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

Optimal. Leaf size=105 \[ -\frac {\sqrt {3 x^2+5 x+2} (x+8)}{2 (2 x+3)}+\frac {43 \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{8 \sqrt {3}}-\frac {57 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{8 \sqrt {5}} \]

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Rubi [A]  time = 0.06, antiderivative size = 105, 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 = {812, 843, 621, 206, 724} \begin {gather*} -\frac {\sqrt {3 x^2+5 x+2} (x+8)}{2 (2 x+3)}+\frac {43 \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{8 \sqrt {3}}-\frac {57 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{8 \sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((8 + x)*Sqrt[2 + 5*x + 3*x^2])/(2*(3 + 2*x)) + (43*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(8*
Sqrt[3]) - (57*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(8*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) \sqrt {2+5 x+3 x^2}}{(3+2 x)^2} \, dx &=-\frac {(8+x) \sqrt {2+5 x+3 x^2}}{2 (3+2 x)}-\frac {1}{8} \int \frac {-72-86 x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {(8+x) \sqrt {2+5 x+3 x^2}}{2 (3+2 x)}+\frac {43}{8} \int \frac {1}{\sqrt {2+5 x+3 x^2}} \, dx-\frac {57}{8} \int \frac {1}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {(8+x) \sqrt {2+5 x+3 x^2}}{2 (3+2 x)}+\frac {43}{4} \operatorname {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {5+6 x}{\sqrt {2+5 x+3 x^2}}\right )+\frac {57}{4} \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,\frac {-7-8 x}{\sqrt {2+5 x+3 x^2}}\right )\\ &=-\frac {(8+x) \sqrt {2+5 x+3 x^2}}{2 (3+2 x)}+\frac {43 \tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{8 \sqrt {3}}-\frac {57 \tanh ^{-1}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{8 \sqrt {5}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 98, normalized size = 0.93 \begin {gather*} -\frac {\sqrt {3 x^2+5 x+2} (x+8)}{4 x+6}+\frac {57 \tanh ^{-1}\left (\frac {-8 x-7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{8 \sqrt {5}}+\frac {43 \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {9 x^2+15 x+6}}\right )}{8 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((8 + x)*Sqrt[2 + 5*x + 3*x^2])/(6 + 4*x)) + (57*ArcTanh[(-7 - 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(8*S
qrt[5]) + (43*ArcTanh[(5 + 6*x)/(2*Sqrt[6 + 15*x + 9*x^2])])/(8*Sqrt[3])

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IntegrateAlgebraic [A]  time = 0.49, size = 101, normalized size = 0.96 \begin {gather*} \frac {\sqrt {3 x^2+5 x+2} (-x-8)}{2 (2 x+3)}+\frac {43 \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {3} (x+1)}\right )}{4 \sqrt {3}}-\frac {57 \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {5} (x+1)}\right )}{4 \sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8 - x)*Sqrt[2 + 5*x + 3*x^2])/(2*(3 + 2*x)) + (43*ArcTanh[Sqrt[2 + 5*x + 3*x^2]/(Sqrt[3]*(1 + x))])/(4*Sqrt
[3]) - (57*ArcTanh[Sqrt[2 + 5*x + 3*x^2]/(Sqrt[5]*(1 + x))])/(4*Sqrt[5])

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fricas [A]  time = 0.42, size = 127, normalized size = 1.21 \begin {gather*} \frac {215 \, \sqrt {3} {\left (2 \, x + 3\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 ) + 171 \, \sqrt {5} {\left (2 \, x + 3\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 ) - 120 \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (x + 8\right )}}{240 \, {\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)^(1/2)/(3+2*x)^2,x, algorithm="fricas")

[Out]

1/240*(215*sqrt(3)*(2*x + 3)*log(4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120*x + 49) + 171*sqrt(5
)*(2*x + 3)*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)) - 120*
sqrt(3*x^2 + 5*x + 2)*(x + 8))/(2*x + 3)

