3.2126 \(\int \frac{(a+b x+c x^2)^2}{(d+e x)^4} \, dx\)

Optimal. Leaf size=139 \[ -\frac{-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{e^5 (d+e x)}+\frac{(2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^5 (d+e x)^2}-\frac{\left (a e^2-b d e+c d^2\right )^2}{3 e^5 (d+e x)^3}-\frac{2 c (2 c d-b e) \log (d+e x)}{e^5}+\frac{c^2 x}{e^4} \]

[Out]

(c^2*x)/e^4 - (c*d^2 - b*d*e + a*e^2)^2/(3*e^5*(d + e*x)^3) + ((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(e^5*(d
+ e*x)^2) - (6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))/(e^5*(d + e*x)) - (2*c*(2*c*d - b*e)*Log[d + e*x])/e^5

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Rubi [A]  time = 0.122614, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {698} \[ -\frac{-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{e^5 (d+e x)}+\frac{(2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^5 (d+e x)^2}-\frac{\left (a e^2-b d e+c d^2\right )^2}{3 e^5 (d+e x)^3}-\frac{2 c (2 c d-b e) \log (d+e x)}{e^5}+\frac{c^2 x}{e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x)/e^4 - (c*d^2 - b*d*e + a*e^2)^2/(3*e^5*(d + e*x)^3) + ((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(e^5*(d
+ e*x)^2) - (6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))/(e^5*(d + e*x)) - (2*c*(2*c*d - b*e)*Log[d + e*x])/e^5

Rule 698

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] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^2}{(d+e x)^4} \, dx &=\int \left (\frac{c^2}{e^4}+\frac{\left (c d^2-b d e+a e^2\right )^2}{e^4 (d+e x)^4}+\frac{2 (-2 c d+b e) \left (c d^2-b d e+a e^2\right )}{e^4 (d+e x)^3}+\frac{6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^4 (d+e x)^2}-\frac{2 c (2 c d-b e)}{e^4 (d+e x)}\right ) \, dx\\ &=\frac{c^2 x}{e^4}-\frac{\left (c d^2-b d e+a e^2\right )^2}{3 e^5 (d+e x)^3}+\frac{(2 c d-b e) \left (c d^2-b d e+a e^2\right )}{e^5 (d+e x)^2}-\frac{6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^5 (d+e x)}-\frac{2 c (2 c d-b e) \log (d+e x)}{e^5}\\ \end{align*}

Mathematica [A]  time = 0.0832742, size = 176, normalized size = 1.27 \[ \frac{-e^2 \left (a^2 e^2+a b e (d+3 e x)+b^2 \left (d^2+3 d e x+3 e^2 x^2\right )\right )+c e \left (b d \left (11 d^2+27 d e x+18 e^2 x^2\right )-2 a e \left (d^2+3 d e x+3 e^2 x^2\right )\right )-6 c (d+e x)^3 (2 c d-b e) \log (d+e x)+c^2 \left (-9 d^2 e^2 x^2-27 d^3 e x-13 d^4+9 d e^3 x^3+3 e^4 x^4\right )}{3 e^5 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(-13*d^4 - 27*d^3*e*x - 9*d^2*e^2*x^2 + 9*d*e^3*x^3 + 3*e^4*x^4) - e^2*(a^2*e^2 + a*b*e*(d + 3*e*x) + b^2
*(d^2 + 3*d*e*x + 3*e^2*x^2)) + c*e*(-2*a*e*(d^2 + 3*d*e*x + 3*e^2*x^2) + b*d*(11*d^2 + 27*d*e*x + 18*e^2*x^2)
) - 6*c*(2*c*d - b*e)*(d + e*x)^3*Log[d + e*x])/(3*e^5*(d + e*x)^3)

