### 3.2322 $$\int \frac{1}{(1+2 x)^{5/2} (2+3 x+5 x^2)^2} \, dx$$

Optimal. Leaf size=296 $\frac{20 x+37}{217 (2 x+1)^{3/2} \left (5 x^2+3 x+2\right )}-\frac{4680}{10633 \sqrt{2 x+1}}-\frac{820}{4557 (2 x+1)^{3/2}}-\frac{5 \sqrt{\frac{1}{434} \left (2632525 \sqrt{35}-12504542\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{10633}+\frac{5 \sqrt{\frac{1}{434} \left (2632525 \sqrt{35}-12504542\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{10633}+\frac{5 \sqrt{\frac{2}{217} \left (12504542+2632525 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{10633}-\frac{5 \sqrt{\frac{2}{217} \left (12504542+2632525 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{10633}$

[Out]

-820/(4557*(1 + 2*x)^(3/2)) - 4680/(10633*Sqrt[1 + 2*x]) + (37 + 20*x)/(217*(1 + 2*x)^(3/2)*(2 + 3*x + 5*x^2))
+ (5*Sqrt[(2*(12504542 + 2632525*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*
(-2 + Sqrt[35])]])/10633 - (5*Sqrt[(2*(12504542 + 2632525*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10
*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/10633 - (5*Sqrt[(-12504542 + 2632525*Sqrt[35])/434]*Log[Sqrt[35] -
Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10633 + (5*Sqrt[(-12504542 + 2632525*Sqrt[35])/434]*Log[
Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10633

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Rubi [A]  time = 0.44318, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.364, Rules used = {740, 828, 826, 1169, 634, 618, 204, 628} $\frac{20 x+37}{217 (2 x+1)^{3/2} \left (5 x^2+3 x+2\right )}-\frac{4680}{10633 \sqrt{2 x+1}}-\frac{820}{4557 (2 x+1)^{3/2}}-\frac{5 \sqrt{\frac{1}{434} \left (2632525 \sqrt{35}-12504542\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{10633}+\frac{5 \sqrt{\frac{1}{434} \left (2632525 \sqrt{35}-12504542\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{10633}+\frac{5 \sqrt{\frac{2}{217} \left (12504542+2632525 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{10633}-\frac{5 \sqrt{\frac{2}{217} \left (12504542+2632525 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{10633}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-820/(4557*(1 + 2*x)^(3/2)) - 4680/(10633*Sqrt[1 + 2*x]) + (37 + 20*x)/(217*(1 + 2*x)^(3/2)*(2 + 3*x + 5*x^2))
+ (5*Sqrt[(2*(12504542 + 2632525*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*
(-2 + Sqrt[35])]])/10633 - (5*Sqrt[(2*(12504542 + 2632525*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10
*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/10633 - (5*Sqrt[(-12504542 + 2632525*Sqrt[35])/434]*Log[Sqrt[35] -
Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10633 + (5*Sqrt[(-12504542 + 2632525*Sqrt[35])/434]*Log[
Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10633

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), 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[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

