3.2328 $$\int \frac{1}{\sqrt{1+2 x} (2+3 x+5 x^2)^3} \, dx$$

Optimal. Leaf size=314 $\frac{\sqrt{2 x+1} (20 x+37)}{434 \left (5 x^2+3 x+2\right )^2}+\frac{\sqrt{2 x+1} (7920 x+9227)}{94178 \left (5 x^2+3 x+2\right )}-\frac{3 \sqrt{\frac{1}{434} \left (64681225 \sqrt{35}-250141922\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{94178}+\frac{3 \sqrt{\frac{1}{434} \left (64681225 \sqrt{35}-250141922\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{94178}-\frac{3 \sqrt{\frac{1}{434} \left (2+\sqrt{35}\right )} \left (7379+264 \sqrt{35}\right ) \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{47089}+\frac{3 \sqrt{\frac{1}{434} \left (2+\sqrt{35}\right )} \left (7379+264 \sqrt{35}\right ) \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{47089}$

[Out]

(Sqrt[1 + 2*x]*(37 + 20*x))/(434*(2 + 3*x + 5*x^2)^2) + (Sqrt[1 + 2*x]*(9227 + 7920*x))/(94178*(2 + 3*x + 5*x^
2)) - (3*Sqrt[(2 + Sqrt[35])/434]*(7379 + 264*Sqrt[35])*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sq
rt[10*(-2 + Sqrt[35])]])/47089 + (3*Sqrt[(2 + Sqrt[35])/434]*(7379 + 264*Sqrt[35])*ArcTan[(Sqrt[10*(2 + Sqrt[3
5])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/47089 - (3*Sqrt[(-250141922 + 64681225*Sqrt[35])/434]*Log[
Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/94178 + (3*Sqrt[(-250141922 + 64681225*Sqrt[3
5])/434]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/94178

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Rubi [A]  time = 0.386157, antiderivative size = 314, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.364, Rules used = {740, 822, 826, 1169, 634, 618, 204, 628} $\frac{\sqrt{2 x+1} (20 x+37)}{434 \left (5 x^2+3 x+2\right )^2}+\frac{\sqrt{2 x+1} (7920 x+9227)}{94178 \left (5 x^2+3 x+2\right )}-\frac{3 \sqrt{\frac{1}{434} \left (64681225 \sqrt{35}-250141922\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{94178}+\frac{3 \sqrt{\frac{1}{434} \left (64681225 \sqrt{35}-250141922\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{94178}-\frac{3 \sqrt{\frac{1}{434} \left (2+\sqrt{35}\right )} \left (7379+264 \sqrt{35}\right ) \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{47089}+\frac{3 \sqrt{\frac{1}{434} \left (2+\sqrt{35}\right )} \left (7379+264 \sqrt{35}\right ) \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{47089}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[1 + 2*x]*(37 + 20*x))/(434*(2 + 3*x + 5*x^2)^2) + (Sqrt[1 + 2*x]*(9227 + 7920*x))/(94178*(2 + 3*x + 5*x^
2)) - (3*Sqrt[(2 + Sqrt[35])/434]*(7379 + 264*Sqrt[35])*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sq
rt[10*(-2 + Sqrt[35])]])/47089 + (3*Sqrt[(2 + Sqrt[35])/434]*(7379 + 264*Sqrt[35])*ArcTan[(Sqrt[10*(2 + Sqrt[3
5])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/47089 - (3*Sqrt[(-250141922 + 64681225*Sqrt[35])/434]*Log[
Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/94178 + (3*Sqrt[(-250141922 + 64681225*Sqrt[3
5])/434]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/94178

