### 3.2325 $$\int \frac{(1+2 x)^{5/2}}{(2+3 x+5 x^2)^3} \, dx$$

Optimal. Leaf size=300 $-\frac{(5-4 x) (2 x+1)^{3/2}}{62 \left (5 x^2+3 x+2\right )^2}+\frac{3 (78 x+11) \sqrt{2 x+1}}{1922 \left (5 x^2+3 x+2\right )}+\frac{3 \sqrt{\frac{1}{310} \left (2705 \sqrt{35}-15082\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{1922}-\frac{3 \sqrt{\frac{1}{310} \left (2705 \sqrt{35}-15082\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{1922}-\frac{3}{961} \sqrt{\frac{1}{310} \left (15082+2705 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{3}{961} \sqrt{\frac{1}{310} \left (15082+2705 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

[Out]

-((5 - 4*x)*(1 + 2*x)^(3/2))/(62*(2 + 3*x + 5*x^2)^2) + (3*Sqrt[1 + 2*x]*(11 + 78*x))/(1922*(2 + 3*x + 5*x^2))
- (3*Sqrt[(15082 + 2705*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt
[35])]])/961 + (3*Sqrt[(15082 + 2705*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[1
0*(-2 + Sqrt[35])]])/961 + (3*Sqrt[(-15082 + 2705*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1
+ 2*x] + 5*(1 + 2*x)])/1922 - (3*Sqrt[(-15082 + 2705*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sq
rt[1 + 2*x] + 5*(1 + 2*x)])/1922

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Rubi [A]  time = 0.435021, 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, 820, 826, 1169, 634, 618, 204, 628} $-\frac{(5-4 x) (2 x+1)^{3/2}}{62 \left (5 x^2+3 x+2\right )^2}+\frac{3 (78 x+11) \sqrt{2 x+1}}{1922 \left (5 x^2+3 x+2\right )}+\frac{3 \sqrt{\frac{1}{310} \left (2705 \sqrt{35}-15082\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{1922}-\frac{3 \sqrt{\frac{1}{310} \left (2705 \sqrt{35}-15082\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{1922}-\frac{3}{961} \sqrt{\frac{1}{310} \left (15082+2705 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{3}{961} \sqrt{\frac{1}{310} \left (15082+2705 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((5 - 4*x)*(1 + 2*x)^(3/2))/(62*(2 + 3*x + 5*x^2)^2) + (3*Sqrt[1 + 2*x]*(11 + 78*x))/(1922*(2 + 3*x + 5*x^2))
- (3*Sqrt[(15082 + 2705*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt
[35])]])/961 + (3*Sqrt[(15082 + 2705*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[1
0*(-2 + Sqrt[35])]])/961 + (3*Sqrt[(-15082 + 2705*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1
+ 2*x] + 5*(1 + 2*x)])/1922 - (3*Sqrt[(-15082 + 2705*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sq
rt[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 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g + (2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/
((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g*(2*a*e*m + b*d*(2*p + 3)) - f*
(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], 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] && LtQ[p, -1] && GtQ[m, 0] && (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)^{5/2}}{\left (2+3 x+5 x^2\right )^3} \, dx &=-\frac{(5-4 x) (1+2 x)^{3/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{1}{62} \int \frac{\sqrt{1+2 x} (27+12 x)}{\left (2+3 x+5 x^2\right )^2} \, dx\\ &=-\frac{(5-4 x) (1+2 x)^{3/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{3 \sqrt{1+2 x} (11+78 x)}{1922 \left (2+3 x+5 x^2\right )}+\frac{\int \frac{201+234 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx}{1922}\\ &=-\frac{(5-4 x) (1+2 x)^{3/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{3 \sqrt{1+2 x} (11+78 x)}{1922 \left (2+3 x+5 x^2\right )}+\frac{1}{961} \operatorname{Subst}\left (\int \frac{168+234 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=-\frac{(5-4 x) (1+2 x)^{3/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{3 \sqrt{1+2 x} (11+78 x)}{1922 \left (2+3 x+5 x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{168 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (168-234 \sqrt{\frac{7}{5}}\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 )}{1922 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{168 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (168-234 \sqrt{\frac{7}{5}}\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 )}{1922 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{(5-4 x) (1+2 x)^{3/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{3 \sqrt{1+2 x} (11+78 x)}{1922 \left (2+3 x+5 x^2\right )}+\frac{\left (3 \left (39+4 \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 )}{9610}+\frac{\left (3 \left (39+4 \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 )}{9610}+\frac{\left (3 \sqrt{\frac{1}{310} \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}{310} \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) (1+2 x)^{3/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{3 \sqrt{1+2 x} (11+78 x)}{1922 \left (2+3 x+5 x^2\right )}+\frac{3 \sqrt{\frac{1}{310} \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}{310} \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 (39+4 \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 )}{4805}-\frac{\left (3 \left (39+4 \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 )}{4805}\\ &=-\frac{(5-4 x) (1+2 x)^{3/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{3 \sqrt{1+2 x} (11+78 x)}{1922 \left (2+3 x+5 x^2\right )}-\frac{3}{961} \sqrt{\frac{7541}{155}+\frac{541 \sqrt{35}}{62}} \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{7541}{155}+\frac{541 \sqrt{35}}{62}} \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}{310} \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}{310} \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 = 0.565355, size = 209, normalized size = 0.7 $\frac{\frac{(480 x+1973) (2 x+1)^{7/2}}{5 x^2+3 x+2}+\frac{217 (20 x+37) (2 x+1)^{7/2}}{\left (5 x^2+3 x+2\right )^2}-192 (2 x+1)^{5/2}-1540 (2 x+1)^{3/2}-2352 \sqrt{2 x+1}+\frac{294 \left (\sqrt{2-i \sqrt{31}} \left (124-47 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+\sqrt{2+i \sqrt{31}} \left (124+47 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{31 \sqrt{5}}}{94178}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2352*Sqrt[1 + 2*x] - 1540*(1 + 2*x)^(3/2) - 192*(1 + 2*x)^(5/2) + (217*(1 + 2*x)^(7/2)*(37 + 20*x))/(2 + 3*x
+ 5*x^2)^2 + ((1 + 2*x)^(7/2)*(1973 + 480*x))/(2 + 3*x + 5*x^2) + (294*(Sqrt[2 - I*Sqrt[31]]*(124 - (47*I)*Sq
rt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]] + Sqrt[2 + I*Sqrt[31]]*(124 + (47*I)*Sqrt[31])*ArcTanh[Sq
rt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]]))/(31*Sqrt[5]))/94178