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 )^2} \, dx &=\frac{37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac{1}{217} \int \frac{255+100 x}{(1+2 x)^{5/2} \left (2+3 x+5 x^2\right )} \, dx\\ &=-\frac{820}{4557 (1+2 x)^{3/2}}+\frac{37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac{\int \frac{145-2050 x}{(1+2 x)^{3/2} \left (2+3 x+5 x^2\right )} \, dx}{1519}\\ &=-\frac{820}{4557 (1+2 x)^{3/2}}-\frac{4680}{10633 \sqrt{1+2 x}}+\frac{37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac{\int \frac{-8345-11700 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx}{10633}\\ &=-\frac{820}{4557 (1+2 x)^{3/2}}-\frac{4680}{10633 \sqrt{1+2 x}}+\frac{37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{-4990-11700 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )}{10633}\\ &=-\frac{820}{4557 (1+2 x)^{3/2}}-\frac{4680}{10633 \sqrt{1+2 x}}+\frac{37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-998 \sqrt{10 \left (2+\sqrt{35}\right )}-\left (-4990+2340 \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 )}{10633 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{-998 \sqrt{10 \left (2+\sqrt{35}\right )}+\left (-4990+2340 \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 )}{10633 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{820}{4557 (1+2 x)^{3/2}}-\frac{4680}{10633 \sqrt{1+2 x}}+\frac{37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac{\left (5 \left (499-234 \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 )}{10633 \sqrt{14 \left (2+\sqrt{35}\right )}}-\frac{\left (5 \left (499-234 \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 )}{10633 \sqrt{14 \left (2+\sqrt{35}\right )}}-\frac{\left (8190+499 \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 )}{74431}-\frac{\left (8190+499 \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 )}{74431}\\ &=-\frac{820}{4557 (1+2 x)^{3/2}}-\frac{4680}{10633 \sqrt{1+2 x}}+\frac{37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}-\frac{5 \sqrt{-\frac{6252271}{217}+\frac{376075 \sqrt{35}}{62}} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{10633}+\frac{5 \sqrt{-\frac{6252271}{217}+\frac{376075 \sqrt{35}}{62}} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{10633}+\frac{\left (2 \left (8190+499 \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 )}{74431}+\frac{\left (2 \left (8190+499 \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 )}{74431}\\ &=-\frac{820}{4557 (1+2 x)^{3/2}}-\frac{4680}{10633 \sqrt{1+2 x}}+\frac{37+20 x}{217 (1+2 x)^{3/2} \left (2+3 x+5 x^2\right )}+\frac{5 \sqrt{\frac{2}{7 \left (-2+\sqrt{35}\right )}} \left (499+234 \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 )}{10633}-\frac{5 \sqrt{\frac{2}{7 \left (-2+\sqrt{35}\right )}} \left (499+234 \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 )}{10633}-\frac{5 \sqrt{-\frac{6252271}{217}+\frac{376075 \sqrt{35}}{62}} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{10633}+\frac{5 \sqrt{-\frac{6252271}{217}+\frac{376075 \sqrt{35}}{62}} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{10633}\\ \end{align*}

Mathematica [C]  time = 0.51755, size = 176, normalized size = 0.59 $\frac{1}{217} \left (\frac{20 x+37}{(2 x+1)^{3/2} \left (5 x^2+3 x+2\right )}-\frac{4680}{49 \sqrt{2 x+1}}-\frac{820}{21 (2 x+1)^{3/2}}+\frac{2 i \sqrt{5} \left (\sqrt{2-i \sqrt{31}} \left (9188 \sqrt{31}+15469 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+\left (-9188 \sqrt{31}+15469 i\right ) \sqrt{2+i \sqrt{31}} \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{10633}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-820/(21*(1 + 2*x)^(3/2)) - 4680/(49*Sqrt[1 + 2*x]) + (37 + 20*x)/((1 + 2*x)^(3/2)*(2 + 3*x + 5*x^2)) + ((2*I
)/10633)*Sqrt[5]*(Sqrt[2 - I*Sqrt[31]]*(15469*I + 9188*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]]
+ (15469*I - 9188*Sqrt[31])*Sqrt[2 + I*Sqrt[31]]*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]]))/217

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Maple [B]  time = 0.081, size = 660, 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/(1+2*x)^(5/2)/(5*x^2+3*x+2)^2,x)

[Out]

-16/343*(89/62*(1+2*x)^(3/2)+233/620*(1+2*x)^(1/2))/((1+2*x)^2-8/5*x+3/5)-4835/659246*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)+22970/2307361*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)+2417
5/329623/(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)-45940/2307361/(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)-9
980/74431/(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)+4835/659246*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)-22970/2307361*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)+24175/329623/(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)-45940/2307361/(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)-9980/74431/(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))*5^(1/2)
*7^(1/2)-16/147/(1+2*x)^(3/2)-128/343/(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 )}^{2}{\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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.62435, size = 3170, normalized size = 10.71 \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)^2,x, algorithm="fricas")

