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]

-2/3*(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]  time = 0.29, 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 = {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 12

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

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 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.10, 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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fricas [B]  time = 0.52, size = 302, normalized size = 3.78 \[ -\frac {1}{8} \cdot 8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 4} {\left (\sqrt {2} - 2\right )} \log \left (2 \cdot 8^{\frac {1}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 4 \, x + 4 \, \sqrt {2} + 4\right ) + \frac {1}{8} \cdot 8^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 4} {\left (\sqrt {2} - 2\right )} \log \left (-2 \cdot 8^{\frac {1}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 4 \, x + 4 \, \sqrt {2} + 4\right ) - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, \sqrt {2} + 4} \arctan \left (\frac {1}{16} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {2 \cdot 8^{\frac {1}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 4 \, x + 4 \, \sqrt {2} + 4} \sqrt {2 \, \sqrt {2} + 4} - \frac {1}{8} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} - \sqrt {2} - 1\right ) - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, \sqrt {2} + 4} \arctan \left (\frac {1}{16} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {-2 \cdot 8^{\frac {1}{4}} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + 4 \, x + 4 \, \sqrt {2} + 4} \sqrt {2 \, \sqrt {2} + 4} - \frac {1}{8} \cdot 8^{\frac {3}{4}} \sqrt {2} \sqrt {x + 1} \sqrt {2 \, \sqrt {2} + 4} + \sqrt {2} + 1\right ) + \frac {2}{15} \, {\left (3 \, x^{2} + x - 17\right )} \sqrt {x + 1} \]

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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giac [B]  time = 1.71, size = 171, normalized size = 2.14 \[ \frac {2}{5} \, {\left (x + 1\right )}^{\frac {5}{2}} - \frac {2}{3} \, {\left (x + 1\right )}^{\frac {3}{2}} + \sqrt {\sqrt {2} + 1} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {x + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) + \sqrt {\sqrt {2} + 1} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {x + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{2} \, \sqrt {\sqrt {2} - 1} \log \left (2^{\frac {1}{4}} \sqrt {x + 1} \sqrt {\sqrt {2} + 2} + x + \sqrt {2} + 1\right ) - \frac {1}{2} \, \sqrt {\sqrt {2} - 1} \log \left (-2^{\frac {1}{4}} \sqrt {x + 1} \sqrt {\sqrt {2} + 2} + x + \sqrt {2} + 1\right ) - 2 \, \sqrt {x + 1} \]

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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maple [B]  time = 0.05, size = 443, normalized size = 5.54 \[ \frac {\left (2+2 \sqrt {2}\right ) \sqrt {2}\, \arctan \left (\frac {2 \sqrt {x +1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}-\frac {\left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {x +1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {2 \sqrt {2}\, \arctan \left (\frac {2 \sqrt {x +1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {\left (2+2 \sqrt {2}\right ) \sqrt {2}\, \arctan \left (\frac {2 \sqrt {x +1}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}-\frac {\left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {x +1}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {2 \sqrt {2}\, \arctan \left (\frac {2 \sqrt {x +1}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (x +1+\sqrt {2}-\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}\right )}{4}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1+\sqrt {2}-\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}\right )}{2}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (x +1+\sqrt {2}+\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}\right )}{4}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x +1+\sqrt {2}+\sqrt {x +1}\, \sqrt {2+2 \sqrt {2}}\right )}{2}+\frac {2 \left (x +1\right )^{\frac {5}{2}}}{5}-\frac {2 \left (x +1\right )^{\frac {3}{2}}}{3}-2 \sqrt {x +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x+1)^(5/2)-2/3*(x+1)^(3/2)-2*(x+1)^(1/2)-1/4*ln(1+x+2^(1/2)+(x+1)^(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)+(x+1)^(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*(x+1)^(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*(x+1)^(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*(x+1)^(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)-(x+1)^(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)-(x+1)^(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*(x+1)^(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*(x+1)^(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*(x+1)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{3} + 1\right )} \sqrt {x + 1}}{x^{2} + 1}\,{d x} \]

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

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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sympy [A]  time = 12.15, size = 56, normalized size = 0.70 \[ \frac {2 \left (x + 1\right )^{\frac {5}{2}}}{5} - \frac {2 \left (x + 1\right )^{\frac {3}{2}}}{3} - 2 \sqrt {x + 1} + 4 \operatorname {RootSum} {\left (512 t^{4} + 32 t^{2} + 1, \left (t \mapsto t \log {\left (- 128 t^{3} + \sqrt {x + 1} \right )} \right )\right )} \]

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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