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

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{1}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )^3} \, dx &=\frac{\sqrt{1+2 x} (37+20 x)}{434 \left (2+3 x+5 x^2\right )^2}+\frac{1}{434} \int \frac{271+100 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )^2} \, dx\\ &=\frac{\sqrt{1+2 x} (37+20 x)}{434 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (9227+7920 x)}{94178 \left (2+3 x+5 x^2\right )}+\frac{\int \frac{26097+7920 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx}{94178}\\ &=\frac{\sqrt{1+2 x} (37+20 x)}{434 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (9227+7920 x)}{94178 \left (2+3 x+5 x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{44274+7920 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )}{47089}\\ &=\frac{\sqrt{1+2 x} (37+20 x)}{434 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (9227+7920 x)}{94178 \left (2+3 x+5 x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{44274 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (44274-1584 \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 )}{94178 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{44274 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (44274-1584 \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 )}{94178 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=\frac{\sqrt{1+2 x} (37+20 x)}{434 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (9227+7920 x)}{94178 \left (2+3 x+5 x^2\right )}+\frac{\left (3 \left (9240+7379 \sqrt{35}\right )\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 )}{3296230}+\frac{\left (3 \left (9240+7379 \sqrt{35}\right )\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 )}{3296230}-\frac{\left (3 \sqrt{\frac{1}{434} \left (-250141922+64681225 \sqrt{35}\right )}\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 )}{94178}+\frac{\left (3 \sqrt{\frac{1}{434} \left (-250141922+64681225 \sqrt{35}\right )}\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 )}{94178}\\ &=\frac{\sqrt{1+2 x} (37+20 x)}{434 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (9227+7920 x)}{94178 \left (2+3 x+5 x^2\right )}-\frac{3 \sqrt{\frac{1}{434} \left (-250141922+64681225 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{94178}+\frac{3 \sqrt{\frac{1}{434} \left (-250141922+64681225 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{94178}-\frac{\left (3 \left (9240+7379 \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 )}{1648115}-\frac{\left (3 \left (9240+7379 \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 )}{1648115}\\ &=\frac{\sqrt{1+2 x} (37+20 x)}{434 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (9227+7920 x)}{94178 \left (2+3 x+5 x^2\right )}-\frac{3 \sqrt{\frac{1}{434} \left (250141922+64681225 \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 )}{47089}+\frac{3 \sqrt{\frac{1}{434} \left (250141922+64681225 \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 )}{47089}-\frac{3 \sqrt{\frac{1}{434} \left (-250141922+64681225 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{94178}+\frac{3 \sqrt{\frac{1}{434} \left (-250141922+64681225 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{94178}\\ \end{align*}

Mathematica [C]  time = 0.651586, size = 151, normalized size = 0.48 $\frac{\frac{1085 \sqrt{2 x+1} \left (39600 x^3+69895 x^2+47861 x+26483\right )}{\left (5 x^2+3 x+2\right )^2}+6 \sqrt{10-5 i \sqrt{31}} \left (228749-23998 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+6 \sqrt{10+5 i \sqrt{31}} \left (228749+23998 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )}{102183130}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((1085*Sqrt[1 + 2*x]*(26483 + 47861*x + 69895*x^2 + 39600*x^3))/(2 + 3*x + 5*x^2)^2 + 6*Sqrt[10 - (5*I)*Sqrt[3
1]]*(228749 - (23998*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]] + 6*Sqrt[10 + (5*I)*Sqrt[31]]*(
228749 + (23998*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]])/102183130

