3.2575 \(\int \frac{5+\sqrt{35}+10 x}{\sqrt{1+2 x} (2+3 x+5 x^2)} \, dx\)

Optimal. Leaf size=105 \[ 2 \sqrt{\frac{10}{\sqrt{35}-2}} \tan ^{-1}\left (\frac{\sqrt{20 x+10}+\sqrt{2+\sqrt{35}}}{\sqrt{\sqrt{35}-2}}\right )-2 \sqrt{\frac{10}{\sqrt{35}-2}} \tan ^{-1}\left (\frac{\sqrt{2+\sqrt{35}}-\sqrt{20 x+10}}{\sqrt{\sqrt{35}-2}}\right ) \]

[Out]

-2*Sqrt[10/(-2 + Sqrt[35])]*ArcTan[(Sqrt[2 + Sqrt[35]] - Sqrt[10 + 20*x])/Sqrt[-2 + Sqrt[35]]] + 2*Sqrt[10/(-2
+ Sqrt[35])]*ArcTan[(Sqrt[2 + Sqrt[35]] + Sqrt[10 + 20*x])/Sqrt[-2 + Sqrt[35]]]

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Rubi [A]  time = 0.243228, antiderivative size = 115, normalized size of antiderivative = 1.1, number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {826, 1161, 618, 204} \[ 2 \sqrt{\frac{10}{\sqrt{35}-2}} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-2 \sqrt{\frac{10}{\sqrt{35}-2}} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(5 + Sqrt[35] + 10*x)/(Sqrt[1 + 2*x]*(2 + 3*x + 5*x^2)),x]

[Out]

-2*Sqrt[10/(-2 + Sqrt[35])]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] - 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]] + 2*
Sqrt[10/(-2 + Sqrt[35])]*ArcTan[(Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]]

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 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

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])

Rubi steps

\begin{align*} \int \frac{5+\sqrt{35}+10 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{-10+2 \left (5+\sqrt{35}\right )+10 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=2 \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 )+2 \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 )\\ &=-\left (4 \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 )\right )-4 \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 )\\ &=-2 \sqrt{\frac{10}{-2+\sqrt{35}}} \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 )+2 \sqrt{\frac{10}{-2+\sqrt{35}}} \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 )\\ \end{align*}

Mathematica [C]  time = 0.294548, size = 141, normalized size = 1.34 \[ \frac{2}{217} \left (\sqrt{2-i \sqrt{31}} \left (31 \sqrt{7}-7 i \sqrt{155}-2 i \sqrt{217}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+\sqrt{2+i \sqrt{31}} \left (31 \sqrt{7}+7 i \sqrt{155}+2 i \sqrt{217}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + Sqrt[35] + 10*x)/(Sqrt[1 + 2*x]*(2 + 3*x + 5*x^2)),x]

[Out]

(2*(Sqrt[2 - I*Sqrt[31]]*(31*Sqrt[7] - (7*I)*Sqrt[155] - (2*I)*Sqrt[217])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2 - I*Sq
rt[31]]] + Sqrt[2 + I*Sqrt[31]]*(31*Sqrt[7] + (7*I)*Sqrt[155] + (2*I)*Sqrt[217])*ArcTanh[Sqrt[5 + 10*x]/Sqrt[2
+ I*Sqrt[31]]]))/217

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Maple [A]  time = 0.168, size = 111, normalized size = 1.1 \begin{align*} 20\,{\frac{1}{\sqrt{10\,\sqrt{5}\sqrt{7}-20}}\arctan \left ({\frac{10\,\sqrt{1+2\,x}-\sqrt{5}\sqrt{2\,\sqrt{5}\sqrt{7}+4}}{\sqrt{10\,\sqrt{5}\sqrt{7}-20}}} \right ) }+20\,{\frac{1}{\sqrt{10\,\sqrt{5}\sqrt{7}-20}}\arctan \left ({\frac{10\,\sqrt{1+2\,x}+\sqrt{5}\sqrt{2\,\sqrt{5}\sqrt{7}+4}}{\sqrt{10\,\sqrt{5}\sqrt{7}-20}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5+10*x+35^(1/2))/(5*x^2+3*x+2)/(1+2*x)^(1/2),x)

[Out]

20/(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))+20/(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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{10 \, x + \sqrt{35} + 5}{{\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((5+10*x+35^(1/2))/(5*x^2+3*x+2)/(1+2*x)^(1/2),x, algorithm="maxima")

[Out]

integrate((10*x + sqrt(35) + 5)/((5*x^2 + 3*x + 2)*sqrt(2*x + 1)), x)

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Fricas [A]  time = 1.66963, size = 181, normalized size = 1.72 \begin{align*} -\frac{2}{31} \, \sqrt{31} \sqrt{10 \, \sqrt{35} + 20} \arctan \left (-\frac{{\left (5 \, \sqrt{31}{\left (2 \, x + 1\right )} - \sqrt{35} \sqrt{31}\right )} \sqrt{10 \, \sqrt{35} + 20}}{310 \, \sqrt{2 \, x + 1}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5+10*x+35^(1/2))/(5*x^2+3*x+2)/(1+2*x)^(1/2),x, algorithm="fricas")

[Out]

-2/31*sqrt(31)*sqrt(10*sqrt(35) + 20)*arctan(-1/310*(5*sqrt(31)*(2*x + 1) - sqrt(35)*sqrt(31))*sqrt(10*sqrt(35
) + 20)/sqrt(2*x + 1))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{10 x + 5 + \sqrt{35}}{\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((5+10*x+35**(1/2))/(5*x**2+3*x+2)/(1+2*x)**(1/2),x)

[Out]

Integral((10*x + 5 + sqrt(35))/(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{10 \, x + \sqrt{35} + 5}{{\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((5+10*x+35^(1/2))/(5*x^2+3*x+2)/(1+2*x)^(1/2),x, algorithm="giac")

[Out]

integrate((10*x + sqrt(35) + 5)/((5*x^2 + 3*x + 2)*sqrt(2*x + 1)), x)