### 3.2324 $$\int \frac{(1+2 x)^{7/2}}{(2+3 x+5 x^2)^3} \, dx$$

Optimal. Leaf size=300 $-\frac{(5-4 x) (2 x+1)^{5/2}}{62 \left (5 x^2+3 x+2\right )^2}-\frac{(957-592 x) \sqrt{2 x+1}}{9610 \left (5 x^2+3 x+2\right )}-\frac{\sqrt{\frac{1}{310} \left (1806875 \sqrt{35}-9651062\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{9610}+\frac{\sqrt{\frac{1}{310} \left (1806875 \sqrt{35}-9651062\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{9610}-\frac{\sqrt{\frac{1}{310} \left (9651062+1806875 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{4805}+\frac{\sqrt{\frac{1}{310} \left (9651062+1806875 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{4805}$

[Out]

-((5 - 4*x)*(1 + 2*x)^(5/2))/(62*(2 + 3*x + 5*x^2)^2) - ((957 - 592*x)*Sqrt[1 + 2*x])/(9610*(2 + 3*x + 5*x^2))
- (Sqrt[(9651062 + 1806875*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + S
qrt[35])]])/4805 + (Sqrt[(9651062 + 1806875*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])
/Sqrt[10*(-2 + Sqrt[35])]])/4805 - (Sqrt[(-9651062 + 1806875*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[3
5])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/9610 + (Sqrt[(-9651062 + 1806875*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 +
Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/9610

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Rubi [A]  time = 0.438287, antiderivative size = 300, 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, 818, 826, 1169, 634, 618, 204, 628} $-\frac{(5-4 x) (2 x+1)^{5/2}}{62 \left (5 x^2+3 x+2\right )^2}-\frac{(957-592 x) \sqrt{2 x+1}}{9610 \left (5 x^2+3 x+2\right )}-\frac{\sqrt{\frac{1}{310} \left (1806875 \sqrt{35}-9651062\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{9610}+\frac{\sqrt{\frac{1}{310} \left (1806875 \sqrt{35}-9651062\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{9610}-\frac{\sqrt{\frac{1}{310} \left (9651062+1806875 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{4805}+\frac{\sqrt{\frac{1}{310} \left (9651062+1806875 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )}{4805}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((5 - 4*x)*(1 + 2*x)^(5/2))/(62*(2 + 3*x + 5*x^2)^2) - ((957 - 592*x)*Sqrt[1 + 2*x])/(9610*(2 + 3*x + 5*x^2))
- (Sqrt[(9651062 + 1806875*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + S
qrt[35])]])/4805 + (Sqrt[(9651062 + 1806875*Sqrt[35])/310]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])
/Sqrt[10*(-2 + Sqrt[35])]])/4805 - (Sqrt[(-9651062 + 1806875*Sqrt[35])/310]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[3
5])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/9610 + (Sqrt[(-9651062 + 1806875*Sqrt[35])/310]*Log[Sqrt[35] + Sqrt[10*(2 +
Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/9610

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 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 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}}{\left (2+3 x+5 x^2\right )^3} \, dx &=-\frac{(5-4 x) (1+2 x)^{5/2}}{62 \left (2+3 x+5 x^2\right )^2}+\frac{1}{62} \int \frac{(1+2 x)^{3/2} (37+4 x)}{\left (2+3 x+5 x^2\right )^2} \, dx\\ &=-\frac{(5-4 x) (1+2 x)^{5/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(957-592 x) \sqrt{1+2 x}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\int \frac{-1797-1088 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx}{9610}\\ &=-\frac{(5-4 x) (1+2 x)^{5/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(957-592 x) \sqrt{1+2 x}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{-2506-1088 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )}{4805}\\ &=-\frac{(5-4 x) (1+2 x)^{5/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(957-592 x) \sqrt{1+2 x}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{-2506 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (-2506+1088 \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 )}{9610 \sqrt{14 \left (2+\sqrt{35}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{-2506 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (-2506+1088 \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 )}{9610 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{(5-4 x) (1+2 x)^{5/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(957-592 x) \sqrt{1+2 x}}{9610 \left (2+3 x+5 x^2\right )}+\frac{\sqrt{1417371+194752 \sqrt{35}} \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 )}{48050}+\frac{\sqrt{1417371+194752 \sqrt{35}} \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 )}{48050}-\frac{\sqrt{\frac{1}{310} \left (-9651062+1806875 \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 )}{9610}+\frac{\sqrt{\frac{1}{310} \left (-9651062+1806875 \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 )}{9610}\\ &=-\frac{(5-4 x) (1+2 x)^{5/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(957-592 x) \sqrt{1+2 x}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\sqrt{\frac{1}{310} \left (-9651062+1806875 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{9610}+\frac{\sqrt{\frac{1}{310} \left (-9651062+1806875 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{9610}-\frac{\sqrt{1417371+194752 \sqrt{35}} \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 )}{24025}-\frac{\sqrt{1417371+194752 \sqrt{35}} \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 )}{24025}\\ &=-\frac{(5-4 x) (1+2 x)^{5/2}}{62 \left (2+3 x+5 x^2\right )^2}-\frac{(957-592 x) \sqrt{1+2 x}}{9610 \left (2+3 x+5 x^2\right )}-\frac{\sqrt{\frac{1}{310} \left (9651062+1806875 \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 )}{4805}+\frac{\sqrt{\frac{1}{310} \left (9651062+1806875 \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 )}{4805}-\frac{\sqrt{\frac{1}{310} \left (-9651062+1806875 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{9610}+\frac{\sqrt{\frac{1}{310} \left (-9651062+1806875 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{9610}\\ \end{align*}

