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

Optimal. Leaf size=126 \[ \frac{5}{32} \sqrt{2 x^2-x+3} (2 x+5)-\frac{243}{64} \sqrt{2 x^2-x+3}-\frac{3667 \sqrt{2 x^2-x+3}}{576 (2 x+5)}+\frac{158527 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{6912 \sqrt{2}}-\frac{2943 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{128 \sqrt{2}} \]

[Out]

(-243*Sqrt[3 - x + 2*x^2])/64 - (3667*Sqrt[3 - x + 2*x^2])/(576*(5 + 2*x)) + (5*(5 + 2*x)*Sqrt[3 - x + 2*x^2])
/32 - (2943*ArcSinh[(1 - 4*x)/Sqrt[23]])/(128*Sqrt[2]) + (158527*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x +
2*x^2])])/(6912*Sqrt[2])

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Rubi [A]  time = 0.202046, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {1650, 1653, 843, 619, 215, 724, 206} \[ \frac{5}{32} \sqrt{2 x^2-x+3} (2 x+5)-\frac{243}{64} \sqrt{2 x^2-x+3}-\frac{3667 \sqrt{2 x^2-x+3}}{576 (2 x+5)}+\frac{158527 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{6912 \sqrt{2}}-\frac{2943 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{128 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-243*Sqrt[3 - x + 2*x^2])/64 - (3667*Sqrt[3 - x + 2*x^2])/(576*(5 + 2*x)) + (5*(5 + 2*x)*Sqrt[3 - x + 2*x^2])
/32 - (2943*ArcSinh[(1 - 4*x)/Sqrt[23]])/(128*Sqrt[2]) + (158527*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x +
2*x^2])])/(6912*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 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)^2 \sqrt{3-x+2 x^2}} \, dx &=-\frac{3667 \sqrt{3-x+2 x^2}}{576 (5+2 x)}-\frac{1}{72} \int \frac{\frac{12007}{16}-1323 x+486 x^2-180 x^3}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx\\ &=-\frac{3667 \sqrt{3-x+2 x^2}}{576 (5+2 x)}+\frac{5}{32} (5+2 x) \sqrt{3-x+2 x^2}-\frac{\int \frac{30314-27216 x+34992 x^2}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{2304}\\ &=-\frac{243}{64} \sqrt{3-x+2 x^2}-\frac{3667 \sqrt{3-x+2 x^2}}{576 (5+2 x)}+\frac{5}{32} (5+2 x) \sqrt{3-x+2 x^2}-\frac{\int \frac{417472-847584 x}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{18432}\\ &=-\frac{243}{64} \sqrt{3-x+2 x^2}-\frac{3667 \sqrt{3-x+2 x^2}}{576 (5+2 x)}+\frac{5}{32} (5+2 x) \sqrt{3-x+2 x^2}+\frac{2943}{128} \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx-\frac{158527 \int \frac{1}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{1152}\\ &=-\frac{243}{64} \sqrt{3-x+2 x^2}-\frac{3667 \sqrt{3-x+2 x^2}}{576 (5+2 x)}+\frac{5}{32} (5+2 x) \sqrt{3-x+2 x^2}+\frac{158527}{576} \operatorname{Subst}\left (\int \frac{1}{288-x^2} \, dx,x,\frac{17-22 x}{\sqrt{3-x+2 x^2}}\right )+\frac{2943 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{128 \sqrt{46}}\\ &=-\frac{243}{64} \sqrt{3-x+2 x^2}-\frac{3667 \sqrt{3-x+2 x^2}}{576 (5+2 x)}+\frac{5}{32} (5+2 x) \sqrt{3-x+2 x^2}-\frac{2943 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{128 \sqrt{2}}+\frac{158527 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{3-x+2 x^2}}\right )}{6912 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.114362, size = 88, normalized size = 0.7 \[ \frac{\frac{48 \sqrt{2 x^2-x+3} \left (180 x^2-1287 x-6176\right )}{2 x+5}+158527 \sqrt{2} \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{4 x^2-2 x+6}}\right )-158922 \sqrt{2} \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{13824} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((48*Sqrt[3 - x + 2*x^2]*(-6176 - 1287*x + 180*x^2))/(5 + 2*x) - 158922*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt[23]] +
158527*Sqrt[2]*ArcTanh[(17 - 22*x)/(12*Sqrt[6 - 2*x + 4*x^2])])/13824

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Maple [A]  time = 0.056, size = 96, normalized size = 0.8 \begin{align*}{\frac{5\,x}{16}\sqrt{2\,{x}^{2}-x+3}}-{\frac{193}{64}\sqrt{2\,{x}^{2}-x+3}}+{\frac{2943\,\sqrt{2}}{256}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) }+{\frac{158527\,\sqrt{2}}{13824}{\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}{1152}\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}} \left ( x+{\frac{5}{2}} \right ) ^{-1}} \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/(2*x^2-x+3)^(1/2),x)

[Out]

5/16*x*(2*x^2-x+3)^(1/2)-193/64*(2*x^2-x+3)^(1/2)+2943/256*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))+158527/13824
*2^(1/2)*arctanh(1/12*(17/2-11*x)*2^(1/2)/(2*(x+5/2)^2-11*x-19/2)^(1/2))-3667/1152/(x+5/2)*(2*(x+5/2)^2-11*x-1
9/2)^(1/2)

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Maxima [A]  time = 1.52076, size = 139, normalized size = 1.1 \begin{align*} \frac{5}{16} \, \sqrt{2 \, x^{2} - x + 3} x + \frac{2943}{256} \, \sqrt{2} \operatorname{arsinh}\left (\frac{4}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) - \frac{158527}{13824} \, \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{193}{64} \, \sqrt{2 \, x^{2} - x + 3} - \frac{3667 \, \sqrt{2 \, x^{2} - x + 3}}{576 \,{\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)^2/(2*x^2-x+3)^(1/2),x, algorithm="maxima")

[Out]

5/16*sqrt(2*x^2 - x + 3)*x + 2943/256*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) - 158527/13824*sqrt(2)*
arcsinh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) - 193/64*sqrt(2*x^2 - x + 3) - 3667/576*s
qrt(2*x^2 - x + 3)/(2*x + 5)

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Fricas [A]  time = 1.44426, size = 389, normalized size = 3.09 \begin{align*} \frac{158922 \, \sqrt{2}{\left (2 \, x + 5\right )} \log \left (-4 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 158527 \, \sqrt{2}{\left (2 \, x + 5\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 ) + 96 \,{\left (180 \, x^{2} - 1287 \, x - 6176\right )} \sqrt{2 \, x^{2} - x + 3}}{27648 \,{\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)^2/(2*x^2-x+3)^(1/2),x, algorithm="fricas")

[Out]

1/27648*(158922*sqrt(2)*(2*x + 5)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) + 158527*
sqrt(2)*(2*x + 5)*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)) + 96*(180*x^2 - 1287*x - 6176)*sqrt(2*x^2 - x + 3))/(2*x + 5)

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{5 \, x^{4} - x^{3} + 3 \, x^{2} + x + 2}{\sqrt{2 \, x^{2} - x + 3}{\left (2 \, x + 5\right )}^{2}}\,{d x} \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/(2*x^2-x+3)^(1/2),x, algorithm="giac")

[Out]

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