3.255 \(\int \frac{1+3 x+4 x^2}{(1+2 x) (2-x+3 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=85 \[ -\frac{4 (691-13668 x)}{268203 \sqrt{3 x^2-x+2}}-\frac{2 (101-77 x)}{897 \left (3 x^2-x+2\right )^{3/2}}-\frac{8 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{169 \sqrt{13}} \]

[Out]

(-2*(101 - 77*x))/(897*(2 - x + 3*x^2)^(3/2)) - (4*(691 - 13668*x))/(268203*Sqrt[2 - x + 3*x^2]) - (8*ArcTanh[
(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(169*Sqrt[13])

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Rubi [A]  time = 0.0944011, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {1646, 822, 12, 724, 206} \[ -\frac{4 (691-13668 x)}{268203 \sqrt{3 x^2-x+2}}-\frac{2 (101-77 x)}{897 \left (3 x^2-x+2\right )^{3/2}}-\frac{8 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{169 \sqrt{13}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 3*x + 4*x^2)/((1 + 2*x)*(2 - x + 3*x^2)^(5/2)),x]

[Out]

(-2*(101 - 77*x))/(897*(2 - x + 3*x^2)^(3/2)) - (4*(691 - 13668*x))/(268203*Sqrt[2 - x + 3*x^2]) - (8*ArcTanh[
(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(169*Sqrt[13])

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 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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{1+3 x+4 x^2}{(1+2 x) \left (2-x+3 x^2\right )^{5/2}} \, dx &=-\frac{2 (101-77 x)}{897 \left (2-x+3 x^2\right )^{3/2}}+\frac{2}{69} \int \frac{\frac{223}{13}+\frac{308 x}{13}}{(1+2 x) \left (2-x+3 x^2\right )^{3/2}} \, dx\\ &=-\frac{2 (101-77 x)}{897 \left (2-x+3 x^2\right )^{3/2}}-\frac{4 (691-13668 x)}{268203 \sqrt{2-x+3 x^2}}+\frac{4 \int \frac{3174}{13 (1+2 x) \sqrt{2-x+3 x^2}} \, dx}{20631}\\ &=-\frac{2 (101-77 x)}{897 \left (2-x+3 x^2\right )^{3/2}}-\frac{4 (691-13668 x)}{268203 \sqrt{2-x+3 x^2}}+\frac{8}{169} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{2 (101-77 x)}{897 \left (2-x+3 x^2\right )^{3/2}}-\frac{4 (691-13668 x)}{268203 \sqrt{2-x+3 x^2}}-\frac{16}{169} \operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )\\ &=-\frac{2 (101-77 x)}{897 \left (2-x+3 x^2\right )^{3/2}}-\frac{4 (691-13668 x)}{268203 \sqrt{2-x+3 x^2}}-\frac{8 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )}{169 \sqrt{13}}\\ \end{align*}

Mathematica [A]  time = 0.0482146, size = 72, normalized size = 0.85 \[ \frac{2 \left (82008 x^3-31482 x^2+79077 x-32963\right )}{268203 \left (3 x^2-x+2\right )^{3/2}}-\frac{8 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{169 \sqrt{13}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 3*x + 4*x^2)/((1 + 2*x)*(2 - x + 3*x^2)^(5/2)),x]

[Out]

(2*(-32963 + 79077*x - 31482*x^2 + 82008*x^3))/(268203*(2 - x + 3*x^2)^(3/2)) - (8*ArcTanh[(9 - 8*x)/(2*Sqrt[1
3]*Sqrt[2 - x + 3*x^2])])/(169*Sqrt[13])

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Maple [B]  time = 0.056, size = 158, normalized size = 1.9 \begin{align*} -{\frac{2}{9} \left ( 3\,{x}^{2}-x+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{-5+30\,x}{207} \left ( 3\,{x}^{2}-x+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{-40+240\,x}{1587}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}}+{\frac{1}{39} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{-{\frac{3}{2}}}}+{\frac{-4+24\,x}{897} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{-{\frac{3}{2}}}}+{\frac{-784+4704\,x}{89401}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}}}}+{\frac{4}{169}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}}}}-{\frac{8\,\sqrt{13}}{2197}{\it Artanh} \left ({\frac{2\,\sqrt{13}}{13} \left ({\frac{9}{2}}-4\,x \right ){\frac{1}{\sqrt{12\, \left ( x+1/2 \right ) ^{2}-16\,x+5}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2+3*x+1)/(1+2*x)/(3*x^2-x+2)^(5/2),x)

[Out]

