### 3.2311 $$\int \frac{(1+2 x)^{3/2}}{2+3 x+5 x^2} \, dx$$

Optimal. Leaf size=253 $\frac{4}{5} \sqrt{2 x+1}+\frac{1}{5} \sqrt{\frac{1}{310} \left (178+35 \sqrt{35}\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{5} \sqrt{\frac{1}{310} \left (178+35 \sqrt{35}\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{5} \sqrt{\frac{2}{155} \left (35 \sqrt{35}-178\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{5} \sqrt{\frac{2}{155} \left (35 \sqrt{35}-178\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

[Out]

(4*Sqrt[1 + 2*x])/5 + (Sqrt[(2*(-178 + 35*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/
Sqrt[10*(-2 + Sqrt[35])]])/5 - (Sqrt[(2*(-178 + 35*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1
+ 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/5 + (Sqrt[(178 + 35*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*
Sqrt[1 + 2*x] + 5*(1 + 2*x)])/5 - (Sqrt[(178 + 35*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1
+ 2*x] + 5*(1 + 2*x)])/5

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Rubi [A]  time = 0.357923, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.318, Rules used = {703, 826, 1169, 634, 618, 204, 628} $\frac{4}{5} \sqrt{2 x+1}+\frac{1}{5} \sqrt{\frac{1}{310} \left (178+35 \sqrt{35}\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{5} \sqrt{\frac{1}{310} \left (178+35 \sqrt{35}\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{5} \sqrt{\frac{2}{155} \left (35 \sqrt{35}-178\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{5} \sqrt{\frac{2}{155} \left (35 \sqrt{35}-178\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*Sqrt[1 + 2*x])/5 + (Sqrt[(2*(-178 + 35*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/
Sqrt[10*(-2 + Sqrt[35])]])/5 - (Sqrt[(2*(-178 + 35*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1
+ 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/5 + (Sqrt[(178 + 35*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*
Sqrt[1 + 2*x] + 5*(1 + 2*x)])/5 - (Sqrt[(178 + 35*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1
+ 2*x] + 5*(1 + 2*x)])/5

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 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)^{3/2}}{2+3 x+5 x^2} \, dx &=\frac{4}{5} \sqrt{1+2 x}+\frac{1}{5} \int \frac{-3+8 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx\\ &=\frac{4}{5} \sqrt{1+2 x}+\frac{2}{5} \operatorname{Subst}\left (\int \frac{-14+8 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=\frac{4}{5} \sqrt{1+2 x}+\frac{\operatorname{Subst}\left (\int \frac{-14 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (-14-8 \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 )}{5 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{-14 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (-14-8 \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 )}{5 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=\frac{4}{5} \sqrt{1+2 x}+\frac{1}{25} \left (4-\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}{25} \left (4-\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}{5} \sqrt{\frac{1}{310} \left (178+35 \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}{5} \sqrt{\frac{1}{310} \left (178+35 \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}{5} \sqrt{1+2 x}+\frac{1}{5} \sqrt{\frac{1}{310} \left (178+35 \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}{5} \sqrt{\frac{1}{310} \left (178+35 \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} \left (2 \left (4-\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}{25} \left (2 \left (4-\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{4}{5} \sqrt{1+2 x}+\frac{1}{5} \sqrt{\frac{2}{155} \left (-178+35 \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}{5} \sqrt{\frac{2}{155} \left (-178+35 \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}{5} \sqrt{\frac{1}{310} \left (178+35 \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}{5} \sqrt{\frac{1}{310} \left (178+35 \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 = 0.265968, size = 128, normalized size = 0.51 $\frac{2}{775} \left (310 \sqrt{2 x+1}-i \sqrt{10-5 i \sqrt{31}} \left (2 \sqrt{31}-31 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+i \sqrt{10+5 i \sqrt{31}} \left (2 \sqrt{31}+31 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(310*Sqrt[1 + 2*x] - I*Sqrt[10 - (5*I)*Sqrt[31]]*(-31*I + 2*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqr
t[31]]] + I*Sqrt[10 + (5*I)*Sqrt[31]]*(31*I + 2*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]]))/775

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Maple [B]  time = 0.073, size = 616, 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)^(3/2)/(5*x^2+3*x+2),x)

[Out]

