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

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

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Rubi [A]  time = 0.24, antiderivative size = 115, normalized size of antiderivative = 1.10, 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} \begin {gather*} 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 ) \end {gather*}

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 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 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 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]))

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*}

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Mathematica [C]  time = 0.27, size = 141, normalized size = 1.34 \begin {gather*} \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 ) \end {gather*}

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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IntegrateAlgebraic [A]  time = 1.38, size = 73, normalized size = 0.70 \begin {gather*} 2 \sqrt {\frac {1}{31} \left (20+10 \sqrt {35}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{434} \left (70+35 \sqrt {35}\right )} (2 x+1)-\sqrt {\frac {1}{62} \left (14+7 \sqrt {35}\right )}}{\sqrt {2 x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[(20 + 10*Sqrt[35])/31]*ArcTan[(-Sqrt[(14 + 7*Sqrt[35])/62] + Sqrt[(70 + 35*Sqrt[35])/434]*(1 + 2*x))/Sq
rt[1 + 2*x]]

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fricas [A]  time = 0.42, size = 52, normalized size = 0.50 \begin {gather*} -\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 {gather*}

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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giac [B]  time = 1.01, size = 313, normalized size = 2.98 \begin {gather*} \frac {1}{7443100} \, \sqrt {31} {\left (210 \, \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (2 \, \sqrt {35} + 35\right )} \sqrt {-140 \, \sqrt {35} + 2450} - \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (-140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 2 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 420 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} \sqrt {140 \, \sqrt {35} + 2450} {\left (2 \, \sqrt {35} - 35\right )} + 980 \, \sqrt {35} \sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450} + 1960 \, \sqrt {35} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450}\right )} \arctan \left (\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} + \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) + \frac {1}{7443100} \, \sqrt {31} {\left (210 \, \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (2 \, \sqrt {35} + 35\right )} \sqrt {-140 \, \sqrt {35} + 2450} - \sqrt {31} \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (-140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 2 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (140 \, \sqrt {35} + 2450\right )}^{\frac {3}{2}} + 420 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} \sqrt {140 \, \sqrt {35} + 2450} {\left (2 \, \sqrt {35} - 35\right )} + 980 \, \sqrt {35} \sqrt {31} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {-140 \, \sqrt {35} + 2450} + 1960 \, \sqrt {35} \left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {140 \, \sqrt {35} + 2450}\right )} \arctan \left (-\frac {5 \, \left (\frac {7}{5}\right )^{\frac {3}{4}} {\left (\left (\frac {7}{5}\right )^{\frac {1}{4}} \sqrt {\frac {1}{35} \, \sqrt {35} + \frac {1}{2}} - \sqrt {2 \, x + 1}\right )}}{7 \, \sqrt {-\frac {1}{35} \, \sqrt {35} + \frac {1}{2}}}\right ) \end {gather*}

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]

1/7443100*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sqrt(35) + 2450) - sqrt(31)*(7/5)^(3/
4)*(-140*sqrt(35) + 2450)^(3/2) + 2*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 420*(7/5)^(3/4)*sqrt(140*sqrt(35
) + 2450)*(2*sqrt(35) - 35) + 980*sqrt(35)*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35) + 2450) + 1960*sqrt(35)*(7/
5)^(1/4)*sqrt(140*sqrt(35) + 2450))*arctan(5/7*(7/5)^(3/4)*((7/5)^(1/4)*sqrt(1/35*sqrt(35) + 1/2) + sqrt(2*x +
1))/sqrt(-1/35*sqrt(35) + 1/2)) + 1/7443100*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-140*sq
rt(35) + 2450) - sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 2*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2)
+ 420*(7/5)^(3/4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 980*sqrt(35)*sqrt(31)*(7/5)^(1/4)*sqrt(-140*s
qrt(35) + 2450) + 1960*sqrt(35)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450))*arctan(-5/7*(7/5)^(3/4)*((7/5)^(1/4)*sq
rt(1/35*sqrt(35) + 1/2) - sqrt(2*x + 1))/sqrt(-1/35*sqrt(35) + 1/2))

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maple [A]  time = 0.33, size = 111, normalized size = 1.06 \begin {gather*} \frac {20 \arctan \left (\frac {10 \sqrt {2 x +1}-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {20 \arctan \left (\frac {10 \sqrt {2 x +1}+\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {10 \, x + \sqrt {35} + 5}{{\left (5 \, x^{2} + 3 \, x + 2\right )} \sqrt {2 \, x + 1}}\,{d x} \end {gather*}

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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mupad [B]  time = 3.27, size = 143, normalized size = 1.36 \begin {gather*} 2\,\sqrt {\frac {10\,\sqrt {35}}{31}+\frac {20}{31}}\,\left (\mathrm {atan}\left (\frac {\sqrt {434}\,\left (39\,\sqrt {35}+140\right )\,\sqrt {2\,x+1}\,{\left (\sqrt {35}-2\right )}^2\,\sqrt {\sqrt {35}+2}}{417074}\right )+\mathrm {atan}\left (\frac {31\,\sqrt {2\,x+1}\,\left (\frac {\sqrt {\frac {10\,\sqrt {35}}{31}+\frac {20}{31}}\,\left (10000\,\sqrt {35}+20000\right )}{39\,\sqrt {35}+140}-\frac {\sqrt {434}\,\left (\frac {390000\,\sqrt {35}}{31}+\frac {1400000}{31}\right )\,{\left (\sqrt {35}-2\right )}^2\,\sqrt {\sqrt {35}+2}}{417074}\right )}{10000}+\frac {\sqrt {434}\,\left (\frac {200000\,\sqrt {35}}{31}+\frac {1950000}{31}\right )\,{\left (2\,x+1\right )}^{3/2}\,{\left (\sqrt {35}-2\right )}^2\,\sqrt {\sqrt {35}+2}}{134540000}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((10*35^(1/2))/31 + 20/31)^(1/2)*(atan((434^(1/2)*(39*35^(1/2) + 140)*(2*x + 1)^(1/2)*(35^(1/2) - 2)^2*(35^(
1/2) + 2)^(1/2))/417074) + atan((31*(2*x + 1)^(1/2)*((((10*35^(1/2))/31 + 20/31)^(1/2)*(10000*35^(1/2) + 20000
))/(39*35^(1/2) + 140) - (434^(1/2)*((390000*35^(1/2))/31 + 1400000/31)*(35^(1/2) - 2)^2*(35^(1/2) + 2)^(1/2))
/417074))/10000 + (434^(1/2)*((200000*35^(1/2))/31 + 1950000/31)*(2*x + 1)^(3/2)*(35^(1/2) - 2)^2*(35^(1/2) +
2)^(1/2))/134540000))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {10 x + 5 + \sqrt {35}}{\sqrt {2 x + 1} \left (5 x^{2} + 3 x + 2\right )}\, dx \end {gather*}

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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