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Maple [B]  time = 0.046, size = 279, normalized size = 2. \begin{align*}{\frac{{c}^{2}x}{{e}^{4}}}-{\frac{{a}^{2}}{3\,e \left ( ex+d \right ) ^{3}}}+{\frac{2\,abd}{3\,{e}^{2} \left ( ex+d \right ) ^{3}}}-{\frac{2\,ac{d}^{2}}{3\,{e}^{3} \left ( ex+d \right ) ^{3}}}-{\frac{{b}^{2}{d}^{2}}{3\,{e}^{3} \left ( ex+d \right ) ^{3}}}+{\frac{2\,{d}^{3}bc}{3\,{e}^{4} \left ( ex+d \right ) ^{3}}}-{\frac{{c}^{2}{d}^{4}}{3\,{e}^{5} \left ( ex+d \right ) ^{3}}}-{\frac{ab}{{e}^{2} \left ( ex+d \right ) ^{2}}}+2\,{\frac{acd}{{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{{b}^{2}d}{{e}^{3} \left ( ex+d \right ) ^{2}}}-3\,{\frac{{d}^{2}bc}{{e}^{4} \left ( ex+d \right ) ^{2}}}+2\,{\frac{{c}^{2}{d}^{3}}{{e}^{5} \left ( ex+d \right ) ^{2}}}+2\,{\frac{c\ln \left ( ex+d \right ) b}{{e}^{4}}}-4\,{\frac{{c}^{2}d\ln \left ( ex+d \right ) }{{e}^{5}}}-2\,{\frac{ac}{{e}^{3} \left ( ex+d \right ) }}-{\frac{{b}^{2}}{{e}^{3} \left ( ex+d \right ) }}+6\,{\frac{bcd}{{e}^{4} \left ( ex+d \right ) }}-6\,{\frac{{c}^{2}{d}^{2}}{{e}^{5} \left ( ex+d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*x/e^4-1/3/e/(e*x+d)^3*a^2+2/3/e^2/(e*x+d)^3*d*a*b-2/3/e^3/(e*x+d)^3*a*c*d^2-1/3/e^3/(e*x+d)^3*d^2*b^2+2/3/
e^4/(e*x+d)^3*d^3*b*c-1/3/e^5/(e*x+d)^3*c^2*d^4-1/e^2/(e*x+d)^2*a*b+2/e^3/(e*x+d)^2*a*d*c+1/e^3/(e*x+d)^2*b^2*
d-3/e^4/(e*x+d)^2*d^2*b*c+2/e^5/(e*x+d)^2*c^2*d^3+2*c/e^4*ln(e*x+d)*b-4*c^2*d*ln(e*x+d)/e^5-2/e^3/(e*x+d)*a*c-
1/e^3/(e*x+d)*b^2+6/e^4/(e*x+d)*b*c*d-6/e^5/(e*x+d)*c^2*d^2

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Maxima [A]  time = 0.993387, size = 262, normalized size = 1.88 \begin{align*} -\frac{13 \, c^{2} d^{4} - 11 \, b c d^{3} e + a b d e^{3} + a^{2} e^{4} +{\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 3 \,{\left (6 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} +{\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 3 \,{\left (10 \, c^{2} d^{3} e - 9 \, b c d^{2} e^{2} + a b e^{4} +{\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x}{3 \,{\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} + \frac{c^{2} x}{e^{4}} - \frac{2 \,{\left (2 \, c^{2} d - b c e\right )} \log \left (e x + d\right )}{e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

-1/3*(13*c^2*d^4 - 11*b*c*d^3*e + a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2 + 3*(6*c^2*d^2*e^2 - 6*b*c*d*e^3
+ (b^2 + 2*a*c)*e^4)*x^2 + 3*(10*c^2*d^3*e - 9*b*c*d^2*e^2 + a*b*e^4 + (b^2 + 2*a*c)*d*e^3)*x)/(e^8*x^3 + 3*d
*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5) + c^2*x/e^4 - 2*(2*c^2*d - b*c*e)*log(e*x + d)/e^5

