### 3.2319 $$\int \frac{\sqrt{1+2 x}}{(2+3 x+5 x^2)^2} \, dx$$

Optimal. Leaf size=270 $\frac{\sqrt{2 x+1} (10 x+3)}{31 \left (5 x^2+3 x+2\right )}+\frac{1}{31} \sqrt{\frac{1}{434} \left (47 \sqrt{35}-218\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{31} \sqrt{\frac{1}{434} \left (47 \sqrt{35}-218\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

[Out]

(Sqrt[1 + 2*x]*(3 + 10*x))/(31*(2 + 3*x + 5*x^2)) - (Sqrt[(2*(218 + 47*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sq
rt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/31 + (Sqrt[(2*(218 + 47*Sqrt[35]))/217]*ArcTan[(Sqrt[1
0*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/31 + (Sqrt[(-218 + 47*Sqrt[35])/434]*Log[Sqrt
[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/31 - (Sqrt[(-218 + 47*Sqrt[35])/434]*Log[Sqrt[35]
+ Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/31

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Rubi [A]  time = 0.34043, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.318, Rules used = {736, 826, 1169, 634, 618, 204, 628} $\frac{\sqrt{2 x+1} (10 x+3)}{31 \left (5 x^2+3 x+2\right )}+\frac{1}{31} \sqrt{\frac{1}{434} \left (47 \sqrt{35}-218\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{31} \sqrt{\frac{1}{434} \left (47 \sqrt{35}-218\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[1 + 2*x]*(3 + 10*x))/(31*(2 + 3*x + 5*x^2)) - (Sqrt[(2*(218 + 47*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sq
rt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/31 + (Sqrt[(2*(218 + 47*Sqrt[35]))/217]*ArcTan[(Sqrt[1
0*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/31 + (Sqrt[(-218 + 47*Sqrt[35])/434]*Log[Sqrt
[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/31 - (Sqrt[(-218 + 47*Sqrt[35])/434]*Log[Sqrt[35]
+ Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/31

Rule 736

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*(b + 2*
c*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m
- 1)*(b*e*m + 2*c*d*(2*p + 3) + 2*c*e*(m + 2*p + 3)*x)*(a + b*x + c*x^2)^(p + 1), 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] && LtQ[p, -1] && GtQ[m
, 0] && (LtQ[m, 1] || (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*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, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + 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+2 x}}{\left (2+3 x+5 x^2\right )^2} \, dx &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}-\frac{1}{31} \int \frac{-7-10 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}-\frac{2}{31} \operatorname{Subst}\left (\int \frac{-4-10 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{-4 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (-4+2 \sqrt{35}\right ) x}{\sqrt{\frac{7}{5}}-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{31 \sqrt{14 \left (2+\sqrt{35}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{-4 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (-4+2 \sqrt{35}\right ) x}{\sqrt{\frac{7}{5}}+\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{31 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}+\frac{\left (35+2 \sqrt{35}\right ) \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 )}{1085}+\frac{\left (35+2 \sqrt{35}\right ) \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 )}{1085}+\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \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 )-\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \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 )\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}+\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )-\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )-\frac{\left (2 \left (35+2 \sqrt{35}\right )\right ) \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 )}{1085}-\frac{\left (2 \left (35+2 \sqrt{35}\right )\right ) \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 )}{1085}\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}-\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \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{1}{31} \sqrt{\frac{2}{217} \left (218+47 \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{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )-\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )\\ \end{align*}

Mathematica [C]  time = 0.369282, size = 145, normalized size = 0.54 $\frac{1}{31} \left (\frac{\sqrt{2 x+1} (10 x+3)}{5 x^2+3 x+2}+\frac{2 \sqrt{10-5 i \sqrt{31}} \left (62-39 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+2 \sqrt{10+5 i \sqrt{31}} \left (62+39 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )}{1085}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((Sqrt[1 + 2*x]*(3 + 10*x))/(2 + 3*x + 5*x^2) + (2*Sqrt[10 - (5*I)*Sqrt[31]]*(62 - (39*I)*Sqrt[31])*ArcTanh[Sq
rt[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]] + 2*Sqrt[10 + (5*I)*Sqrt[31]]*(62 + (39*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]
/Sqrt[2 + I*Sqrt[31]]])/1085)/31

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Maple [B]  time = 0.243, size = 972, normalized size = 3.6 \begin{align*} \text{result too large to display} \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)^2,x)

[Out]

