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

Optimal. Leaf size=102 \[ -\frac {\sqrt {2 x+3} (35 x+29)}{2 \left (3 x^2+5 x+2\right )^2}+\frac {3 \sqrt {2 x+3} (1063 x+878)}{10 \left (3 x^2+5 x+2\right )}+730 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {4713}{5} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {820, 822, 826, 1166, 207} \begin {gather*} -\frac {\sqrt {2 x+3} (35 x+29)}{2 \left (3 x^2+5 x+2\right )^2}+\frac {3 \sqrt {2 x+3} (1063 x+878)}{10 \left (3 x^2+5 x+2\right )}+730 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {4713}{5} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[3 + 2*x]*(29 + 35*x))/(2*(2 + 5*x + 3*x^2)^2) + (3*Sqrt[3 + 2*x]*(878 + 1063*x))/(10*(2 + 5*x + 3*x^2))
+ 730*ArcTanh[Sqrt[3 + 2*x]] - (4713*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/5

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 820

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

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 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {(5-x) \sqrt {3+2 x}}{\left (2+5 x+3 x^2\right )^3} \, dx &=-\frac {\sqrt {3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}-\frac {1}{2} \int \frac {286+175 x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac {\sqrt {3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac {3 \sqrt {3+2 x} (878+1063 x)}{10 \left (2+5 x+3 x^2\right )}+\frac {1}{10} \int \frac {6839+3189 x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac {\sqrt {3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac {3 \sqrt {3+2 x} (878+1063 x)}{10 \left (2+5 x+3 x^2\right )}+\frac {1}{5} \operatorname {Subst}\left (\int \frac {4111+3189 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt {3+2 x}\right )\\ &=-\frac {\sqrt {3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac {3 \sqrt {3+2 x} (878+1063 x)}{10 \left (2+5 x+3 x^2\right )}-2190 \operatorname {Subst}\left (\int \frac {1}{-3+3 x^2} \, dx,x,\sqrt {3+2 x}\right )+\frac {14139}{5} \operatorname {Subst}\left (\int \frac {1}{-5+3 x^2} \, dx,x,\sqrt {3+2 x}\right )\\ &=-\frac {\sqrt {3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac {3 \sqrt {3+2 x} (878+1063 x)}{10 \left (2+5 x+3 x^2\right )}+730 \tanh ^{-1}\left (\sqrt {3+2 x}\right )-\frac {4713}{5} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 81, normalized size = 0.79 \begin {gather*} \frac {1}{50} \left (\frac {5 \sqrt {2 x+3} \left (9567 x^3+23847 x^2+19373 x+5123\right )}{\left (3 x^2+5 x+2\right )^2}-9426 \sqrt {15} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right )+730 \tanh ^{-1}\left (\sqrt {2 x+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

730*ArcTanh[Sqrt[3 + 2*x]] + ((5*Sqrt[3 + 2*x]*(5123 + 19373*x + 23847*x^2 + 9567*x^3))/(2 + 5*x + 3*x^2)^2 -
9426*Sqrt[15]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/50

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IntegrateAlgebraic [A]  time = 0.26, size = 102, normalized size = 1.00 \begin {gather*} \frac {\sqrt {2 x+3} \left (9567 (2 x+3)^3-38409 (2 x+3)^2+49637 (2 x+3)-20555\right )}{5 \left (3 (2 x+3)^2-8 (2 x+3)+5\right )^2}+730 \tanh ^{-1}\left (\sqrt {2 x+3}\right )-\frac {4713}{5} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 + 2*x]*(-20555 + 49637*(3 + 2*x) - 38409*(3 + 2*x)^2 + 9567*(3 + 2*x)^3))/(5*(5 - 8*(3 + 2*x) + 3*(3 +
2*x)^2)^2) + 730*ArcTanh[Sqrt[3 + 2*x]] - (4713*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/5

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fricas [B]  time = 0.42, size = 170, normalized size = 1.67 \begin {gather*} \frac {4713 \, \sqrt {5} \sqrt {3} {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 18250 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 18250 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) + 5 \, {\left (9567 \, x^{3} + 23847 \, x^{2} + 19373 \, x + 5123\right )} \sqrt {2 \, x + 3}}{50 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50*(4713*sqrt(5)*sqrt(3)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)
/(3*x + 2)) + 18250*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(sqrt(2*x + 3) + 1) - 18250*(9*x^4 + 30*x^3 + 37*x
^2 + 20*x + 4)*log(sqrt(2*x + 3) - 1) + 5*(9567*x^3 + 23847*x^2 + 19373*x + 5123)*sqrt(2*x + 3))/(9*x^4 + 30*x
^3 + 37*x^2 + 20*x + 4)

