### 3.2312 $$\int \frac{\sqrt{1+2 x}}{2+3 x+5 x^2} \, dx$$

Optimal. Leaf size=222 $\frac{\log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}-\frac{\log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}-\sqrt{\frac{2}{5 \left (\sqrt{35}-2\right )}} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\sqrt{\frac{2}{5 \left (\sqrt{35}-2\right )}} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

[Out]

-(Sqrt[2/(5*(-2 + Sqrt[35]))]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]]) +
Sqrt[2/(5*(-2 + Sqrt[35]))]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] + L
og[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/Sqrt[10*(2 + Sqrt[35])] - Log[Sqrt[35] + Sq
rt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/Sqrt[10*(2 + Sqrt[35])]

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Rubi [A]  time = 0.247047, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.318, Rules used = {699, 1127, 1161, 618, 204, 1164, 628} $\frac{\log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}-\frac{\log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}-\sqrt{\frac{2}{5 \left (\sqrt{35}-2\right )}} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\sqrt{\frac{2}{5 \left (\sqrt{35}-2\right )}} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[1 + 2*x]/(2 + 3*x + 5*x^2),x]

[Out]

-(Sqrt[2/(5*(-2 + Sqrt[35]))]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]]) +
Sqrt[2/(5*(-2 + Sqrt[35]))]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] + L
og[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/Sqrt[10*(2 + Sqrt[35])] - Log[Sqrt[35] + Sq
rt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/Sqrt[10*(2 + Sqrt[35])]

Rule 699

Int[Sqrt[(d_.) + (e_.)*(x_)]/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(c*d^2
- b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*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] && NeQ[2*c*d - b*e, 0]

Rule 1127

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

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 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 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 1164

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

Rubi steps

\begin{align*} \int \frac{\sqrt{1+2 x}}{2+3 x+5 x^2} \, dx &=4 \operatorname{Subst}\left (\int \frac{x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{\sqrt{\frac{7}{5}}-x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\right )+2 \operatorname{Subst}\left (\int \frac{\sqrt{\frac{7}{5}}+x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=\frac{1}{5} \operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{7}{5}}-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )+\frac{1}{5} \operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{7}{5}}+\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 x}{-\sqrt{\frac{7}{5}}-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x-x^2} \, dx,x,\sqrt{1+2 x}\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-2 x}{-\sqrt{\frac{7}{5}}+\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x-x^2} \, dx,x,\sqrt{1+2 x}\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}\\ &=\frac{\log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}-\frac{\log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}-\frac{2}{5} \operatorname{Subst}\left (\int \frac{1}{\frac{2}{5} \left (2-\sqrt{35}\right )-x^2} \, dx,x,-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 \sqrt{1+2 x}\right )-\frac{2}{5} \operatorname{Subst}\left (\int \frac{1}{\frac{2}{5} \left (2-\sqrt{35}\right )-x^2} \, dx,x,\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 \sqrt{1+2 x}\right )\\ &=-\sqrt{\frac{2}{5 \left (-2+\sqrt{35}\right )}} \tan ^{-1}\left (\sqrt{\frac{5}{2 \left (-2+\sqrt{35}\right )}} \left (\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-2 \sqrt{1+2 x}\right )\right )+\sqrt{\frac{2}{5 \left (-2+\sqrt{35}\right )}} \tan ^{-1}\left (\sqrt{\frac{5}{2 \left (-2+\sqrt{35}\right )}} \left (\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 \sqrt{1+2 x}\right )\right )+\frac{\log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}-\frac{\log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{\sqrt{10 \left (2+\sqrt{35}\right )}}\\ \end{align*}

Mathematica [C]  time = 0.202769, size = 112, normalized size = 0.5 $\frac{2 \left (\sqrt{-2+i \sqrt{31}} \left (\sqrt{31}-2 i\right ) \tan ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{-2-i \sqrt{31}}}\right )+\sqrt{-2-i \sqrt{31}} \left (\sqrt{31}+2 i\right ) \tan ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{-2+i \sqrt{31}}}\right )\right )}{5 \sqrt{217}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[1 + 2*x]/(2 + 3*x + 5*x^2),x]

[Out]

(2*(Sqrt[-2 + I*Sqrt[31]]*(-2*I + Sqrt[31])*ArcTan[Sqrt[5 + 10*x]/Sqrt[-2 - I*Sqrt[31]]] + Sqrt[-2 - I*Sqrt[31
]]*(2*I + Sqrt[31])*ArcTan[Sqrt[5 + 10*x]/Sqrt[-2 + I*Sqrt[31]]]))/(5*Sqrt[217])

