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

Optimal. Leaf size=126 \[ \frac{5}{48} \sqrt{2 x^2-x+3} (2 x+5)^2-\frac{337}{192} \sqrt{2 x^2-x+3} (2 x+5)+\frac{1669}{128} \sqrt{2 x^2-x+3}-\frac{3667 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{96 \sqrt{2}}+\frac{9657 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{256 \sqrt{2}} \]

[Out]

(1669*Sqrt[3 - x + 2*x^2])/128 - (337*(5 + 2*x)*Sqrt[3 - x + 2*x^2])/192 + (5*(5 + 2*x)^2*Sqrt[3 - x + 2*x^2])
/48 + (9657*ArcSinh[(1 - 4*x)/Sqrt[23]])/(256*Sqrt[2]) - (3667*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*
x^2])])/(96*Sqrt[2])

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Rubi [A]  time = 0.210483, antiderivative size = 126, 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 = {1653, 843, 619, 215, 724, 206} \[ \frac{5}{48} \sqrt{2 x^2-x+3} (2 x+5)^2-\frac{337}{192} \sqrt{2 x^2-x+3} (2 x+5)+\frac{1669}{128} \sqrt{2 x^2-x+3}-\frac{3667 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{96 \sqrt{2}}+\frac{9657 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{256 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1669*Sqrt[3 - x + 2*x^2])/128 - (337*(5 + 2*x)*Sqrt[3 - x + 2*x^2])/192 + (5*(5 + 2*x)^2*Sqrt[3 - x + 2*x^2])
/48 + (9657*ArcSinh[(1 - 4*x)/Sqrt[23]])/(256*Sqrt[2]) - (3667*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*
x^2])])/(96*Sqrt[2])

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

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) \sqrt{3-x+2 x^2}} \, dx &=\frac{5}{48} (5+2 x)^2 \sqrt{3-x+2 x^2}+\frac{1}{96} \int \frac{-2183-3054 x-4092 x^2-2696 x^3}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx\\ &=-\frac{337}{192} (5+2 x) \sqrt{3-x+2 x^2}+\frac{5}{48} (5+2 x)^2 \sqrt{3-x+2 x^2}+\frac{\int \frac{24504+128736 x+160224 x^2}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{3072}\\ &=\frac{1669}{128} \sqrt{3-x+2 x^2}-\frac{337}{192} (5+2 x) \sqrt{3-x+2 x^2}+\frac{5}{48} (5+2 x)^2 \sqrt{3-x+2 x^2}+\frac{\int \frac{997152-1854144 x}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{24576}\\ &=\frac{1669}{128} \sqrt{3-x+2 x^2}-\frac{337}{192} (5+2 x) \sqrt{3-x+2 x^2}+\frac{5}{48} (5+2 x)^2 \sqrt{3-x+2 x^2}-\frac{9657}{256} \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx+\frac{3667}{16} \int \frac{1}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx\\ &=\frac{1669}{128} \sqrt{3-x+2 x^2}-\frac{337}{192} (5+2 x) \sqrt{3-x+2 x^2}+\frac{5}{48} (5+2 x)^2 \sqrt{3-x+2 x^2}-\frac{3667}{8} \operatorname{Subst}\left (\int \frac{1}{288-x^2} \, dx,x,\frac{17-22 x}{\sqrt{3-x+2 x^2}}\right )-\frac{9657 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{256 \sqrt{46}}\\ &=\frac{1669}{128} \sqrt{3-x+2 x^2}-\frac{337}{192} (5+2 x) \sqrt{3-x+2 x^2}+\frac{5}{48} (5+2 x)^2 \sqrt{3-x+2 x^2}+\frac{9657 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{256 \sqrt{2}}-\frac{3667 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{3-x+2 x^2}}\right )}{96 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.103197, size = 81, normalized size = 0.64 \[ \frac{4 \sqrt{2 x^2-x+3} \left (160 x^2-548 x+2637\right )-29336 \sqrt{2} \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{4 x^2-2 x+6}}\right )+28971 \sqrt{2} \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{1536} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[3 - x + 2*x^2]*(2637 - 548*x + 160*x^2) + 28971*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt[23]] - 29336*Sqrt[2]*Ar
cTanh[(17 - 22*x)/(12*Sqrt[6 - 2*x + 4*x^2])])/1536

