3.17.17 \(\int \frac {d+e x}{\sqrt {9+12 x+4 x^2}} \, dx\) [1617]

Optimal. Leaf size=56 \[ \frac {1}{4} e \sqrt {9+12 x+4 x^2}+\frac {(2 d-3 e) (3+2 x) \log (3+2 x)}{4 \sqrt {9+12 x+4 x^2}} \]

[Out]

1/4*(2*d-3*e)*(3+2*x)*ln(3+2*x)/((3+2*x)^2)^(1/2)+1/4*e*((3+2*x)^2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {654, 622, 31} \begin {gather*} \frac {(2 x+3) (2 d-3 e) \log (2 x+3)}{4 \sqrt {4 x^2+12 x+9}}+\frac {1}{4} e \sqrt {4 x^2+12 x+9} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[9 + 12*x + 4*x^2])/4 + ((2*d - 3*e)*(3 + 2*x)*Log[3 + 2*x])/(4*Sqrt[9 + 12*x + 4*x^2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 622

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

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}{\sqrt {9+12 x+4 x^2}} \, dx &=\frac {1}{4} e \sqrt {9+12 x+4 x^2}+\frac {1}{2} (2 d-3 e) \int \frac {1}{\sqrt {9+12 x+4 x^2}} \, dx\\ &=\frac {1}{4} e \sqrt {9+12 x+4 x^2}+\frac {((2 d-3 e) (6+4 x)) \int \frac {1}{6+4 x} \, dx}{2 \sqrt {9+12 x+4 x^2}}\\ &=\frac {1}{4} e \sqrt {9+12 x+4 x^2}+\frac {(2 d-3 e) (3+2 x) \log (3+2 x)}{4 \sqrt {9+12 x+4 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 42, normalized size = 0.75 \begin {gather*} \frac {(3+2 x) (e (3+2 x)+(2 d-3 e) \log (3+2 x))}{4 \sqrt {(3+2 x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3 + 2*x)*(e*(3 + 2*x) + (2*d - 3*e)*Log[3 + 2*x]))/(4*Sqrt[(3 + 2*x)^2])

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Maple [A]
time = 0.56, size = 40, normalized size = 0.71

method	result	size
meijerg	\(\frac {3 e \left (\frac {2 x}{3}-\ln \left (1+\frac {2 x}{3}\right )\right )}{4}+\frac {d \ln \left (1+\frac {2 x}{3}\right )}{2}\)	\(26\)
default	\(\frac {\left (2 x +3\right ) \left (2 \ln \left (2 x +3\right ) d -3 e \ln \left (2 x +3\right )+2 e x \right )}{4 \sqrt {\left (2 x +3\right )^{2}}}\)	\(40\)
risch	\(\frac {\sqrt {\left (2 x +3\right )^{2}}\, e x}{4 x +6}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (-\frac {3 e}{2}+d \right ) \ln \left (2 x +3\right )}{4 x +6}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)/(4*x^2+12*x+9)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/4*(2*x+3)*(2*ln(2*x+3)*d-3*e*ln(2*x+3)+2*e*x)/((2*x+3)^2)^(1/2)

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Maxima [A]
time = 0.51, size = 32, normalized size = 0.57 \begin {gather*} \frac {1}{2} \, d \log \left (x + \frac {3}{2}\right ) - \frac {3}{4} \, e \log \left (x + \frac {3}{2}\right ) + \frac {1}{4} \, \sqrt {4 \, x^{2} + 12 \, x + 9} e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(4*x^2+12*x+9)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.24, size = 22, normalized size = 0.39 \begin {gather*} \frac {1}{2} \, x e + \frac {1}{4} \, {\left (2 \, d - 3 \, e\right )} \log \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)^(1/2),x, algorithm="fricas")

[Out]

1/2*x*e + 1/4*(2*d - 3*e)*log(2*x + 3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d + e x}{\sqrt {\left (2 x + 3\right )^{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)**(1/2),x)

[Out]

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

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Giac [A]
time = 0.62, size = 39, normalized size = 0.70 \begin {gather*} \frac {x e}{2 \, \mathrm {sgn}\left (2 \, x + 3\right )} + \frac {{\left (2 \, d - 3 \, e\right )} \log \left ({\left | 2 \, x + 3 \right |}\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)^(1/2),x, algorithm="giac")

[Out]

1/2*x*e/sgn(2*x + 3) + 1/4*(2*d - 3*e)*log(abs(2*x + 3))/sgn(2*x + 3)

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Mupad [B]
time = 1.50, size = 46, normalized size = 0.82 \begin {gather*} \frac {e\,\sqrt {4\,x^2+12\,x+9}}{4}-\frac {3\,e\,\ln \left (x+\frac {\left |2\,x+3\right |}{2}+\frac {3}{2}\right )}{4}+\frac {d\,\ln \left (4\,x+6\right )\,\mathrm {sign}\left (8\,x+12\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(12*x + 4*x^2 + 9)^(1/2),x)

[Out]

(e*(12*x + 4*x^2 + 9)^(1/2))/4 - (3*e*log(x + abs(2*x + 3)/2 + 3/2))/4 + (d*log(4*x + 6)*sign(8*x + 12))/2

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