3.2326 $$\int \frac{(1+2 x)^{3/2}}{(2+3 x+5 x^2)^3} \, dx$$

Optimal. Leaf size=300 $-\frac{\sqrt{2 x+1} (5-4 x)}{62 \left (5 x^2+3 x+2\right )^2}+\frac{\sqrt{2 x+1} (120 x+67)}{1922 \left (5 x^2+3 x+2\right )}-\frac{3 \sqrt{\frac{1}{434} \left (2705 \sqrt{35}-15082\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{1922}+\frac{3 \sqrt{\frac{1}{434} \left (2705 \sqrt{35}-15082\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{1922}-\frac{3}{961} \sqrt{\frac{1}{434} \left (15082+2705 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{3}{961} \sqrt{\frac{1}{434} \left (15082+2705 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

[Out]

-((5 - 4*x)*Sqrt[1 + 2*x])/(62*(2 + 3*x + 5*x^2)^2) + (Sqrt[1 + 2*x]*(67 + 120*x))/(1922*(2 + 3*x + 5*x^2)) -
(3*Sqrt[(15082 + 2705*Sqrt[35])/434]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35
])]])/961 + (3*Sqrt[(15082 + 2705*Sqrt[35])/434]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(
-2 + Sqrt[35])]])/961 - (3*Sqrt[(-15082 + 2705*Sqrt[35])/434]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 +
2*x] + 5*(1 + 2*x)])/1922 + (3*Sqrt[(-15082 + 2705*Sqrt[35])/434]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[
1 + 2*x] + 5*(1 + 2*x)])/1922

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Rubi [A]  time = 0.421929, antiderivative size = 300, 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 = {738, 822, 826, 1169, 634, 618, 204, 628} $-\frac{\sqrt{2 x+1} (5-4 x)}{62 \left (5 x^2+3 x+2\right )^2}+\frac{\sqrt{2 x+1} (120 x+67)}{1922 \left (5 x^2+3 x+2\right )}-\frac{3 \sqrt{\frac{1}{434} \left (2705 \sqrt{35}-15082\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{1922}+\frac{3 \sqrt{\frac{1}{434} \left (2705 \sqrt{35}-15082\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{1922}-\frac{3}{961} \sqrt{\frac{1}{434} \left (15082+2705 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{3}{961} \sqrt{\frac{1}{434} \left (15082+2705 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((5 - 4*x)*Sqrt[1 + 2*x])/(62*(2 + 3*x + 5*x^2)^2) + (Sqrt[1 + 2*x]*(67 + 120*x))/(1922*(2 + 3*x + 5*x^2)) -
(3*Sqrt[(15082 + 2705*Sqrt[35])/434]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35
])]])/961 + (3*Sqrt[(15082 + 2705*Sqrt[35])/434]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(
-2 + Sqrt[35])]])/961 - (3*Sqrt[(-15082 + 2705*Sqrt[35])/434]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 +
2*x] + 5*(1 + 2*x)])/1922 + (3*Sqrt[(-15082 + 2705*Sqrt[35])/434]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[
1 + 2*x] + 5*(1 + 2*x)])/1922

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*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 - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, 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, 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+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^3} \, dx &=-\frac{(5-4 x) \sqrt{1+2 x}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{1}{62} \int \frac{17+20 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )^2} \, dx\\ &=-\frac{(5-4 x) \sqrt{1+2 x}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (67+120 x)}{1922 \left (2+3 x+5 x^2\right )}+\frac{\int \frac{1239+840 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx}{13454}\\ &=-\frac{(5-4 x) \sqrt{1+2 x}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (67+120 x)}{1922 \left (2+3 x+5 x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{1638+840 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )}{6727}\\ &=-\frac{(5-4 x) \sqrt{1+2 x}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (67+120 x)}{1922 \left (2+3 x+5 x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{1638 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (1638-168 \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 )}{13454 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{1638 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (1638-168 \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 )}{13454 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{(5-4 x) \sqrt{1+2 x}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (67+120 x)}{1922 \left (2+3 x+5 x^2\right )}+\frac{\left (3 \left (140+39 \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 )}{67270}+\frac{\left (3 \left (140+39 \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 )}{67270}-\frac{\left (3 \sqrt{\frac{1}{434} \left (-15082+2705 \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 )}{1922}+\frac{\left (3 \sqrt{\frac{1}{434} \left (-15082+2705 \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 )}{1922}\\ &=-\frac{(5-4 x) \sqrt{1+2 x}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (67+120 x)}{1922 \left (2+3 x+5 x^2\right )}-\frac{3 \sqrt{\frac{1}{434} \left (-15082+2705 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{1922}+\frac{3 \sqrt{\frac{1}{434} \left (-15082+2705 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{1922}-\frac{\left (3 \left (140+39 \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 )}{33635}-\frac{\left (3 \left (140+39 \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 )}{33635}\\ &=-\frac{(5-4 x) \sqrt{1+2 x}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{\sqrt{1+2 x} (67+120 x)}{1922 \left (2+3 x+5 x^2\right )}-\frac{3}{961} \sqrt{\frac{1}{434} \left (15082+2705 \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{3}{961} \sqrt{\frac{1}{434} \left (15082+2705 \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{3 \sqrt{\frac{1}{434} \left (-15082+2705 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{1922}+\frac{3 \sqrt{\frac{1}{434} \left (-15082+2705 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{1922}\\ \end{align*}

