3.17.19 \(\int \frac {d+e x}{(9+12 x+4 x^2)^{5/2}} \, dx\) [1619]

Optimal. Leaf size=52 \[ -\frac {e}{12 \left (9+12 x+4 x^2\right )^{3/2}}-\frac {2 d-3 e}{16 (3+2 x) \left (9+12 x+4 x^2\right )^{3/2}} \]

[Out]

-1/12*e/(4*x^2+12*x+9)^(3/2)+1/16*(-2*d+3*e)/(3+2*x)/(4*x^2+12*x+9)^(3/2)

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Rubi [A]
time = 0.01, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {654, 621} \begin {gather*} -\frac {2 d-3 e}{16 (2 x+3) \left (4 x^2+12 x+9\right )^{3/2}}-\frac {e}{12 \left (4 x^2+12 x+9\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(9 + 12*x + 4*x^2)^(5/2),x]

[Out]

-1/12*e/(9 + 12*x + 4*x^2)^(3/2) - (2*d - 3*e)/(16*(3 + 2*x)*(9 + 12*x + 4*x^2)^(3/2))

Rule 621

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[2*((a + b*x + c*x^2)^(p + 1)/((2*p + 1)*(b + 2*
c*x))), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {d+e x}{\left (9+12 x+4 x^2\right )^{5/2}} \, dx &=-\frac {e}{12 \left (9+12 x+4 x^2\right )^{3/2}}+\frac {1}{2} (2 d-3 e) \int \frac {1}{\left (9+12 x+4 x^2\right )^{5/2}} \, dx\\ &=-\frac {e}{12 \left (9+12 x+4 x^2\right )^{3/2}}-\frac {2 d-3 e}{16 (3+2 x) \left (9+12 x+4 x^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 34, normalized size = 0.65 \begin {gather*} \frac {-6 d-e (3+8 x)}{48 (3+2 x)^3 \sqrt {(3+2 x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(9 + 12*x + 4*x^2)^(5/2),x]

[Out]

(-6*d - e*(3 + 8*x))/(48*(3 + 2*x)^3*Sqrt[(3 + 2*x)^2])

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Maple [A]
time = 0.56, size = 28, normalized size = 0.54

method	result	size
gosper	\(-\frac {\left (2 x +3\right ) \left (8 e x +6 d +3 e \right )}{48 \left (\left (2 x +3\right )^{2}\right )^{\frac {5}{2}}}\)	\(28\)
default	\(-\frac {\left (2 x +3\right ) \left (8 e x +6 d +3 e \right )}{48 \left (\left (2 x +3\right )^{2}\right )^{\frac {5}{2}}}\)	\(28\)
risch	\(\frac {16 \sqrt {\left (2 x +3\right )^{2}}\, \left (-\frac {1}{96} e x -\frac {1}{256} e -\frac {1}{128} d \right )}{\left (2 x +3\right )^{5}}\)	\(30\)
meijerg	\(\frac {e \,x^{2} \left (\frac {4}{9} x^{2}+\frac {8}{3} x +6\right )}{2916 \left (1+\frac {2 x}{3}\right )^{4}}+\frac {d x \left (\frac {8}{27} x^{3}+\frac {16}{9} x^{2}+4 x +4\right )}{972 \left (1+\frac {2 x}{3}\right )^{4}}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)/(4*x^2+12*x+9)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/48*(2*x+3)*(8*e*x+6*d+3*e)/((2*x+3)^2)^(5/2)

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Maxima [A]
time = 0.49, size = 38, normalized size = 0.73 \begin {gather*} -\frac {e}{12 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}}} - \frac {d}{8 \, {\left (2 \, x + 3\right )}^{4}} + \frac {3 \, e}{16 \, {\left (2 \, x + 3\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(4*x^2+12*x+9)^(5/2),x, algorithm="maxima")

[Out]

-1/12*e/(4*x^2 + 12*x + 9)^(3/2) - 1/8*d/(2*x + 3)^4 + 3/16*e/(2*x + 3)^4

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Fricas [A]
time = 2.43, size = 36, normalized size = 0.69 \begin {gather*} -\frac {{\left (8 \, x + 3\right )} e + 6 \, d}{48 \, {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(4*x^2+12*x+9)^(5/2),x, algorithm="fricas")

[Out]

-1/48*((8*x + 3)*e + 6*d)/(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d + e x}{\left (\left (2 x + 3\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(4*x**2+12*x+9)**(5/2),x)

[Out]

Integral((d + e*x)/((2*x + 3)**2)**(5/2), x)

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Giac [A]
time = 0.80, size = 30, normalized size = 0.58 \begin {gather*} -\frac {8 \, x e + 6 \, d + 3 \, e}{48 \, {\left (2 \, x + 3\right )}^{4} \mathrm {sgn}\left (2 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(4*x^2+12*x+9)^(5/2),x, algorithm="giac")

[Out]

-1/48*(8*x*e + 6*d + 3*e)/((2*x + 3)^4*sgn(2*x + 3))

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Mupad [B]
time = 0.10, size = 32, normalized size = 0.62 \begin {gather*} -\frac {\left (6\,d+3\,e+8\,e\,x\right )\,\sqrt {4\,x^2+12\,x+9}}{48\,{\left (2\,x+3\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(12*x + 4*x^2 + 9)^(5/2),x)

[Out]

-((6*d + 3*e + 8*e*x)*(12*x + 4*x^2 + 9)^(1/2))/(48*(2*x + 3)^5)

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