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

Optimal. Leaf size=137 \[ \frac {(352 x+333) \left (3 x^2+5 x+2\right )^{3/2}}{240 (2 x+3)^4}+\frac {(1528 x+2087) \sqrt {3 x^2+5 x+2}}{3200 (2 x+3)^2}-\frac {3}{32} \sqrt {3} \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )+\frac {2359 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{6400 \sqrt {5}} \]

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Rubi [A]  time = 0.08, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {810, 843, 621, 206, 724} \begin {gather*} \frac {(352 x+333) \left (3 x^2+5 x+2\right )^{3/2}}{240 (2 x+3)^4}+\frac {(1528 x+2087) \sqrt {3 x^2+5 x+2}}{3200 (2 x+3)^2}-\frac {3}{32} \sqrt {3} \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )+\frac {2359 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{6400 \sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2087 + 1528*x)*Sqrt[2 + 5*x + 3*x^2])/(3200*(3 + 2*x)^2) + ((333 + 352*x)*(2 + 5*x + 3*x^2)^(3/2))/(240*(3 +
2*x)^4) - (3*Sqrt[3]*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/32 + (2359*ArcTanh[(7 + 8*x)/(2*Sq
rt[5]*Sqrt[2 + 5*x + 3*x^2])])/(6400*Sqrt[5])

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])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*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] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 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]

Rubi steps

\begin {align*} \int \frac {(5-x) \left (2+5 x+3 x^2\right )^{3/2}}{(3+2 x)^5} \, dx &=\frac {(333+352 x) \left (2+5 x+3 x^2\right )^{3/2}}{240 (3+2 x)^4}-\frac {1}{160} \int \frac {(139+120 x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^3} \, dx\\ &=\frac {(2087+1528 x) \sqrt {2+5 x+3 x^2}}{3200 (3+2 x)^2}+\frac {(333+352 x) \left (2+5 x+3 x^2\right )^{3/2}}{240 (3+2 x)^4}+\frac {\int \frac {-6082-7200 x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx}{12800}\\ &=\frac {(2087+1528 x) \sqrt {2+5 x+3 x^2}}{3200 (3+2 x)^2}+\frac {(333+352 x) \left (2+5 x+3 x^2\right )^{3/2}}{240 (3+2 x)^4}-\frac {9}{32} \int \frac {1}{\sqrt {2+5 x+3 x^2}} \, dx+\frac {2359 \int \frac {1}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx}{6400}\\ &=\frac {(2087+1528 x) \sqrt {2+5 x+3 x^2}}{3200 (3+2 x)^2}+\frac {(333+352 x) \left (2+5 x+3 x^2\right )^{3/2}}{240 (3+2 x)^4}-\frac {9}{16} \operatorname {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {5+6 x}{\sqrt {2+5 x+3 x^2}}\right )-\frac {2359 \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,\frac {-7-8 x}{\sqrt {2+5 x+3 x^2}}\right )}{3200}\\ &=\frac {(2087+1528 x) \sqrt {2+5 x+3 x^2}}{3200 (3+2 x)^2}+\frac {(333+352 x) \left (2+5 x+3 x^2\right )^{3/2}}{240 (3+2 x)^4}-\frac {3}{32} \sqrt {3} \tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )+\frac {2359 \tanh ^{-1}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{6400 \sqrt {5}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 110, normalized size = 0.80 \begin {gather*} \frac {-7077 \sqrt {5} \tanh ^{-1}\left (\frac {-8 x-7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )-9000 \sqrt {3} \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {9 x^2+15 x+6}}\right )+\frac {10 \sqrt {3 x^2+5 x+2} \left (60576 x^3+190412 x^2+211148 x+82989\right )}{(2 x+3)^4}}{96000} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((10*Sqrt[2 + 5*x + 3*x^2]*(82989 + 211148*x + 190412*x^2 + 60576*x^3))/(3 + 2*x)^4 - 7077*Sqrt[5]*ArcTanh[(-7
- 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])] - 9000*Sqrt[3]*ArcTanh[(5 + 6*x)/(2*Sqrt[6 + 15*x + 9*x^2])])/96000