Mathematica [C]  time = 0.48492, size = 198, normalized size = 0.66 $\frac{\frac{(2960 x+4391) (2 x+1)^{5/2}}{5 x^2+3 x+2}+\frac{217 (20 x+37) (2 x+1)^{5/2}}{\left (5 x^2+3 x+2\right )^2}-1184 (2 x+1)^{3/2}-3276 \sqrt{2 x+1}+\frac{42 \left (\sqrt{2-i \sqrt{31}} \left (1209-218 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+\sqrt{2+i \sqrt{31}} \left (1209+218 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{31 \sqrt{5}}}{94178}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3276*Sqrt[1 + 2*x] - 1184*(1 + 2*x)^(3/2) + (217*(1 + 2*x)^(5/2)*(37 + 20*x))/(2 + 3*x + 5*x^2)^2 + ((1 + 2*
x)^(5/2)*(4391 + 2960*x))/(2 + 3*x + 5*x^2) + (42*(Sqrt[2 - I*Sqrt[31]]*(1209 - (218*I)*Sqrt[31])*ArcTanh[Sqrt
[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]] + Sqrt[2 + I*Sqrt[31]]*(1209 + (218*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[
2 + I*Sqrt[31]]]))/(31*Sqrt[5]))/94178

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Maple [B]  time = 0.078, size = 662, normalized size = 2.2 \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)^(3/2)/(5*x^2+3*x+2)^3,x)

[Out]

1600*(3/7688*(1+2*x)^(7/2)-41/153760*(1+2*x)^(5/2)+4/4805*(1+2*x)^(3/2)-819/768800*(1+2*x)^(1/2))/(5*(1+2*x)^2
-8*x+3)^2+141/119164*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)-327/417074*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)-705/59582/(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)+327/208537/(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)+234/6727/(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)*7^(1/2)-141/119164*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)+327/417074*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)-705/595
82/(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)+327/208537/(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)+234/6727
/(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)*7^(1/2)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.84599, size = 2778, normalized size = 9.26 \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)^(3/2)/(5*x^2+3*x+2)^3,x, algorithm="fricas")

[Out]

-1/2352683507962900*(357492*256095875^(1/4)*sqrt(217)*sqrt(35)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(8159
3620*sqrt(35) + 512191750)*arctan(1/971794421886819908125*256095875^(3/4)*sqrt(3787)*sqrt(217)*sqrt(256095875^
(1/4)*sqrt(217)*(4*sqrt(35)*sqrt(31) - 39*sqrt(31))*sqrt(2*x + 1)*sqrt(81593620*sqrt(35) + 512191750) + 564092
5850*x + 564092585*sqrt(35) + 2820462925)*(39*sqrt(35)*sqrt(31) - 140*sqrt(31))*sqrt(81593620*sqrt(35) + 51219
1750) - 1/53405465484875*256095875^(3/4)*sqrt(217)*sqrt(2*x + 1)*sqrt(81593620*sqrt(35) + 512191750)*(39*sqrt(
35) - 140) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 357492*256095875^(1/4)*sqrt(217)*sqrt(35)*(25*x^4 + 30*
x^3 + 29*x^2 + 12*x + 4)*sqrt(81593620*sqrt(35) + 512191750)*arctan(1/14576916328302298621875*256095875^(3/4)*
sqrt(217)*sqrt(-852075*256095875^(1/4)*sqrt(217)*(4*sqrt(35)*sqrt(31) - 39*sqrt(31))*sqrt(2*x + 1)*sqrt(815936
20*sqrt(35) + 512191750) + 4806491893638750*x + 480649189363875*sqrt(35) + 2403245946819375)*(39*sqrt(35)*sqrt
(31) - 140*sqrt(31))*sqrt(81593620*sqrt(35) + 512191750) - 1/53405465484875*256095875^(3/4)*sqrt(217)*sqrt(2*x
+ 1)*sqrt(81593620*sqrt(35) + 512191750)*(39*sqrt(35) - 140) - 1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) - 3*25
6095875^(1/4)*sqrt(217)*(15082*sqrt(35)*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4) - 94675*sqrt(31)*(25*x^
4 + 30*x^3 + 29*x^2 + 12*x + 4))*sqrt(81593620*sqrt(35) + 512191750)*log(852075/31*256095875^(1/4)*sqrt(217)*(
4*sqrt(35)*sqrt(31) - 39*sqrt(31))*sqrt(2*x + 1)*sqrt(81593620*sqrt(35) + 512191750) + 155048125601250*x + 155
04812560125*sqrt(35) + 77524062800625) + 3*256095875^(1/4)*sqrt(217)*(15082*sqrt(35)*sqrt(31)*(25*x^4 + 30*x^3
+ 29*x^2 + 12*x + 4) - 94675*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4))*sqrt(81593620*sqrt(35) + 5121917
50)*log(-852075/31*256095875^(1/4)*sqrt(217)*(4*sqrt(35)*sqrt(31) - 39*sqrt(31))*sqrt(2*x + 1)*sqrt(81593620*s
qrt(35) + 512191750) + 155048125601250*x + 15504812560125*sqrt(35) + 77524062800625) - 1224080909450*(600*x^3
+ 695*x^2 + 565*x - 21)*sqrt(2*x + 1))/(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)