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Maple [A]  time = 0.056, size = 92, normalized size = 0.7 \begin{align*}{\frac{5\,{x}^{2}}{12}\sqrt{2\,{x}^{2}-x+3}}-{\frac{137\,x}{96}\sqrt{2\,{x}^{2}-x+3}}+{\frac{879}{128}\sqrt{2\,{x}^{2}-x+3}}-{\frac{9657\,\sqrt{2}}{512}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) }-{\frac{3667\,\sqrt{2}}{192}{\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 ) } \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)/(2*x^2-x+3)^(1/2),x)

[Out]

5/12*x^2*(2*x^2-x+3)^(1/2)-137/96*x*(2*x^2-x+3)^(1/2)+879/128*(2*x^2-x+3)^(1/2)-9657/512*2^(1/2)*arcsinh(4/23*
23^(1/2)*(x-1/4))-3667/192*2^(1/2)*arctanh(1/12*(17/2-11*x)*2^(1/2)/(2*(x+5/2)^2-11*x-19/2)^(1/2))

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Maxima [A]  time = 1.52173, size = 134, normalized size = 1.06 \begin{align*} \frac{5}{12} \, \sqrt{2 \, x^{2} - x + 3} x^{2} - \frac{137}{96} \, \sqrt{2 \, x^{2} - x + 3} x - \frac{9657}{512} \, \sqrt{2} \operatorname{arsinh}\left (\frac{4}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) + \frac{3667}{192} \, \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{879}{128} \, \sqrt{2 \, x^{2} - x + 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)/(2*x^2-x+3)^(1/2),x, algorithm="maxima")

[Out]

5/12*sqrt(2*x^2 - x + 3)*x^2 - 137/96*sqrt(2*x^2 - x + 3)*x - 9657/512*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*
sqrt(23)) + 3667/192*sqrt(2)*arcsinh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) + 879/128*sq
rt(2*x^2 - x + 3)

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Fricas [A]  time = 1.4016, size = 344, normalized size = 2.73 \begin{align*} \frac{1}{384} \,{\left (160 \, x^{2} - 548 \, x + 2637\right )} \sqrt{2 \, x^{2} - x + 3} + \frac{9657}{1024} \, \sqrt{2} \log \left (4 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + \frac{3667}{384} \, \sqrt{2} \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 ) \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)/(2*x^2-x+3)^(1/2),x, algorithm="fricas")

[Out]

1/384*(160*x^2 - 548*x + 2637)*sqrt(2*x^2 - x + 3) + 9657/1024*sqrt(2)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x
- 1) - 32*x^2 + 16*x - 25) + 3667/384*sqrt(2)*log(-(24*sqrt(2)*sqrt(2*x^2 - x + 3)*(22*x - 17) + 1060*x^2 - 10
36*x + 1153)/(4*x^2 + 20*x + 25))

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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 ) \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)/(2*x**2-x+3)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.16402, size = 161, normalized size = 1.28 \begin{align*} \frac{1}{384} \,{\left (4 \,{\left (40 \, x - 137\right )} x + 2637\right )} \sqrt{2 \, x^{2} - x + 3} + \frac{9657}{512} \, \sqrt{2} \log \left (-4 \, \sqrt{2} x + \sqrt{2} + 4 \, \sqrt{2 \, x^{2} - x + 3}\right ) - \frac{3667}{192} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x + \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\right ) + \frac{3667}{192} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x - 11 \, \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\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)/(2*x^2-x+3)^(1/2),x, algorithm="giac")

[Out]

1/384*(4*(40*x - 137)*x + 2637)*sqrt(2*x^2 - x + 3) + 9657/512*sqrt(2)*log(-4*sqrt(2)*x + sqrt(2) + 4*sqrt(2*x
^2 - x + 3)) - 3667/192*sqrt(2)*log(abs(-2*sqrt(2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) + 3667/192*sqrt(2)*lo
g(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt(2*x^2 - x + 3)))