[Out]

-1/10765101069366070602*(620294748*161637035^(1/4)*sqrt(4298)*sqrt(217)*sqrt(35)*(20*x^4 + 32*x^3 + 25*x^2 + 1
1*x + 2)*sqrt(12504542*sqrt(35) + 92138375)*arctan(1/55152316249116723757744225*161637035^(3/4)*sqrt(4298)*sqr
t(1535)*sqrt(217)*sqrt(149)*sqrt(161637035^(1/4)*sqrt(4298)*sqrt(217)*(234*sqrt(35)*sqrt(31) - 499*sqrt(31))*s
qrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375) + 7775911578470*x + 777591157847*sqrt(35) + 3887955789235)*sqr
t(12504542*sqrt(35) + 92138375)*(499*sqrt(35) - 8190) - 1/58486518937462105*161637035^(3/4)*sqrt(4298)*sqrt(21
7)*sqrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375)*(499*sqrt(35) - 8190) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt
(31)) + 620294748*161637035^(1/4)*sqrt(4298)*sqrt(217)*sqrt(35)*(20*x^4 + 32*x^3 + 25*x^2 + 11*x + 2)*sqrt(125
04542*sqrt(35) + 92138375)*arctan(1/4729311118361759062226567293750*161637035^(3/4)*sqrt(4298)*sqrt(217)*sqrt(
149)*sqrt(-11286950937500*161637035^(1/4)*sqrt(4298)*sqrt(217)*(234*sqrt(35)*sqrt(31) - 499*sqrt(31))*sqrt(2*x
+ 1)*sqrt(12504542*sqrt(35) + 92138375) + 87766332480529071315625000*x + 8776633248052907131562500*sqrt(35) +
43883166240264535657812500)*sqrt(12504542*sqrt(35) + 92138375)*(499*sqrt(35) - 8190) - 1/58486518937462105*16
1637035^(3/4)*sqrt(4298)*sqrt(217)*sqrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375)*(499*sqrt(35) - 8190) - 1/
31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 3*161637035^(1/4)*sqrt(4298)*sqrt(217)*(12504542*sqrt(35)*sqrt(31)*(20
*x^4 + 32*x^3 + 25*x^2 + 11*x + 2) - 92138375*sqrt(31)*(20*x^4 + 32*x^3 + 25*x^2 + 11*x + 2))*sqrt(12504542*sq
rt(35) + 92138375)*log(11286950937500/53789*161637035^(1/4)*sqrt(4298)*sqrt(217)*(234*sqrt(35)*sqrt(31) - 499*
sqrt(31))*sqrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375) + 1631678084376528125000*x + 163167808437652812500*
sqrt(35) + 815839042188264062500) - 3*161637035^(1/4)*sqrt(4298)*sqrt(217)*(12504542*sqrt(35)*sqrt(31)*(20*x^4
+ 32*x^3 + 25*x^2 + 11*x + 2) - 92138375*sqrt(31)*(20*x^4 + 32*x^3 + 25*x^2 + 11*x + 2))*sqrt(12504542*sqrt(3
5) + 92138375)*log(-11286950937500/53789*161637035^(1/4)*sqrt(4298)*sqrt(217)*(234*sqrt(35)*sqrt(31) - 499*sqr
t(31))*sqrt(2*x + 1)*sqrt(12504542*sqrt(35) + 92138375) + 1631678084376528125000*x + 163167808437652812500*sqr
t(35) + 815839042188264062500) + 337474562505598*(140400*x^3 + 183140*x^2 + 112560*x + 34121)*sqrt(2*x + 1))/(
20*x^4 + 32*x^3 + 25*x^2 + 11*x + 2)

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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 )^{2}}\, 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)**2,x)

[Out]

Integral(1/((2*x + 1)**(5/2)*(5*x**2 + 3*x + 2)**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 )}^{2}{\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)^2,x, algorithm="giac")

[Out]

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