3.349 \(\int \frac{2+x+3 x^2-x^3+5 x^4}{(5+2 x)^4 \sqrt{3-x+2 x^2}} \, dx\)

Optimal. Leaf size=135 \[ -\frac{3163415 \sqrt{2 x^2-x+3}}{5971968 (2 x+5)}+\frac{394907 \sqrt{2 x^2-x+3}}{248832 (2 x+5)^2}-\frac{3667 \sqrt{2 x^2-x+3}}{1728 (2 x+5)^3}+\frac{22389491 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{71663616 \sqrt{2}}-\frac{5 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{16 \sqrt{2}} \]

[Out]

(-3667*Sqrt[3 - x + 2*x^2])/(1728*(5 + 2*x)^3) + (394907*Sqrt[3 - x + 2*x^2])/(248832*(5 + 2*x)^2) - (3163415*
Sqrt[3 - x + 2*x^2])/(5971968*(5 + 2*x)) - (5*ArcSinh[(1 - 4*x)/Sqrt[23]])/(16*Sqrt[2]) + (22389491*ArcTanh[(1
7 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(71663616*Sqrt[2])

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Rubi [A]  time = 0.204716, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {1650, 843, 619, 215, 724, 206} \[ -\frac{3163415 \sqrt{2 x^2-x+3}}{5971968 (2 x+5)}+\frac{394907 \sqrt{2 x^2-x+3}}{248832 (2 x+5)^2}-\frac{3667 \sqrt{2 x^2-x+3}}{1728 (2 x+5)^3}+\frac{22389491 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{71663616 \sqrt{2}}-\frac{5 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{16 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + x + 3*x^2 - x^3 + 5*x^4)/((5 + 2*x)^4*Sqrt[3 - x + 2*x^2]),x]

[Out]

(-3667*Sqrt[3 - x + 2*x^2])/(1728*(5 + 2*x)^3) + (394907*Sqrt[3 - x + 2*x^2])/(248832*(5 + 2*x)^2) - (3163415*
Sqrt[3 - x + 2*x^2])/(5971968*(5 + 2*x)) - (5*ArcSinh[(1 - 4*x)/Sqrt[23]])/(16*Sqrt[2]) + (22389491*ArcTanh[(1
7 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(71663616*Sqrt[2])

Rule 1650

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomia
lQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*
x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^
(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Q + c*d*R*(m + 1) - b*e*R*(m + p + 2)
- c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{2+x+3 x^2-x^3+5 x^4}{(5+2 x)^4 \sqrt{3-x+2 x^2}} \, dx &=-\frac{3667 \sqrt{3-x+2 x^2}}{1728 (5+2 x)^3}-\frac{1}{216} \int \frac{\frac{28687}{16}-\frac{4271 x}{2}+1458 x^2-540 x^3}{(5+2 x)^3 \sqrt{3-x+2 x^2}} \, dx\\ &=-\frac{3667 \sqrt{3-x+2 x^2}}{1728 (5+2 x)^3}+\frac{394907 \sqrt{3-x+2 x^2}}{248832 (5+2 x)^2}+\frac{\int \frac{\frac{1464275}{16}-\frac{413797 x}{4}+38880 x^2}{(5+2 x)^2 \sqrt{3-x+2 x^2}} \, dx}{31104}\\ &=-\frac{3667 \sqrt{3-x+2 x^2}}{1728 (5+2 x)^3}+\frac{394907 \sqrt{3-x+2 x^2}}{248832 (5+2 x)^2}-\frac{3163415 \sqrt{3-x+2 x^2}}{5971968 (5+2 x)}-\frac{\int \frac{\frac{11181273}{16}-1399680 x}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{2239488}\\ &=-\frac{3667 \sqrt{3-x+2 x^2}}{1728 (5+2 x)^3}+\frac{394907 \sqrt{3-x+2 x^2}}{248832 (5+2 x)^2}-\frac{3163415 \sqrt{3-x+2 x^2}}{5971968 (5+2 x)}+\frac{5}{16} \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx-\frac{22389491 \int \frac{1}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{11943936}\\ &=-\frac{3667 \sqrt{3-x+2 x^2}}{1728 (5+2 x)^3}+\frac{394907 \sqrt{3-x+2 x^2}}{248832 (5+2 x)^2}-\frac{3163415 \sqrt{3-x+2 x^2}}{5971968 (5+2 x)}+\frac{22389491 \operatorname{Subst}\left (\int \frac{1}{288-x^2} \, dx,x,\frac{17-22 x}{\sqrt{3-x+2 x^2}}\right )}{5971968}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{16 \sqrt{46}}\\ &=-\frac{3667 \sqrt{3-x+2 x^2}}{1728 (5+2 x)^3}+\frac{394907 \sqrt{3-x+2 x^2}}{248832 (5+2 x)^2}-\frac{3163415 \sqrt{3-x+2 x^2}}{5971968 (5+2 x)}-\frac{5 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{16 \sqrt{2}}+\frac{22389491 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{3-x+2 x^2}}\right )}{71663616 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.151497, size = 88, normalized size = 0.65 \[ \frac{-\frac{24 \sqrt{2 x^2-x+3} \left (12653660 x^2+44312764 x+44369687\right )}{(2 x+5)^3}[phone] \sqrt{2} \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{4 x^2-2 x+6}}\right )-22394880 \sqrt{2} \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{143327232} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + x + 3*x^2 - x^3 + 5*x^4)/((5 + 2*x)^4*Sqrt[3 - x + 2*x^2]),x]

