256\,\sqrt {\frac {\sqrt {2}}{4}-\frac {1}{4}}\,\sqrt {-\frac {\sqrt {2}}{4}-\frac {1}{4}}-64}-\frac {\sqrt {2}\,\sqrt {\frac {\sqrt {2}}{4}-\frac {1}{4}}\,\sqrt {x+1}\,64{}\mathrm {i}}{256\,\sqrt {\frac {\sqrt {2}}{4}-\frac {1}{4}}\,\sqrt {-\frac {\sqrt {2}}{4}-\frac {1}{4}}-64}\right )\,\left (\sqrt {-\frac {\sqrt {2}}{4}-\frac {1}{4}}\,2{}\mathrm {i}+\sqrt {\frac {\sqrt {2}}{4}-\frac {1}{4}}\,2{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-\frac {\sqrt {2}}{4}-\frac {1}{4}}\,\sqrt {x+1}\,64{}\mathrm {i}}{256\,\sqrt {\frac {\sqrt {2}}{4}-\frac {1}{4}}\,\sqrt {-\frac {\sqrt {2}}{4}-\frac {1}{4}}+64}+\frac {\sqrt {2}\,\sqrt {\frac {\sqrt {2}}{4}-\frac {1}{4}}\,\sqrt {x+1}\,64{}\mathrm {i}}{256\,\sqrt {\frac {\sqrt {2}}{4}-\frac {1}{4}}\,\sqrt {-\frac {\sqrt {2}}{4}-\frac {1}{4}}+64}\right )\,\left (\sqrt {-\frac {\sqrt {2}}{4}-\frac {1}{4}}\,2{}\mathrm {i}-\sqrt {\frac {\sqrt {2}}{4}-\frac {1}{4}}\,2{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(x + 1)^(5/2))/5 - (2*(x + 1)^(3/2))/3 - 2*(x + 1)^(1/2) - atan((2^(1/2)*(- 2^(1/2)/4 - 1/4)^(1/2)*(x + 1)^
(1/2)*64i)/(256*(2^(1/2)/4 - 1/4)^(1/2)*(- 2^(1/2)/4 - 1/4)^(1/2) - 64) - (2^(1/2)*(2^(1/2)/4 - 1/4)^(1/2)*(x
+ 1)^(1/2)*64i)/(256*(2^(1/2)/4 - 1/4)^(1/2)*(- 2^(1/2)/4 - 1/4)^(1/2) - 64))*((- 2^(1/2)/4 - 1/4)^(1/2)*2i +
(2^(1/2)/4 - 1/4)^(1/2)*2i) + atan((2^(1/2)*(- 2^(1/2)/4 - 1/4)^(1/2)*(x + 1)^(1/2)*64i)/(256*(2^(1/2)/4 - 1/4
)^(1/2)*(- 2^(1/2)/4 - 1/4)^(1/2) + 64) + (2^(1/2)*(2^(1/2)/4 - 1/4)^(1/2)*(x + 1)^(1/2)*64i)/(256*(2^(1/2)/4
- 1/4)^(1/2)*(- 2^(1/2)/4 - 1/4)^(1/2) + 64))*((- 2^(1/2)/4 - 1/4)^(1/2)*2i - (2^(1/2)/4 - 1/4)^(1/2)*2i)

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