3.2323 \(\int \frac{(1+2 x)^{9/2}}{(2+3 x+5 x^2)^3} \, dx\)

Optimal. Leaf size=313 \[ -\frac{(5-4 x) (2 x+1)^{7/2}}{62 \left (5 x^2+3 x+2\right )^2}-\frac{(1143-1088 x) (2 x+1)^{3/2}}{9610 \left (5 x^2+3 x+2\right )}-\frac{1584 \sqrt{2 x+1}}{24025}+\frac{3 \sqrt{\frac{1}{310} \left (64681225 \sqrt{35}-250141922\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{48050}-\frac{3 \sqrt{\frac{1}{310} \left (64681225 \sqrt{35}-250141922\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{48050}-\frac{3 \sqrt{\frac{1}{310} \left (250141922+64681225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{24025}+\frac{3 \sqrt{\frac{1}{310} \left (250141922+64681225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{24025} \]

[Out]

(-1584*Sqrt[1 + 2*x])/24025 - ((5 - 4*x)*(1 + 2*x)^(7/2))/(62*(2 + 3*x + 5*x^2)^2) - ((1143 - 1088*x)*(1 + 2*x
)^(3/2))/(9610*(2 + 3*x + 5*x^2)) - (3*Sqrt[(250141922 + 64681225*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35]
)] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/24025 + (3*Sqrt[(250141922 + 64681225*Sqrt[35])/310]*ArcTan[
(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/24025 + (3*Sqrt[(-250141922 + 64681225
*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/48050 - (3*Sqrt[(-2501419
22 + 64681225*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/48050

________________________________________________________________________________________

Rubi [A]  time = 0.521533, antiderivative size = 313, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {738, 818, 824, 826, 1169, 634, 618, 204, 628} \[ -\frac{(5-4 x) (2 x+1)^{7/2}}{62 \left (5 x^2+3 x+2\right )^2}-\frac{(1143-1088 x) (2 x+1)^{3/2}}{9610 \left (5 x^2+3 x+2\right )}-\frac{1584 \sqrt{2 x+1}}{24025}+\frac{3 \sqrt{\frac{1}{310} \left (64681225 \sqrt{35}-250141922\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{48050}-\frac{3 \sqrt{\frac{1}{310} \left (64681225 \sqrt{35}-250141922\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{48050}-\frac{3 \sqrt{\frac{1}{310} \left (250141922+64681225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{24025}+\frac{3 \sqrt{\frac{1}{310} \left (250141922+64681225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{24025} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x)^(9/2)/(2 + 3*x + 5*x^2)^3,x]

[Out]

(-1584*Sqrt[1 + 2*x])/24025 - ((5 - 4*x)*(1 + 2*x)^(7/2))/(62*(2 + 3*x + 5*x^2)^2) - ((1143 - 1088*x)*(1 + 2*x
)^(3/2))/(9610*(2 + 3*x + 5*x^2)) - (3*Sqrt[(250141922 + 64681225*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35]
)] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/24025 + (3*Sqrt[(250141922 + 64681225*Sqrt[35])/310]*ArcTan[
(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/24025 + (3*Sqrt[(-250141922 + 64681225
*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/48050 - (3*Sqrt[(-2501419
22 + 64681225*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/48050

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 818

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

Rule 824

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), 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] && FractionQ[m] && GtQ[m, 0]