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Maple [B]  time = 0.087, size = 486, normalized size = 2.2 \begin{align*}{\frac{\sqrt{5}\sqrt{2\,\sqrt{5}\sqrt{7}+4}}{155}\ln \left ( \sqrt{5}\sqrt{7}+10\,x+5+\sqrt{2\,\sqrt{5}\sqrt{7}+4}\sqrt{5}\sqrt{1+2\,x} \right ) }-{\frac{4\,\sqrt{5}\sqrt{7}+8}{31\,\sqrt{10\,\sqrt{5}\sqrt{7}-20}}\arctan \left ({\frac{1}{\sqrt{10\,\sqrt{5}\sqrt{7}-20}} \left ( 10\,\sqrt{1+2\,x}+\sqrt{5}\sqrt{2\,\sqrt{5}\sqrt{7}+4} \right ) } \right ) }-{\frac{\sqrt{7}\sqrt{2\,\sqrt{5}\sqrt{7}+4}}{62}\ln \left ( \sqrt{5}\sqrt{7}+10\,x+5+\sqrt{2\,\sqrt{5}\sqrt{7}+4}\sqrt{5}\sqrt{1+2\,x} \right ) }+{\frac{\sqrt{5} \left ( 2\,\sqrt{5}\sqrt{7}+4 \right ) \sqrt{7}}{31\,\sqrt{10\,\sqrt{5}\sqrt{7}-20}}\arctan \left ({\frac{1}{\sqrt{10\,\sqrt{5}\sqrt{7}-20}} \left ( 10\,\sqrt{1+2\,x}+\sqrt{5}\sqrt{2\,\sqrt{5}\sqrt{7}+4} \right ) } \right ) }-{\frac{\sqrt{5}\sqrt{2\,\sqrt{5}\sqrt{7}+4}}{155}\ln \left ( -\sqrt{2\,\sqrt{5}\sqrt{7}+4}\sqrt{5}\sqrt{1+2\,x}+\sqrt{5}\sqrt{7}+10\,x+5 \right ) }-{\frac{4\,\sqrt{5}\sqrt{7}+8}{31\,\sqrt{10\,\sqrt{5}\sqrt{7}-20}}\arctan \left ({\frac{1}{\sqrt{10\,\sqrt{5}\sqrt{7}-20}} \left ( -\sqrt{5}\sqrt{2\,\sqrt{5}\sqrt{7}+4}+10\,\sqrt{1+2\,x} \right ) } \right ) }+{\frac{\sqrt{7}\sqrt{2\,\sqrt{5}\sqrt{7}+4}}{62}\ln \left ( -\sqrt{2\,\sqrt{5}\sqrt{7}+4}\sqrt{5}\sqrt{1+2\,x}+\sqrt{5}\sqrt{7}+10\,x+5 \right ) }+{\frac{\sqrt{5} \left ( 2\,\sqrt{5}\sqrt{7}+4 \right ) \sqrt{7}}{31\,\sqrt{10\,\sqrt{5}\sqrt{7}-20}}\arctan \left ({\frac{1}{\sqrt{10\,\sqrt{5}\sqrt{7}-20}} \left ( -\sqrt{5}\sqrt{2\,\sqrt{5}\sqrt{7}+4}+10\,\sqrt{1+2\,x} \right ) } \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*x)^(1/2)/(5*x^2+3*x+2),x)

[Out]

1/155*ln(5^(1/2)*7^(1/2)+10*x+5+(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(1+2*x)^(1/2))*5^(1/2)*(2*5^(1/2)*7^(1/2)+
4)^(1/2)-2/31/(10*5^(1/2)*7^(1/2)-20)^(1/2)*arctan((10*(1+2*x)^(1/2)+5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2))/(10*
5^(1/2)*7^(1/2)-20)^(1/2))*(2*5^(1/2)*7^(1/2)+4)-1/62*ln(5^(1/2)*7^(1/2)+10*x+5+(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^
(1/2)*(1+2*x)^(1/2))*7^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+1/31/(10*5^(1/2)*7^(1/2)-20)^(1/2)*arctan((10*(1+2*x)
^(1/2)+5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))*5^(1/2)*(2*5^(1/2)*7^(1/2)+4)*7^(1/
2)-1/155*ln(-(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(1+2*x)^(1/2)+5^(1/2)*7^(1/2)+10*x+5)*5^(1/2)*(2*5^(1/2)*7^(1
/2)+4)^(1/2)-2/31/(10*5^(1/2)*7^(1/2)-20)^(1/2)*arctan((-5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+10*(1+2*x)^(1/2))
/(10*5^(1/2)*7^(1/2)-20)^(1/2))*(2*5^(1/2)*7^(1/2)+4)+1/62*ln(-(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(1+2*x)^(1/
2)+5^(1/2)*7^(1/2)+10*x+5)*7^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+1/31/(10*5^(1/2)*7^(1/2)-20)^(1/2)*arctan((-5^(
1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+10*(1+2*x)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))*5^(1/2)*(2*5^(1/2)*7^(1/2)+4
)*7^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 \, x + 1}}{5 \, x^{2} + 3 \, x + 2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^(1/2)/(5*x^2+3*x+2),x, algorithm="maxima")

[Out]

integrate(sqrt(2*x + 1)/(5*x^2 + 3*x + 2), x)