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Sympy [A]  time = 92.2554, size = 490, normalized size = 1.63 \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)**(3/2)/(5*x**2+3*x+2)**3,x)

[Out]

229120*(2*x + 1)**(7/2)/(-24109568*x + 5381600*(2*x + 1)**4 - 8610560*(2*x + 1)**3 + 18512704*(2*x + 1)**2 - 1
506848) - 1774080*(2*x + 1)**(7/2)/(-168766976*x + 37671200*(2*x + 1)**4 - 60273920*(2*x + 1)**3 + 129588928*(
2*x + 1)**2 - 10547936) - 259072*(2*x + 1)**(5/2)/(-24109568*x + 5381600*(2*x + 1)**4 - 8610560*(2*x + 1)**3 +
18512704*(2*x + 1)**2 - 1506848) - 940352*(2*x + 1)**(5/2)/(-168766976*x + 37671200*(2*x + 1)**4 - 60273920*(
2*x + 1)**3 + 129588928*(2*x + 1)**2 - 10547936) + 3017984*(2*x + 1)**(3/2)/(5*(-24109568*x + 5381600*(2*x + 1
)**4 - 8610560*(2*x + 1)**3 + 18512704*(2*x + 1)**2 - 1506848)) - 6868736*(2*x + 1)**(3/2)/(5*(-168766976*x +
37671200*(2*x + 1)**4 - 60273920*(2*x + 1)**3 + 129588928*(2*x + 1)**2 - 10547936)) + 128*(2*x + 1)**(3/2)/(-6
944*x + 4340*(2*x + 1)**2 + 2604) - 974848*sqrt(2*x + 1)/(5*(-24109568*x + 5381600*(2*x + 1)**4 - 8610560*(2*x
+ 1)**3 + 18512704*(2*x + 1)**2 - 1506848)) - 5403328*sqrt(2*x + 1)/(-168766976*x + 37671200*(2*x + 1)**4 - 6
0273920*(2*x + 1)**3 + 129588928*(2*x + 1)**2 - 10547936) + 1728*sqrt(2*x + 1)/(5*(-6944*x + 4340*(2*x + 1)**2
+ 2604)) + 256*RootSum(75465931487403231630327808*_t**4 + 9053854476152406016*_t**2 + 333142578125, Lambda(_t
, _t*log(21632117045402271744*_t**3/158378125 + 10865340674816*_t/1108646875 + sqrt(2*x + 1))))/5 - 448*RootSu
m(3697830642882758349886062592*_t**4 + 2111968303753265086464*_t**2 + 705698730253125, Lambda(_t, _t*log(-3459
438283411209322496*_t**3/1377792122625 + 251494140770688*_t/357205365125 + sqrt(2*x + 1))))/5 + 64*RootSum(199
50060344639488*_t**4 + 498437272576*_t**2 + 10878125, Lambda(_t, _t*log(-11049511452672*_t**3/2205125 + 307918
256*_t/2205125 + sqrt(2*x + 1))))/5

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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