3.2313 \(\int \frac{1}{\sqrt{1+2 x} (2+3 x+5 x^2)} \, dx\)

Optimal. Leaf size=218 \[ -\frac{\log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{\sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{\sqrt{14 \left (2+\sqrt{35}\right )}}-\sqrt{\frac{2}{217} \left (2+\sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\sqrt{\frac{2}{217} \left (2+\sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]

[Out]

-(Sqrt[(2*(2 + Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])
+ Sqrt[(2*(2 + Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] -
Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/Sqrt[14*(2 + Sqrt[35])] + Log[Sqrt[35] +
Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/Sqrt[14*(2 + Sqrt[35])]

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Rubi [A]  time = 0.271973, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {707, 1094, 634, 618, 204, 628} \[ -\frac{\log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{\sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )}{\sqrt{14 \left (2+\sqrt{35}\right )}}-\sqrt{\frac{2}{217} \left (2+\sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\sqrt{\frac{2}{217} \left (2+\sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[(2*(2 + Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])
+ Sqrt[(2*(2 + Sqrt[35]))/217]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] -
Log[Sqrt[35] - Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/Sqrt[14*(2 + Sqrt[35])] + Log[Sqrt[35] +
Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)]/Sqrt[14*(2 + Sqrt[35])]

Rule 707

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(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}, 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]

Rule 1094

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}
, Dist[1/(2*c*q*r), Int[(r - x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(r + x)/(q + r*x + x^2), x], x
]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx &=4 \operatorname{Subst}\left (\int \frac{1}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=\sqrt{\frac{2}{7 \left (2+\sqrt{35}\right )}} \operatorname{Subst}\left (\int \frac{\sqrt{\frac{2}{5} \left (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 )+\sqrt{\frac{2}{7 \left (2+\sqrt{35}\right )}} \operatorname{Subst}\left (\int \frac{\sqrt{\frac{2}{5} \left (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 )\\ &=\frac{\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 )}{\sqrt{35}}+\frac{\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 )}{\sqrt{35}}-\frac{\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 )}{\sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\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 )}{\sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{\log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{\sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{\sqrt{14 \left (2+\sqrt{35}\right )}}-\frac{2 \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 )}{\sqrt{35}}-\frac{2 \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 )}{\sqrt{35}}\\ &=-\sqrt{\frac{2}{7 \left (-2+\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 )+\sqrt{\frac{2}{7 \left (-2+\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{\log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{\sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )}{\sqrt{14 \left (2+\sqrt{35}\right )}}\\ \end{align*}

Mathematica [C]  time = 0.246324, size = 112, normalized size = 0.51 \[ \frac{2 \left (\sqrt{2-i \sqrt{31}} \left (\sqrt{31}-2 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+\sqrt{2+i \sqrt{31}} \left (\sqrt{31}+2 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{7 \sqrt{155}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Sqrt[2 - I*Sqrt[31]]*(-2*I + Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sqrt[31]]] + Sqrt[2 + I*Sqrt[31]]
*(2*I + Sqrt[31])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 + I*Sqrt[31]]]))/(7*Sqrt[155])

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Maple [B]  time = 0.071, size = 607, normalized size = 2.8 \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)^(1/2)/(5*x^2+3*x+2),x)

[Out]

