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

Optimal. Leaf size=108 \[ \frac{2203 x+9897}{119232 \sqrt{2 x^2-x+3}}-\frac{3667 \sqrt{2 x^2-x+3}}{10368 (2 x+5)}+\frac{25951 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{41472 \sqrt{2}}-\frac{5 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{8 \sqrt{2}} \]

[Out]

(9897 + 2203*x)/(119232*Sqrt[3 - x + 2*x^2]) - (3667*Sqrt[3 - x + 2*x^2])/(10368*(5 + 2*x)) - (5*ArcSinh[(1 -
4*x)/Sqrt[23]])/(8*Sqrt[2]) + (25951*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(41472*Sqrt[2])

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Rubi [A]  time = 0.152743, antiderivative size = 108, 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, 1650, 843, 619, 215, 724, 206} \[ \frac{2203 x+9897}{119232 \sqrt{2 x^2-x+3}}-\frac{3667 \sqrt{2 x^2-x+3}}{10368 (2 x+5)}+\frac{25951 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{41472 \sqrt{2}}-\frac{5 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{8 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9897 + 2203*x)/(119232*Sqrt[3 - x + 2*x^2]) - (3667*Sqrt[3 - x + 2*x^2])/(10368*(5 + 2*x)) - (5*ArcSinh[(1 -
4*x)/Sqrt[23]])/(8*Sqrt[2]) + (25951*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(41472*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 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)^2 \left (3-x+2 x^2\right )^{3/2}} \, dx &=\frac{9897+2203 x}{119232 \sqrt{3-x+2 x^2}}+\frac{2}{23} \int \frac{-\frac{33649}{20736}+\frac{131215 x}{10368}+\frac{115 x^2}{4}}{(5+2 x)^2 \sqrt{3-x+2 x^2}} \, dx\\ &=\frac{9897+2203 x}{119232 \sqrt{3-x+2 x^2}}-\frac{3667 \sqrt{3-x+2 x^2}}{10368 (5+2 x)}-\frac{1}{828} \int \frac{\frac{100073}{192}-1035 x}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx\\ &=\frac{9897+2203 x}{119232 \sqrt{3-x+2 x^2}}-\frac{3667 \sqrt{3-x+2 x^2}}{10368 (5+2 x)}+\frac{5}{8} \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx-\frac{25951 \int \frac{1}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{6912}\\ &=\frac{9897+2203 x}{119232 \sqrt{3-x+2 x^2}}-\frac{3667 \sqrt{3-x+2 x^2}}{10368 (5+2 x)}+\frac{25951 \operatorname{Subst}\left (\int \frac{1}{288-x^2} \, dx,x,\frac{17-22 x}{\sqrt{3-x+2 x^2}}\right )}{3456}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{8 \sqrt{46}}\\ &=\frac{9897+2203 x}{119232 \sqrt{3-x+2 x^2}}-\frac{3667 \sqrt{3-x+2 x^2}}{10368 (5+2 x)}-\frac{5 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{8 \sqrt{2}}+\frac{25951 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{3-x+2 x^2}}\right )}{41472 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.317964, size = 104, normalized size = 0.96 \[ \frac{\frac{8 (2203 x+9897)}{23 \sqrt{x^2-\frac{x}{2}+\frac{3}{2}}}-\frac{14668 \sqrt{4 x^2-2 x+6}}{2 x+5}+25951 \log \left (12 \sqrt{4 x^2-2 x+6}-22 x+17\right )-25951 \log (2 x+5)+25920 \sinh ^{-1}\left (\frac{4 x-1}{\sqrt{23}}\right )}{41472 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*(9897 + 2203*x))/(23*Sqrt[3/2 - x/2 + x^2]) - (14668*Sqrt[6 - 2*x + 4*x^2])/(5 + 2*x) + 25920*ArcSinh[(-1
+ 4*x)/Sqrt[23]] - 25951*Log[5 + 2*x] + 25951*Log[17 - 22*x + 12*Sqrt[6 - 2*x + 4*x^2]])/(41472*Sqrt[2])

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Maple [A]  time = 0.057, size = 152, normalized size = 1.4 \begin{align*} -{\frac{5\,x}{8}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}+{\frac{99}{32}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}+{\frac{-1529+6116\,x}{736}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}+{\frac{5\,\sqrt{2}}{16}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) }-{\frac{25951}{13824}{\frac{1}{\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}}}}}-{\frac{-637493+2549972\,x}{317952}{\frac{1}{\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}}}}}+{\frac{25951\,\sqrt{2}}{82944}{\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} \left ( x+{\frac{5}{2}} \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}}}}} \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)^(3/2),x)

[Out]

-5/8*x/(2*x^2-x+3)^(1/2)+99/32/(2*x^2-x+3)^(1/2)+1529/736*(-1+4*x)/(2*x^2-x+3)^(1/2)+5/16*2^(1/2)*arcsinh(4/23
*23^(1/2)*(x-1/4))-25951/13824/(2*(x+5/2)^2-11*x-19/2)^(1/2)-637493/317952*(-1+4*x)/(2*(x+5/2)^2-11*x-19/2)^(1
/2)+25951/82944*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-19/2)^(1/2)

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Maxima [A]  time = 1.52622, size = 157, normalized size = 1.45 \begin{align*} \frac{5}{16} \, \sqrt{2} \operatorname{arsinh}\left (\frac{4}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) - \frac{25951}{82944} \, \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{26645 \, x}{79488 \, \sqrt{2 \, x^{2} - x + 3}} + \frac{30313}{26496 \, \sqrt{2 \, x^{2} - x + 3}} - \frac{3667}{576 \,{\left (2 \, \sqrt{2 \, x^{2} - x + 3} x + 5 \, \sqrt{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/(2*x^2-x+3)^(3/2),x, algorithm="maxima")

[Out]

5/16*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) - 25951/82944*sqrt(2)*arcsinh(22/23*sqrt(23)*x/abs(2*x +
5) - 17/23*sqrt(23)/abs(2*x + 5)) - 26645/79488*x/sqrt(2*x^2 - x + 3) + 30313/26496/sqrt(2*x^2 - x + 3) - 366
7/576/(2*sqrt(2*x^2 - x + 3)*x + 5*sqrt(2*x^2 - x + 3))

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Fricas [A]  time = 1.37733, size = 458, normalized size = 4.24 \begin{align*} \frac{596160 \, \sqrt{2}{\left (4 \, x^{3} + 8 \, x^{2} + x + 15\right )} \log \left (-4 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 596873 \, \sqrt{2}{\left (4 \, x^{3} + 8 \, x^{2} + x + 15\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 (53290 \, x^{2} - 48653 \, x + 51351\right )} \sqrt{2 \, x^{2} - x + 3}}{3815424 \,{\left (4 \, x^{3} + 8 \, x^{2} + x + 15\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)^(3/2),x, algorithm="fricas")

[Out]

1/3815424*(596160*sqrt(2)*(4*x^3 + 8*x^2 + x + 15)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*
x - 25) + 596873*sqrt(2)*(4*x^3 + 8*x^2 + x + 15)*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*(53290*x^2 - 48653*x + 51351)*sqrt(2*x^2 - x + 3))/(4*x^3 + 8*x^2 +
x + 15)

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

[Out]

Integral((5*x**4 - x**3 + 3*x**2 + x + 2)/((2*x + 5)**2*(2*x**2 - x + 3)**(3/2)), 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}{{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}}{\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)^(3/2),x, algorithm="giac")

[Out]

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