\(\int \frac {a+b x+c x^2}{(b d+2 c d x)^3} \, dx\) [1115]

Optimal result

Integrand size = 22, antiderivative size = 44 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^3} \, dx=\frac {b^2-4 a c}{16 c^2 d^3 (b+2 c x)^2}+\frac {\log (b+2 c x)}{8 c^2 d^3} \]

[Out]

1/16*(-4*a*c+b^2)/c^2/d^3/(2*c*x+b)^2+1/8*ln(2*c*x+b)/c^2/d^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, 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} \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^3} \, dx=\frac {b^2-4 a c}{16 c^2 d^3 (b+2 c x)^2}+\frac {\log (b+2 c x)}{8 c^2 d^3} \]

[In]

Int[(a + b*x + c*x^2)/(b*d + 2*c*d*x)^3,x]

[Out]

(b^2 - 4*a*c)/(16*c^2*d^3*(b + 2*c*x)^2) + Log[b + 2*c*x]/(8*c^2*d^3)

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^2+4 a c}{4 c d^3 (b+2 c x)^3}+\frac {1}{4 c d^3 (b+2 c x)}\right ) \, dx \\ & = \frac {b^2-4 a c}{16 c^2 d^3 (b+2 c x)^2}+\frac {\log (b+2 c x)}{8 c^2 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^3} \, dx=\frac {\frac {b^2-4 a c}{(b+2 c x)^2}+2 \log (b+2 c x)}{16 c^2 d^3} \]

[In]

Integrate[(a + b*x + c*x^2)/(b*d + 2*c*d*x)^3,x]

[Out]

((b^2 - 4*a*c)/(b + 2*c*x)^2 + 2*Log[b + 2*c*x])/(16*c^2*d^3)

Maple [A] (verified)

Time = 2.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\frac {\ln \left (2 c x +b \right )}{8 c^{2}}-\frac {4 a c -b^{2}}{16 c^{2} \left (2 c x +b \right )^{2}}}{d^{3}}\)	\(41\)
norman	\(-\frac {4 a c -b^{2}}{16 c^{2} d^{3} \left (2 c x +b \right )^{2}}+\frac {\ln \left (2 c x +b \right )}{8 c^{2} d^{3}}\)	\(43\)
risch	\(-\frac {a}{4 c \,d^{3} \left (2 c x +b \right )^{2}}+\frac {b^{2}}{16 c^{2} d^{3} \left (2 c x +b \right )^{2}}+\frac {\ln \left (2 c x +b \right )}{8 c^{2} d^{3}}\)	\(53\)
parallelrisch	\(\frac {4 \ln \left (\frac {b}{2}+c x \right ) x^{2} b^{2} c^{2}+4 \ln \left (\frac {b}{2}+c x \right ) x \,b^{3} c +8 x^{2} a \,c^{3}-2 b^{2} c^{2} x^{2}+\ln \left (\frac {b}{2}+c x \right ) b^{4}+8 a b \,c^{2} x -2 b^{3} c x}{8 b^{2} c^{2} d^{3} \left (2 c x +b \right )^{2}}\)	\(102\)

[In]

int((c*x^2+b*x+a)/(2*c*d*x+b*d)^3,x,method=_RETURNVERBOSE)

[Out]

1/d^3*(1/8/c^2*ln(2*c*x+b)-1/16*(4*a*c-b^2)/c^2/(2*c*x+b)^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.59 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^3} \, dx=\frac {b^{2} - 4 \, a c + 2 \, {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (2 \, c x + b\right )}{16 \, {\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\right )}} \]

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d)^3,x, algorithm="fricas")

[Out]

1/16*(b^2 - 4*a*c + 2*(4*c^2*x^2 + 4*b*c*x + b^2)*log(2*c*x + b))/(4*c^4*d^3*x^2 + 4*b*c^3*d^3*x + b^2*c^2*d^3
)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.36 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^3} \, dx=\frac {- 4 a c + b^{2}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac {\log {\left (b + 2 c x \right )}}{8 c^{2} d^{3}} \]

[In]

integrate((c*x**2+b*x+a)/(2*c*d*x+b*d)**3,x)

[Out]

(-4*a*c + b**2)/(16*b**2*c**2*d**3 + 64*b*c**3*d**3*x + 64*c**4*d**3*x**2) + log(b + 2*c*x)/(8*c**2*d**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.36 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^3} \, dx=\frac {b^{2} - 4 \, a c}{16 \, {\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\right )}} + \frac {\log \left (2 \, c x + b\right )}{8 \, c^{2} d^{3}} \]

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d)^3,x, algorithm="maxima")

[Out]

1/16*(b^2 - 4*a*c)/(4*c^4*d^3*x^2 + 4*b*c^3*d^3*x + b^2*c^2*d^3) + 1/8*log(2*c*x + b)/(c^2*d^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^3} \, dx=\frac {\log \left ({\left | 2 \, c x + b \right |}\right )}{8 \, c^{2} d^{3}} + \frac {b^{2} - 4 \, a c}{16 \, {\left (2 \, c x + b\right )}^{2} c^{2} d^{3}} \]

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d)^3,x, algorithm="giac")

[Out]

1/8*log(abs(2*c*x + b))/(c^2*d^3) + 1/16*(b^2 - 4*a*c)/((2*c*x + b)^2*c^2*d^3)

Mupad [B] (verification not implemented)

Time = 9.96 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^3} \, dx=\frac {\ln \left (b+2\,c\,x\right )}{8\,c^2\,d^3}-\frac {\frac {a\,c}{4}-\frac {b^2}{16}}{c^2\,d^3\,{\left (b+2\,c\,x\right )}^2} \]

[In]

int((a + b*x + c*x^2)/(b*d + 2*c*d*x)^3,x)

[Out]

log(b + 2*c*x)/(8*c^2*d^3) - ((a*c)/4 - b^2/16)/(c^2*d^3*(b + 2*c*x)^2)