1/62*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)-1/217*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))*7^(1/2)*(2*5^(1/2)*
7^(1/2)+4)^(1/2)-5/31/(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)+2/217/(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))*5^(1/2)*(2*5^(1/2)*7^(1/2)+4)*7^(1
/2)+4/7/(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)-1/62*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)+1/217*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)*7^(1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)-5/31/(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)+2/217/(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)+4/7/(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{1}{{\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt{2 \, x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.61098, size = 1427, normalized size = 6.55 \begin{align*} -\frac{1}{470890} \, \sqrt{217} 35^{\frac{1}{4}}{\left (2 \, \sqrt{35} \sqrt{31} - 35 \, \sqrt{31}\right )} \sqrt{4 \, \sqrt{35} + 70} \log \left (4340 \, \sqrt{217} 35^{\frac{1}{4}} \sqrt{31} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + 9417800 \, x + 941780 \, \sqrt{35} + 4708900\right ) + \frac{1}{470890} \, \sqrt{217} 35^{\frac{1}{4}}{\left (2 \, \sqrt{35} \sqrt{31} - 35 \, \sqrt{31}\right )} \sqrt{4 \, \sqrt{35} + 70} \log \left (-4340 \, \sqrt{217} 35^{\frac{1}{4}} \sqrt{31} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + 9417800 \, x + 941780 \, \sqrt{35} + 4708900\right ) - \frac{2}{7595} \, \sqrt{217} 35^{\frac{3}{4}} \sqrt{4 \, \sqrt{35} + 70} \arctan \left (\frac{1}{235445} \, \sqrt{1085} \sqrt{217} 35^{\frac{1}{4}} \sqrt{\sqrt{217} 35^{\frac{1}{4}} \sqrt{31} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + 2170 \, x + 217 \, \sqrt{35} + 1085} \sqrt{4 \, \sqrt{35} + 70} - \frac{1}{217} \, \sqrt{217} 35^{\frac{1}{4}} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} - \frac{1}{31} \, \sqrt{35} \sqrt{31} - \frac{2}{31} \, \sqrt{31}\right ) - \frac{2}{7595} \, \sqrt{217} 35^{\frac{3}{4}} \sqrt{4 \, \sqrt{35} + 70} \arctan \left (\frac{1}{470890} \, \sqrt{217} 35^{\frac{1}{4}} \sqrt{-4340 \, \sqrt{217} 35^{\frac{1}{4}} \sqrt{31} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + 9417800 \, x + 941780 \, \sqrt{35} + 4708900} \sqrt{4 \, \sqrt{35} + 70} - \frac{1}{217} \, \sqrt{217} 35^{\frac{1}{4}} \sqrt{2 \, x + 1} \sqrt{4 \, \sqrt{35} + 70} + \frac{1}{31} \, \sqrt{35} \sqrt{31} + \frac{2}{31} \, \sqrt{31}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/470890*sqrt(217)*35^(1/4)*(2*sqrt(35)*sqrt(31) - 35*sqrt(31))*sqrt(4*sqrt(35) + 70)*log(4340*sqrt(217)*35^(
1/4)*sqrt(31)*sqrt(2*x + 1)*sqrt(4*sqrt(35) + 70) + 9417800*x + 941780*sqrt(35) + 4708900) + 1/470890*sqrt(217
)*35^(1/4)*(2*sqrt(35)*sqrt(31) - 35*sqrt(31))*sqrt(4*sqrt(35) + 70)*log(-4340*sqrt(217)*35^(1/4)*sqrt(31)*sqr
t(2*x + 1)*sqrt(4*sqrt(35) + 70) + 9417800*x + 941780*sqrt(35) + 4708900) - 2/7595*sqrt(217)*35^(3/4)*sqrt(4*s
qrt(35) + 70)*arctan(1/235445*sqrt(1085)*sqrt(217)*35^(1/4)*sqrt(sqrt(217)*35^(1/4)*sqrt(31)*sqrt(2*x + 1)*sqr
t(4*sqrt(35) + 70) + 2170*x + 217*sqrt(35) + 1085)*sqrt(4*sqrt(35) + 70) - 1/217*sqrt(217)*35^(1/4)*sqrt(2*x +
1)*sqrt(4*sqrt(35) + 70) - 1/31*sqrt(35)*sqrt(31) - 2/31*sqrt(31)) - 2/7595*sqrt(217)*35^(3/4)*sqrt(4*sqrt(35
) + 70)*arctan(1/470890*sqrt(217)*35^(1/4)*sqrt(-4340*sqrt(217)*35^(1/4)*sqrt(31)*sqrt(2*x + 1)*sqrt(4*sqrt(35
) + 70) + 9417800*x + 941780*sqrt(35) + 4708900)*sqrt(4*sqrt(35) + 70) - 1/217*sqrt(217)*35^(1/4)*sqrt(2*x + 1
)*sqrt(4*sqrt(35) + 70) + 1/31*sqrt(35)*sqrt(31) + 2/31*sqrt(31))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 x + 1} \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)**(1/2)/(5*x**2+3*x+2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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