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

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

[Out]

-2*Sqrt[1 + x] - (2*(1 + x)^(3/2))/3 + (2*(1 + x)^(5/2))/5 + (1 - I)^(3/2)*ArcTanh[Sqrt[1 + x]/Sqrt[1 - I]] +
(1 + I)^(3/2)*ArcTanh[Sqrt[1 + x]/Sqrt[1 + I]]

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Rubi [B]  time = 0.285622, antiderivative size = 224, normalized size of antiderivative = 2.8, number of steps used = 16, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {1625, 1629, 825, 12, 708, 1094, 634, 618, 204, 628} \[ \frac{2}{5} (x+1)^{5/2}-\frac{2}{3} (x+1)^{3/2}-2 \sqrt{x+1}-\frac{\log \left (x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}}}+\frac{\log \left (x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}}}-\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{x+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )+\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{x+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right ) \]

Warning: Unable to verify antiderivative.

[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 1625

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 1629

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]

Rule 825

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 12

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

Rule 708

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 1094

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 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^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 \operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\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 )}{2 \sqrt{1+\sqrt{2}}}+\frac{\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 )}{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} \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 )-\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 )\\ &=-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*}

Mathematica [A]  time = 0.103947, size = 68, normalized size = 0.85 \[ \frac{2}{15} \sqrt{x+1} \left (3 x^2+x-17\right )+(1-i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{1-i}}\right )+(1+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{1+i}}\right ) \]

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 + (1 - I)^(3/2)*ArcTanh[Sqrt[1 + x]/Sqrt[1 - I]] + (1 + I)^(3/2)*ArcTanh[
Sqrt[1 + x]/Sqrt[1 + I]]

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Maple [B]  time = 0.033, size = 443, normalized size = 5.5 \begin{align*}{\frac{2}{5} \left ( 1+x \right ) ^{{\frac{5}{2}}}}-{\frac{2}{3} \left ( 1+x \right ) ^{{\frac{3}{2}}}}-2\,\sqrt{1+x}-{\frac{\sqrt{2+2\,\sqrt{2}}\sqrt{2}}{4}\ln \left ( 1+x+\sqrt{2}+\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }+{\frac{\sqrt{2+2\,\sqrt{2}}}{2}\ln \left ( 1+x+\sqrt{2}+\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }+{\frac{ \left ( 2+2\,\sqrt{2} \right ) \sqrt{2}}{2\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}+\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }-{\frac{2+2\,\sqrt{2}}{\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}+\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }+2\,{\frac{\sqrt{2}}{\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,\sqrt{1+x}+\sqrt{2+2\,\sqrt{2}}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) }+{\frac{\sqrt{2+2\,\sqrt{2}}\sqrt{2}}{4}\ln \left ( 1+x+\sqrt{2}-\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }-{\frac{\sqrt{2+2\,\sqrt{2}}}{2}\ln \left ( 1+x+\sqrt{2}-\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }+{\frac{ \left ( 2+2\,\sqrt{2} \right ) \sqrt{2}}{2\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}-\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }-{\frac{2+2\,\sqrt{2}}{\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}-\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }+2\,{\frac{\sqrt{2}}{\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,\sqrt{1+x}-\sqrt{2+2\,\sqrt{2}}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(1+x)^(5/2)-2/3*(1+x)^(3/2)-2*(1+x)^(1/2)-1/4*ln(1+x+2^(1/2)+(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))*(2+2*2^(1/2)
)^(1/2)*2^(1/2)+1/2*ln(1+x+2^(1/2)+(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))*(2+2*2^(1/2))^(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))*(2+2*2^(1/2))*2^(1/2)-1/(-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+2*2^(1/2))+2/(-2+2*2^(1/2))^(1/2)*arcta
n((2*(1+x)^(1/2)+(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)+1/4*ln(1+x+2^(1/2)-(1+x)^(1/2)*(2+2*2^(1/2
))^(1/2))*(2+2*2^(1/2))^(1/2)*2^(1/2)-1/2*ln(1+x+2^(1/2)-(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))*(2+2*2^(1/2))^(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))*(2+2*2^(1/2))*2^(1/2
)-1/(-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+2*2^(1/2))+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)

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

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]  time = 1.60087, size = 986, normalized size = 12.32 \begin{align*} -\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{align*}

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 = 7.36695, size = 56, normalized size = 0.7 \begin{align*} \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{align*}

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

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]

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