5/6727*(2/25*(-5425*7^(1/2)+2170*5^(1/2))/(2*5^(1/2)-5*7^(1/2))*(1+2*x)^(1/2)+1/25*7^(1/2)*(2*5^(1/2)*7^(1/2)+
4)^(1/2)*(-1085*5^(1/2)+310*7^(1/2))/(2*5^(1/2)-5*7^(1/2)))/(1/5*5^(1/2)*7^(1/2)+2*x+1+1/5*(2*5^(1/2)*7^(1/2)+
4)^(1/2)*5^(1/2)*(1+2*x)^(1/2))-109/6727/(2*5^(1/2)-5*7^(1/2))*ln(5+10*x+35^(1/2)+(1+2*x)^(1/2)*(20+10*35^(1/2
))^(1/2))*(2*35^(1/2)+4)^(1/2)*35^(1/2)+235/1922/(2*5^(1/2)-5*7^(1/2))*ln(5+10*x+35^(1/2)+(1+2*x)^(1/2)*(20+10
*35^(1/2))^(1/2))*(2*35^(1/2)+4)^(1/2)+218/6727/(2*5^(1/2)-5*7^(1/2))/(-20+10*35^(1/2))^(1/2)*arctan((10*(1+2*
x)^(1/2)+(20+10*35^(1/2))^(1/2))/(-20+10*35^(1/2))^(1/2))*(20+10*35^(1/2))^(1/2)*(2*35^(1/2)+4)^(1/2)*35^(1/2)
-235/961/(2*5^(1/2)-5*7^(1/2))/(-20+10*35^(1/2))^(1/2)*arctan((10*(1+2*x)^(1/2)+(20+10*35^(1/2))^(1/2))/(-20+1
0*35^(1/2))^(1/2))*(20+10*35^(1/2))^(1/2)*(2*35^(1/2)+4)^(1/2)-40/31/(2*5^(1/2)-5*7^(1/2))/(-20+10*35^(1/2))^(
1/2)*arctan((10*(1+2*x)^(1/2)+(20+10*35^(1/2))^(1/2))/(-20+10*35^(1/2))^(1/2))*5^(1/2)+80/217/(2*5^(1/2)-5*7^(
1/2))/(-20+10*35^(1/2))^(1/2)*arctan((10*(1+2*x)^(1/2)+(20+10*35^(1/2))^(1/2))/(-20+10*35^(1/2))^(1/2))*7^(1/2
)-5/6727*(-2/25*(-5425*7^(1/2)+2170*5^(1/2))/(2*5^(1/2)-5*7^(1/2))*(1+2*x)^(1/2)+1/25*7^(1/2)*(2*5^(1/2)*7^(1/
2)+4)^(1/2)*(-1085*5^(1/2)+310*7^(1/2))/(2*5^(1/2)-5*7^(1/2)))/(1/5*5^(1/2)*7^(1/2)+2*x+1-1/5*(2*5^(1/2)*7^(1/
2)+4)^(1/2)*5^(1/2)*(1+2*x)^(1/2))+109/6727/(2*5^(1/2)-5*7^(1/2))*ln(5+10*x+35^(1/2)-(1+2*x)^(1/2)*(20+10*35^(
1/2))^(1/2))*(2*35^(1/2)+4)^(1/2)*35^(1/2)-235/1922/(2*5^(1/2)-5*7^(1/2))*ln(5+10*x+35^(1/2)-(1+2*x)^(1/2)*(20
+10*35^(1/2))^(1/2))*(2*35^(1/2)+4)^(1/2)+218/6727/(2*5^(1/2)-5*7^(1/2))/(-20+10*35^(1/2))^(1/2)*arctan((-(20+
10*35^(1/2))^(1/2)+10*(1+2*x)^(1/2))/(-20+10*35^(1/2))^(1/2))*(20+10*35^(1/2))^(1/2)*(2*35^(1/2)+4)^(1/2)*35^(
1/2)-235/961/(2*5^(1/2)-5*7^(1/2))/(-20+10*35^(1/2))^(1/2)*arctan((-(20+10*35^(1/2))^(1/2)+10*(1+2*x)^(1/2))/(
-20+10*35^(1/2))^(1/2))*(20+10*35^(1/2))^(1/2)*(2*35^(1/2)+4)^(1/2)-40/31/(2*5^(1/2)-5*7^(1/2))/(-20+10*35^(1/
2))^(1/2)*arctan((-(20+10*35^(1/2))^(1/2)+10*(1+2*x)^(1/2))/(-20+10*35^(1/2))^(1/2))*5^(1/2)+80/217/(2*5^(1/2)
-5*7^(1/2))/(-20+10*35^(1/2))^(1/2)*arctan((-(20+10*35^(1/2))^(1/2)+10*(1+2*x)^(1/2))/(-20+10*35^(1/2))^(1/2))
*7^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 \, x + 1}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.09446, size = 2176, normalized size = 8.06 \begin{align*} \text{result too large to display} \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)^2,x, algorithm="fricas")

