3.2310 \(\int \frac{(1+2 x)^{5/2}}{2+3 x+5 x^2} \, dx\)

Optimal. Leaf size=266 \[ \frac{4}{15} (2 x+1)^{3/2}+\frac{16}{25} \sqrt{2 x+1}+\frac{1}{25} \sqrt{\frac{1}{310} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{25} \sqrt{\frac{1}{310} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{25} \sqrt{\frac{2}{155} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{25} \sqrt{\frac{2}{155} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]

[Out]

(16*Sqrt[1 + 2*x])/25 + (4*(1 + 2*x)^(3/2))/15 + (Sqrt[(2*(7162 + 1225*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sq
rt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/25 - (Sqrt[(2*(7162 + 1225*Sqrt[35]))/155]*ArcTan[(Sqr
t[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/25 + (Sqrt[(-7162 + 1225*Sqrt[35])/310]*Lo
g[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/25 - (Sqrt[(-7162 + 1225*Sqrt[35])/310]*Log
[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/25

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Rubi [A]  time = 0.436201, antiderivative size = 266, 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 = {703, 824, 826, 1169, 634, 618, 204, 628} \[ \frac{4}{15} (2 x+1)^{3/2}+\frac{16}{25} \sqrt{2 x+1}+\frac{1}{25} \sqrt{\frac{1}{310} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{25} \sqrt{\frac{1}{310} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{25} \sqrt{\frac{2}{155} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{25} \sqrt{\frac{2}{155} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*Sqrt[1 + 2*x])/25 + (4*(1 + 2*x)^(3/2))/15 + (Sqrt[(2*(7162 + 1225*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sq
rt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/25 - (Sqrt[(2*(7162 + 1225*Sqrt[35]))/155]*ArcTan[(Sqr
t[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/25 + (Sqrt[(-7162 + 1225*Sqrt[35])/310]*Lo
g[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/25 - (Sqrt[(-7162 + 1225*Sqrt[35])/310]*Log
[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/25

Rule 703

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

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)^{5/2}}{2+3 x+5 x^2} \, dx &=\frac{4}{15} (1+2 x)^{3/2}+\frac{1}{5} \int \frac{\sqrt{1+2 x} (-3+8 x)}{2+3 x+5 x^2} \, dx\\ &=\frac{16}{25} \sqrt{1+2 x}+\frac{4}{15} (1+2 x)^{3/2}+\frac{1}{25} \int \frac{-47-38 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx\\ &=\frac{16}{25} \sqrt{1+2 x}+\frac{4}{15} (1+2 x)^{3/2}+\frac{2}{25} \operatorname{Subst}\left (\int \frac{-56-38 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=\frac{16}{25} \sqrt{1+2 x}+\frac{4}{15} (1+2 x)^{3/2}+\frac{\operatorname{Subst}\left (\int \frac{-56 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (-56+38 \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 )}{25 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{-56 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (-56+38 \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 )}{25 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=\frac{16}{25} \sqrt{1+2 x}+\frac{4}{15} (1+2 x)^{3/2}-\frac{1}{125} \sqrt{921+152 \sqrt{35}} \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 )-\frac{1}{125} \sqrt{921+152 \sqrt{35}} \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 )+\frac{1}{25} \sqrt{\frac{1}{310} \left (-7162+1225 \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}{25} \sqrt{\frac{1}{310} \left (-7162+1225 \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{16}{25} \sqrt{1+2 x}+\frac{4}{15} (1+2 x)^{3/2}+\frac{1}{25} \sqrt{\frac{1}{310} \left (-7162+1225 \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}{25} \sqrt{\frac{1}{310} \left (-7162+1225 \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}{125} \left (2 \sqrt{921+152 \sqrt{35}}\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 )+\frac{1}{125} \left (2 \sqrt{921+152 \sqrt{35}}\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 )\\ &=\frac{16}{25} \sqrt{1+2 x}+\frac{4}{15} (1+2 x)^{3/2}+\frac{1}{25} \sqrt{\frac{2}{155} \left (7162+1225 \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}{25} \sqrt{\frac{2}{155} \left (7162+1225 \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}{25} \sqrt{\frac{1}{310} \left (-7162+1225 \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}{25} \sqrt{\frac{1}{310} \left (-7162+1225 \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 = 0.276248, size = 133, normalized size = 0.5 \[ \frac{2 \left (310 \sqrt{2 x+1} (10 x+17)+3 i \sqrt{10-5 i \sqrt{31}} \left (27 \sqrt{31}+124 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )-3 i \sqrt{10+5 i \sqrt{31}} \left (27 \sqrt{31}-124 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{11625} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(310*Sqrt[1 + 2*x]*(17 + 10*x) + (3*I)*Sqrt[10 - (5*I)*Sqrt[31]]*(124*I + 27*Sqrt[31])*ArcTanh[Sqrt[5 + 10*
x]/Sqrt[2 - I*Sqrt[31]]] - (3*I)*Sqrt[10 + (5*I)*Sqrt[31]]*(-124*I + 27*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[
2 + I*Sqrt[31]]]))/11625

