3.2314 \(\int \frac{1}{(1+2 x)^{3/2} (2+3 x+5 x^2)} \, dx\)

Optimal. Leaf size=253 \[ -\frac{4}{7 \sqrt{2 x+1}}-\frac{1}{7} \sqrt{\frac{1}{434} \left (178+35 \sqrt{35}\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{7} \sqrt{\frac{1}{434} \left (178+35 \sqrt{35}\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{7} \sqrt{\frac{2}{217} \left (35 \sqrt{35}-178\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{7} \sqrt{\frac{2}{217} \left (35 \sqrt{35}-178\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]

[Out]

-4/(7*Sqrt[1 + 2*x]) + (Sqrt[(2*(-178 + 35*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])
/Sqrt[10*(-2 + Sqrt[35])]])/7 - (Sqrt[(2*(-178 + 35*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[
1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/7 - (Sqrt[(178 + 35*Sqrt[35])/434]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]
*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/7 + (Sqrt[(178 + 35*Sqrt[35])/434]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[
1 + 2*x] + 5*(1 + 2*x)])/7

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Rubi [A]  time = 0.337402, 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 = {709, 826, 1169, 634, 618, 204, 628} \[ -\frac{4}{7 \sqrt{2 x+1}}-\frac{1}{7} \sqrt{\frac{1}{434} \left (178+35 \sqrt{35}\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{7} \sqrt{\frac{1}{434} \left (178+35 \sqrt{35}\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{7} \sqrt{\frac{2}{217} \left (35 \sqrt{35}-178\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{7} \sqrt{\frac{2}{217} \left (35 \sqrt{35}-178\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-4/(7*Sqrt[1 + 2*x]) + (Sqrt[(2*(-178 + 35*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])
/Sqrt[10*(-2 + Sqrt[35])]])/7 - (Sqrt[(2*(-178 + 35*Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[
1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/7 - (Sqrt[(178 + 35*Sqrt[35])/434]*Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]
*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/7 + (Sqrt[(178 + 35*Sqrt[35])/434]*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[
1 + 2*x] + 5*(1 + 2*x)])/7

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 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)^{3/2} \left (2+3 x+5 x^2\right )} \, dx &=-\frac{4}{7 \sqrt{1+2 x}}+\frac{1}{7} \int \frac{-1-10 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx\\ &=-\frac{4}{7 \sqrt{1+2 x}}+\frac{2}{7} \operatorname{Subst}\left (\int \frac{8-10 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=-\frac{4}{7 \sqrt{1+2 x}}+\frac{\operatorname{Subst}\left (\int \frac{8 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (8+2 \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 )}{7 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{8 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (8+2 \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 )}{7 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{4}{7 \sqrt{1+2 x}}-\frac{\left (4+\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 )}{7 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\left (4+\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 )}{7 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{1}{245} \left (-35+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}{245} \left (-35+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{4}{7 \sqrt{1+2 x}}-\frac{1}{7} \sqrt{\frac{89}{217}+\frac{5 \sqrt{35}}{62}} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )+\frac{1}{7} \sqrt{\frac{89}{217}+\frac{5 \sqrt{35}}{62}} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )+\frac{1}{245} \left (2 \left (35-4 \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}{245} \left (2 \left (35-4 \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}{7 \sqrt{1+2 x}}+\frac{1}{7} \sqrt{\frac{2}{217} \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}{7} \sqrt{\frac{2}{217} \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}{7} \sqrt{\frac{89}{217}+\frac{5 \sqrt{35}}{62}} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )+\frac{1}{7} \sqrt{\frac{89}{217}+\frac{5 \sqrt{35}}{62}} \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 = 0.45037, size = 122, normalized size = 0.48 \[ \frac{2 \left (-\frac{2170}{\sqrt{2 x+1}}+\sqrt{10-5 i \sqrt{31}} \left (124+27 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+\sqrt{10+5 i \sqrt{31}} \left (124-27 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{7595} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-2170/Sqrt[1 + 2*x] + Sqrt[10 - (5*I)*Sqrt[31]]*(124 + (27*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*
Sqrt[31]]] + Sqrt[10 + (5*I)*Sqrt[31]]*(124 - (27*I)*Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]]))/
7595

