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

Optimal. Leaf size=119 \[ -\frac {6 (47 x+37)}{5 (2 x+3)^2 \sqrt {3 x^2+5 x+2}}-\frac {864 \sqrt {3 x^2+5 x+2}}{25 (2 x+3)}-\frac {166 \sqrt {3 x^2+5 x+2}}{5 (2 x+3)^2}+\frac {483 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{25 \sqrt {5}} \]

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Rubi [A]  time = 0.07, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {822, 834, 806, 724, 206} \begin {gather*} -\frac {6 (47 x+37)}{5 (2 x+3)^2 \sqrt {3 x^2+5 x+2}}-\frac {864 \sqrt {3 x^2+5 x+2}}{25 (2 x+3)}-\frac {166 \sqrt {3 x^2+5 x+2}}{5 (2 x+3)^2}+\frac {483 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{25 \sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(37 + 47*x))/(5*(3 + 2*x)^2*Sqrt[2 + 5*x + 3*x^2]) - (166*Sqrt[2 + 5*x + 3*x^2])/(5*(3 + 2*x)^2) - (864*Sq
rt[2 + 5*x + 3*x^2])/(25*(3 + 2*x)) + (483*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(25*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 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 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[Sim
plify[m + 2*p + 3], 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 834

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

Rubi steps

\begin {align*} \int \frac {5-x}{(3+2 x)^3 \left (2+5 x+3 x^2\right )^{3/2}} \, dx &=-\frac {6 (37+47 x)}{5 (3+2 x)^2 \sqrt {2+5 x+3 x^2}}-\frac {2}{5} \int \frac {431+564 x}{(3+2 x)^3 \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {6 (37+47 x)}{5 (3+2 x)^2 \sqrt {2+5 x+3 x^2}}-\frac {166 \sqrt {2+5 x+3 x^2}}{5 (3+2 x)^2}+\frac {1}{25} \int \frac {-1575-2490 x}{(3+2 x)^2 \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {6 (37+47 x)}{5 (3+2 x)^2 \sqrt {2+5 x+3 x^2}}-\frac {166 \sqrt {2+5 x+3 x^2}}{5 (3+2 x)^2}-\frac {864 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}+\frac {483}{25} \int \frac {1}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {6 (37+47 x)}{5 (3+2 x)^2 \sqrt {2+5 x+3 x^2}}-\frac {166 \sqrt {2+5 x+3 x^2}}{5 (3+2 x)^2}-\frac {864 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}-\frac {966}{25} \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,\frac {-7-8 x}{\sqrt {2+5 x+3 x^2}}\right )\\ &=-\frac {6 (37+47 x)}{5 (3+2 x)^2 \sqrt {2+5 x+3 x^2}}-\frac {166 \sqrt {2+5 x+3 x^2}}{5 (3+2 x)^2}-\frac {864 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}+\frac {483 \tanh ^{-1}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{25 \sqrt {5}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 79, normalized size = 0.66 \begin {gather*} -\frac {483 \tanh ^{-1}\left (\frac {-8 x-7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{25 \sqrt {5}}-\frac {2 \left (2592 x^3+9453 x^2+10988 x+3977\right )}{25 (2 x+3)^2 \sqrt {3 x^2+5 x+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3977 + 10988*x + 9453*x^2 + 2592*x^3))/(25*(3 + 2*x)^2*Sqrt[2 + 5*x + 3*x^2]) - (483*ArcTanh[(-7 - 8*x)/(
2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(25*Sqrt[5])

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IntegrateAlgebraic [A]  time = 0.53, size = 88, normalized size = 0.74 \begin {gather*} \frac {966 \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {5} (x+1)}\right )}{25 \sqrt {5}}-\frac {2 \sqrt {3 x^2+5 x+2} \left (2592 x^3+9453 x^2+10988 x+3977\right )}{25 (x+1) (2 x+3)^2 (3 x+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2 + 5*x + 3*x^2]*(3977 + 10988*x + 9453*x^2 + 2592*x^3))/(25*(1 + x)*(3 + 2*x)^2*(2 + 3*x)) + (966*Ar
cTanh[Sqrt[2 + 5*x + 3*x^2]/(Sqrt[5]*(1 + x))])/(25*Sqrt[5])