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Fricas [B]  time = 2.65576, size = 1442, normalized size = 6.5 \begin{align*} \frac{1}{336350} \, \sqrt{155} 35^{\frac{1}{4}}{\left (2 \, \sqrt{35} \sqrt{31} - 35 \, \sqrt{31}\right )} \sqrt{4 \, \sqrt{35} + 70} \log \left (\frac{124}{7} \, \sqrt{155} 35^{\frac{3}{4}} \sqrt{31} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + 192200 \, x + 19220 \, \sqrt{35} + 96100\right ) - \frac{1}{336350} \, \sqrt{155} 35^{\frac{1}{4}}{\left (2 \, \sqrt{35} \sqrt{31} - 35 \, \sqrt{31}\right )} \sqrt{4 \, \sqrt{35} + 70} \log \left (-\frac{124}{7} \, \sqrt{155} 35^{\frac{3}{4}} \sqrt{31} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + 192200 \, x + 19220 \, \sqrt{35} + 96100\right ) - \frac{2}{5425} \, \sqrt{155} 35^{\frac{3}{4}} \sqrt{4 \, \sqrt{35} + 70} \arctan \left (\frac{1}{1177225} \, \sqrt{155} 35^{\frac{3}{4}} \sqrt{31} \sqrt{7} \sqrt{\sqrt{155} 35^{\frac{3}{4}} \sqrt{31} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + 10850 \, x + 1085 \, \sqrt{35} + 5425} \sqrt{4 \, \sqrt{35} + 70} - \frac{1}{1085} \, \sqrt{155} 35^{\frac{3}{4}} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} - \frac{1}{31} \, \sqrt{35} \sqrt{31} - \frac{2}{31} \, \sqrt{31}\right ) - \frac{2}{5425} \, \sqrt{155} 35^{\frac{3}{4}} \sqrt{4 \, \sqrt{35} + 70} \arctan \left (\frac{1}{2354450} \, \sqrt{155} 35^{\frac{3}{4}} \sqrt{7} \sqrt{-124 \, \sqrt{155} 35^{\frac{3}{4}} \sqrt{31} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + 1345400 \, x + 134540 \, \sqrt{35} + 672700} \sqrt{4 \, \sqrt{35} + 70} - \frac{1}{1085} \, \sqrt{155} 35^{\frac{3}{4}} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + \frac{1}{31} \, \sqrt{35} \sqrt{31} + \frac{2}{31} \, \sqrt{31}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^(1/2)/(5*x^2+3*x+2),x, algorithm="fricas")

[Out]

3/4)*sqrt(31)*sqrt(2*x + 1)*sqrt(4*sqrt(35) + 70) + 192200*x + 19220*sqrt(35) + 96100) - 1/336350*sqrt(155)*35
*x + 1)*sqrt(4*sqrt(35) + 70) + 192200*x + 19220*sqrt(35) + 96100) - 2/5425*sqrt(155)*35^(3/4)*sqrt(4*sqrt(35)
+ 70)*arctan(1/1177225*sqrt(155)*35^(3/4)*sqrt(31)*sqrt(7)*sqrt(sqrt(155)*35^(3/4)*sqrt(31)*sqrt(2*x + 1)*sqr
t(4*sqrt(35) + 70) + 10850*x + 1085*sqrt(35) + 5425)*sqrt(4*sqrt(35) + 70) - 1/1085*sqrt(155)*35^(3/4)*sqrt(2*
x + 1)*sqrt(4*sqrt(35) + 70) - 1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) - 2/5425*sqrt(155)*35^(3/4)*sqrt(4*sqrt
(35) + 70)*arctan(1/2354450*sqrt(155)*35^(3/4)*sqrt(7)*sqrt(-124*sqrt(155)*35^(3/4)*sqrt(31)*sqrt(2*x + 1)*sqr
t(4*sqrt(35) + 70) + 1345400*x + 134540*sqrt(35) + 672700)*sqrt(4*sqrt(35) + 70) - 1/1085*sqrt(155)*35^(3/4)*s
qrt(2*x + 1)*sqrt(4*sqrt(35) + 70) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31))

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Sympy [A]  time = 2.5134, size = 32, normalized size = 0.14 \begin{align*} 4 \operatorname{RootSum}{\left (1230080 t^{4} + 1984 t^{2} + 7, \left ( t \mapsto t \log{\left (9920 t^{3} + 8 t + \sqrt{2 x + 1} \right )} \right )\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)**(1/2)/(5*x**2+3*x+2),x)

[Out]

4*RootSum(1230080*_t**4 + 1984*_t**2 + 7, Lambda(_t, _t*log(9920*_t**3 + 8*_t + sqrt(2*x + 1))))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 \, x + 1}}{5 \, x^{2} + 3 \, x + 2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^(1/2)/(5*x^2+3*x+2),x, algorithm="giac")

[Out]

integrate(sqrt(2*x + 1)/(5*x^2 + 3*x + 2), x)