-2/9/(3*x^2-x+2)^(3/2)+5/207*(-1+6*x)/(3*x^2-x+2)^(3/2)+40/1587*(-1+6*x)/(3*x^2-x+2)^(1/2)+1/39/(3*(x+1/2)^2-4
*x+5/4)^(3/2)+4/897*(-1+6*x)/(3*(x+1/2)^2-4*x+5/4)^(3/2)+784/89401*(-1+6*x)/(3*(x+1/2)^2-4*x+5/4)^(1/2)+4/169/
(3*(x+1/2)^2-4*x+5/4)^(1/2)-8/2197*13^(1/2)*arctanh(2/13*(9/2-4*x)*13^(1/2)/(12*(x+1/2)^2-16*x+5)^(1/2))

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Maxima [A]  time = 1.47288, size = 126, normalized size = 1.48 \begin{align*} \frac{8}{2197} \, \sqrt{13} \operatorname{arsinh}\left (\frac{8 \, \sqrt{23} x}{23 \,{\left | 2 \, x + 1 \right |}} - \frac{9 \, \sqrt{23}}{23 \,{\left | 2 \, x + 1 \right |}}\right ) + \frac{18224 \, x}{89401 \, \sqrt{3 \, x^{2} - x + 2}} - \frac{2764}{268203 \, \sqrt{3 \, x^{2} - x + 2}} + \frac{154 \, x}{897 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} - \frac{202}{897 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+3*x+1)/(1+2*x)/(3*x^2-x+2)^(5/2),x, algorithm="maxima")

[Out]

8/2197*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9/23*sqrt(23)/abs(2*x + 1)) + 18224/89401*x/sqrt(3*x^2
- x + 2) - 2764/268203/sqrt(3*x^2 - x + 2) + 154/897*x/(3*x^2 - x + 2)^(3/2) - 202/897/(3*x^2 - x + 2)^(3/2)

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Fricas [A]  time = 1.10805, size = 344, normalized size = 4.05 \begin{align*} \frac{2 \,{\left (3174 \, \sqrt{13}{\left (9 \, x^{4} - 6 \, x^{3} + 13 \, x^{2} - 4 \, x + 4\right )} \log \left (-\frac{4 \, \sqrt{13} \sqrt{3 \, x^{2} - x + 2}{\left (8 \, x - 9\right )} + 220 \, x^{2} - 196 \, x + 185}{4 \, x^{2} + 4 \, x + 1}\right ) + 13 \,{\left (82008 \, x^{3} - 31482 \, x^{2} + 79077 \, x - 32963\right )} \sqrt{3 \, x^{2} - x + 2}\right )}}{3486639 \,{\left (9 \, x^{4} - 6 \, x^{3} + 13 \, x^{2} - 4 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+3*x+1)/(1+2*x)/(3*x^2-x+2)^(5/2),x, algorithm="fricas")

[Out]

2/3486639*(3174*sqrt(13)*(9*x^4 - 6*x^3 + 13*x^2 - 4*x + 4)*log(-(4*sqrt(13)*sqrt(3*x^2 - x + 2)*(8*x - 9) + 2
20*x^2 - 196*x + 185)/(4*x^2 + 4*x + 1)) + 13*(82008*x^3 - 31482*x^2 + 79077*x - 32963)*sqrt(3*x^2 - x + 2))/(
9*x^4 - 6*x^3 + 13*x^2 - 4*x + 4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{4 x^{2} + 3 x + 1}{\left (2 x + 1\right ) \left (3 x^{2} - x + 2\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2+3*x+1)/(1+2*x)/(3*x**2-x+2)**(5/2),x)

[Out]

Integral((4*x**2 + 3*x + 1)/((2*x + 1)*(3*x**2 - x + 2)**(5/2)), x)

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Giac [A]  time = 1.20905, size = 136, normalized size = 1.6 \begin{align*} \frac{8}{2197} \, \sqrt{13} \log \left (-\frac{{\left | -4 \, \sqrt{3} x - 2 \, \sqrt{13} - 2 \, \sqrt{3} + 4 \, \sqrt{3 \, x^{2} - x + 2} \right |}}{2 \,{\left (2 \, \sqrt{3} x - \sqrt{13} + \sqrt{3} - 2 \, \sqrt{3 \, x^{2} - x + 2}\right )}}\right ) + \frac{2 \,{\left (3 \,{\left (6 \,{\left (4556 \, x - 1749\right )} x + 26359\right )} x - 32963\right )}}{268203 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+3*x+1)/(1+2*x)/(3*x^2-x+2)^(5/2),x, algorithm="giac")

[Out]

8/2197*sqrt(13)*log(-1/2*abs(-4*sqrt(3)*x - 2*sqrt(13) - 2*sqrt(3) + 4*sqrt(3*x^2 - x + 2))/(2*sqrt(3)*x - sqr
t(13) + sqrt(3) - 2*sqrt(3*x^2 - x + 2))) + 2/268203*(3*(6*(4556*x - 1749)*x + 26359)*x - 32963)/(3*x^2 - x +
2)^(3/2)