3.2503 \(\int \frac {5-x}{(3+2 x)^4 \sqrt {2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=114 \[ -\frac {72 \sqrt {3 x^2+5 x+2}}{25 (2 x+3)}-\frac {49 \sqrt {3 x^2+5 x+2}}{30 (2 x+3)^2}-\frac {13 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^3}+\frac {331 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{100 \sqrt {5}} \]

[Out]

331/500*arctanh(1/10*(7+8*x)*5^(1/2)/(3*x^2+5*x+2)^(1/2))*5^(1/2)-13/15*(3*x^2+5*x+2)^(1/2)/(3+2*x)^3-49/30*(3
*x^2+5*x+2)^(1/2)/(3+2*x)^2-72/25*(3*x^2+5*x+2)^(1/2)/(3+2*x)

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Rubi [A]  time = 0.07, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {834, 806, 724, 206} \[ -\frac {72 \sqrt {3 x^2+5 x+2}}{25 (2 x+3)}-\frac {49 \sqrt {3 x^2+5 x+2}}{30 (2 x+3)^2}-\frac {13 \sqrt {3 x^2+5 x+2}}{15 (2 x+3)^3}+\frac {331 \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{100 \sqrt {5}} \]

Antiderivative was successfully verified.

[In]

Int[(5 - x)/((3 + 2*x)^4*Sqrt[2 + 5*x + 3*x^2]),x]

[Out]

(-13*Sqrt[2 + 5*x + 3*x^2])/(15*(3 + 2*x)^3) - (49*Sqrt[2 + 5*x + 3*x^2])/(30*(3 + 2*x)^2) - (72*Sqrt[2 + 5*x
+ 3*x^2])/(25*(3 + 2*x)) + (331*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(100*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 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)^4 \sqrt {2+5 x+3 x^2}} \, dx &=-\frac {13 \sqrt {2+5 x+3 x^2}}{15 (3+2 x)^3}-\frac {1}{15} \int \frac {-\frac {11}{2}+78 x}{(3+2 x)^3 \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {13 \sqrt {2+5 x+3 x^2}}{15 (3+2 x)^3}-\frac {49 \sqrt {2+5 x+3 x^2}}{30 (3+2 x)^2}+\frac {1}{150} \int \frac {-\frac {45}{2}-735 x}{(3+2 x)^2 \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {13 \sqrt {2+5 x+3 x^2}}{15 (3+2 x)^3}-\frac {49 \sqrt {2+5 x+3 x^2}}{30 (3+2 x)^2}-\frac {72 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}+\frac {331}{100} \int \frac {1}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {13 \sqrt {2+5 x+3 x^2}}{15 (3+2 x)^3}-\frac {49 \sqrt {2+5 x+3 x^2}}{30 (3+2 x)^2}-\frac {72 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}-\frac {331}{50} \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,\frac {-7-8 x}{\sqrt {2+5 x+3 x^2}}\right )\\ &=-\frac {13 \sqrt {2+5 x+3 x^2}}{15 (3+2 x)^3}-\frac {49 \sqrt {2+5 x+3 x^2}}{30 (3+2 x)^2}-\frac {72 \sqrt {2+5 x+3 x^2}}{25 (3+2 x)}+\frac {331 \tanh ^{-1}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )}{100 \sqrt {5}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 74, normalized size = 0.65 \[ \frac {-\frac {10 \sqrt {3 x^2+5 x+2} \left (1728 x^2+5674 x+4753\right )}{(2 x+3)^3}-993 \sqrt {5} \tanh ^{-1}\left (\frac {-8 x-7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )}{1500} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 - x)/((3 + 2*x)^4*Sqrt[2 + 5*x + 3*x^2]),x]

[Out]

((-10*Sqrt[2 + 5*x + 3*x^2]*(4753 + 5674*x + 1728*x^2))/(3 + 2*x)^3 - 993*Sqrt[5]*ArcTanh[(-7 - 8*x)/(2*Sqrt[5
]*Sqrt[2 + 5*x + 3*x^2])])/1500

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fricas [A]  time = 0.76, size = 110, normalized size = 0.96 \[ \frac {993 \, \sqrt {5} {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )} \log \left (\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{4 \, x^{2} + 12 \, x + 9}\right ) - 20 \, {\left (1728 \, x^{2} + 5674 \, x + 4753\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{3000 \, {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3+2*x)^4/(3*x^2+5*x+2)^(1/2),x, algorithm="fricas")

[Out]

