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

Optimal. Leaf size=94 \[ -\frac {6 (47 x+37)}{5 (2 x+3) \sqrt {3 x^2+5 x+2}}-\frac {856 \sqrt {3 x^2+5 x+2}}{25 (2 x+3)}+\frac {302 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{25 \sqrt {5}} \]

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Rubi [A]  time = 0.05, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {822, 806, 724, 206} \begin {gather*} -\frac {6 (47 x+37)}{5 (2 x+3) \sqrt {3 x^2+5 x+2}}-\frac {856 \sqrt {3 x^2+5 x+2}}{25 (2 x+3)}+\frac {302 \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)^2*(2 + 5*x + 3*x^2)^(3/2)),x]

[Out]

(-6*(37 + 47*x))/(5*(3 + 2*x)*Sqrt[2 + 5*x + 3*x^2]) - (856*Sqrt[2 + 5*x + 3*x^2])/(25*(3 + 2*x)) + (302*ArcTa
nh[(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])

Rubi steps

\begin {align*} \int \frac {5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )^{3/2}} \, dx &=-\frac {6 (37+47 x)}{5 (3+2 x) \sqrt {2+5 x+3 x^2}}-\frac {2}{5} \int \frac {209+282 x}{(3+2 x)^2 \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {6 (37+47 x)}{5 (3+2 x) \sqrt {2+5 x+3 x^2}}-\frac {856 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}+\frac {302}{25} \int \frac {1}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {6 (37+47 x)}{5 (3+2 x) \sqrt {2+5 x+3 x^2}}-\frac {856 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}-\frac {604}{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) \sqrt {2+5 x+3 x^2}}-\frac {856 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}+\frac {302 \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 = 0.03, size = 90, normalized size = 0.96 \begin {gather*} -\frac {2 \left (6420 x^2+151 \sqrt {5} (2 x+3) \sqrt {3 x^2+5 x+2} \tanh ^{-1}\left (\frac {-8 x-7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )+14225 x+7055\right )}{125 (2 x+3) \sqrt {3 x^2+5 x+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(7055 + 14225*x + 6420*x^2 + 151*Sqrt[5]*(3 + 2*x)*Sqrt[2 + 5*x + 3*x^2]*ArcTanh[(-7 - 8*x)/(2*Sqrt[5]*Sqr
t[2 + 5*x + 3*x^2])]))/(125*(3 + 2*x)*Sqrt[2 + 5*x + 3*x^2])

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IntegrateAlgebraic [A]  time = 0.49, size = 83, normalized size = 0.88 \begin {gather*} \frac {604 \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 (1284 x^2+2845 x+1411\right )}{25 (x+1) (2 x+3) (3 x+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2 + 5*x + 3*x^2]*(1411 + 2845*x + 1284*x^2))/(25*(1 + x)*(3 + 2*x)*(2 + 3*x)) + (604*ArcTanh[Sqrt[2 +
5*x + 3*x^2]/(Sqrt[5]*(1 + x))])/(25*Sqrt[5])

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fricas [A]  time = 0.41, size = 110, normalized size = 1.17 \begin {gather*} \frac {151 \, \sqrt {5} {\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\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 ) - 10 \, {\left (1284 \, x^{2} + 2845 \, x + 1411\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{125 \, {\left (6 \, x^{3} + 19 \, x^{2} + 19 \, x + 6\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/125*(151*sqrt(5)*(6*x^3 + 19*x^2 + 19*x + 6)*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)) - 10*(1284*x^2 + 2845*x + 1411)*sqrt(3*x^2 + 5*x + 2))/(6*x^3 + 19*x^2 + 19*x + 6)

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giac [B]  time = 0.32, size = 170, normalized size = 1.81 \begin {gather*} \frac {2}{125} \, \sqrt {5} {\left (214 \, \sqrt {5} \sqrt {3} + 151 \, \log \left (-\sqrt {5} \sqrt {3} + 4\right )\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right ) + \frac {2 \, {\left (\frac {\frac {1007}{\mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )} - \frac {65}{{\left (2 \, x + 3\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )}}{2 \, x + 3} - \frac {642}{\mathrm {sgn}\left (\frac {1}{2 \, x + 3}\right )}\right )}}{25 \, \sqrt {-\frac {8}{2 \, x + 3} + \frac {5}{{\left (2 \, x + 3\right )}^{2}} + 3}} - \frac {302 \, \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 )}{125 \, \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+2*x)^2/(3*x^2+5*x+2)^(3/2),x, algorithm="giac")

[Out]

2/125*sqrt(5)*(214*sqrt(5)*sqrt(3) + 151*log(-sqrt(5)*sqrt(3) + 4))*sgn(1/(2*x + 3)) + 2/25*((1007/sgn(1/(2*x
+ 3)) - 65/((2*x + 3)*sgn(1/(2*x + 3))))/(2*x + 3) - 642/sgn(1/(2*x + 3)))/sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 +
3) - 302/125*sqrt(5)*log(abs(sqrt(5)*(sqrt(-8/(2*x + 3) + 5/(2*x + 3)^2 + 3) + sqrt(5)/(2*x + 3)) - 4))/sgn(1
/(2*x + 3))

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maple [A]  time = 0.01, size = 90, normalized size = 0.96 \begin {gather*} -\frac {302 \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 {13}{10 \left (x +\frac {3}{2}\right ) \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}+\frac {151}{25 \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}-\frac {214 \left (6 x +5\right )}{25 \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)^2/(3*x^2+5*x+2)^(3/2),x)

[Out]

-13/10/(x+3/2)/(-4*x+3*(x+3/2)^2-19/4)^(1/2)+151/25/(-4*x+3*(x+3/2)^2-19/4)^(1/2)-214/25*(6*x+5)/(-4*x+3*(x+3/
2)^2-19/4)^(1/2)-302/125*5^(1/2)*arctanh(2/5*(-4*x-7/2)*5^(1/2)/(-16*x+12*(x+3/2)^2-19)^(1/2))

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maxima [A]  time = 1.36, size = 106, normalized size = 1.13 \begin {gather*} -\frac {302}{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 {1284 \, x}{25 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {919}{25 \, \sqrt {3 \, x^{2} + 5 \, x + 2}} - \frac {13}{5 \, {\left (2 \, \sqrt {3 \, x^{2} + 5 \, x + 2} x + 3 \, \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-302/125*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) - 1284/25*x/sqrt(3*x^2
+ 5*x + 2) - 919/25/sqrt(3*x^2 + 5*x + 2) - 13/5/(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 )}^2\,{\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)^2*(5*x + 3*x^2 + 2)^(3/2)),x)

[Out]

-int((x - 5)/((2*x + 3)^2*(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}{12 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 56 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 95 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 69 x \sqrt {3 x^{2} + 5 x + 2} + 18 \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {5}{12 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 56 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 95 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 69 x \sqrt {3 x^{2} + 5 x + 2} + 18 \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)**2/(3*x**2+5*x+2)**(3/2),x)

[Out]

-Integral(x/(12*x**4*sqrt(3*x**2 + 5*x + 2) + 56*x**3*sqrt(3*x**2 + 5*x + 2) + 95*x**2*sqrt(3*x**2 + 5*x + 2)
+ 69*x*sqrt(3*x**2 + 5*x + 2) + 18*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-5/(12*x**4*sqrt(3*x**2 + 5*x + 2) +
56*x**3*sqrt(3*x**2 + 5*x + 2) + 95*x**2*sqrt(3*x**2 + 5*x + 2) + 69*x*sqrt(3*x**2 + 5*x + 2) + 18*sqrt(3*x**
2 + 5*x + 2)), x)

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