[Out]

((-24*Sqrt[3 - x + 2*x^2]*(44369687 + 44312764*x + 12653660*x^2))/(5 + 2*x)^3 - 22394880*Sqrt[2]*ArcSinh[(1 -
4*x)/Sqrt[23]] + 22389491*Sqrt[2]*ArcTanh[(17 - 22*x)/(12*Sqrt[6 - 2*x + 4*x^2])])/143327232

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Maple [A]  time = 0.059, size = 109, normalized size = 0.8 \begin{align*}{\frac{5\,\sqrt{2}}{32}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) }+{\frac{394907}{995328}\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}} \left ( x+{\frac{5}{2}} \right ) ^{-2}}-{\frac{3163415}{11943936}\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}} \left ( x+{\frac{5}{2}} \right ) ^{-1}}+{\frac{22389491\,\sqrt{2}}{143327232}{\it Artanh} \left ({\frac{\sqrt{2}}{12} \left ({\frac{17}{2}}-11\,x \right ){\frac{1}{\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}}}}} \right ) }-{\frac{3667}{13824}\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}} \left ( x+{\frac{5}{2}} \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^4-x^3+3*x^2+x+2)/(5+2*x)^4/(2*x^2-x+3)^(1/2),x)

[Out]

5/32*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))+394907/995328/(x+5/2)^2*(2*(x+5/2)^2-11*x-19/2)^(1/2)-3163415/1194
3936/(x+5/2)*(2*(x+5/2)^2-11*x-19/2)^(1/2)[phone]/143327232*2^(1/2)*arctanh(1/12*(17/2-11*x)*2^(1/2)/(2*(x+5
/2)^2-11*x-19/2)^(1/2))-3667/13824/(x+5/2)^3*(2*(x+5/2)^2-11*x-19/2)^(1/2)

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Maxima [A]  time = 1.49446, size = 177, normalized size = 1.31 \begin{align*} \frac{5}{32} \, \sqrt{2} \operatorname{arsinh}\left (\frac{4}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) - \frac{22389491}{143327232} \, \sqrt{2} \operatorname{arsinh}\left (\frac{22 \, \sqrt{23} x}{23 \,{\left | 2 \, x + 5 \right |}} - \frac{17 \, \sqrt{23}}{23 \,{\left | 2 \, x + 5 \right |}}\right ) - \frac{3667 \, \sqrt{2 \, x^{2} - x + 3}}{1728 \,{\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )}} + \frac{394907 \, \sqrt{2 \, x^{2} - x + 3}}{248832 \,{\left (4 \, x^{2} + 20 \, x + 25\right )}} - \frac{3163415 \, \sqrt{2 \, x^{2} - x + 3}}{5971968 \,{\left (2 \, x + 5\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^4-x^3+3*x^2+x+2)/(5+2*x)^4/(2*x^2-x+3)^(1/2),x, algorithm="maxima")