[Out]

1/21268688630*(3844*77315^(1/4)*sqrt(217)*sqrt(35)*(5*x^2 + 3*x + 2)*sqrt(20492*sqrt(35) + 154630)*arctan(1/26
522397764975*77315^(3/4)*sqrt(1645)*sqrt(217)*sqrt(77315^(1/4)*sqrt(217)*(sqrt(35)*sqrt(31) - 2*sqrt(31))*sqrt
(2*x + 1)*sqrt(20492*sqrt(35) + 154630) + 3161690*x + 316169*sqrt(35) + 1580845)*sqrt(20492*sqrt(35) + 154630)
*(2*sqrt(35) - 35) - 1/520098005*77315^(3/4)*sqrt(217)*sqrt(2*x + 1)*sqrt(20492*sqrt(35) + 154630)*(2*sqrt(35)
- 35) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 3844*77315^(1/4)*sqrt(217)*sqrt(35)*(5*x^2 + 3*x + 2)*sqrt(
20492*sqrt(35) + 154630)*arctan(1/53044795529950*77315^(3/4)*sqrt(217)*sqrt(-6580*77315^(1/4)*sqrt(217)*(sqrt(
35)*sqrt(31) - 2*sqrt(31))*sqrt(2*x + 1)*sqrt(20492*sqrt(35) + 154630) + 20803920200*x + 2080392020*sqrt(35) +
10401960100)*sqrt(20492*sqrt(35) + 154630)*(2*sqrt(35) - 35) - 1/520098005*77315^(3/4)*sqrt(217)*sqrt(2*x + 1
)*sqrt(20492*sqrt(35) + 154630)*(2*sqrt(35) - 35) - 1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 77315^(1/4)*sqrt
(217)*(218*sqrt(35)*sqrt(31)*(5*x^2 + 3*x + 2) - 1645*sqrt(31)*(5*x^2 + 3*x + 2))*sqrt(20492*sqrt(35) + 154630
)*log(6580*77315^(1/4)*sqrt(217)*(sqrt(35)*sqrt(31) - 2*sqrt(31))*sqrt(2*x + 1)*sqrt(20492*sqrt(35) + 154630)
+ 20803920200*x + 2080392020*sqrt(35) + 10401960100) - 77315^(1/4)*sqrt(217)*(218*sqrt(35)*sqrt(31)*(5*x^2 + 3
*x + 2) - 1645*sqrt(31)*(5*x^2 + 3*x + 2))*sqrt(20492*sqrt(35) + 154630)*log(-6580*77315^(1/4)*sqrt(217)*(sqrt
(35)*sqrt(31) - 2*sqrt(31))*sqrt(2*x + 1)*sqrt(20492*sqrt(35) + 154630) + 20803920200*x + 2080392020*sqrt(35)
+ 10401960100) + 686086730*(10*x + 3)*sqrt(2*x + 1))/(5*x^2 + 3*x + 2)

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Sympy [A]  time = 5.22867, size = 83, normalized size = 0.31 \begin{align*} \frac{80 \left (2 x + 1\right )^{\frac{3}{2}}}{- 992 x + 620 \left (2 x + 1\right )^{2} + 372} - \frac{32 \sqrt{2 x + 1}}{- 992 x + 620 \left (2 x + 1\right )^{2} + 372} + 16 \operatorname{RootSum}{\left (407144088666112 t^{4} + 3325152256 t^{2} + 11045, \left ( t \mapsto t \log{\left (\frac{33312534528 t^{3}}{235} + \frac{166784 t}{235} + \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)**2,x)

[Out]

80*(2*x + 1)**(3/2)/(-992*x + 620*(2*x + 1)**2 + 372) - 32*sqrt(2*x + 1)/(-992*x + 620*(2*x + 1)**2 + 372) + 1
6*RootSum(407144088666112*_t**4 + 3325152256*_t**2 + 11045, Lambda(_t, _t*log(33312534528*_t**3/235 + 166784*_
t/235 + 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}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{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)^2,x, algorithm="giac")

[Out]

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