### 3.2309 $$\int \frac{(1+2 x)^{7/2}}{2+3 x+5 x^2} \, dx$$

Optimal. Leaf size=279 $\frac{4}{25} (2 x+1)^{5/2}+\frac{16}{75} (2 x+1)^{3/2}-\frac{76}{125} \sqrt{2 x+1}-\frac{1}{125} \sqrt{\frac{1}{310} \left (168698+42875 \sqrt{35}\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{125} \sqrt{\frac{1}{310} \left (168698+42875 \sqrt{35}\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{125} \sqrt{\frac{2}{155} \left (42875 \sqrt{35}-168698\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{125} \sqrt{\frac{2}{155} \left (42875 \sqrt{35}-168698\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

[Out]

(-76*Sqrt[1 + 2*x])/125 + (16*(1 + 2*x)^(3/2))/75 + (4*(1 + 2*x)^(5/2))/25 + (Sqrt[(2*(-168698 + 42875*Sqrt[35
]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/125 - (Sqrt[(2*(-16869
8 + 42875*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/125 -
(Sqrt[(168698 + 42875*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/125
+ (Sqrt[(168698 + 42875*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/1
25

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Rubi [A]  time = 0.60758, antiderivative size = 279, 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 = {703, 824, 826, 1169, 634, 618, 204, 628} $\frac{4}{25} (2 x+1)^{5/2}+\frac{16}{75} (2 x+1)^{3/2}-\frac{76}{125} \sqrt{2 x+1}-\frac{1}{125} \sqrt{\frac{1}{310} \left (168698+42875 \sqrt{35}\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{125} \sqrt{\frac{1}{310} \left (168698+42875 \sqrt{35}\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{125} \sqrt{\frac{2}{155} \left (42875 \sqrt{35}-168698\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{125} \sqrt{\frac{2}{155} \left (42875 \sqrt{35}-168698\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[(1 + 2*x)^(7/2)/(2 + 3*x + 5*x^2),x]

[Out]

(-76*Sqrt[1 + 2*x])/125 + (16*(1 + 2*x)^(3/2))/75 + (4*(1 + 2*x)^(5/2))/25 + (Sqrt[(2*(-168698 + 42875*Sqrt[35
]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/125 - (Sqrt[(2*(-16869
8 + 42875*Sqrt[35]))/155]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/125 -
(Sqrt[(168698 + 42875*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/125
+ (Sqrt[(168698 + 42875*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/1
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)^{7/2}}{2+3 x+5 x^2} \, dx &=\frac{4}{25} (1+2 x)^{5/2}+\frac{1}{5} \int \frac{(1+2 x)^{3/2} (-3+8 x)}{2+3 x+5 x^2} \, dx\\ &=\frac{16}{75} (1+2 x)^{3/2}+\frac{4}{25} (1+2 x)^{5/2}+\frac{1}{25} \int \frac{(-47-38 x) \sqrt{1+2 x}}{2+3 x+5 x^2} \, dx\\ &=-\frac{76}{125} \sqrt{1+2 x}+\frac{16}{75} (1+2 x)^{3/2}+\frac{4}{25} (1+2 x)^{5/2}+\frac{1}{125} \int \frac{-83-432 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx\\ &=-\frac{76}{125} \sqrt{1+2 x}+\frac{16}{75} (1+2 x)^{3/2}+\frac{4}{25} (1+2 x)^{5/2}+\frac{2}{125} \operatorname{Subst}\left (\int \frac{266-432 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=-\frac{76}{125} \sqrt{1+2 x}+\frac{16}{75} (1+2 x)^{3/2}+\frac{4}{25} (1+2 x)^{5/2}+\frac{\operatorname{Subst}\left (\int \frac{266 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (266+432 \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 )}{125 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{266 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (266+432 \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 )}{125 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{76}{125} \sqrt{1+2 x}+\frac{16}{75} (1+2 x)^{3/2}+\frac{4}{25} (1+2 x)^{5/2}+\frac{1}{625} \left (-216+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 )+\frac{1}{625} \left (-216+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 )-\frac{1}{125} \sqrt{\frac{1}{310} \left (168698+42875 \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}{125} \sqrt{\frac{1}{310} \left (168698+42875 \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{76}{125} \sqrt{1+2 x}+\frac{16}{75} (1+2 x)^{3/2}+\frac{4}{25} (1+2 x)^{5/2}-\frac{1}{125} \sqrt{\frac{1}{310} \left (168698+42875 \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} \sqrt{\frac{1}{310} \left (168698+42875 \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}{625} \left (2 \left (216-19 \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 )+\frac{1}{625} \left (2 \left (216-19 \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 )\\ &=-\frac{76}{125} \sqrt{1+2 x}+\frac{16}{75} (1+2 x)^{3/2}+\frac{4}{25} (1+2 x)^{5/2}+\frac{1}{125} \sqrt{\frac{2}{155} \left (-168698+42875 \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}{125} \sqrt{\frac{2}{155} \left (-168698+42875 \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}{125} \sqrt{\frac{1}{310} \left (168698+42875 \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} \sqrt{\frac{1}{310} \left (168698+42875 \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 = 0.315421, size = 134, normalized size = 0.48 $\frac{2 \left (620 \sqrt{2 x+1} \left (30 x^2+50 x-11\right )+3 \sqrt{10-5 i \sqrt{31}} \left (589+178 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+3 \sqrt{10+5 i \sqrt{31}} \left (589-178 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{58125}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(1 + 2*x)^(7/2)/(2 + 3*x + 5*x^2),x]

[Out]

