### 3.2315 $$\int \frac{1}{(1+2 x)^{5/2} (2+3 x+5 x^2)} \, dx$$

Optimal. Leaf size=266 $-\frac{16}{49 \sqrt{2 x+1}}-\frac{4}{21 (2 x+1)^{3/2}}-\frac{1}{49} \sqrt{\frac{1}{434} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{49} \sqrt{\frac{1}{434} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

[Out]

-4/(21*(1 + 2*x)^(3/2)) - 16/(49*Sqrt[1 + 2*x]) + (Sqrt[(2*(7162 + 1225*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + S
qrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/49 - (Sqrt[(2*(7162 + 1225*Sqrt[35]))/217]*ArcTan[(Sq
rt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/49 - (Sqrt[(-7162 + 1225*Sqrt[35])/434]*L
og[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/49 + (Sqrt[(-7162 + 1225*Sqrt[35])/434]*Lo
g[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/49

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Rubi [A]  time = 0.382157, 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 = {709, 828, 826, 1169, 634, 618, 204, 628} $-\frac{16}{49 \sqrt{2 x+1}}-\frac{4}{21 (2 x+1)^{3/2}}-\frac{1}{49} \sqrt{\frac{1}{434} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{49} \sqrt{\frac{1}{434} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-4/(21*(1 + 2*x)^(3/2)) - 16/(49*Sqrt[1 + 2*x]) + (Sqrt[(2*(7162 + 1225*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + S
qrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/49 - (Sqrt[(2*(7162 + 1225*Sqrt[35]))/217]*ArcTan[(Sq
rt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/49 - (Sqrt[(-7162 + 1225*Sqrt[35])/434]*L
og[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/49 + (Sqrt[(-7162 + 1225*Sqrt[35])/434]*Lo
g[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/49

Rule 709

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

Rule 828

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

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}{(1+2 x)^{5/2} \left (2+3 x+5 x^2\right )} \, dx &=-\frac{4}{21 (1+2 x)^{3/2}}+\frac{1}{7} \int \frac{-1-10 x}{(1+2 x)^{3/2} \left (2+3 x+5 x^2\right )} \, dx\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}+\frac{1}{49} \int \frac{-39-40 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}+\frac{2}{49} \operatorname{Subst}\left (\int \frac{-38-40 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}+\frac{\operatorname{Subst}\left (\int \frac{-38 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (-38+8 \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 )}{49 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{-38 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (-38+8 \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 )}{49 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}-\frac{\left (140+19 \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 )}{1715}-\frac{\left (140+19 \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 )}{1715}-\frac{1}{49} \sqrt{\frac{1}{434} \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}{49} \sqrt{\frac{1}{434} \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{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}-\frac{1}{49} \sqrt{\frac{1}{434} \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}{49} \sqrt{\frac{1}{434} \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{\left (2 \left (140+19 \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 )}{1715}+\frac{\left (2 \left (140+19 \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 )}{1715}\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}+\frac{1}{49} \sqrt{\frac{2}{217} \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}{49} \sqrt{\frac{2}{217} \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}{49} \sqrt{\frac{1}{434} \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}{49} \sqrt{\frac{1}{434} \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 = 0.527477, size = 133, normalized size = 0.5 $\frac{2 \left (-\frac{2170 (24 x+19)}{(2 x+1)^{3/2}}+3 i \sqrt{10-5 i \sqrt{31}} \left (178 \sqrt{31}+589 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )-3 i \sqrt{10+5 i \sqrt{31}} \left (178 \sqrt{31}-589 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{159495}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*((-2170*(19 + 24*x))/(1 + 2*x)^(3/2) + (3*I)*Sqrt[10 - (5*I)*Sqrt[31]]*(589*I + 178*Sqrt[31])*ArcTanh[Sqrt[
5 + 10*x]/Sqrt[2 - I*Sqrt[31]]] - (3*I)*Sqrt[10 + (5*I)*Sqrt[31]]*(-589*I + 178*Sqrt[31])*ArcTanh[Sqrt[5 + 10*
x]/Sqrt[2 + I*Sqrt[31]]]))/159495

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Maple [B]  time = 0.075, 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/(1+2*x)^(5/2)/(5*x^2+3*x+2),x)

[Out]

-27/3038*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)+89/10633*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)+135/1519/(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)-178/10633/(10*5^(1/2)*7^(1/2)-20)^(1/2)*ar
ctan((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)-76/343/(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)+27/3038*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)-89/10633*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)+135/1519/(10*5^(1/2)*7^(1/2)-2
0)^(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)-178/10633/(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)-76/343/(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)-4/21/(1+2*x)^(3/2)-16/49/(1+2*x)^(1/2)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.7599, size = 2337, normalized size = 8.79 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/13774945170*(74028*sqrt(217)*35^(3/4)*sqrt(2)*(4*x^2 + 4*x + 1)*sqrt(7162*sqrt(35) + 42875)*arctan(1/326335
010575*sqrt(1085)*sqrt(217)*sqrt(199)*35^(3/4)*sqrt(2)*sqrt(sqrt(217)*35^(1/4)*sqrt(2)*(4*sqrt(35)*sqrt(31) -
19*sqrt(31))*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875) + 431830*x + 43183*sqrt(35) + 215915)*sqrt(7162*sqrt(35
) + 42875)*(19*sqrt(35) - 140) - 1/1511405*sqrt(217)*35^(3/4)*sqrt(2)*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875
)*(19*sqrt(35) - 140) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 74028*sqrt(217)*35^(3/4)*sqrt(2)*(4*x^2 + 4*
x + 1)*sqrt(7162*sqrt(35) + 42875)*arctan(1/799520775908750*sqrt(217)*sqrt(199)*35^(3/4)*sqrt(2)*sqrt(-6512712
500*sqrt(217)*35^(1/4)*sqrt(2)*(4*sqrt(35)*sqrt(31) - 19*sqrt(31))*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875) +
2812384638875000*x + 281238463887500*sqrt(35) + 1406192319437500)*sqrt(7162*sqrt(35) + 42875)*(19*sqrt(35) -
140) - 1/1511405*sqrt(217)*35^(3/4)*sqrt(2)*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875)*(19*sqrt(35) - 140) - 1/
31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 3*sqrt(217)*35^(1/4)*sqrt(2)*(7162*sqrt(35)*sqrt(31)*(4*x^2 + 4*x + 1)
- 42875*sqrt(31)*(4*x^2 + 4*x + 1))*sqrt(7162*sqrt(35) + 42875)*log(6512712500/199*sqrt(217)*35^(1/4)*sqrt(2)
*(4*sqrt(35)*sqrt(31) - 19*sqrt(31))*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875) + 14132586125000*x + 1413258612
500*sqrt(35) + 7066293062500) - 3*sqrt(217)*35^(1/4)*sqrt(2)*(7162*sqrt(35)*sqrt(31)*(4*x^2 + 4*x + 1) - 42875
*sqrt(31)*(4*x^2 + 4*x + 1))*sqrt(7162*sqrt(35) + 42875)*log(-6512712500/199*sqrt(217)*35^(1/4)*sqrt(2)*(4*sqr
t(35)*sqrt(31) - 19*sqrt(31))*sqrt(2*x + 1)*sqrt(7162*sqrt(35) + 42875) + 14132586125000*x + 1413258612500*sqr
t(35) + 7066293062500) + 374828440*(24*x + 19)*sqrt(2*x + 1))/(4*x^2 + 4*x + 1)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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