### 3.348 $$\int \frac{2+x+3 x^2-x^3+5 x^4}{(5+2 x)^3 \sqrt{3-x+2 x^2}} \, dx$$

Optimal. Leaf size=128 $\frac{92239 \sqrt{2 x^2-x+3}}{27648 (2 x+5)}-\frac{3667 \sqrt{2 x^2-x+3}}{1152 (2 x+5)^2}+\frac{5}{16} \sqrt{2 x^2-x+3}-\frac{1546507 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{331776 \sqrt{2}}+\frac{149 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{32 \sqrt{2}}$

[Out]

(5*Sqrt[3 - x + 2*x^2])/16 - (3667*Sqrt[3 - x + 2*x^2])/(1152*(5 + 2*x)^2) + (92239*Sqrt[3 - x + 2*x^2])/(2764
8*(5 + 2*x)) + (149*ArcSinh[(1 - 4*x)/Sqrt[23]])/(32*Sqrt[2]) - (1546507*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[
3 - x + 2*x^2])])/(331776*Sqrt[2])

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Rubi [A]  time = 0.207631, antiderivative size = 128, 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} $\frac{92239 \sqrt{2 x^2-x+3}}{27648 (2 x+5)}-\frac{3667 \sqrt{2 x^2-x+3}}{1152 (2 x+5)^2}+\frac{5}{16} \sqrt{2 x^2-x+3}-\frac{1546507 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{331776 \sqrt{2}}+\frac{149 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{32 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(5*Sqrt[3 - x + 2*x^2])/16 - (3667*Sqrt[3 - x + 2*x^2])/(1152*(5 + 2*x)^2) + (92239*Sqrt[3 - x + 2*x^2])/(2764
8*(5 + 2*x)) + (149*ArcSinh[(1 - 4*x)/Sqrt[23]])/(32*Sqrt[2]) - (1546507*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[
3 - x + 2*x^2])])/(331776*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)^3 \sqrt{3-x+2 x^2}} \, dx &=-\frac{3667 \sqrt{3-x+2 x^2}}{1152 (5+2 x)^2}-\frac{1}{144} \int \frac{\frac{20347}{16}-\frac{6917 x}{4}+972 x^2-360 x^3}{(5+2 x)^2 \sqrt{3-x+2 x^2}} \, dx\\ &=-\frac{3667 \sqrt{3-x+2 x^2}}{1152 (5+2 x)^2}+\frac{92239 \sqrt{3-x+2 x^2}}{27648 (5+2 x)}+\frac{\int \frac{\frac{647841}{16}-67392 x+12960 x^2}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{10368}\\ &=\frac{5}{16} \sqrt{3-x+2 x^2}-\frac{3667 \sqrt{3-x+2 x^2}}{1152 (5+2 x)^2}+\frac{92239 \sqrt{3-x+2 x^2}}{27648 (5+2 x)}+\frac{\int \frac{\frac{777441}{2}-772416 x}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{82944}\\ &=\frac{5}{16} \sqrt{3-x+2 x^2}-\frac{3667 \sqrt{3-x+2 x^2}}{1152 (5+2 x)^2}+\frac{92239 \sqrt{3-x+2 x^2}}{27648 (5+2 x)}-\frac{149}{32} \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx+\frac{1546507 \int \frac{1}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{55296}\\ &=\frac{5}{16} \sqrt{3-x+2 x^2}-\frac{3667 \sqrt{3-x+2 x^2}}{1152 (5+2 x)^2}+\frac{92239 \sqrt{3-x+2 x^2}}{27648 (5+2 x)}-\frac{1546507 \operatorname{Subst}\left (\int \frac{1}{288-x^2} \, dx,x,\frac{17-22 x}{\sqrt{3-x+2 x^2}}\right )}{27648}-\frac{149 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{32 \sqrt{46}}\\ &=\frac{5}{16} \sqrt{3-x+2 x^2}-\frac{3667 \sqrt{3-x+2 x^2}}{1152 (5+2 x)^2}+\frac{92239 \sqrt{3-x+2 x^2}}{27648 (5+2 x)}+\frac{149 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{32 \sqrt{2}}-\frac{1546507 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{3-x+2 x^2}}\right )}{331776 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.129727, size = 88, normalized size = 0.69 $\frac{\frac{24 \sqrt{2 x^2-x+3} \left (34560 x^2+357278 x+589187\right )}{(2 x+5)^2}-1546507 \sqrt{2} \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{4 x^2-2 x+6}}\right )+1544832 \sqrt{2} \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{663552}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((24*Sqrt[3 - x + 2*x^2]*(589187 + 357278*x + 34560*x^2))/(5 + 2*x)^2 + 1544832*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt
[23]] - 1546507*Sqrt[2]*ArcTanh[(17 - 22*x)/(12*Sqrt[6 - 2*x + 4*x^2])])/663552

