\(\int \frac {d+e x}{9+12 x+4 x^2} \, dx\) [1540]

Optimal result

Integrand size = 18, antiderivative size = 30 \[ \int \frac {d+e x}{9+12 x+4 x^2} \, dx=-\frac {2 d-3 e}{4 (3+2 x)}+\frac {1}{4} e \log (3+2 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {27, 45} \[ \int \frac {d+e x}{9+12 x+4 x^2} \, dx=\frac {1}{4} e \log (2 x+3)-\frac {2 d-3 e}{4 (2 x+3)} \]

[In]

Int[(d + e*x)/(9 + 12*x + 4*x^2),x]

[Out]

-1/4*(2*d - 3*e)/(3 + 2*x) + (e*Log[3 + 2*x])/4

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x}{(3+2 x)^2} \, dx \\ & = \int \left (\frac {2 d-3 e}{2 (3+2 x)^2}+\frac {e}{2 (3+2 x)}\right ) \, dx \\ & = -\frac {2 d-3 e}{4 (3+2 x)}+\frac {1}{4} e \log (3+2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {d+e x}{9+12 x+4 x^2} \, dx=\frac {-2 d+3 e}{4 (3+2 x)}+\frac {1}{4} e \log (3+2 x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87

method	result	size
norman	\(\frac {-\frac {d}{2}+\frac {3 e}{4}}{2 x +3}+\frac {e \ln \left (2 x +3\right )}{4}\)	\(26\)
default	\(-\frac {\frac {d}{2}-\frac {3 e}{4}}{2 x +3}+\frac {e \ln \left (2 x +3\right )}{4}\)	\(27\)
risch	\(-\frac {d}{4 \left (x +\frac {3}{2}\right )}+\frac {3 e}{8 \left (x +\frac {3}{2}\right )}+\frac {e \ln \left (2 x +3\right )}{4}\)	\(27\)
parallelrisch	\(\frac {2 e \ln \left (x +\frac {3}{2}\right ) x +3 e \ln \left (x +\frac {3}{2}\right )-2 d +3 e}{8 x +12}\)	\(32\)
meijerg	\(\frac {e \left (-\frac {2 x}{3 \left (1+\frac {2 x}{3}\right )}+\ln \left (1+\frac {2 x}{3}\right )\right )}{4}+\frac {d x}{9+6 x}\)	\(33\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {d+e x}{9+12 x+4 x^2} \, dx=\frac {{\left (2 \, e x + 3 \, e\right )} \log \left (2 \, x + 3\right ) - 2 \, d + 3 \, e}{4 \, {\left (2 \, x + 3\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {d+e x}{9+12 x+4 x^2} \, dx=\frac {e \log {\left (2 x + 3 \right )}}{4} + \frac {- 2 d + 3 e}{8 x + 12} \]

[In]

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

[Out]

e*log(2*x + 3)/4 + (-2*d + 3*e)/(8*x + 12)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {d+e x}{9+12 x+4 x^2} \, dx=\frac {1}{4} \, e \log \left (2 \, x + 3\right ) - \frac {2 \, d - 3 \, e}{4 \, {\left (2 \, x + 3\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {d+e x}{9+12 x+4 x^2} \, dx=\frac {1}{4} \, e \log \left ({\left | 2 \, x + 3 \right |}\right ) - \frac {2 \, d - 3 \, e}{4 \, {\left (2 \, x + 3\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {d+e x}{9+12 x+4 x^2} \, dx=\frac {e\,\ln \left (x+\frac {3}{2}\right )}{4}-\frac {\frac {d}{2}-\frac {3\,e}{4}}{2\,x+3} \]

[In]

int((d + e*x)/(12*x + 4*x^2 + 9),x)

[Out]

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