[Out]

5/32*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) - 22389491/143327232*sqrt(2)*arcsinh(22/23*sqrt(23)*x/ab
s(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) - 3667/1728*sqrt(2*x^2 - x + 3)/(8*x^3 + 60*x^2 + 150*x + 125) + 394
907/248832*sqrt(2*x^2 - x + 3)/(4*x^2 + 20*x + 25) - 3163415/5971968*sqrt(2*x^2 - x + 3)/(2*x + 5)

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Fricas [A]  time = 1.34501, size = 502, normalized size = 3.72 \begin{align*} \frac{22394880 \, \sqrt{2}{\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )} \log \left (-4 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 22389491 \, \sqrt{2}{\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )} \log \left (\frac{24 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (22 \, x - 17\right )} - 1060 \, x^{2} + 1036 \, x - 1153}{4 \, x^{2} + 20 \, x + 25}\right ) - 48 \,{\left (12653660 \, x^{2} + 44312764 \, x + 44369687\right )} \sqrt{2 \, x^{2} - x + 3}}{286654464 \,{\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^4-x^3+3*x^2+x+2)/(5+2*x)^4/(2*x^2-x+3)^(1/2),x, algorithm="fricas")

[Out]

1/286654464*(22394880*sqrt(2)*(8*x^3 + 60*x^2 + 150*x + 125)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32
*x^2 + 16*x - 25) + 22389491*sqrt(2)*(8*x^3 + 60*x^2 + 150*x + 125)*log((24*sqrt(2)*sqrt(2*x^2 - x + 3)*(22*x
- 17) - 1060*x^2 + 1036*x - 1153)/(4*x^2 + 20*x + 25)) - 48*(12653660*x^2 + 44312764*x + 44369687)*sqrt(2*x^2
- x + 3))/(8*x^3 + 60*x^2 + 150*x + 125)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{5 x^{4} - x^{3} + 3 x^{2} + x + 2}{\left (2 x + 5\right )^{4} \sqrt{2 x^{2} - x + 3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**4-x**3+3*x**2+x+2)/(5+2*x)**4/(2*x**2-x+3)**(1/2),x)

[Out]

Integral((5*x**4 - x**3 + 3*x**2 + x + 2)/((2*x + 5)**4*sqrt(2*x**2 - x + 3)), x)

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Giac [B]  time = 1.20272, size = 385, normalized size = 2.85 \begin{align*} -\frac{5}{32} \, \sqrt{2} \log \left (-2 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac{22389491}{143327232} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x + \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\right ) - \frac{22389491}{143327232} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x - 11 \, \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\right ) - \frac{\sqrt{2}{\left (215012404 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )}^{5} + 3010410772 \,{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )}^{4} + 2740802468 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )}^{3} - 21459328844 \,{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )}^{2} + 14434519361 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )} - 5957650879\right )}}{11943936 \,{\left (2 \,{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )}^{2} + 10 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )} - 11\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^4-x^3+3*x^2+x+2)/(5+2*x)^4/(2*x^2-x+3)^(1/2),x, algorithm="giac")

[Out]

-5/32*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 22389491/143327232*sqrt(2)*log(abs(-2*sq
rt(2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) - 22389491/143327232*sqrt(2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2
*sqrt(2*x^2 - x + 3))) - 1/11943936*sqrt(2)*(215012404*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^5 + 301041077
2*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^4 + 2740802468*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^3 - 21459328844*(
sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 + 14434519361*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) - 5957650879)/(2*(s
qrt(2)*x - sqrt(2*x^2 - x + 3))^2 + 10*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) - 11)^3