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Maple [B]  time = 0.083, 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)^(5/2)/(5*x^2+3*x+2)^3,x)

[Out]

1600*(117/153760*(1+2*x)^(7/2)-4/4805*(1+2*x)^(5/2)+287/768800*(1+2*x)^(3/2)-147/192200*(1+2*x)^(1/2))/(5*(1+2
*x)^2-8*x+3)^2+327/297910*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)-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))*7^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)-327/29791/(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)+141/59582/(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))
*5^(1/2)*(2*5^(1/2)*7^(1/2)+4)*7^(1/2)+24/961/(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)-327/297910*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)+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)*7^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)-327/2
9791/(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)+141/59582/(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)+24/961/
(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{5}{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)^(5/2)/(5*x^2+3*x+2)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.73719, size = 2738, normalized size = 9.13 \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)^(5/2)/(5*x^2+3*x+2)^3,x, algorithm="fricas")

[Out]

1/1680488219973500*(357492*256095875^(1/4)*sqrt(155)*sqrt(35)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(81593
620*sqrt(35) + 512191750)*arctan(1/694138872776299934375*256095875^(3/4)*sqrt(2705)*sqrt(217)*sqrt(155)*sqrt(2
56095875^(1/4)*sqrt(155)*(39*sqrt(35)*sqrt(31) - 140*sqrt(31))*sqrt(2*x + 1)*sqrt(81593620*sqrt(35) + 51219175
0) + 28204629250*x + 2820462925*sqrt(35) + 14102314625)*sqrt(81593620*sqrt(35) + 512191750)*(4*sqrt(35) - 39)
- 1/7629352212125*256095875^(3/4)*sqrt(155)*sqrt(2*x + 1)*sqrt(81593620*sqrt(35) + 512191750)*(4*sqrt(35) - 39
) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 357492*256095875^(1/4)*sqrt(155)*sqrt(35)*(25*x^4 + 30*x^3 + 29*
x^2 + 12*x + 4)*sqrt(81593620*sqrt(35) + 512191750)*arctan(1/2082416618328899803125*256095875^(3/4)*sqrt(217)*
sqrt(155)*sqrt(-24345*256095875^(1/4)*sqrt(155)*(39*sqrt(35)*sqrt(31) - 140*sqrt(31))*sqrt(2*x + 1)*sqrt(81593
620*sqrt(35) + 512191750) + 686641699091250*x + 68664169909125*sqrt(35) + 343320849545625)*sqrt(81593620*sqrt(
35) + 512191750)*(4*sqrt(35) - 39) - 1/7629352212125*256095875^(3/4)*sqrt(155)*sqrt(2*x + 1)*sqrt(81593620*sqr
t(35) + 512191750)*(4*sqrt(35) - 39) - 1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 3*256095875^(1/4)*sqrt(155)*(
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(24345/217*256095875^(1/4)*sqrt(155)*(39*sqrt(35)*sqrt(31) - 14
0*sqrt(31))*sqrt(2*x + 1)*sqrt(81593620*sqrt(35) + 512191750) + 3164247461250*x + 316424746125*sqrt(35) + 1582
123730625) - 3*256095875^(1/4)*sqrt(155)*(15082*sqrt(35)*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4) - 9467
5*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4))*sqrt(81593620*sqrt(35) + 512191750)*log(-24345/217*256095875
^(1/4)*sqrt(155)*(39*sqrt(35)*sqrt(31) - 140*sqrt(31))*sqrt(2*x + 1)*sqrt(81593620*sqrt(35) + 512191750) + 316
4247461250*x + 316424746125*sqrt(35) + 1582123730625) + 874343506750*(1170*x^3 + 1115*x^2 + 381*x - 89)*sqrt(2
*x + 1))/(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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