4/5*(1+2*x)^(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))*5^(1/2)
*(2*5^(1/2)*7^(1/2)+4)^(1/2)-1/155*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)+27/155/(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))*(2*5^(1/2)*7^(1/2)+4)+2/155/(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)*(2*
5^(1/2)*7^(1/2)+4)*7^(1/2)-4/5/(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)+27/1550*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)*5^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+1/155*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)+27/155/(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)+2/155/(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)*(2*5^(1/2)*7^(1/2)+4)*7^(1/2)-4/5/(10*5^(1/2)*7^(1/2)-20)^(1/2)*arc
tan((-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{3}{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)^(3/2)/(5*x^2+3*x+2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.55124, size = 2006, normalized size = 7.93 \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)^(3/2)/(5*x^2+3*x+2),x, algorithm="fricas")

[Out]

-1/1118363750*42875^(1/4)*sqrt(155)*(178*sqrt(35)*sqrt(31) + 1225*sqrt(31))*sqrt(-12460*sqrt(35) + 85750)*log(
620/19*42875^(1/4)*sqrt(155)*(4*sqrt(35)*sqrt(31) + 35*sqrt(31))*sqrt(2*x + 1)*sqrt(-12460*sqrt(35) + 85750) +
235445000*x + 23544500*sqrt(35) + 117722500) + 1/1118363750*42875^(1/4)*sqrt(155)*(178*sqrt(35)*sqrt(31) + 12
25*sqrt(31))*sqrt(-12460*sqrt(35) + 85750)*log(-620/19*42875^(1/4)*sqrt(155)*(4*sqrt(35)*sqrt(31) + 35*sqrt(31
))*sqrt(2*x + 1)*sqrt(-12460*sqrt(35) + 85750) + 235445000*x + 23544500*sqrt(35) + 117722500) + 2/949375*42875
^(1/4)*sqrt(155)*sqrt(35)*sqrt(-12460*sqrt(35) + 85750)*arctan(1/5205983256250*42875^(3/4)*sqrt(155)*sqrt(-620
*42875^(1/4)*sqrt(155)*(4*sqrt(35)*sqrt(31) + 35*sqrt(31))*sqrt(2*x + 1)*sqrt(-12460*sqrt(35) + 85750) + 44734
55000*x + 447345500*sqrt(35) + 2236727500)*(sqrt(35)*sqrt(19) + 4*sqrt(19))*sqrt(-12460*sqrt(35) + 85750) - 1/
25253375*42875^(3/4)*sqrt(155)*sqrt(2*x + 1)*(sqrt(35) + 4)*sqrt(-12460*sqrt(35) + 85750) + 1/31*sqrt(35)*sqrt
(31) + 2/31*sqrt(31)) + 2/949375*42875^(1/4)*sqrt(155)*sqrt(35)*sqrt(-12460*sqrt(35) + 85750)*arctan(-1/252533
75*42875^(3/4)*sqrt(155)*sqrt(2*x + 1)*(sqrt(35) + 4)*sqrt(-12460*sqrt(35) + 85750) + 1/16793494375*42875^(3/4
)*sqrt(42875^(1/4)*sqrt(155)*(4*sqrt(35)*sqrt(31) + 35*sqrt(31))*sqrt(2*x + 1)*sqrt(-12460*sqrt(35) + 85750) +
7215250*x + 721525*sqrt(35) + 3607625)*(sqrt(35)*sqrt(19) + 4*sqrt(19))*sqrt(-12460*sqrt(35) + 85750) - 1/31*
sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 4/5*sqrt(2*x + 1)

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Sympy [A]  time = 17.6129, size = 85, normalized size = 0.34 \begin{align*} \frac{4 \sqrt{2 x + 1}}{5} + \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} - \frac{28 \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 )}}{5} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sqrt(2*x + 1)/5 + 16*RootSum(1230080*_t**4 + 1984*_t**2 + 7, Lambda(_t, _t*log(9920*_t**3 + 8*_t + sqrt(2*x
+ 1))))/5 - 28*RootSum(1722112*_t**4 + 1984*_t**2 + 5, Lambda(_t, _t*log(-27776*_t**3/5 + 108*_t/5 + 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{3}{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)^(3/2)/(5*x^2+3*x+2),x, algorithm="giac")

[Out]

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