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giac [A]  time = 0.19, size = 120, normalized size = 1.18 \begin {gather*} \frac {4713}{50} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) + \frac {9567 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 38409 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 49637 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 20555 \, \sqrt {2 \, x + 3}}{5 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 365 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 365 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4713/50*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/5*(9567*(2*x + 3
)^(7/2) - 38409*(2*x + 3)^(5/2) + 49637*(2*x + 3)^(3/2) - 20555*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19)^2 +
365*log(sqrt(2*x + 3) + 1) - 365*log(abs(sqrt(2*x + 3) - 1))

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maple [A]  time = 0.02, size = 124, normalized size = 1.22 \begin {gather*} -\frac {4713 \sqrt {15}\, \arctanh \left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{25}-365 \ln \left (-1+\sqrt {2 x +3}\right )+365 \ln \left (\sqrt {2 x +3}+1\right )+\frac {\frac {4527 \left (2 x +3\right )^{\frac {3}{2}}}{5}-1611 \sqrt {2 x +3}}{\left (6 x +4\right )^{2}}-\frac {3}{\left (\sqrt {2 x +3}+1\right )^{2}}+\frac {56}{\sqrt {2 x +3}+1}+\frac {3}{\left (-1+\sqrt {2 x +3}\right )^{2}}+\frac {56}{-1+\sqrt {2 x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

162*(503/90*(2*x+3)^(3/2)-179/18*(2*x+3)^(1/2))/(6*x+4)^2-4713/25*arctanh(1/5*15^(1/2)*(2*x+3)^(1/2))*15^(1/2)
-3/((2*x+3)^(1/2)+1)^2+56/((2*x+3)^(1/2)+1)+365*ln((2*x+3)^(1/2)+1)+3/(-1+(2*x+3)^(1/2))^2+56/(-1+(2*x+3)^(1/2
))-365*ln(-1+(2*x+3)^(1/2))

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maxima [A]  time = 1.06, size = 134, normalized size = 1.31 \begin {gather*} \frac {4713}{50} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) + \frac {9567 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 38409 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 49637 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 20555 \, \sqrt {2 \, x + 3}}{5 \, {\left (9 \, {\left (2 \, x + 3\right )}^{4} - 48 \, {\left (2 \, x + 3\right )}^{3} + 94 \, {\left (2 \, x + 3\right )}^{2} - 160 \, x - 215\right )}} + 365 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 365 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4713/50*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/5*(9567*(2*x + 3)^(7/2) -
38409*(2*x + 3)^(5/2) + 49637*(2*x + 3)^(3/2) - 20555*sqrt(2*x + 3))/(9*(2*x + 3)^4 - 48*(2*x + 3)^3 + 94*(2*
x + 3)^2 - 160*x - 215) + 365*log(sqrt(2*x + 3) + 1) - 365*log(sqrt(2*x + 3) - 1)

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mupad [B]  time = 2.39, size = 101, normalized size = 0.99 \begin {gather*} 730\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )+\frac {\frac {4111\,\sqrt {2\,x+3}}{9}-\frac {49637\,{\left (2\,x+3\right )}^{3/2}}{45}+\frac {12803\,{\left (2\,x+3\right )}^{5/2}}{15}-\frac {1063\,{\left (2\,x+3\right )}^{7/2}}{5}}{\frac {160\,x}{9}-\frac {94\,{\left (2\,x+3\right )}^2}{9}+\frac {16\,{\left (2\,x+3\right )}^3}{3}-{\left (2\,x+3\right )}^4+\frac {215}{9}}-\frac {4713\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

730*atanh((2*x + 3)^(1/2)) + ((4111*(2*x + 3)^(1/2))/9 - (49637*(2*x + 3)^(3/2))/45 + (12803*(2*x + 3)^(5/2))/
15 - (1063*(2*x + 3)^(7/2))/5)/((160*x)/9 - (94*(2*x + 3)^2)/9 + (16*(2*x + 3)^3)/3 - (2*x + 3)^4 + 215/9) - (
4713*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/2))/5))/25

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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