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Maple [B]  time = 0.444, size = 1100, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/20436626*(6/1353025*(-6045943503600+620096769600*5^(1/2)*7^(1/2))/(-390+40*5^(1/2)*7^(1/2))*(1+2*x)^(3/2)+1/
6765125/(-390+40*5^(1/2)*7^(1/2))*(-91048526818200*5^(1/2)+65791327714000*7^(1/2))*(2*5^(1/2)*7^(1/2)+4)^(1/2)
*(1+2*x)+2/6765125*(-59423591568600*5^(1/2)*7^(1/2)+320925328420550)/(-390+40*5^(1/2)*7^(1/2))*(1+2*x)^(1/2)+1
/13530250*(-123371070933600*7^(1/2)+152992435939000*5^(1/2))*(2*5^(1/2)*7^(1/2)+4)^(1/2)/(-390+40*5^(1/2)*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))^2-6065475/5839036/(20*
5^(1/2)*7^(1/2)-195)*ln(5+10*x+35^(1/2)+(1+2*x)^(1/2)*(20+10*35^(1/2))^(1/2))*5^(1/2)*(2*35^(1/2)+4)^(1/2)+153
21765/20436626/(20*5^(1/2)*7^(1/2)-195)*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)*7^(1/2)+6065475/2919518/(20*5^(1/2)*7^(1/2)-195)/(-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)*5^(1/2)*(2*35^(1/2)+4)^(1/2)-15321765/1021
8313/(20*5^(1/2)*7^(1/2)-195)/(-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)*7^(1/2)-8633430/329623/(20*5^(1/2)*7^(1/2)-195)/
(-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))*35^(1/2)+442
7400/47089/(20*5^(1/2)*7^(1/2)-195)/(-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))-5/20436626*(-6/1353025*(-6045943503600+620096769600*5^(1/2)*7^(1/2))/(-390+40*5^(1/2)*
7^(1/2))*(1+2*x)^(3/2)+1/6765125/(-390+40*5^(1/2)*7^(1/2))*(-91048526818200*5^(1/2)+65791327714000*7^(1/2))*(2
*5^(1/2)*7^(1/2)+4)^(1/2)*(1+2*x)-2/6765125*(-59423591568600*5^(1/2)*7^(1/2)+320925328420550)/(-390+40*5^(1/2)
*7^(1/2))*(1+2*x)^(1/2)+1/13530250*(-123371070933600*7^(1/2)+152992435939000*5^(1/2))*(2*5^(1/2)*7^(1/2)+4)^(1
/2)/(-390+40*5^(1/2)*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
))^2+6065475/5839036/(20*5^(1/2)*7^(1/2)-195)*ln(5+10*x+35^(1/2)-(1+2*x)^(1/2)*(20+10*35^(1/2))^(1/2))*5^(1/2)
*(2*35^(1/2)+4)^(1/2)-15321765/20436626/(20*5^(1/2)*7^(1/2)-195)*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)*7^(1/2)+6065475/2919518/(20*5^(1/2)*7^(1/2)-195)/(-20+10*35^(1/2))^(1/2)*arct
an((-(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)*5^(1/2)*(2*35^(1
/2)+4)^(1/2)-15321765/10218313/(20*5^(1/2)*7^(1/2)-195)/(-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)*7^(1/2)-8633430/32962
3/(20*5^(1/2)*7^(1/2)-195)/(-20+10*35^(1/2))^(1/2)*arctan((-(20+10*35^(1/2))^(1/2)+10*(1+2*x)^(1/2))/(-20+10*3
5^(1/2))^(1/2))*35^(1/2)+4427400/47089/(20*5^(1/2)*7^(1/2)-195)/(-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))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.83957, size = 3406, normalized size = 10.85 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/121786816718039847492020*(19347824532*97578096035^(1/4)*sqrt(105602)*sqrt(217)*sqrt(35)*(25*x^4 + 30*x^3 +
29*x^2 + 12*x + 4)*sqrt(250141922*sqrt(35) + 2263842875)*arctan(1/15121769925583791519919258475683975*97578096
035^(3/4)*sqrt(1677751)*sqrt(105602)*sqrt(37715)*sqrt(217)*sqrt(97578096035^(1/4)*sqrt(105602)*sqrt(217)*(264*
sqrt(35)*sqrt(31) - 7379*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875) + 5959242818165770*x +
595924281816577*sqrt(35) + 2979621409082885)*sqrt(250141922*sqrt(35) + 2263842875)*(7379*sqrt(35) - 9240) - 1/
1101288930146897876195*97578096035^(3/4)*sqrt(105602)*sqrt(217)*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 226384
2875)*(7379*sqrt(35) - 9240) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 19347824532*97578096035^(1/4)*sqrt(10
5602)*sqrt(217)*sqrt(35)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(250141922*sqrt(35) + 2263842875)*arctan(1/
389007531335643036849922924286970256875*97578096035^(3/4)*sqrt(1677751)*sqrt(105602)*sqrt(217)*sqrt(-249588676
96875*97578096035^(1/4)*sqrt(105602)*sqrt(217)*(264*sqrt(35)*sqrt(31) - 7379*sqrt(31))*sqrt(2*x + 1)*sqrt(2501
41922*sqrt(35) + 2263842875) + 148735953072151976291860968750*x + 14873595307215197629186096875*sqrt(35) + 743
67976536075988145930484375)*sqrt(250141922*sqrt(35) + 2263842875)*(7379*sqrt(35) - 9240) - 1/11012889301468978
76195*97578096035^(3/4)*sqrt(105602)*sqrt(217)*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875)*(7379*sqrt(
35) - 9240) - 1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) - 3*97578096035^(1/4)*sqrt(105602)*sqrt(217)*(250141922*
sqrt(35)*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4) - 2263842875*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x
+ 4))*sqrt(250141922*sqrt(35) + 2263842875)*log(24958867696875/1677751*97578096035^(1/4)*sqrt(105602)*sqrt(21
7)*(264*sqrt(35)*sqrt(31) - 7379*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875) + 8865198296538
1618781250*x + 8865198296538161878125*sqrt(35) + 44325991482690809390625) + 3*97578096035^(1/4)*sqrt(105602)*s
qrt(217)*(250141922*sqrt(35)*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4) - 2263842875*sqrt(31)*(25*x^4 + 30
*x^3 + 29*x^2 + 12*x + 4))*sqrt(250141922*sqrt(35) + 2263842875)*log(-24958867696875/1677751*97578096035^(1/4)
*sqrt(105602)*sqrt(217)*(264*sqrt(35)*sqrt(31) - 7379*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 226384
2875) + 88651982965381618781250*x + 8865198296538161878125*sqrt(35) + 44325991482690809390625) - 1293155691541
972090*(39600*x^3 + 69895*x^2 + 47861*x + 26483)*sqrt(2*x + 1))/(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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