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

Optimal. Leaf size=135 \[ -\frac {(x+8) \left (3 x^2+5 x+2\right )^{3/2}}{4 (2 x+3)^2}+\frac {3 (43 x+93) \sqrt {3 x^2+5 x+2}}{16 (2 x+3)}-\frac {343}{64} \sqrt {3} \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )+\frac {1329 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{64 \sqrt {5}} \]

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Rubi [A]  time = 0.08, antiderivative size = 135, 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 = {812, 843, 621, 206, 724} \begin {gather*} -\frac {(x+8) \left (3 x^2+5 x+2\right )^{3/2}}{4 (2 x+3)^2}+\frac {3 (43 x+93) \sqrt {3 x^2+5 x+2}}{16 (2 x+3)}-\frac {343}{64} \sqrt {3} \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )+\frac {1329 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{64 \sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(93 + 43*x)*Sqrt[2 + 5*x + 3*x^2])/(16*(3 + 2*x)) - ((8 + x)*(2 + 5*x + 3*x^2)^(3/2))/(4*(3 + 2*x)^2) - (34
3*Sqrt[3]*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/64 + (1329*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2
+ 5*x + 3*x^2])])/(64*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 812

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

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)^3} \, dx &=-\frac {(8+x) \left (2+5 x+3 x^2\right )^{3/2}}{4 (3+2 x)^2}-\frac {3}{32} \int \frac {(-144-172 x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2} \, dx\\ &=\frac {3 (93+43 x) \sqrt {2+5 x+3 x^2}}{16 (3+2 x)}-\frac {(8+x) \left (2+5 x+3 x^2\right )^{3/2}}{4 (3+2 x)^2}+\frac {3}{256} \int \frac {-2344-2744 x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {3 (93+43 x) \sqrt {2+5 x+3 x^2}}{16 (3+2 x)}-\frac {(8+x) \left (2+5 x+3 x^2\right )^{3/2}}{4 (3+2 x)^2}-\frac {1029}{64} \int \frac {1}{\sqrt {2+5 x+3 x^2}} \, dx+\frac {1329}{64} \int \frac {1}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {3 (93+43 x) \sqrt {2+5 x+3 x^2}}{16 (3+2 x)}-\frac {(8+x) \left (2+5 x+3 x^2\right )^{3/2}}{4 (3+2 x)^2}-\frac {1029}{32} \operatorname {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {5+6 x}{\sqrt {2+5 x+3 x^2}}\right )-\frac {1329}{32} \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,\frac {-7-8 x}{\sqrt {2+5 x+3 x^2}}\right )\\ &=\frac {3 (93+43 x) \sqrt {2+5 x+3 x^2}}{16 (3+2 x)}-\frac {(8+x) \left (2+5 x+3 x^2\right )^{3/2}}{4 (3+2 x)^2}-\frac {343}{64} \sqrt {3} \tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )+\frac {1329 \tanh ^{-1}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{64 \sqrt {5}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 110, normalized size = 0.81 \begin {gather*} \frac {1}{320} \left (-1329 \sqrt {5} \tanh ^{-1}\left (\frac {-8 x-7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )-1715 \sqrt {3} \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {9 x^2+15 x+6}}\right )-\frac {20 \sqrt {3 x^2+5 x+2} \left (12 x^3-142 x^2-777 x-773\right )}{(2 x+3)^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-20*Sqrt[2 + 5*x + 3*x^2]*(-773 - 777*x - 142*x^2 + 12*x^3))/(3 + 2*x)^2 - 1329*Sqrt[5]*ArcTanh[(-7 - 8*x)/(
2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])] - 1715*Sqrt[3]*ArcTanh[(5 + 6*x)/(2*Sqrt[6 + 15*x + 9*x^2])])/320

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IntegrateAlgebraic [A]  time = 0.59, size = 111, normalized size = 0.82 \begin {gather*} -\frac {343}{32} \sqrt {3} \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {3} (x+1)}\right )+\frac {1329 \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {5} (x+1)}\right )}{32 \sqrt {5}}+\frac {\sqrt {3 x^2+5 x+2} \left (-12 x^3+142 x^2+777 x+773\right )}{16 (2 x+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2 + 5*x + 3*x^2]*(773 + 777*x + 142*x^2 - 12*x^3))/(16*(3 + 2*x)^2) - (343*Sqrt[3]*ArcTanh[Sqrt[2 + 5*x
+ 3*x^2]/(Sqrt[3]*(1 + x))])/32 + (1329*ArcTanh[Sqrt[2 + 5*x + 3*x^2]/(Sqrt[5]*(1 + x))])/(32*Sqrt[5])

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fricas [A]  time = 0.42, size = 153, normalized size = 1.13 \begin {gather*} \frac {1715 \, \sqrt {3} {\left (4 \, x^{2} + 12 \, x + 9\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 ) + 1329 \, \sqrt {5} {\left (4 \, x^{2} + 12 \, x + 9\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 ) - 40 \, {\left (12 \, x^{3} - 142 \, x^{2} - 777 \, x - 773\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{640 \, {\left (4 \, x^{2} + 12 \, x + 9\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)^3,x, algorithm="fricas")

[Out]