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Maple [B]  time = 0.077, 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/(1+2*x)^(3/2)/(5*x^2+3*x+2),x)

[Out]

1/217*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)+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))*7^(1/2)*(2*5^(1/
2)*7^(1/2)+4)^(1/2)-10/217/(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)-27/1519/(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))*5^(1/2)*(2*5^(1/2)*7^(1/2)+
4)*7^(1/2)+16/49/(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)-1/217*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)-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)*7^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)-10/217/(10*5^(1/2)*7^(1/2)-20)^(1/2)*arcta
n((-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)
-27/1519/(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)+16/49/(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)-4/7/(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{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.77233, size = 2148, normalized size = 8.49 \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)^(3/2)/(5*x^2+3*x+2),x, algorithm="fricas")

[Out]

1/2191992950*(2356*42875^(1/4)*sqrt(217)*sqrt(35)*(2*x + 1)*sqrt(-12460*sqrt(35) + 85750)*arctan(1/36441882793
75*42875^(3/4)*sqrt(217)*sqrt(31)*sqrt(42875^(1/4)*sqrt(217)*(sqrt(35)*sqrt(31) + 4*sqrt(31))*sqrt(2*x + 1)*sq
rt(-12460*sqrt(35) + 85750) + 1443050*x + 144305*sqrt(35) + 721525)*(4*sqrt(35)*sqrt(19) + 35*sqrt(19))*sqrt(-
12460*sqrt(35) + 85750) - 1/176773625*42875^(3/4)*sqrt(217)*sqrt(2*x + 1)*(4*sqrt(35) + 35)*sqrt(-12460*sqrt(3
5) + 85750) - 1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) + 2356*42875^(1/4)*sqrt(217)*sqrt(35)*(2*x + 1)*sqrt(-12
460*sqrt(35) + 85750)*arctan(1/255093179556250*42875^(3/4)*sqrt(217)*sqrt(-151900*42875^(1/4)*sqrt(217)*(sqrt(
35)*sqrt(31) + 4*sqrt(31))*sqrt(2*x + 1)*sqrt(-12460*sqrt(35) + 85750) + 219199295000*x + 21919929500*sqrt(35)
+ 109599647500)*(4*sqrt(35)*sqrt(19) + 35*sqrt(19))*sqrt(-12460*sqrt(35) + 85750) - 1/176773625*42875^(3/4)*s
qrt(217)*sqrt(2*x + 1)*(4*sqrt(35) + 35)*sqrt(-12460*sqrt(35) + 85750) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31
)) + 42875^(1/4)*sqrt(217)*(178*sqrt(35)*sqrt(31)*(2*x + 1) + 1225*sqrt(31)*(2*x + 1))*sqrt(-12460*sqrt(35) +
85750)*log(151900/19*42875^(1/4)*sqrt(217)*(sqrt(35)*sqrt(31) + 4*sqrt(31))*sqrt(2*x + 1)*sqrt(-12460*sqrt(35)
+ 85750) + 11536805000*x + 1153680500*sqrt(35) + 5768402500) - 42875^(1/4)*sqrt(217)*(178*sqrt(35)*sqrt(31)*(
2*x + 1) + 1225*sqrt(31)*(2*x + 1))*sqrt(-12460*sqrt(35) + 85750)*log(-151900/19*42875^(1/4)*sqrt(217)*(sqrt(3
5)*sqrt(31) + 4*sqrt(31))*sqrt(2*x + 1)*sqrt(-12460*sqrt(35) + 85750) + 11536805000*x + 1153680500*sqrt(35) +
5768402500) - 1252567400*sqrt(2*x + 1))/(2*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{3}{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)**(3/2)/(5*x**2+3*x+2),x)

[Out]

Integral(1/((2*x + 1)**(3/2)*(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{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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