3.10.41 \(\int \frac {(a+b x+c x^2)^2}{(b d+2 c d x)^4} \, dx\)

Optimal. Leaf size=66 \[ -\frac {\left (b^2-4 a c\right )^2}{96 c^3 d^4 (b+2 c x)^3}+\frac {b^2-4 a c}{16 c^3 d^4 (b+2 c x)}+\frac {x}{16 c^2 d^4} \]

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Rubi [A]  time = 0.05, antiderivative size = 66, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {\left (b^2-4 a c\right )^2}{96 c^3 d^4 (b+2 c x)^3}+\frac {b^2-4 a c}{16 c^3 d^4 (b+2 c x)}+\frac {x}{16 c^2 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^4,x]

[Out]

x/(16*c^2*d^4) - (b^2 - 4*a*c)^2/(96*c^3*d^4*(b + 2*c*x)^3) + (b^2 - 4*a*c)/(16*c^3*d^4*(b + 2*c*x))

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)^4} \, dx &=\int \left (\frac {1}{16 c^2 d^4}+\frac {\left (-b^2+4 a c\right )^2}{16 c^2 d^4 (b+2 c x)^4}+\frac {-b^2+4 a c}{8 c^2 d^4 (b+2 c x)^2}\right ) \, dx\\ &=\frac {x}{16 c^2 d^4}-\frac {\left (b^2-4 a c\right )^2}{96 c^3 d^4 (b+2 c x)^3}+\frac {b^2-4 a c}{16 c^3 d^4 (b+2 c x)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 53, normalized size = 0.80 \begin {gather*} \frac {-\frac {\left (b^2-4 a c\right )^2}{(b+2 c x)^3}+\frac {6 \left (b^2-4 a c\right )}{b+2 c x}+6 c x}{96 c^3 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^4,x]

[Out]

(6*c*x - (b^2 - 4*a*c)^2/(b + 2*c*x)^3 + (6*(b^2 - 4*a*c))/(b + 2*c*x))/(96*c^3*d^4)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^4,x]

[Out]

IntegrateAlgebraic[(a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^4, x]

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fricas [B]  time = 0.39, size = 125, normalized size = 1.89 \begin {gather*} \frac {48 \, c^{4} x^{4} + 72 \, b c^{3} x^{3} + 5 \, b^{4} - 16 \, a b^{2} c - 16 \, a^{2} c^{2} + 12 \, {\left (5 \, b^{2} c^{2} - 8 \, a c^{3}\right )} x^{2} + 6 \, {\left (5 \, b^{3} c - 16 \, a b c^{2}\right )} x}{96 \, {\left (8 \, c^{6} d^{4} x^{3} + 12 \, b c^{5} d^{4} x^{2} + 6 \, b^{2} c^{4} d^{4} x + b^{3} c^{3} d^{4}\right )}} \end {gather*}

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)^4,x, algorithm="fricas")

[Out]

1/96*(48*c^4*x^4 + 72*b*c^3*x^3 + 5*b^4 - 16*a*b^2*c - 16*a^2*c^2 + 12*(5*b^2*c^2 - 8*a*c^3)*x^2 + 6*(5*b^3*c
- 16*a*b*c^2)*x)/(8*c^6*d^4*x^3 + 12*b*c^5*d^4*x^2 + 6*b^2*c^4*d^4*x + b^3*c^3*d^4)

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giac [A]  time = 0.16, size = 82, normalized size = 1.24 \begin {gather*} \frac {x}{16 \, c^{2} d^{4}} + \frac {24 \, b^{2} c^{2} x^{2} - 96 \, a c^{3} x^{2} + 24 \, b^{3} c x - 96 \, a b c^{2} x + 5 \, b^{4} - 16 \, a b^{2} c - 16 \, a^{2} c^{2}}{96 \, {\left (2 \, c x + b\right )}^{3} c^{3} d^{4}} \end {gather*}

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)^4,x, algorithm="giac")

[Out]

1/16*x/(c^2*d^4) + 1/96*(24*b^2*c^2*x^2 - 96*a*c^3*x^2 + 24*b^3*c*x - 96*a*b*c^2*x + 5*b^4 - 16*a*b^2*c - 16*a
^2*c^2)/((2*c*x + b)^3*c^3*d^4)

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maple [A]  time = 0.05, size = 67, normalized size = 1.02 \begin {gather*} \frac {\frac {x}{16 c^{2}}-\frac {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}{96 \left (2 c x +b \right )^{3} c^{3}}-\frac {4 a c -b^{2}}{16 \left (2 c x +b \right ) c^{3}}}{d^{4}} \end {gather*}

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)^4,x)

[Out]

1/d^4*(1/16/c^2*x-1/96/c^3*(16*a^2*c^2-8*a*b^2*c+b^4)/(2*c*x+b)^3-1/16/c^3*(4*a*c-b^2)/(2*c*x+b))

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maxima [A]  time = 1.44, size = 116, normalized size = 1.76 \begin {gather*} \frac {5 \, b^{4} - 16 \, a b^{2} c - 16 \, a^{2} c^{2} + 24 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 24 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x}{96 \, {\left (8 \, c^{6} d^{4} x^{3} + 12 \, b c^{5} d^{4} x^{2} + 6 \, b^{2} c^{4} d^{4} x + b^{3} c^{3} d^{4}\right )}} + \frac {x}{16 \, c^{2} d^{4}} \end {gather*}

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)^4,x, algorithm="maxima")

[Out]

1/96*(5*b^4 - 16*a*b^2*c - 16*a^2*c^2 + 24*(b^2*c^2 - 4*a*c^3)*x^2 + 24*(b^3*c - 4*a*b*c^2)*x)/(8*c^6*d^4*x^3
+ 12*b*c^5*d^4*x^2 + 6*b^2*c^4*d^4*x + b^3*c^3*d^4) + 1/16*x/(c^2*d^4)

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mupad [B]  time = 0.46, size = 67, normalized size = 1.02 \begin {gather*} -\frac {\frac {a^2\,c^2}{6}-b\,\left (\frac {c^3\,x^3}{3}-a\,c^2\,x\right )-\frac {c^4\,x^4}{2}+a\,c^3\,x^2+\frac {a\,b^2\,c}{6}}{c^3\,d^4\,{\left (b+2\,c\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^2/(b*d + 2*c*d*x)^4,x)

[Out]

-((a^2*c^2)/6 - b*((c^3*x^3)/3 - a*c^2*x) - (c^4*x^4)/2 + a*c^3*x^2 + (a*b^2*c)/6)/(c^3*d^4*(b + 2*c*x)^3)

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sympy [A]  time = 0.86, size = 117, normalized size = 1.77 \begin {gather*} \frac {- 16 a^{2} c^{2} - 16 a b^{2} c + 5 b^{4} + x^{2} \left (- 96 a c^{3} + 24 b^{2} c^{2}\right ) + x \left (- 96 a b c^{2} + 24 b^{3} c\right )}{96 b^{3} c^{3} d^{4} + 576 b^{2} c^{4} d^{4} x + 1152 b c^{5} d^{4} x^{2} + 768 c^{6} d^{4} x^{3}} + \frac {x}{16 c^{2} d^{4}} \end {gather*}

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)**4,x)

[Out]

(-16*a**2*c**2 - 16*a*b**2*c + 5*b**4 + x**2*(-96*a*c**3 + 24*b**2*c**2) + x*(-96*a*b*c**2 + 24*b**3*c))/(96*b
**3*c**3*d**4 + 576*b**2*c**4*d**4*x + 1152*b*c**5*d**4*x**2 + 768*c**6*d**4*x**3) + x/(16*c**2*d**4)

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