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

Optimal. Leaf size=101 \[ \frac{917 x+1191}{3312 \sqrt{2 x^2-x+3}}+\frac{5}{8} \sqrt{2 x^2-x+3}-\frac{3667 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{1728 \sqrt{2}}+\frac{39 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{16 \sqrt{2}} \]

[Out]

(1191 + 917*x)/(3312*Sqrt[3 - x + 2*x^2]) + (5*Sqrt[3 - x + 2*x^2])/8 + (39*ArcSinh[(1 - 4*x)/Sqrt[23]])/(16*S
qrt[2]) - (3667*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(1728*Sqrt[2])

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Rubi [A]  time = 0.151246, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {1646, 1653, 843, 619, 215, 724, 206} \[ \frac{917 x+1191}{3312 \sqrt{2 x^2-x+3}}+\frac{5}{8} \sqrt{2 x^2-x+3}-\frac{3667 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{1728 \sqrt{2}}+\frac{39 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{16 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1191 + 917*x)/(3312*Sqrt[3 - x + 2*x^2]) + (5*Sqrt[3 - x + 2*x^2])/8 + (39*ArcSinh[(1 - 4*x)/Sqrt[23]])/(16*S
qrt[2]) - (3667*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(1728*Sqrt[2])

Rule 1646

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

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) \left (3-x+2 x^2\right )^{3/2}} \, dx &=\frac{1191+917 x}{3312 \sqrt{3-x+2 x^2}}+\frac{2}{23} \int \frac{-\frac{6739}{576}+\frac{69 x}{8}+\frac{115 x^2}{4}}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx\\ &=\frac{1191+917 x}{3312 \sqrt{3-x+2 x^2}}+\frac{5}{8} \sqrt{3-x+2 x^2}+\frac{1}{92} \int \frac{\frac{3611}{72}-\frac{897 x}{2}}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx\\ &=\frac{1191+917 x}{3312 \sqrt{3-x+2 x^2}}+\frac{5}{8} \sqrt{3-x+2 x^2}-\frac{39}{16} \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx+\frac{3667}{288} \int \frac{1}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx\\ &=\frac{1191+917 x}{3312 \sqrt{3-x+2 x^2}}+\frac{5}{8} \sqrt{3-x+2 x^2}-\frac{3667}{144} \operatorname{Subst}\left (\int \frac{1}{288-x^2} \, dx,x,\frac{17-22 x}{\sqrt{3-x+2 x^2}}\right )-\frac{39 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{16 \sqrt{46}}\\ &=\frac{1191+917 x}{3312 \sqrt{3-x+2 x^2}}+\frac{5}{8} \sqrt{3-x+2 x^2}+\frac{39 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{16 \sqrt{2}}-\frac{3667 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{3-x+2 x^2}}\right )}{1728 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.349433, size = 86, normalized size = 0.85 \[ \frac{\frac{12 \left (4140 x^2-1153 x+7401\right )}{23 \sqrt{x^2-\frac{x}{2}+\frac{3}{2}}}-3667 \log \left (12 \sqrt{4 x^2-2 x+6}-22 x+17\right )+3667 \log (2 x+5)-4212 \sinh ^{-1}\left (\frac{4 x-1}{\sqrt{23}}\right )}{1728 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*(7401 - 1153*x + 4140*x^2))/(23*Sqrt[3/2 - x/2 + x^2]) - 4212*ArcSinh[(-1 + 4*x)/Sqrt[23]] + 3667*Log[5 +
2*x] - 3667*Log[17 - 22*x + 12*Sqrt[6 - 2*x + 4*x^2]])/(1728*Sqrt[2])

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Maple [A]  time = 0.061, size = 148, normalized size = 1.5 \begin{align*}{\frac{5\,{x}^{2}}{4}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}+{\frac{39\,x}{16}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}-{\frac{309}{64}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}-{\frac{-5507+22028\,x}{1472}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}-{\frac{39\,\sqrt{2}}{32}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) }+{\frac{3667}{576}{\frac{1}{\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}}}}}+{\frac{-40337+161348\,x}{13248}{\frac{1}{\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}}}}}-{\frac{3667\,\sqrt{2}}{3456}{\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)^(3/2),x)

[Out]

5/4*x^2/(2*x^2-x+3)^(1/2)+39/16*x/(2*x^2-x+3)^(1/2)-309/64/(2*x^2-x+3)^(1/2)-5507/1472*(-1+4*x)/(2*x^2-x+3)^(1
/2)-39/32*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))+3667/576/(2*(x+5/2)^2-11*x-19/2)^(1/2)+40337/13248*(-1+4*x)/(
2*(x+5/2)^2-11*x-19/2)^(1/2)-3667/3456*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 = 1.52708, size = 134, normalized size = 1.33 \begin{align*} \frac{5 \, x^{2}}{4 \, \sqrt{2 \, x^{2} - x + 3}} - \frac{39}{32} \, \sqrt{2} \operatorname{arsinh}\left (\frac{4}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) + \frac{3667}{3456} \, \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{1153 \, x}{3312 \, \sqrt{2 \, x^{2} - x + 3}} + \frac{2467}{1104 \, \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)^(3/2),x, algorithm="maxima")

[Out]

5/4*x^2/sqrt(2*x^2 - x + 3) - 39/32*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) + 3667/3456*sqrt(2)*arcsi
nh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) - 1153/3312*x/sqrt(2*x^2 - x + 3) + 2467/1104/
sqrt(2*x^2 - x + 3)

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Fricas [A]  time = 1.38289, size = 413, normalized size = 4.09 \begin{align*} \frac{96876 \, \sqrt{2}{\left (2 \, x^{2} - x + 3\right )} \log \left (4 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 84341 \, \sqrt{2}{\left (2 \, x^{2} - x + 3\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 (4140 \, x^{2} - 1153 \, x + 7401\right )} \sqrt{2 \, x^{2} - x + 3}}{158976 \,{\left (2 \, x^{2} - x + 3\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)^(3/2),x, algorithm="fricas")

[Out]

1/158976*(96876*sqrt(2)*(2*x^2 - x + 3)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) + 84
341*sqrt(2)*(2*x^2 - x + 3)*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*(4140*x^2 - 1153*x + 7401)*sqrt(2*x^2 - x + 3))/(2*x^2 - x + 3)

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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 ) \left (2 x^{2} - x + 3\right )^{\frac{3}{2}}}\, 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)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.16943, size = 159, normalized size = 1.57 \begin{align*} \frac{39}{32} \, \sqrt{2} \log \left (-4 \, \sqrt{2} x + \sqrt{2} + 4 \, \sqrt{2 \, x^{2} - x + 3}\right ) - \frac{3667}{3456} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x + \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\right ) + \frac{3667}{3456} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x - 11 \, \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\right ) + \frac{{\left (4140 \, x - 1153\right )} x + 7401}{3312 \, \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)^(3/2),x, algorithm="giac")

[Out]

39/32*sqrt(2)*log(-4*sqrt(2)*x + sqrt(2) + 4*sqrt(2*x^2 - x + 3)) - 3667/3456*sqrt(2)*log(abs(-2*sqrt(2)*x + s
qrt(2) + 2*sqrt(2*x^2 - x + 3))) + 3667/3456*sqrt(2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt(2*x^2 - x + 3)
)) + 1/3312*((4140*x - 1153)*x + 7401)/sqrt(2*x^2 - x + 3)