3.2317 \(\int \frac{(1+2 x)^{5/2}}{(2+3 x+5 x^2)^2} \, dx\)

Optimal. Leaf size=283 \[ -\frac{(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}-\frac{8}{155} \sqrt{2 x+1}+\frac{1}{155} \sqrt{\frac{1}{310} \left (10325 \sqrt{35}-32678\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{155} \sqrt{\frac{1}{310} \left (10325 \sqrt{35}-32678\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{155} \sqrt{\frac{2}{155} \left (32678+10325 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{1}{155} \sqrt{\frac{2}{155} \left (32678+10325 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]

[Out]

(-8*Sqrt[1 + 2*x])/155 - ((5 - 4*x)*(1 + 2*x)^(3/2))/(31*(2 + 3*x + 5*x^2)) - (Sqrt[(2*(32678 + 10325*Sqrt[35]
))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/155 + (Sqrt[(2*(32678 +
10325*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/155 + (S
qrt[(-32678 + 10325*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/155 -
(Sqrt[(-32678 + 10325*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/155

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Rubi [A]  time = 0.410117, antiderivative size = 283, 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, 824, 826, 1169, 634, 618, 204, 628} \[ -\frac{(5-4 x) (2 x+1)^{3/2}}{31 \left (5 x^2+3 x+2\right )}-\frac{8}{155} \sqrt{2 x+1}+\frac{1}{155} \sqrt{\frac{1}{310} \left (10325 \sqrt{35}-32678\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{155} \sqrt{\frac{1}{310} \left (10325 \sqrt{35}-32678\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{155} \sqrt{\frac{2}{155} \left (32678+10325 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{1}{155} \sqrt{\frac{2}{155} \left (32678+10325 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*Sqrt[1 + 2*x])/155 - ((5 - 4*x)*(1 + 2*x)^(3/2))/(31*(2 + 3*x + 5*x^2)) - (Sqrt[(2*(32678 + 10325*Sqrt[35]
))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/155 + (Sqrt[(2*(32678 +
10325*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/155 + (S
qrt[(-32678 + 10325*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/155 -
(Sqrt[(-32678 + 10325*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/155

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 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}}{\left (2+3 x+5 x^2\right )^2} \, dx &=-\frac{(5-4 x) (1+2 x)^{3/2}}{31 \left (2+3 x+5 x^2\right )}+\frac{1}{31} \int \frac{(19-4 x) \sqrt{1+2 x}}{2+3 x+5 x^2} \, dx\\ &=-\frac{8}{155} \sqrt{1+2 x}-\frac{(5-4 x) (1+2 x)^{3/2}}{31 \left (2+3 x+5 x^2\right )}+\frac{1}{155} \int \frac{111+194 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx\\ &=-\frac{8}{155} \sqrt{1+2 x}-\frac{(5-4 x) (1+2 x)^{3/2}}{31 \left (2+3 x+5 x^2\right )}+\frac{2}{155} \operatorname{Subst}\left (\int \frac{28+194 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=-\frac{8}{155} \sqrt{1+2 x}-\frac{(5-4 x) (1+2 x)^{3/2}}{31 \left (2+3 x+5 x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{28 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (28-194 \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 )}{155 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{28 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (28-194 \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 )}{155 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{8}{155} \sqrt{1+2 x}-\frac{(5-4 x) (1+2 x)^{3/2}}{31 \left (2+3 x+5 x^2\right )}+\frac{1}{775} \left (97+2 \sqrt{35}\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 )+\frac{1}{775} \left (97+2 \sqrt{35}\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 )+\frac{1}{155} \sqrt{\frac{1}{310} \left (-32678+10325 \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}{155} \sqrt{\frac{1}{310} \left (-32678+10325 \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{8}{155} \sqrt{1+2 x}-\frac{(5-4 x) (1+2 x)^{3/2}}{31 \left (2+3 x+5 x^2\right )}+\frac{1}{155} \sqrt{\frac{1}{310} \left (-32678+10325 \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}{155} \sqrt{\frac{1}{310} \left (-32678+10325 \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}{775} \left (2 \left (97+2 \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 )-\frac{1}{775} \left (2 \left (97+2 \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 )\\ &=-\frac{8}{155} \sqrt{1+2 x}-\frac{(5-4 x) (1+2 x)^{3/2}}{31 \left (2+3 x+5 x^2\right )}-\frac{1}{155} \sqrt{\frac{2}{155} \left (32678+10325 \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}{155} \sqrt{\frac{2}{155} \left (32678+10325 \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}{155} \sqrt{\frac{1}{310} \left (-32678+10325 \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}{155} \sqrt{\frac{1}{310} \left (-32678+10325 \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 = 0.672765, size = 141, normalized size = 0.5 \[ \frac{-\frac{155 \sqrt{2 x+1} (54 x+41)}{5 x^2+3 x+2}+2 \sqrt{10-5 i \sqrt{31}} \left (62-101 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+2 \sqrt{10+5 i \sqrt{31}} \left (62+101 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )}{24025} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-155*Sqrt[1 + 2*x]*(41 + 54*x))/(2 + 3*x + 5*x^2) + 2*Sqrt[10 - (5*I)*Sqrt[31]]*(62 - (101*I)*Sqrt[31])*ArcT
anh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]] + 2*Sqrt[10 + (5*I)*Sqrt[31]]*(62 + (101*I)*Sqrt[31])*ArcTanh[Sqrt[5
+ 10*x]/Sqrt[2 + I*Sqrt[31]]])/24025