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Maple [A]  time = 0.065, size = 102, normalized size = 0.8 \begin{align*}{\frac{5}{16}\sqrt{2\,{x}^{2}-x+3}}-{\frac{149\,\sqrt{2}}{64}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) }-{\frac{3667}{4608}\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}} \left ( x+{\frac{5}{2}} \right ) ^{-2}}+{\frac{92239}{55296}\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}} \left ( x+{\frac{5}{2}} \right ) ^{-1}}-{\frac{1546507\,\sqrt{2}}{663552}{\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)^3/(2*x^2-x+3)^(1/2),x)

[Out]

5/16*(2*x^2-x+3)^(1/2)-149/64*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))-3667/4608/(x+5/2)^2*(2*(x+5/2)^2-11*x-19/
2)^(1/2)+92239/55296/(x+5/2)*(2*(x+5/2)^2-11*x-19/2)^(1/2)-1546507/663552*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 = 1.50934, size = 154, normalized size = 1.2 \begin{align*} -\frac{149}{64} \, \sqrt{2} \operatorname{arsinh}\left (\frac{4}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) + \frac{1546507}{663552} \, \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{5}{16} \, \sqrt{2 \, x^{2} - x + 3} - \frac{3667 \, \sqrt{2 \, x^{2} - x + 3}}{1152 \,{\left (4 \, x^{2} + 20 \, x + 25\right )}} + \frac{92239 \, \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)^3/(2*x^2-x+3)^(1/2),x, algorithm="maxima")

[Out]

-149/64*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) + 1546507/663552*sqrt(2)*arcsinh(22/23*sqrt(23)*x/abs
(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) + 5/16*sqrt(2*x^2 - x + 3) - 3667/1152*sqrt(2*x^2 - x + 3)/(4*x^2 + 2
0*x + 25) + 92239/27648*sqrt(2*x^2 - x + 3)/(2*x + 5)

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Fricas [A]  time = 1.42837, size = 443, normalized size = 3.46 \begin{align*} \frac{1544832 \, \sqrt{2}{\left (4 \, x^{2} + 20 \, x + 25\right )} \log \left (4 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 1546507 \, \sqrt{2}{\left (4 \, x^{2} + 20 \, x + 25\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 (34560 \, x^{2} + 357278 \, x + 589187\right )} \sqrt{2 \, x^{2} - x + 3}}{1327104 \,{\left (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)^3/(2*x^2-x+3)^(1/2),x, algorithm="fricas")

[Out]

1/1327104*(1544832*sqrt(2)*(4*x^2 + 20*x + 25)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 2
5) + 1546507*sqrt(2)*(4*x^2 + 20*x + 25)*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*(34560*x^2 + 357278*x + 589187)*sqrt(2*x^2 - x + 3))/(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 )^{3} \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)**3/(2*x**2-x+3)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.20277, size = 335, normalized size = 2.62 \begin{align*} \frac{149}{64} \, \sqrt{2} \log \left (-2 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )} + 1\right ) - \frac{1546507}{663552} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x + \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\right ) + \frac{1546507}{663552} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x - 11 \, \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\right ) + \frac{5}{16} \, \sqrt{2 \, x^{2} - x + 3} + \frac{\sqrt{2}{\left (2381290 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )}^{3} + 16628406 \,{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )}^{2} - 25697445 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )} + 16720645\right )}}{55296 \,{\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 )}^{2}} \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)^3/(2*x^2-x+3)^(1/2),x, algorithm="giac")

[Out]

149/64*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) - 1546507/663552*sqrt(2)*log(abs(-2*sqrt(
2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) + 1546507/663552*sqrt(2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt(2
*x^2 - x + 3))) + 5/16*sqrt(2*x^2 - x + 3) + 1/55296*sqrt(2)*(2381290*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)
)^3 + 16628406*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 - 25697445*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1672
0645)/(2*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 + 10*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) - 11)^2