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)^{9/2}}{\left (2+3 x+5 x^2\right )^3} \, dx &=-\frac{(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{1}{62} \int \frac{(47-4 x) (1+2 x)^{5/2}}{\left (2+3 x+5 x^2\right )^2} \, dx\\ &=-\frac{(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\int \frac{\sqrt{1+2 x} (-4269+1584 x)}{2+3 x+5 x^2} \, dx}{9610}\\ &=-\frac{1584 \sqrt{1+2 x}}{24025}-\frac{(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\int \frac{-27681-44274 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx}{48050}\\ &=-\frac{1584 \sqrt{1+2 x}}{24025}-\frac{(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{-11088-44274 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )}{24025}\\ &=-\frac{1584 \sqrt{1+2 x}}{24025}-\frac{(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{-11088 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (-11088+44274 \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 )}{48050 \sqrt{14 \left (2+\sqrt{35}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{-11088 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (-11088+44274 \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 )}{48050 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{1584 \sqrt{1+2 x}}{24025}-\frac{(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\left (3 \left (9240-7379 \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 )}{240250 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\left (3 \left (9240-7379 \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 )}{240250 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\left (3 \left (7379+264 \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 )}{240250}+\frac{\left (3 \left (7379+264 \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 )}{240250}\\ &=-\frac{1584 \sqrt{1+2 x}}{24025}-\frac{(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}+\frac{3 \left (7379-264 \sqrt{35}\right ) \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{48050 \sqrt{10 \left (2+\sqrt{35}\right )}}-\frac{3 \left (7379-264 \sqrt{35}\right ) \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{48050 \sqrt{10 \left (2+\sqrt{35}\right )}}-\frac{\left (3 \left (7379+264 \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 )}{120125}-\frac{\left (3 \left (7379+264 \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 )}{120125}\\ &=-\frac{1584 \sqrt{1+2 x}}{24025}-\frac{(5-4 x) (1+2 x)^{7/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(1143-1088 x) (1+2 x)^{3/2}}{9610 \left (2+3 x+5 x^2\right )}-\frac{3 \sqrt{\frac{1}{310} \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 )}{24025}+\frac{3 \sqrt{\frac{1}{310} \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 )}{24025}+\frac{3 \left (7379-264 \sqrt{35}\right ) \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{48050 \sqrt{10 \left (2+\sqrt{35}\right )}}-\frac{3 \left (7379-264 \sqrt{35}\right ) \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{48050 \sqrt{10 \left (2+\sqrt{35}\right )}}\\ \end{align*}

Mathematica [C]  time = 0.697732, size = 236, normalized size = 0.75 \[ \frac{-\frac{7 (640 x+409) (2 x+1)^{11/2}}{5 x^2+3 x+2}+\frac{217 (20 x+37) (2 x+1)^{11/2}}{\left (5 x^2+3 x+2\right )^2}+1792 (2 x+1)^{9/2}+1932 (2 x+1)^{7/2}-2352 (2 x+1)^{5/2}-\frac{47236}{5} (2 x+1)^{3/2}-\frac{155232}{25} \sqrt{2 x+1}+\frac{294 \left (\sqrt{2-i \sqrt{31}} \left (8184-7907 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+\sqrt{2+i \sqrt{31}} \left (8184+7907 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{775 \sqrt{5}}}{94178} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x)^(9/2)/(2 + 3*x + 5*x^2)^3,x]

[Out]

((-155232*Sqrt[1 + 2*x])/25 - (47236*(1 + 2*x)^(3/2))/5 - 2352*(1 + 2*x)^(5/2) + 1932*(1 + 2*x)^(7/2) + 1792*(
1 + 2*x)^(9/2) + (217*(1 + 2*x)^(11/2)*(37 + 20*x))/(2 + 3*x + 5*x^2)^2 - (7*(1 + 2*x)^(11/2)*(409 + 640*x))/(
2 + 3*x + 5*x^2) + (294*(Sqrt[2 - I*Sqrt[31]]*(8184 - (7907*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqr
t[31]]] + Sqrt[2 + I*Sqrt[31]]*(8184 + (7907*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]]))/(775*
Sqrt[5]))/94178

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Maple [B]  time = 0.077, size = 662, normalized size = 2.1 \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)^(9/2)/(5*x^2+3*x+2)^3,x)

[Out]

