\(\int \frac {a+b x+c x^2}{b d+2 c d x} \, dx\) [1113]

Optimal result

Integrand size = 22, antiderivative size = 48 \[ \int \frac {a+b x+c x^2}{b d+2 c d x} \, dx=\frac {b x}{4 c d}+\frac {x^2}{4 d}-\frac {\left (b^2-4 a c\right ) \log (b+2 c x)}{8 c^2 d} \]

[Out]

1/4*b*x/c/d+1/4*x^2/d-1/8*(-4*a*c+b^2)*ln(2*c*x+b)/c^2/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {697} \[ \int \frac {a+b x+c x^2}{b d+2 c d x} \, dx=-\frac {\left (b^2-4 a c\right ) \log (b+2 c x)}{8 c^2 d}+\frac {b x}{4 c d}+\frac {x^2}{4 d} \]

[In]

Int[(a + b*x + c*x^2)/(b*d + 2*c*d*x),x]

[Out]

(b*x)/(4*c*d) + x^2/(4*d) - ((b^2 - 4*a*c)*Log[b + 2*c*x])/(8*c^2*d)

Rule 697

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {b}{4 c d}+\frac {x}{2 d}+\frac {-b^2+4 a c}{4 c d (b+2 c x)}\right ) \, dx \\ & = \frac {b x}{4 c d}+\frac {x^2}{4 d}-\frac {\left (b^2-4 a c\right ) \log (b+2 c x)}{8 c^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77 \[ \int \frac {a+b x+c x^2}{b d+2 c d x} \, dx=\frac {2 c x (b+c x)-\left (b^2-4 a c\right ) \log (b+2 c x)}{8 c^2 d} \]

[In]

Integrate[(a + b*x + c*x^2)/(b*d + 2*c*d*x),x]

[Out]

(2*c*x*(b + c*x) - (b^2 - 4*a*c)*Log[b + 2*c*x])/(8*c^2*d)

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {\frac {c \,x^{2}+b x}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (2 c x +b \right )}{8 c^{2}}}{d}\)	\(42\)
norman	\(\frac {x^{2}}{4 d}+\frac {b x}{4 c d}+\frac {\left (4 a c -b^{2}\right ) \ln \left (2 c x +b \right )}{8 c^{2} d}\)	\(45\)
parallelrisch	\(\frac {2 c^{2} x^{2}+4 \ln \left (\frac {b}{2}+c x \right ) a c -\ln \left (\frac {b}{2}+c x \right ) b^{2}+2 b c x}{8 c^{2} d}\)	\(48\)
risch	\(\frac {x^{2}}{4 d}+\frac {b x}{4 c d}+\frac {\ln \left (2 c x +b \right ) a}{2 c d}-\frac {\ln \left (2 c x +b \right ) b^{2}}{8 c^{2} d}\)	\(54\)

[In]

int((c*x^2+b*x+a)/(2*c*d*x+b*d),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/4/c*(c*x^2+b*x)+1/8*(4*a*c-b^2)/c^2*ln(2*c*x+b))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.81 \[ \int \frac {a+b x+c x^2}{b d+2 c d x} \, dx=\frac {2 \, c^{2} x^{2} + 2 \, b c x - {\left (b^{2} - 4 \, a c\right )} \log \left (2 \, c x + b\right )}{8 \, c^{2} d} \]

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d),x, algorithm="fricas")

[Out]

1/8*(2*c^2*x^2 + 2*b*c*x - (b^2 - 4*a*c)*log(2*c*x + b))/(c^2*d)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77 \[ \int \frac {a+b x+c x^2}{b d+2 c d x} \, dx=\frac {b x}{4 c d} + \frac {x^{2}}{4 d} + \frac {\left (4 a c - b^{2}\right ) \log {\left (b + 2 c x \right )}}{8 c^{2} d} \]

[In]

integrate((c*x**2+b*x+a)/(2*c*d*x+b*d),x)

[Out]

b*x/(4*c*d) + x**2/(4*d) + (4*a*c - b**2)*log(b + 2*c*x)/(8*c**2*d)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.85 \[ \int \frac {a+b x+c x^2}{b d+2 c d x} \, dx=\frac {c x^{2} + b x}{4 \, c d} - \frac {{\left (b^{2} - 4 \, a c\right )} \log \left (2 \, c x + b\right )}{8 \, c^{2} d} \]

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d),x, algorithm="maxima")

[Out]

1/4*(c*x^2 + b*x)/(c*d) - 1/8*(b^2 - 4*a*c)*log(2*c*x + b)/(c^2*d)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int \frac {a+b x+c x^2}{b d+2 c d x} \, dx=-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, c x + b \right |}\right )}{8 \, c^{2} d} + \frac {c^{2} d x^{2} + b c d x}{4 \, c^{2} d^{2}} \]

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d),x, algorithm="giac")

[Out]

-1/8*(b^2 - 4*a*c)*log(abs(2*c*x + b))/(c^2*d) + 1/4*(c^2*d*x^2 + b*c*d*x)/(c^2*d^2)

Mupad [B] (verification not implemented)

Time = 9.83 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92 \[ \int \frac {a+b x+c x^2}{b d+2 c d x} \, dx=\frac {x^2}{4\,d}+\frac {b\,x}{4\,c\,d}+\frac {\ln \left (b+2\,c\,x\right )\,\left (4\,a\,c-b^2\right )}{8\,c^2\,d} \]

[In]

int((a + b*x + c*x^2)/(b*d + 2*c*d*x),x)

[Out]

x^2/(4*d) + (b*x)/(4*c*d) + (log(b + 2*c*x)*(4*a*c - b^2))/(8*c^2*d)