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fricas [A]  time = 0.41, size = 125, normalized size = 1.05 \begin {gather*} \frac {483 \, \sqrt {5} {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\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 (2592 \, x^{3} + 9453 \, x^{2} + 10988 \, x + 3977\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{250 \, {\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/250*(483*sqrt(5)*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*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*(2592*x^3 + 9453*x^2 + 10988*x + 3977)*sqrt(3*x^2 + 5*x + 2))/(12*
x^4 + 56*x^3 + 95*x^2 + 69*x + 18)

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giac [B]  time = 0.36, size = 225, normalized size = 1.89 \begin {gather*} \frac {483}{125} \, \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 {6 \, {\left (903 \, x + 653\right )}}{125 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {2 \, {\left (2442 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{3} + 9999 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 35473 \, \sqrt {3} x + 12979 \, \sqrt {3} - 35473 \, \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}}{125 \, {\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+2*x)^3/(3*x^2+5*x+2)^(3/2),x, algorithm="giac")

[Out]

483/125*sqrt(5)*log(abs(-4*sqrt(3)*x - 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))/abs(-4*sqrt(3)*x + 2*s
qrt(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))) - 6/125*(903*x + 653)/sqrt(3*x^2 + 5*x + 2) - 2/125*(2442*(sqrt
(3)*x - sqrt(3*x^2 + 5*x + 2))^3 + 9999*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^2 + 35473*sqrt(3)*x + 1297
9*sqrt(3) - 35473*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 - sqr
t(3*x^2 + 5*x + 2)) + 11)^2

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maple [A]  time = 0.01, size = 111, normalized size = 0.93 \begin {gather*} -\frac {483 \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 )}{125}-\frac {5}{2 \left (x +\frac {3}{2}\right ) \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}+\frac {483}{50 \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}-\frac {216 \left (6 x +5\right )}{25 \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}-\frac {13}{40 \left (x +\frac {3}{2}\right )^{2} \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2/(x+3/2)/(-4*x+3*(x+3/2)^2-19/4)^(1/2)+483/50/(-4*x+3*(x+3/2)^2-19/4)^(1/2)-216/25*(6*x+5)/(-4*x+3*(x+3/2)
^2-19/4)^(1/2)-483/125*5^(1/2)*arctanh(2/5*(-4*x-7/2)*5^(1/2)/(-16*x+12*(x+3/2)^2-19)^(1/2))-13/40/(x+3/2)^2/(
-4*x+3*(x+3/2)^2-19/4)^(1/2)

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maxima [A]  time = 1.27, size = 157, normalized size = 1.32 \begin {gather*} -\frac {483}{125} \, \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 {1296 \, x}{25 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {1677}{50 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {13}{10 \, {\left (4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} x^{2} + 12 \, \sqrt {3 \, x^{2} + 5 \, x + 2} x + 9 \, \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}} - \frac {5}{2 \, \sqrt {3 \, x^{2} + 5 \, x + 2} x + 3 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-483/125*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) - 1296/25*x/sqrt(3*x^2
+ 5*x + 2) - 1677/50/sqrt(3*x^2 + 5*x + 2) - 13/10/(4*sqrt(3*x^2 + 5*x + 2)*x^2 + 12*sqrt(3*x^2 + 5*x + 2)*x
+ 9*sqrt(3*x^2 + 5*x + 2)) - 5/(2*sqrt(3*x^2 + 5*x + 2)*x + 3*sqrt(3*x^2 + 5*x + 2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x}{24 x^{5} \sqrt {3 x^{2} + 5 x + 2} + 148 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 358 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 423 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 243 x \sqrt {3 x^{2} + 5 x + 2} + 54 \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {5}{24 x^{5} \sqrt {3 x^{2} + 5 x + 2} + 148 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 358 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 423 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 243 x \sqrt {3 x^{2} + 5 x + 2} + 54 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x/(24*x**5*sqrt(3*x**2 + 5*x + 2) + 148*x**4*sqrt(3*x**2 + 5*x + 2) + 358*x**3*sqrt(3*x**2 + 5*x + 2
) + 423*x**2*sqrt(3*x**2 + 5*x + 2) + 243*x*sqrt(3*x**2 + 5*x + 2) + 54*sqrt(3*x**2 + 5*x + 2)), x) - Integral
(-5/(24*x**5*sqrt(3*x**2 + 5*x + 2) + 148*x**4*sqrt(3*x**2 + 5*x + 2) + 358*x**3*sqrt(3*x**2 + 5*x + 2) + 423*
x**2*sqrt(3*x**2 + 5*x + 2) + 243*x*sqrt(3*x**2 + 5*x + 2) + 54*sqrt(3*x**2 + 5*x + 2)), x)

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