1600*(-1723/768800*(1+2*x)^(7/2)-3833/4805000*(1+2*x)^(5/2)-14693/19220000*(1+2*x)^(3/2)-4851/2402500*(1+2*x)^
(1/2))/(5*(1+2*x)^2-8*x+3)^2+35997/7447750*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)-23721/2979100*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)-35997/744775/(10*5^(1/2)*7^(1/2)-20)^(1/2)*arct
an((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)
+23721/1489550/(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)+1584/24025/(10*5^(1/2)*7^(1/2)-20)^(1/2)*arc
tan((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)-3599
7/7447750*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)+23721/2979100*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)-35997/744775/(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)+23721/1489550/(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)+1584/24025/(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{9}{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)^(9/2)/(5*x^2+3*x+2)^3,x, algorithm="maxima")

[Out]

integrate((2*x + 1)^(9/2)/(5*x^2 + 3*x + 2)^3, x)

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Fricas [B]  time = 2.8415, size = 3374, normalized size = 10.78 \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)^(9/2)/(5*x^2+3*x+2)^3,x, algorithm="fricas")

[Out]

1/44382950698994113517500*(19347824532*97578096035^(1/4)*sqrt(105602)*sqrt(155)*sqrt(35)*(25*x^4 + 30*x^3 + 29
*x^2 + 12*x + 4)*sqrt(250141922*sqrt(35) + 2263842875)*arctan(1/2160252846511970217131322639383425*97578096035
^(3/4)*sqrt(1677751)*sqrt(105602)*sqrt(7543)*sqrt(155)*sqrt(97578096035^(1/4)*sqrt(105602)*sqrt(155)*(7379*sqr
t(35)*sqrt(31) - 9240*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875) + 29796214090828850*x + 29
79621409082885*sqrt(35) + 14898107045414425)*sqrt(250141922*sqrt(35) + 2263842875)*(264*sqrt(35) - 7379) - 1/1
57326990020985410885*97578096035^(3/4)*sqrt(105602)*sqrt(155)*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 22638428
75)*(264*sqrt(35) - 7379) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 19347824532*97578096035^(1/4)*sqrt(10560
2)*sqrt(155)*sqrt(35)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(250141922*sqrt(35) + 2263842875)*arctan(1/793
8929210931490547957610699734086875*97578096035^(3/4)*sqrt(1677751)*sqrt(105602)*sqrt(155)*sqrt(-101872929375*9
7578096035^(1/4)*sqrt(105602)*sqrt(155)*(7379*sqrt(35)*sqrt(31) - 9240*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*
sqrt(35) + 2263842875) + 3035427613717387271262468750*x + 303542761371738727126246875*sqrt(35) + 1517713806858
693635631234375)*sqrt(250141922*sqrt(35) + 2263842875)*(264*sqrt(35) - 7379) - 1/157326990020985410885*9757809
6035^(3/4)*sqrt(105602)*sqrt(155)*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875)*(264*sqrt(35) - 7379) -
1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 3*97578096035^(1/4)*sqrt(105602)*sqrt(155)*(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(25
0141922*sqrt(35) + 2263842875)*log(101872929375/1677751*97578096035^(1/4)*sqrt(105602)*sqrt(155)*(7379*sqrt(35
)*sqrt(31) - 9240*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875) + 1809224142150645281250*x + 1
80922414215064528125*sqrt(35) + 904612071075322640625) - 3*97578096035^(1/4)*sqrt(105602)*sqrt(155)*(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(-101872929375/1677751*97578096035^(1/4)*sqrt(105602)*sqrt(15
5)*(7379*sqrt(35)*sqrt(31) - 9240*sqrt(31))*sqrt(2*x + 1)*sqrt(250141922*sqrt(35) + 2263842875) + 180922414215
0645281250*x + 180922414215064528125*sqrt(35) + 904612071075322640625) - 923682636815694350*(86150*x^3 + 14455
7*x^2 + 87291*x + 27977)*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)**(9/2)/(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{9}{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)^(9/2)/(5*x^2+3*x+2)^3,x, algorithm="giac")

[Out]

integrate((2*x + 1)^(9/2)/(5*x^2 + 3*x + 2)^3, x)