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giac [B]  time = 0.75, size = 291, normalized size = 2.77 \begin {gather*} -\frac {43}{24} \, \sqrt {3} \log \left (\frac {{\left | -2 \, \sqrt {3} + 2 \, \sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {2 \, \sqrt {5}}{2 \, x + 3} \right |}}{{\left | 2 \, \sqrt {3} + 2 \, \sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {2 \, \sqrt {5}}{2 \, x + 3} \right |}}\right ) \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) + \frac {57}{40} \, \sqrt {5} \log \left ({\left | \sqrt {5} {\left (\sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {\sqrt {5}}{2 \, x + 3}\right )} - 4 \right |}\right ) \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) - \frac {13}{8} \, \sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) + \frac {4 \, {\left (\sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {\sqrt {5}}{2 \, x + 3}\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) - 3 \, \sqrt {5} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )}{4 \, {\left ({\left (\sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac {\sqrt {5}}{2 \, x + 3}\right )}^{2} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-43/24*sqrt(3)*log(abs(-2*sqrt(3) + 2*sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + 2*sqrt(5)/(2*x + 3))/abs(2*sqrt
(3) + 2*sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + 2*sqrt(5)/(2*x + 3)))*sgn(1/(2*x + 3)) + 57/40*sqrt(5)*log(ab
s(sqrt(5)*(sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + sqrt(5)/(2*x + 3)) - 4))*sgn(1/(2*x + 3)) - 13/8*sqrt(-8/(
2*x + 3) + 5/(2*x + 3)^2 + 3)*sgn(1/(2*x + 3)) + 1/4*(4*(sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + sqrt(5)/(2*x
+ 3))*sgn(1/(2*x + 3)) - 3*sqrt(5)*sgn(1/(2*x + 3)))/((sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + sqrt(5)/(2*x
+ 3))^2 - 3)

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maple [A]  time = 0.06, size = 121, normalized size = 1.15 \begin {gather*} \frac {57 \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 )}{40}+\frac {43 \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 )}{24}-\frac {13 \left (-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {3}{2}}}{10 \left (x +\frac {3}{2}\right )}-\frac {57 \sqrt {-16 x +12 \left (x +\frac {3}{2}\right )^{2}-19}}{40}+\frac {13 \left (6 x +5\right ) \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-13/10/(x+3/2)*(-4*x+3*(x+3/2)^2-19/4)^(3/2)-57/40*(-16*x+12*(x+3/2)^2-19)^(1/2)+43/24*3^(1/2)*ln(1/3*(3*x+5/2
)*3^(1/2)+(-4*x+3*(x+3/2)^2-19/4)^(1/2))+57/40*5^(1/2)*arctanh(2/5*(-4*x-7/2)*5^(1/2)/(-16*x+12*(x+3/2)^2-19)^
(1/2))+13/20*(6*x+5)*(-4*x+3*(x+3/2)^2-19/4)^(1/2)

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maxima [A]  time = 1.22, size = 105, normalized size = 1.00 \begin {gather*} \frac {43}{24} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac {5}{2}\right ) + \frac {57}{40} \, \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 {1}{4} \, \sqrt {3 \, x^{2} + 5 \, x + 2} - \frac {13 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{4 \, {\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)^(1/2)/(3+2*x)^2,x, algorithm="maxima")

[Out]

43/24*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) + 57/40*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)
/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) - 1/4*sqrt(3*x^2 + 5*x + 2) - 13/4*sqrt(3*x^2 + 5*x + 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 )\,\sqrt {3\,x^2+5\,x+2}}{{\left (2\,x+3\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \left (- \frac {5 \sqrt {3 x^{2} + 5 x + 2}}{4 x^{2} + 12 x + 9}\right )\, dx - \int \frac {x \sqrt {3 x^{2} + 5 x + 2}}{4 x^{2} + 12 x + 9}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-5*sqrt(3*x**2 + 5*x + 2)/(4*x**2 + 12*x + 9), x) - Integral(x*sqrt(3*x**2 + 5*x + 2)/(4*x**2 + 12*x
+ 9), x)

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