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Maple [B]  time = 0.076, size = 642, normalized size = 2.3 \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)^2,x)

[Out]

16*(-27/3100*(1+2*x)^(3/2)-7/1550*(1+2*x)^(1/2))/((1+2*x)^2-8/5*x+3/5)-101/9610*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)+132/24025*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)+101/4805/(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)-264/4805/(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)+8/155/(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)+101/9610*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)-132/24025*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)+101/4805/(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)-264/480
5/(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)+8/155/(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 )}^{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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.89147, size = 2557, normalized size = 9.04 \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)^2,x, algorithm="fricas")

[Out]

1/199574966022250*(1149356*5969915^(1/4)*sqrt(826)*sqrt(155)*sqrt(35)*(5*x^2 + 3*x + 2)*sqrt(32678*sqrt(35) +
361375)*arctan(1/32833385198242899725*5969915^(3/4)*sqrt(826)*sqrt(299)*sqrt(155)*sqrt(59)*sqrt(5969915^(1/4)*
sqrt(826)*sqrt(155)*(97*sqrt(35)*sqrt(31) - 70*sqrt(31))*sqrt(2*x + 1)*sqrt(32678*sqrt(35) + 361375) + 4153485
2450*x + 4153485245*sqrt(35) + 20767426225)*sqrt(32678*sqrt(35) + 361375)*(2*sqrt(35) - 97) - 1/1715389406185*
5969915^(3/4)*sqrt(826)*sqrt(155)*sqrt(2*x + 1)*sqrt(32678*sqrt(35) + 361375)*(2*sqrt(35) - 97) + 1/31*sqrt(35
)*sqrt(31) + 2/31*sqrt(31)) + 1149356*5969915^(1/4)*sqrt(826)*sqrt(155)*sqrt(35)*(5*x^2 + 3*x + 2)*sqrt(32678*
sqrt(35) + 361375)*arctan(1/1641669259912144986250*5969915^(3/4)*sqrt(826)*sqrt(299)*sqrt(155)*sqrt(-147500*59
69915^(1/4)*sqrt(826)*sqrt(155)*(97*sqrt(35)*sqrt(31) - 70*sqrt(31))*sqrt(2*x + 1)*sqrt(32678*sqrt(35) + 36137
5) + 6126390736375000*x + 612639073637500*sqrt(35) + 3063195368187500)*sqrt(32678*sqrt(35) + 361375)*(2*sqrt(3
5) - 97) - 1/1715389406185*5969915^(3/4)*sqrt(826)*sqrt(155)*sqrt(2*x + 1)*sqrt(32678*sqrt(35) + 361375)*(2*sq
rt(35) - 97) - 1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 5969915^(1/4)*sqrt(826)*sqrt(155)*(32678*sqrt(35)*sqr
t(31)*(5*x^2 + 3*x + 2) - 361375*sqrt(31)*(5*x^2 + 3*x + 2))*sqrt(32678*sqrt(35) + 361375)*log(147500/299*5969
915^(1/4)*sqrt(826)*sqrt(155)*(97*sqrt(35)*sqrt(31) - 70*sqrt(31))*sqrt(2*x + 1)*sqrt(32678*sqrt(35) + 361375)
+ 20489601125000*x + 2048960112500*sqrt(35) + 10244800562500) - 5969915^(1/4)*sqrt(826)*sqrt(155)*(32678*sqrt
(35)*sqrt(31)*(5*x^2 + 3*x + 2) - 361375*sqrt(31)*(5*x^2 + 3*x + 2))*sqrt(32678*sqrt(35) + 361375)*log(-147500
/299*5969915^(1/4)*sqrt(826)*sqrt(155)*(97*sqrt(35)*sqrt(31) - 70*sqrt(31))*sqrt(2*x + 1)*sqrt(32678*sqrt(35)
+ 361375) + 20489601125000*x + 2048960112500*sqrt(35) + 10244800562500) - 1287580425950*(54*x + 41)*sqrt(2*x +
1))/(5*x^2 + 3*x + 2)