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IntegrateAlgebraic [A]  time = 0.61, size = 111, normalized size = 0.81 \begin {gather*} -\frac {3}{16} \sqrt {3} \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {3} (x+1)}\right )+\frac {2359 \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {5} (x+1)}\right )}{3200 \sqrt {5}}+\frac {\sqrt {3 x^2+5 x+2} \left (60576 x^3+190412 x^2+211148 x+82989\right )}{9600 (2 x+3)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2 + 5*x + 3*x^2]*(82989 + 211148*x + 190412*x^2 + 60576*x^3))/(9600*(3 + 2*x)^4) - (3*Sqrt[3]*ArcTanh[Sq
rt[2 + 5*x + 3*x^2]/(Sqrt[3]*(1 + x))])/16 + (2359*ArcTanh[Sqrt[2 + 5*x + 3*x^2]/(Sqrt[5]*(1 + x))])/(3200*Sqr
t[5])

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fricas [A]  time = 0.43, size = 183, normalized size = 1.34 \begin {gather*} \frac {9000 \, \sqrt {3} {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) + 7077 \, \sqrt {5} {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )} \log \left (\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{4 \, x^{2} + 12 \, x + 9}\right ) + 20 \, {\left (60576 \, x^{3} + 190412 \, x^{2} + 211148 \, x + 82989\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{192000 \, {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192000*(9000*sqrt(3)*(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5)
+ 72*x^2 + 120*x + 49) + 7077*sqrt(5)*(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)*log((4*sqrt(5)*sqrt(3*x^2 + 5*
x + 2)*(8*x + 7) + 124*x^2 + 212*x + 89)/(4*x^2 + 12*x + 9)) + 20*(60576*x^3 + 190412*x^2 + 211148*x + 82989)*
sqrt(3*x^2 + 5*x + 2))/(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)

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giac [A]  time = 0.29, size = 106, normalized size = 0.77 \begin {gather*} -\frac {1}{19200} \, {\left (\frac {5 \, {\left (\frac {10 \, {\left (\frac {195 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )}{2 \, x + 3} - 488 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )\right )}}{2 \, x + 3} + 4109 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )\right )}}{2 \, x + 3} - 7572 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )\right )} \sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} - \frac {631}{1600} \, \sqrt {3} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/19200*(5*(10*(195*sgn(1/(2*x + 3))/(2*x + 3) - 488*sgn(1/(2*x + 3)))/(2*x + 3) + 4109*sgn(1/(2*x + 3)))/(2*
x + 3) - 7572*sgn(1/(2*x + 3)))*sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) - 631/1600*sqrt(3)*sgn(1/(2*x + 3))

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maple [B]  time = 0.06, size = 221, normalized size = 1.61 \begin {gather*} -\frac {2359 \sqrt {5}\, \arctanh \left (\frac {2 \left (-4 x -\frac {7}{2}\right ) \sqrt {5}}{5 \sqrt {-16 x +12 \left (x +\frac {3}{2}\right )^{2}-19}}\right )}{32000}-\frac {3 \sqrt {3}\, \ln \left (\frac {\left (3 x +\frac {5}{2}\right ) \sqrt {3}}{3}+\sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}\right )}{32}-\frac {17 \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {5}{2}}}{300 \left (x +\frac {3}{2}\right )^{3}}-\frac {1129 \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {5}{2}}}{12000 \left (x +\frac {3}{2}\right )^{2}}-\frac {911 \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {5}{2}}}{7500 \left (x +\frac {3}{2}\right )}+\frac {2359 \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {3}{2}}}{60000}-\frac {109 \left (6 x +5\right ) \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}{4000}+\frac {2359 \sqrt {-16 x +12 \left (x +\frac {3}{2}\right )^{2}-19}}{32000}+\frac {911 \left (6 x +5\right ) \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {3}{2}}}{15000}-\frac {13 \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {5}{2}}}{320 \left (x +\frac {3}{2}\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-17/300/(x+3/2)^3*(-4*x+3*(x+3/2)^2-19/4)^(5/2)-1129/12000/(x+3/2)^2*(-4*x+3*(x+3/2)^2-19/4)^(5/2)-911/7500/(x
+3/2)*(-4*x+3*(x+3/2)^2-19/4)^(5/2)+2359/60000*(-4*x+3*(x+3/2)^2-19/4)^(3/2)-109/4000*(6*x+5)*(-4*x+3*(x+3/2)^
2-19/4)^(1/2)-3/32*3^(1/2)*ln(1/3*(3*x+5/2)*3^(1/2)+(-4*x+3*(x+3/2)^2-19/4)^(1/2))+2359/32000*(-16*x+12*(x+3/2
)^2-19)^(1/2)-2359/32000*5^(1/2)*arctanh(2/5*(-4*x-7/2)*5^(1/2)/(-16*x+12*(x+3/2)^2-19)^(1/2))+911/15000*(6*x+
5)*(-4*x+3*(x+3/2)^2-19/4)^(3/2)-13/320/(x+3/2)^4*(-4*x+3*(x+3/2)^2-19/4)^(5/2)