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Maple [B]  time = 0.078, size = 625, normalized size = 2.4 \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),x)

[Out]

4/15*(1+2*x)^(3/2)+16/25*(1+2*x)^(1/2)-89/3875*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)+27/1550*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)+178/775/(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)-27/775
/(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)-16/25/(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)+89/3875*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)-27/1550*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)
+178/775/(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)-27/775/(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)-16/25/
(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}}}{5 \, x^{2} + 3 \, x + 2}\,{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),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.71444, size = 2144, 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)^(5/2)/(5*x^2+3*x+2),x, algorithm="fricas")

[Out]

-1/1673341250*sqrt(155)*35^(1/4)*sqrt(2)*(7162*sqrt(35)*sqrt(31) - 42875*sqrt(31))*sqrt(7162*sqrt(35) + 42875)
*log(26582500/199*sqrt(155)*35^(1/4)*sqrt(2)*(19*sqrt(35)*sqrt(31) - 140*sqrt(31))*sqrt(2*x + 1)*sqrt(7162*sqr
t(35) + 42875) + 288420125000*x + 28842012500*sqrt(35) + 144210062500) + 1/1673341250*sqrt(155)*35^(1/4)*sqrt(
2)*(7162*sqrt(35)*sqrt(31) - 42875*sqrt(31))*sqrt(7162*sqrt(35) + 42875)*log(-26582500/199*sqrt(155)*35^(1/4)*
sqrt(2)*(19*sqrt(35)*sqrt(31) - 140*sqrt(31))*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875) + 288420125000*x + 288
42012500*sqrt(35) + 144210062500) + 2/135625*sqrt(155)*35^(3/4)*sqrt(2)*sqrt(7162*sqrt(35) + 42875)*arctan(1/4
6619287225*sqrt(217)*sqrt(199)*sqrt(155)*35^(3/4)*sqrt(2)*sqrt(sqrt(155)*35^(1/4)*sqrt(2)*(19*sqrt(35)*sqrt(31
) - 140*sqrt(31))*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875) + 2159150*x + 215915*sqrt(35) + 1079575)*sqrt(7162
*sqrt(35) + 42875)*(4*sqrt(35) - 19) - 1/215915*sqrt(155)*35^(3/4)*sqrt(2)*sqrt(2*x + 1)*sqrt(7162*sqrt(35) +
42875)*(4*sqrt(35) - 19) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 2/135625*sqrt(155)*35^(3/4)*sqrt(2)*sqrt(
7162*sqrt(35) + 42875)*arctan(1/16316750528750*sqrt(199)*sqrt(155)*35^(3/4)*sqrt(2)*sqrt(-26582500*sqrt(155)*3
5^(1/4)*sqrt(2)*(19*sqrt(35)*sqrt(31) - 140*sqrt(31))*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875) + 573956048750
00*x + 5739560487500*sqrt(35) + 28697802437500)*sqrt(7162*sqrt(35) + 42875)*(4*sqrt(35) - 19) - 1/215915*sqrt(
155)*35^(3/4)*sqrt(2)*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875)*(4*sqrt(35) - 19) - 1/31*sqrt(35)*sqrt(31) - 2
/31*sqrt(31)) + 4/75*(10*x + 17)*sqrt(2*x + 1)

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Sympy [A]  time = 33.9032, size = 97, normalized size = 0.36 \begin{align*} \frac{4 \left (2 x + 1\right )^{\frac{3}{2}}}{15} + \frac{16 \sqrt{2 x + 1}}{25} - \frac{76 \operatorname{RootSum}{\left (1230080 t^{4} + 1984 t^{2} + 7, \left ( t \mapsto t \log{\left (9920 t^{3} + 8 t + \sqrt{2 x + 1} \right )} \right )\right )}}{25} - \frac{112 \operatorname{RootSum}{\left (1722112 t^{4} + 1984 t^{2} + 5, \left ( t \mapsto t \log{\left (- \frac{27776 t^{3}}{5} + \frac{108 t}{5} + \sqrt{2 x + 1} \right )} \right )\right )}}{25} \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),x)

[Out]

4*(2*x + 1)**(3/2)/15 + 16*sqrt(2*x + 1)/25 - 76*RootSum(1230080*_t**4 + 1984*_t**2 + 7, Lambda(_t, _t*log(992
0*_t**3 + 8*_t + sqrt(2*x + 1))))/25 - 112*RootSum(1722112*_t**4 + 1984*_t**2 + 5, Lambda(_t, _t*log(-27776*_t
**3/5 + 108*_t/5 + sqrt(2*x + 1))))/25

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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}}}{5 \, x^{2} + 3 \, x + 2}\,{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),x, algorithm="giac")

[Out]

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