1/3000*(993*sqrt(5)*(8*x^3 + 36*x^2 + 54*x + 27)*log((4*sqrt(5)*sqrt(3*x^2 + 5*x + 2)*(8*x + 7) + 124*x^2 + 21
2*x + 89)/(4*x^2 + 12*x + 9)) - 20*(1728*x^2 + 5674*x + 4753)*sqrt(3*x^2 + 5*x + 2))/(8*x^3 + 36*x^2 + 54*x +
27)

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giac [B]  time = 0.31, size = 257, normalized size = 2.25 \[ \frac {331}{500} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) - \frac {3972 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{5} + 29790 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{4} + 255470 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{3} + 338835 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 632175 \, \sqrt {3} x + 149502 \, \sqrt {3} - 632175 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{150 \, {\left (2 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )}^{2} + 6 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} + 11\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3+2*x)^4/(3*x^2+5*x+2)^(1/2),x, algorithm="giac")

[Out]

331/500*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))) - 1/150*(3972*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^5 + 29790*sqr
t(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^4 + 255470*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2))^3 + 338835*sqrt(3)*(sq
rt(3)*x - sqrt(3*x^2 + 5*x + 2))^2 + 632175*sqrt(3)*x + 149502*sqrt(3) - 632175*sqrt(3*x^2 + 5*x + 2))/(2*(sqr
t(3)*x - sqrt(3*x^2 + 5*x + 2))^2 + 6*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) + 11)^3

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maple [A]  time = 0.01, size = 95, normalized size = 0.83 \[ -\frac {331 \sqrt {5}\, \arctanh \left (\frac {2 \left (-4 x -\frac {7}{2}\right ) \sqrt {5}}{5 \sqrt {-16 x +12 \left (x +\frac {3}{2}\right )^{2}-19}}\right )}{500}-\frac {49 \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}{120 \left (x +\frac {3}{2}\right )^{2}}-\frac {36 \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}{25 \left (x +\frac {3}{2}\right )}-\frac {13 \sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}}{120 \left (x +\frac {3}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-x)/(3+2*x)^4/(3*x^2+5*x+2)^(1/2),x)

[Out]

-49/120/(x+3/2)^2*(-4*x+3*(x+3/2)^2-19/4)^(1/2)-36/25/(x+3/2)*(-4*x+3*(x+3/2)^2-19/4)^(1/2)-331/500*5^(1/2)*ar
ctanh(2/5*(-4*x-7/2)*5^(1/2)/(-16*x+12*(x+3/2)^2-19)^(1/2))-13/120/(x+3/2)^3*(-4*x+3*(x+3/2)^2-19/4)^(1/2)

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maxima [A]  time = 1.21, size = 121, normalized size = 1.06 \[ -\frac {331}{500} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) - \frac {13 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{15 \, {\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )}} - \frac {49 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{30 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac {72 \, \sqrt {3 \, x^{2} + 5 \, x + 2}}{25 \, {\left (2 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3+2*x)^4/(3*x^2+5*x+2)^(1/2),x, algorithm="maxima")

[Out]

-331/500*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) - 13/15*sqrt(3*x^2 + 5
*x + 2)/(8*x^3 + 36*x^2 + 54*x + 27) - 49/30*sqrt(3*x^2 + 5*x + 2)/(4*x^2 + 12*x + 9) - 72/25*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 \[ -\int \frac {x-5}{{\left (2\,x+3\right )}^4\,\sqrt {3\,x^2+5\,x+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - 5)/((2*x + 3)^4*(5*x + 3*x^2 + 2)^(1/2)),x)

[Out]

-int((x - 5)/((2*x + 3)^4*(5*x + 3*x^2 + 2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x}{16 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 96 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 216 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 216 x \sqrt {3 x^{2} + 5 x + 2} + 81 \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {5}{16 x^{4} \sqrt {3 x^{2} + 5 x + 2} + 96 x^{3} \sqrt {3 x^{2} + 5 x + 2} + 216 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 216 x \sqrt {3 x^{2} + 5 x + 2} + 81 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3+2*x)**4/(3*x**2+5*x+2)**(1/2),x)

[Out]

-Integral(x/(16*x**4*sqrt(3*x**2 + 5*x + 2) + 96*x**3*sqrt(3*x**2 + 5*x + 2) + 216*x**2*sqrt(3*x**2 + 5*x + 2)
+ 216*x*sqrt(3*x**2 + 5*x + 2) + 81*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-5/(16*x**4*sqrt(3*x**2 + 5*x + 2)
+ 96*x**3*sqrt(3*x**2 + 5*x + 2) + 216*x**2*sqrt(3*x**2 + 5*x + 2) + 216*x*sqrt(3*x**2 + 5*x + 2) + 81*sqrt(3
*x**2 + 5*x + 2)), x)

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