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maxima [B]  time = 1.34, size = 227, normalized size = 1.66 \begin {gather*} \frac {1129}{4000} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} - \frac {13 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}}{20 \, {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} - \frac {34 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}}{75 \, {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )}} - \frac {1129 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}}{3000 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac {327}{2000} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x - \frac {3}{32} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac {5}{2}\right ) - \frac {2359}{32000} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) + \frac {179}{16000} \, \sqrt {3 \, x^{2} + 5 \, x + 2} - \frac {911 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}}{3000 \, {\left (2 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1129/4000*(3*x^2 + 5*x + 2)^(3/2) - 13/20*(3*x^2 + 5*x + 2)^(5/2)/(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81) - 3
4/75*(3*x^2 + 5*x + 2)^(5/2)/(8*x^3 + 36*x^2 + 54*x + 27) - 1129/3000*(3*x^2 + 5*x + 2)^(5/2)/(4*x^2 + 12*x +
9) - 327/2000*sqrt(3*x^2 + 5*x + 2)*x - 3/32*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) - 2359/320
00*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) + 179/16000*sqrt(3*x^2 + 5*x
+ 2) - 911/3000*(3*x^2 + 5*x + 2)^(3/2)/(2*x + 3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\left (x-5\right )\,{\left (3\,x^2+5\,x+2\right )}^{3/2}}{{\left (2\,x+3\right )}^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \left (- \frac {10 \sqrt {3 x^{2} + 5 x + 2}}{32 x^{5} + 240 x^{4} + 720 x^{3} + 1080 x^{2} + 810 x + 243}\right )\, dx - \int \left (- \frac {23 x \sqrt {3 x^{2} + 5 x + 2}}{32 x^{5} + 240 x^{4} + 720 x^{3} + 1080 x^{2} + 810 x + 243}\right )\, dx - \int \left (- \frac {10 x^{2} \sqrt {3 x^{2} + 5 x + 2}}{32 x^{5} + 240 x^{4} + 720 x^{3} + 1080 x^{2} + 810 x + 243}\right )\, dx - \int \frac {3 x^{3} \sqrt {3 x^{2} + 5 x + 2}}{32 x^{5} + 240 x^{4} + 720 x^{3} + 1080 x^{2} + 810 x + 243}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-10*sqrt(3*x**2 + 5*x + 2)/(32*x**5 + 240*x**4 + 720*x**3 + 1080*x**2 + 810*x + 243), x) - Integral(
-23*x*sqrt(3*x**2 + 5*x + 2)/(32*x**5 + 240*x**4 + 720*x**3 + 1080*x**2 + 810*x + 243), x) - Integral(-10*x**2
*sqrt(3*x**2 + 5*x + 2)/(32*x**5 + 240*x**4 + 720*x**3 + 1080*x**2 + 810*x + 243), x) - Integral(3*x**3*sqrt(3
*x**2 + 5*x + 2)/(32*x**5 + 240*x**4 + 720*x**3 + 1080*x**2 + 810*x + 243), x)

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