(2*(620*Sqrt[1 + 2*x]*(-11 + 50*x + 30*x^2) + 3*Sqrt[10 - (5*I)*Sqrt[31]]*(589 + (178*I)*Sqrt[31])*ArcTanh[Sqr
t[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]] + 3*Sqrt[10 + (5*I)*Sqrt[31]]*(589 - (178*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x
]/Sqrt[2 + I*Sqrt[31]]]))/58125

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Maple [B]  time = 0.303, size = 634, 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)^(7/2)/(5*x^2+3*x+2),x)

[Out]

4/25*(1+2*x)^(5/2)+16/75*(1+2*x)^(3/2)-76/125*(1+2*x)^(1/2)+233/38750*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)+89/3875*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))*7^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)-233/3875/(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)-178/3875/(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)+76/125/(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)-233/38750*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)-89/3875*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)*7^(1/2
)*(2*5^(1/2)*7^(1/2)+4)^(1/2)-233/3875/(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)-178/3875/(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)+76/125/(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{7}{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)^(7/2)/(5*x^2+3*x+2),x, algorithm="maxima")

[Out]

integrate((2*x + 1)^(7/2)/(5*x^2 + 3*x + 2), x)

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Fricas [B]  time = 2.67357, size = 2519, normalized size = 9.03 \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)^(7/2)/(5*x^2+3*x+2),x, algorithm="fricas")

[Out]

1/1752203762968750*42875^(1/4)*sqrt(155)*(168698*sqrt(35)*sqrt(31) + 1500625*sqrt(31))*sqrt(-14465853500*sqrt(
35) + 128678593750)*log(26582500/34021*42875^(1/4)*sqrt(155)*(216*sqrt(35)*sqrt(31) + 665*sqrt(31))*sqrt(2*x +
1)*sqrt(-14465853500*sqrt(35) + 128678593750) + 353314653125000*x + 35331465312500*sqrt(35) + 176657326562500
) - 1/1752203762968750*42875^(1/4)*sqrt(155)*(168698*sqrt(35)*sqrt(31) + 1500625*sqrt(31))*sqrt(-14465853500*s
qrt(35) + 128678593750)*log(-26582500/34021*42875^(1/4)*sqrt(155)*(216*sqrt(35)*sqrt(31) + 665*sqrt(31))*sqrt(
2*x + 1)*sqrt(-14465853500*sqrt(35) + 128678593750) + 353314653125000*x + 35331465312500*sqrt(35) + 1766573265
62500) + 2/830703125*42875^(1/4)*sqrt(155)*sqrt(35)*sqrt(-14465853500*sqrt(35) + 128678593750)*arctan(1/204468
2140744622640625*42875^(3/4)*sqrt(34021)*sqrt(217)*sqrt(155)*sqrt(42875^(1/4)*sqrt(155)*(216*sqrt(35)*sqrt(31)
+ 665*sqrt(31))*sqrt(2*x + 1)*sqrt(-14465853500*sqrt(35) + 128678593750) + 452181616250*x + 45218161625*sqrt(
35) + 226090808125)*(19*sqrt(35) + 216)*sqrt(-14465853500*sqrt(35) + 128678593750) - 1/1582635656875*42875^(3/
4)*sqrt(155)*sqrt(2*x + 1)*(19*sqrt(35) + 216)*sqrt(-14465853500*sqrt(35) + 128678593750) - 1/31*sqrt(35)*sqrt
(31) - 2/31*sqrt(31)) + 2/830703125*42875^(1/4)*sqrt(155)*sqrt(35)*sqrt(-14465853500*sqrt(35) + 128678593750)*
arctan(1/715638749260617924218750*42875^(3/4)*sqrt(34021)*sqrt(155)*sqrt(-26582500*42875^(1/4)*sqrt(155)*(216*
sqrt(35)*sqrt(31) + 665*sqrt(31))*sqrt(2*x + 1)*sqrt(-14465853500*sqrt(35) + 128678593750) + 12020117813965625
000*x + 1202011781396562500*sqrt(35) + 6010058906982812500)*(19*sqrt(35) + 216)*sqrt(-14465853500*sqrt(35) + 1
28678593750) - 1/1582635656875*42875^(3/4)*sqrt(155)*sqrt(2*x + 1)*(19*sqrt(35) + 216)*sqrt(-14465853500*sqrt(
35) + 128678593750) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 8/375*(30*x^2 + 50*x - 11)*sqrt(2*x + 1)

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Sympy [A]  time = 61.4254, size = 109, normalized size = 0.39 \begin{align*} \frac{4 \left (2 x + 1\right )^{\frac{5}{2}}}{25} + \frac{16 \left (2 x + 1\right )^{\frac{3}{2}}}{75} - \frac{76 \sqrt{2 x + 1}}{125} - \frac{864 \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 )}}{125} + \frac{532 \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 )}}{125} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)**(7/2)/(5*x**2+3*x+2),x)

[Out]

4*(2*x + 1)**(5/2)/25 + 16*(2*x + 1)**(3/2)/75 - 76*sqrt(2*x + 1)/125 - 864*RootSum(1230080*_t**4 + 1984*_t**2
+ 7, Lambda(_t, _t*log(9920*_t**3 + 8*_t + sqrt(2*x + 1))))/125 + 532*RootSum(1722112*_t**4 + 1984*_t**2 + 5,
Lambda(_t, _t*log(-27776*_t**3/5 + 108*_t/5 + sqrt(2*x + 1))))/125

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, x + 1\right )}^{\frac{7}{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)^(7/2)/(5*x^2+3*x+2),x, algorithm="giac")

[Out]

integrate((2*x + 1)^(7/2)/(5*x^2 + 3*x + 2), x)