Mathematica [C]  time = 0.626067, size = 223, normalized size = 0.74 $\frac{-\frac{5 (400 x+89) (2 x+1)^{9/2}}{5 x^2+3 x+2}+\frac{217 (20 x+37) (2 x+1)^{9/2}}{\left (5 x^2+3 x+2\right )^2}+800 (2 x+1)^{7/2}+196 (2 x+1)^{5/2}-2352 (2 x+1)^{3/2}-\frac{35084}{5} \sqrt{2 x+1}+\frac{98 \left (\sqrt{2-i \sqrt{31}} \left (5549-902 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+\sqrt{2+i \sqrt{31}} \left (5549+902 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{155 \sqrt{5}}}{94178}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((-35084*Sqrt[1 + 2*x])/5 - 2352*(1 + 2*x)^(3/2) + 196*(1 + 2*x)^(5/2) + 800*(1 + 2*x)^(7/2) + (217*(1 + 2*x)^
(9/2)*(37 + 20*x))/(2 + 3*x + 5*x^2)^2 - (5*(1 + 2*x)^(9/2)*(89 + 400*x))/(2 + 3*x + 5*x^2) + (98*(Sqrt[2 - I*
Sqrt[31]]*(5549 - (902*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]] + Sqrt[2 + I*Sqrt[31]]*(5549
+ (902*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]]))/(155*Sqrt[5]))/94178

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Maple [B]  time = 0.076, size = 662, 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+2*x)^(7/2)/(5*x^2+3*x+2)^3,x)

[Out]

1600*(17/24025*(1+2*x)^(7/2)-11789/3844000*(1+2*x)^(5/2)+1771/961000*(1+2*x)^(3/2)-8771/3844000*(1+2*x)^(1/2))
/(5*(1+2*x)^2-8*x+3)^2+7353/2979100*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)-451/297910*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)-7353/297910/(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)+451/148955
/(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)+358/4805/(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)*7^(1/2)-7353/2979100*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)+451
/297910*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)-7353/297910/(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)+451/148955/(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)+358/4805/(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)*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}}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{3}}\,{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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.71219, size = 2979, normalized size = 9.93 \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)^(7/2)/(5*x^2+3*x+2)^3,x, algorithm="fricas")

[Out]

-1/157428509198663500*(102361876*121835^(1/4)*sqrt(155)*sqrt(118)*sqrt(35)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x +
4)*sqrt(9651062*sqrt(35) + 63240625)*arctan(1/5314799928145246742525*121835^(3/4)*sqrt(26629)*sqrt(413)*sqrt(1
55)*sqrt(118)*sqrt(121835^(1/4)*sqrt(155)*sqrt(118)*(544*sqrt(35)*sqrt(31) - 6265*sqrt(31))*sqrt(2*x + 1)*sqrt
(9651062*sqrt(35) + 63240625) + 528443184850*x + 52844318485*sqrt(35) + 264221592425)*sqrt(9651062*sqrt(35) +
63240625)*(179*sqrt(35) - 544) - 1/3117814790615*121835^(3/4)*sqrt(155)*sqrt(118)*sqrt(2*x + 1)*sqrt(9651062*s
qrt(35) + 63240625)*(179*sqrt(35) - 544) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31)) + 102361876*121835^(1/4)*sq
rt(155)*sqrt(118)*sqrt(35)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(9651062*sqrt(35) + 63240625)*arctan(1/23
252249685635454498546875*121835^(3/4)*sqrt(26629)*sqrt(155)*sqrt(118)*sqrt(-7905078125*121835^(1/4)*sqrt(155)*
sqrt(118)*(544*sqrt(35)*sqrt(31) - 6265*sqrt(31))*sqrt(2*x + 1)*sqrt(9651062*sqrt(35) + 63240625) + 4177384660
863066406250*x + 417738466086306640625*sqrt(35) + 2088692330431533203125)*sqrt(9651062*sqrt(35) + 63240625)*(1
79*sqrt(35) - 544) - 1/3117814790615*121835^(3/4)*sqrt(155)*sqrt(118)*sqrt(2*x + 1)*sqrt(9651062*sqrt(35) + 63
240625)*(179*sqrt(35) - 544) - 1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) - 121835^(1/4)*sqrt(155)*sqrt(118)*(965
1062*sqrt(35)*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4) - 63240625*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 1
2*x + 4))*sqrt(9651062*sqrt(35) + 63240625)*log(7905078125/26629*121835^(1/4)*sqrt(155)*sqrt(118)*(544*sqrt(35
)*sqrt(31) - 6265*sqrt(31))*sqrt(2*x + 1)*sqrt(9651062*sqrt(35) + 63240625) + 156873508613281250*x + 156873508
61328125*sqrt(35) + 78436754306640625) + 121835^(1/4)*sqrt(155)*sqrt(118)*(9651062*sqrt(35)*sqrt(31)*(25*x^4 +
30*x^3 + 29*x^2 + 12*x + 4) - 63240625*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4))*sqrt(9651062*sqrt(35)
+ 63240625)*log(-7905078125/26629*121835^(1/4)*sqrt(155)*sqrt(118)*(544*sqrt(35)*sqrt(31) - 6265*sqrt(31))*sqr
t(2*x + 1)*sqrt(9651062*sqrt(35) + 63240625) + 156873508613281250*x + 15687350861328125*sqrt(35) + 78436754306
640625) - 16381738730350*(5440*x^3 - 3629*x^2 - 4167*x - 2689)*sqrt(2*x + 1))/(25*x^4 + 30*x^3 + 29*x^2 + 12*x
+ 4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**3,x)

[Out]

Timed out

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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}}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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