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

Optimal result

Integrand size = 20, antiderivative size = 56 \[ \int \frac {d+e x}{\sqrt {9+12 x+4 x^2}} \, dx=\frac {1}{4} e \sqrt {9+12 x+4 x^2}+\frac {(2 d-3 e) (3+2 x) \log (3+2 x)}{4 \sqrt {9+12 x+4 x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {654, 622, 31} \[ \int \frac {d+e x}{\sqrt {9+12 x+4 x^2}} \, dx=\frac {(2 x+3) (2 d-3 e) \log (2 x+3)}{4 \sqrt {4 x^2+12 x+9}}+\frac {1}{4} e \sqrt {4 x^2+12 x+9} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.73 \[ \int \frac {d+e x}{\sqrt {9+12 x+4 x^2}} \, dx=\frac {(3+2 x) \left (\frac {e x}{2}+\frac {1}{4} (2 d-3 e) \log (3+2 x)\right )}{\sqrt {(3+2 x)^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.46

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	\(\frac {\sqrt {\left (2 x +3\right )^{2}}\, e x}{4 x +6}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (-\frac {3 e}{2}+d \right ) \ln \left (2 x +3\right )}{4 x +6}\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.55 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.36 \[ \int \frac {d+e x}{\sqrt {9+12 x+4 x^2}} \, dx=\frac {1}{2} \, e x + \frac {1}{4} \, {\left (2 \, d - 3 \, e\right )} \log \left (2 \, x + 3\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {d+e x}{\sqrt {9+12 x+4 x^2}} \, dx=\frac {e \sqrt {4 x^{2} + 12 x + 9}}{4} + \frac {\left (d - \frac {3 e}{2}\right ) \left (x + \frac {3}{2}\right ) \log {\left (x + \frac {3}{2} \right )}}{2 \sqrt {\left (x + \frac {3}{2}\right )^{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.54 \[ \int \frac {d+e x}{\sqrt {9+12 x+4 x^2}} \, dx=\frac {1}{2} \, d \log \left (x + \frac {3}{2}\right ) - \frac {3}{4} \, e \log \left (x + \frac {3}{2}\right ) + \frac {1}{4} \, \sqrt {4 \, x^{2} + 12 \, x + 9} e \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.66 \[ \int \frac {d+e x}{\sqrt {9+12 x+4 x^2}} \, dx=\frac {e x}{2 \, \mathrm {sgn}\left (2 \, x + 3\right )} + \frac {{\left (2 \, d - 3 \, e\right )} \log \left ({\left | 2 \, x + 3 \right |}\right )}{4 \, \mathrm {sgn}\left (2 \, x + 3\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.48 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {d+e x}{\sqrt {9+12 x+4 x^2}} \, dx=\frac {e\,\sqrt {4\,x^2+12\,x+9}}{4}-\frac {3\,e\,\ln \left (x+\frac {\left |2\,x+3\right |}{2}+\frac {3}{2}\right )}{4}+\frac {d\,\ln \left (4\,x+6\right )\,\mathrm {sign}\left (8\,x+12\right )}{2} \]

[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