1/640*(1715*sqrt(3)*(4*x^2 + 12*x + 9)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120*x + 49) +
1329*sqrt(5)*(4*x^2 + 12*x + 9)*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)) - 40*(12*x^3 - 142*x^2 - 777*x - 773)*sqrt(3*x^2 + 5*x + 2))/(4*x^2 + 12*x + 9)

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giac [B]  time = 0.34, size = 259, normalized size = 1.92 \begin {gather*} -\frac {1}{32} \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x - 89\right )} + \frac {1329}{320} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) + \frac {343}{64} \, \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) + \frac {5 \, {\left (510 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{3} + 1869 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 6259 \, \sqrt {3} x + 2209 \, \sqrt {3} - 6259 \, \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}}{32 \, {\left (2 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 6 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} + 11\right )}^{2}} \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)^3,x, algorithm="giac")

[Out]

-1/32*sqrt(3*x^2 + 5*x + 2)*(6*x - 89) + 1329/320*sqrt(5)*log(abs(-4*sqrt(3)*x - 2*sqrt(5) - 6*sqrt(3) + 4*sqr
t(3*x^2 + 5*x + 2))/abs(-4*sqrt(3)*x + 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))) + 343/64*sqrt(3)*log(
abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) - 5)) + 5/32*(510*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^3 + 1
869*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^2 + 6259*sqrt(3)*x + 2209*sqrt(3) - 6259*sqrt(3*x^2 + 5*x + 2)
)/(2*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^2 + 6*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) + 11)^2

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maple [A]  time = 0.06, size = 179, normalized size = 1.33 \begin {gather*} -\frac {1329 \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 )}{320}-\frac {343 \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 )}{64}-\frac {13 \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {5}{2}}}{40 \left (x +\frac {3}{2}\right )^{2}}+\frac {31 \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {5}{2}}}{50 \left (x +\frac {3}{2}\right )}+\frac {443 \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {3}{2}}}{200}-\frac {171 \left (6 x +5\right ) \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}{160}+\frac {1329 \sqrt {-16 x +12 \left (x +\frac {3}{2}\right )^{2}-19}}{320}-\frac {31 \left (6 x +5\right ) \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {3}{2}}}{100} \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)^3,x)

[Out]

-13/40/(x+3/2)^2*(-4*x+3*(x+3/2)^2-19/4)^(5/2)+31/50/(x+3/2)*(-4*x+3*(x+3/2)^2-19/4)^(5/2)+443/200*(-4*x+3*(x+
3/2)^2-19/4)^(3/2)-171/160*(6*x+5)*(-4*x+3*(x+3/2)^2-19/4)^(1/2)-343/64*3^(1/2)*ln(1/3*(3*x+5/2)*3^(1/2)+(-4*x
+3*(x+3/2)^2-19/4)^(1/2))+1329/320*(-16*x+12*(x+3/2)^2-19)^(1/2)-1329/320*5^(1/2)*arctanh(2/5*(-4*x-7/2)*5^(1/
2)/(-16*x+12*(x+3/2)^2-19)^(1/2))-31/100*(6*x+5)*(-4*x+3*(x+3/2)^2-19/4)^(3/2)

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maxima [A]  time = 1.19, size = 160, normalized size = 1.19 \begin {gather*} \frac {39}{40} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} - \frac {13 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}}{10 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac {513}{80} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x - \frac {343}{64} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac {5}{2}\right ) - \frac {1329}{320} \, \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 {237}{80} \, \sqrt {3 \, x^{2} + 5 \, x + 2} + \frac {31 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}}{20 \, {\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)^3,x, algorithm="maxima")

[Out]

39/40*(3*x^2 + 5*x + 2)^(3/2) - 13/10*(3*x^2 + 5*x + 2)^(5/2)/(4*x^2 + 12*x + 9) - 513/80*sqrt(3*x^2 + 5*x + 2
)*x - 343/64*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) - 1329/320*sqrt(5)*log(sqrt(5)*sqrt(3*x^2
+ 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) + 237/80*sqrt(3*x^2 + 5*x + 2) + 31/20*(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 )}^3} \,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)^3,x)

[Out]

-int(((x - 5)*(5*x + 3*x^2 + 2)^(3/2))/(2*x + 3)^3, 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}}{8 x^{3} + 36 x^{2} + 54 x + 27}\right )\, dx - \int \left (- \frac {23 x \sqrt {3 x^{2} + 5 x + 2}}{8 x^{3} + 36 x^{2} + 54 x + 27}\right )\, dx - \int \left (- \frac {10 x^{2} \sqrt {3 x^{2} + 5 x + 2}}{8 x^{3} + 36 x^{2} + 54 x + 27}\right )\, dx - \int \frac {3 x^{3} \sqrt {3 x^{2} + 5 x + 2}}{8 x^{3} + 36 x^{2} + 54 x + 27}\, 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)**3,x)

[Out]

-Integral(-10*sqrt(3*x**2 + 5*x + 2)/(8*x**3 + 36*x**2 + 54*x + 27), x) - Integral(-23*x*sqrt(3*x**2 + 5*x + 2
)/(8*x**3 + 36*x**2 + 54*x + 27), x) - Integral(-10*x**2*sqrt(3*x**2 + 5*x + 2)/(8*x**3 + 36*x**2 + 54*x + 27)
, x) - Integral(3*x**3*sqrt(3*x**2 + 5*x + 2)/(8*x**3 + 36*x**2 + 54*x + 27), x)

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