3.1128 \(\int \frac{(a+b x+c x^2)^2}{(b d+2 c d x)^3} \, dx\)

Optimal. Leaf size=79 \[ -\frac{\left (b^2-4 a c\right )^2}{64 c^3 d^3 (b+2 c x)^2}-\frac{\left (b^2-4 a c\right ) \log (b+2 c x)}{16 c^3 d^3}+\frac{b x}{16 c^2 d^3}+\frac{x^2}{16 c d^3} \]

[Out]

(b*x)/(16*c^2*d^3) + x^2/(16*c*d^3) - (b^2 - 4*a*c)^2/(64*c^3*d^3*(b + 2*c*x)^2) - ((b^2 - 4*a*c)*Log[b + 2*c*
x])/(16*c^3*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.0635056, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {683} \[ -\frac{\left (b^2-4 a c\right )^2}{64 c^3 d^3 (b+2 c x)^2}-\frac{\left (b^2-4 a c\right ) \log (b+2 c x)}{16 c^3 d^3}+\frac{b x}{16 c^2 d^3}+\frac{x^2}{16 c d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x)/(16*c^2*d^3) + x^2/(16*c*d^3) - (b^2 - 4*a*c)^2/(64*c^3*d^3*(b + 2*c*x)^2) - ((b^2 - 4*a*c)*Log[b + 2*c*
x])/(16*c^3*d^3)

Rule 683

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*} \int \frac{\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^3} \, dx &=\int \left (\frac{b}{16 c^2 d^3}+\frac{x}{8 c d^3}+\frac{\left (-b^2+4 a c\right )^2}{16 c^2 d^3 (b+2 c x)^3}+\frac{-b^2+4 a c}{8 c^2 d^3 (b+2 c x)}\right ) \, dx\\ &=\frac{b x}{16 c^2 d^3}+\frac{x^2}{16 c d^3}-\frac{\left (b^2-4 a c\right )^2}{64 c^3 d^3 (b+2 c x)^2}-\frac{\left (b^2-4 a c\right ) \log (b+2 c x)}{16 c^3 d^3}\\ \end{align*}

Mathematica [A]  time = 0.0406992, size = 61, normalized size = 0.77 \[ \frac{-\frac{\left (b^2-4 a c\right )^2}{(b+2 c x)^2}-4 \left (b^2-4 a c\right ) \log (b+2 c x)+4 b c x+4 c^2 x^2}{64 c^3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*c*x + 4*c^2*x^2 - (b^2 - 4*a*c)^2/(b + 2*c*x)^2 - 4*(b^2 - 4*a*c)*Log[b + 2*c*x])/(64*c^3*d^3)

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 115, normalized size = 1.5 \begin{align*}{\frac{{x}^{2}}{16\,c{d}^{3}}}+{\frac{bx}{16\,{c}^{2}{d}^{3}}}-{\frac{{a}^{2}}{4\,c{d}^{3} \left ( 2\,cx+b \right ) ^{2}}}+{\frac{{b}^{2}a}{8\,{c}^{2}{d}^{3} \left ( 2\,cx+b \right ) ^{2}}}-{\frac{{b}^{4}}{64\,{c}^{3}{d}^{3} \left ( 2\,cx+b \right ) ^{2}}}+{\frac{\ln \left ( 2\,cx+b \right ) a}{4\,{c}^{2}{d}^{3}}}-{\frac{\ln \left ( 2\,cx+b \right ){b}^{2}}{16\,{c}^{3}{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x^2/c/d^3+1/16*b*x/c^2/d^3-1/4/d^3/c/(2*c*x+b)^2*a^2+1/8/d^3/c^2/(2*c*x+b)^2*a*b^2-1/64/d^3/c^3/(2*c*x+b)
^2*b^4+1/4/d^3/c^2*ln(2*c*x+b)*a-1/16/d^3/c^3*ln(2*c*x+b)*b^2

________________________________________________________________________________________

Maxima [A]  time = 1.13658, size = 130, normalized size = 1.65 \begin{align*} -\frac{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}{64 \,{\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}} + \frac{c x^{2} + b x}{16 \, c^{2} d^{3}} - \frac{{\left (b^{2} - 4 \, a c\right )} \log \left (2 \, c x + b\right )}{16 \, c^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/(4*c^5*d^3*x^2 + 4*b*c^4*d^3*x + b^2*c^3*d^3) + 1/16*(c*x^2 + b*x)/(c^2*d
^3) - 1/16*(b^2 - 4*a*c)*log(2*c*x + b)/(c^3*d^3)

________________________________________________________________________________________

Fricas [B]  time = 2.02038, size = 312, normalized size = 3.95 \begin{align*} \frac{16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 20 \, b^{2} c^{2} x^{2} + 4 \, b^{3} c x - b^{4} + 8 \, a b^{2} c - 16 \, a^{2} c^{2} - 4 \,{\left (b^{4} - 4 \, a b^{2} c + 4 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 4 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x\right )} \log \left (2 \, c x + b\right )}{64 \,{\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(16*c^4*x^4 + 32*b*c^3*x^3 + 20*b^2*c^2*x^2 + 4*b^3*c*x - b^4 + 8*a*b^2*c - 16*a^2*c^2 - 4*(b^4 - 4*a*b^2
*c + 4*(b^2*c^2 - 4*a*c^3)*x^2 + 4*(b^3*c - 4*a*b*c^2)*x)*log(2*c*x + b))/(4*c^5*d^3*x^2 + 4*b*c^4*d^3*x + b^2
*c^3*d^3)

________________________________________________________________________________________

Sympy [A]  time = 1.08025, size = 102, normalized size = 1.29 \begin{align*} \frac{b x}{16 c^{2} d^{3}} - \frac{16 a^{2} c^{2} - 8 a b^{2} c + b^{4}}{64 b^{2} c^{3} d^{3} + 256 b c^{4} d^{3} x + 256 c^{5} d^{3} x^{2}} + \frac{x^{2}}{16 c d^{3}} + \frac{\left (4 a c - b^{2}\right ) \log{\left (b + 2 c x \right )}}{16 c^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/(16*c**2*d**3) - (16*a**2*c**2 - 8*a*b**2*c + b**4)/(64*b**2*c**3*d**3 + 256*b*c**4*d**3*x + 256*c**5*d**3
*x**2) + x**2/(16*c*d**3) + (4*a*c - b**2)*log(b + 2*c*x)/(16*c**3*d**3)

________________________________________________________________________________________

Giac [A]  time = 1.13045, size = 119, normalized size = 1.51 \begin{align*} -\frac{{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, c x + b \right |}\right )}{16 \, c^{3} d^{3}} - \frac{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}{64 \,{\left (2 \, c x + b\right )}^{2} c^{3} d^{3}} + \frac{c^{5} d^{3} x^{2} + b c^{4} d^{3} x}{16 \, c^{6} d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(b^2 - 4*a*c)*log(abs(2*c*x + b))/(c^3*d^3) - 1/64*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/((2*c*x + b)^2*c^3*d^3
) + 1/16*(c^5*d^3*x^2 + b*c^4*d^3*x)/(c^6*d^6)