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Fricas [B]  time = 1.94339, size = 578, normalized size = 4.16 \begin{align*} \frac{3 \, c^{2} e^{4} x^{4} + 9 \, c^{2} d e^{3} x^{3} - 13 \, c^{2} d^{4} + 11 \, b c d^{3} e - a b d e^{3} - a^{2} e^{4} -{\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} - 3 \,{\left (3 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} +{\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} - 3 \,{\left (9 \, c^{2} d^{3} e - 9 \, b c d^{2} e^{2} + a b e^{4} +{\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x - 6 \,{\left (2 \, c^{2} d^{4} - b c d^{3} e +{\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x^{3} + 3 \,{\left (2 \, c^{2} d^{2} e^{2} - b c d e^{3}\right )} x^{2} + 3 \,{\left (2 \, c^{2} d^{3} e - b c d^{2} e^{2}\right )} x\right )} \log \left (e x + d\right )}{3 \,{\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/3*(3*c^2*e^4*x^4 + 9*c^2*d*e^3*x^3 - 13*c^2*d^4 + 11*b*c*d^3*e - a*b*d*e^3 - a^2*e^4 - (b^2 + 2*a*c)*d^2*e^2
- 3*(3*c^2*d^2*e^2 - 6*b*c*d*e^3 + (b^2 + 2*a*c)*e^4)*x^2 - 3*(9*c^2*d^3*e - 9*b*c*d^2*e^2 + a*b*e^4 + (b^2 +
2*a*c)*d*e^3)*x - 6*(2*c^2*d^4 - b*c*d^3*e + (2*c^2*d*e^3 - b*c*e^4)*x^3 + 3*(2*c^2*d^2*e^2 - b*c*d*e^3)*x^2
+ 3*(2*c^2*d^3*e - b*c*d^2*e^2)*x)*log(e*x + d))/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)

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Sympy [A]  time = 7.45517, size = 218, normalized size = 1.57 \begin{align*} \frac{c^{2} x}{e^{4}} + \frac{2 c \left (b e - 2 c d\right ) \log{\left (d + e x \right )}}{e^{5}} - \frac{a^{2} e^{4} + a b d e^{3} + 2 a c d^{2} e^{2} + b^{2} d^{2} e^{2} - 11 b c d^{3} e + 13 c^{2} d^{4} + x^{2} \left (6 a c e^{4} + 3 b^{2} e^{4} - 18 b c d e^{3} + 18 c^{2} d^{2} e^{2}\right ) + x \left (3 a b e^{4} + 6 a c d e^{3} + 3 b^{2} d e^{3} - 27 b c d^{2} e^{2} + 30 c^{2} d^{3} e\right )}{3 d^{3} e^{5} + 9 d^{2} e^{6} x + 9 d e^{7} x^{2} + 3 e^{8} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**2/(e*x+d)**4,x)

[Out]

c**2*x/e**4 + 2*c*(b*e - 2*c*d)*log(d + e*x)/e**5 - (a**2*e**4 + a*b*d*e**3 + 2*a*c*d**2*e**2 + b**2*d**2*e**2
- 11*b*c*d**3*e + 13*c**2*d**4 + x**2*(6*a*c*e**4 + 3*b**2*e**4 - 18*b*c*d*e**3 + 18*c**2*d**2*e**2) + x*(3*a
*b*e**4 + 6*a*c*d*e**3 + 3*b**2*d*e**3 - 27*b*c*d**2*e**2 + 30*c**2*d**3*e))/(3*d**3*e**5 + 9*d**2*e**6*x + 9*
d*e**7*x**2 + 3*e**8*x**3)

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Giac [A]  time = 1.10542, size = 230, normalized size = 1.65 \begin{align*} c^{2} x e^{\left (-4\right )} - 2 \,{\left (2 \, c^{2} d - b c e\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) - \frac{{\left (13 \, c^{2} d^{4} - 11 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} + a b d e^{3} + 3 \,{\left (6 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} + b^{2} e^{4} + 2 \, a c e^{4}\right )} x^{2} + a^{2} e^{4} + 3 \,{\left (10 \, c^{2} d^{3} e - 9 \, b c d^{2} e^{2} + b^{2} d e^{3} + 2 \, a c d e^{3} + a b e^{4}\right )} x\right )} e^{\left (-5\right )}}{3 \,{\left (x e + d\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^4,x, algorithm="giac")

[Out]

c^2*x*e^(-4) - 2*(2*c^2*d - b*c*e)*e^(-5)*log(abs(x*e + d)) - 1/3*(13*c^2*d^4 - 11*b*c*d^3*e + b^2*d^2*e^2 + 2
*a*c*d^2*e^2 + a*b*d*e^3 + 3*(6*c^2*d^2*e^2 - 6*b*c*d*e^3 + b^2*e^4 + 2*a*c*e^4)*x^2 + a^2*e^4 + 3*(10*c^2*d^3
*e - 9*b*c*d^2*e^2 + b^2*d*e^3 + 2*a*c*d*e^3 + a*b*e^4)*x)*e^(-5)/(x*e + d)^3