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Sympy [A]  time = 127.154, size = 246, normalized size = 0.87 \begin{align*} - \frac{304 \left (2 x + 1\right )^{\frac{3}{2}}}{5 \left (- 992 x + 620 \left (2 x + 1\right )^{2} + 372\right )} - \frac{896 \left (2 x + 1\right )^{\frac{3}{2}}}{5 \left (- 6944 x + 4340 \left (2 x + 1\right )^{2} + 2604\right )} + \frac{608 \sqrt{2 x + 1}}{25 \left (- 992 x + 620 \left (2 x + 1\right )^{2} + 372\right )} - \frac{12096 \sqrt{2 x + 1}}{25 \left (- 6944 x + 4340 \left (2 x + 1\right )^{2} + 2604\right )} - \frac{304 \operatorname{RootSum}{\left (407144088666112 t^{4} + 3325152256 t^{2} + 11045, \left ( t \mapsto t \log{\left (\frac{33312534528 t^{3}}{235} + \frac{166784 t}{235} + \sqrt{2 x + 1} \right )} \right )\right )}}{25} - \frac{448 \operatorname{RootSum}{\left (19950060344639488 t^{4} + 498437272576 t^{2} + 10878125, \left ( t \mapsto t \log{\left (- \frac{11049511452672 t^{3}}{2205125} + \frac{307918256 t}{2205125} + \sqrt{2 x + 1} \right )} \right )\right )}}{25} + \frac{64 \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} + \frac{16 \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 )}}{5} \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)**2,x)

[Out]

-304*(2*x + 1)**(3/2)/(5*(-992*x + 620*(2*x + 1)**2 + 372)) - 896*(2*x + 1)**(3/2)/(5*(-6944*x + 4340*(2*x + 1
)**2 + 2604)) + 608*sqrt(2*x + 1)/(25*(-992*x + 620*(2*x + 1)**2 + 372)) - 12096*sqrt(2*x + 1)/(25*(-6944*x +
4340*(2*x + 1)**2 + 2604)) - 304*RootSum(407144088666112*_t**4 + 3325152256*_t**2 + 11045, Lambda(_t, _t*log(3
3312534528*_t**3/235 + 166784*_t/235 + sqrt(2*x + 1))))/25 - 448*RootSum(19950060344639488*_t**4 + 49843727257
6*_t**2 + 10878125, Lambda(_t, _t*log(-11049511452672*_t**3/2205125 + 307918256*_t/2205125 + sqrt(2*x + 1))))/
25 + 64*RootSum(1722112*_t**4 + 1984*_t**2 + 5, Lambda(_t, _t*log(-27776*_t**3/5 + 108*_t/5 + sqrt(2*x + 1))))
/25 + 16*RootSum(1230080*_t**4 + 1984*_t**2 + 7, Lambda(_t, _t*log(9920*_t**3 + 8*_t + 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{5}{2}}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{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)^2,x, algorithm="giac")

[Out]

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