3.2125 \(\int \frac{(a+b x+c x^2)^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=138 \[ \frac{\log (d+e x) \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^5}-\frac{\left (a e^2-b d e+c d^2\right )^2}{2 e^5 (d+e x)^2}+\frac{2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^5 (d+e x)}-\frac{c x (3 c d-2 b e)}{e^4}+\frac{c^2 x^2}{2 e^3} \]

[Out]

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

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Rubi [A]  time = 0.135546, antiderivative size = 138, 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} \[ \frac{\log (d+e x) \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^5}-\frac{\left (a e^2-b d e+c d^2\right )^2}{2 e^5 (d+e x)^2}+\frac{2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^5 (d+e x)}-\frac{c x (3 c d-2 b e)}{e^4}+\frac{c^2 x^2}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0741781, size = 176, normalized size = 1.28 \[ \frac{2 (d+e x)^2 \log (d+e x) \left (2 c e (a e-3 b d)+b^2 e^2+6 c^2 d^2\right )+2 c e \left (a d e (3 d+4 e x)+b \left (-4 d^2 e x-5 d^3+4 d e^2 x^2+2 e^3 x^3\right )\right )+e^2 (b d-a e) (a e+3 b d+4 b e x)+c^2 \left (-11 d^2 e^2 x^2+2 d^3 e x+7 d^4-4 d e^3 x^3+e^4 x^4\right )}{2 e^5 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*(b*d - a*e)*(3*b*d + a*e + 4*b*e*x) + c^2*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4) +
2*c*e*(a*d*e*(3*d + 4*e*x) + b*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + 2*(6*c^2*d^2 + b^2*e^2 + 2*c*
e*(-3*b*d + a*e))*(d + e*x)^2*Log[d + e*x])/(2*e^5*(d + e*x)^2)

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Maple [A]  time = 0.048, size = 266, normalized size = 1.9 \begin{align*}{\frac{{c}^{2}{x}^{2}}{2\,{e}^{3}}}+2\,{\frac{bcx}{{e}^{3}}}-3\,{\frac{{c}^{2}dx}{{e}^{4}}}-{\frac{{a}^{2}}{2\,e \left ( ex+d \right ) ^{2}}}+{\frac{abd}{{e}^{2} \left ( ex+d \right ) ^{2}}}-{\frac{ac{d}^{2}}{{e}^{3} \left ( ex+d \right ) ^{2}}}-{\frac{{b}^{2}{d}^{2}}{2\,{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{{d}^{3}bc}{{e}^{4} \left ( ex+d \right ) ^{2}}}-{\frac{{c}^{2}{d}^{4}}{2\,{e}^{5} \left ( ex+d \right ) ^{2}}}+2\,{\frac{\ln \left ( ex+d \right ) ac}{{e}^{3}}}+{\frac{{b}^{2}\ln \left ( ex+d \right ) }{{e}^{3}}}-6\,{\frac{\ln \left ( ex+d \right ) bcd}{{e}^{4}}}+6\,{\frac{\ln \left ( ex+d \right ){c}^{2}{d}^{2}}{{e}^{5}}}-2\,{\frac{ab}{{e}^{2} \left ( ex+d \right ) }}+4\,{\frac{acd}{{e}^{3} \left ( ex+d \right ) }}+2\,{\frac{{b}^{2}d}{{e}^{3} \left ( ex+d \right ) }}-6\,{\frac{{d}^{2}bc}{{e}^{4} \left ( ex+d \right ) }}+4\,{\frac{{c}^{2}{d}^{3}}{{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)^3,x)

[Out]

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

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Maxima [A]  time = 1.02102, size = 250, normalized size = 1.81 \begin{align*} \frac{7 \, c^{2} d^{4} - 10 \, b c d^{3} e - 2 \, a b d e^{3} - a^{2} e^{4} + 3 \,{\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 4 \,{\left (2 \, c^{2} d^{3} e - 3 \, b c d^{2} e^{2} - a b e^{4} +{\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x}{2 \,{\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} + \frac{c^{2} e x^{2} - 2 \,{\left (3 \, c^{2} d - 2 \, b c e\right )} x}{2 \, e^{4}} + \frac{{\left (6 \, c^{2} d^{2} - 6 \, b c d e +{\left (b^{2} + 2 \, a c\right )} e^{2}\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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.10841, size = 602, normalized size = 4.36 \begin{align*} \frac{c^{2} e^{4} x^{4} + 7 \, c^{2} d^{4} - 10 \, b c d^{3} e - 2 \, a b d e^{3} - a^{2} e^{4} + 3 \,{\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} - 4 \,{\left (c^{2} d e^{3} - b c e^{4}\right )} x^{3} -{\left (11 \, c^{2} d^{2} e^{2} - 8 \, b c d e^{3}\right )} x^{2} + 2 \,{\left (c^{2} d^{3} e - 4 \, b c d^{2} e^{2} - 2 \, a b e^{4} + 2 \,{\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x + 2 \,{\left (6 \, c^{2} d^{4} - 6 \, b c d^{3} e +{\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} +{\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} + 2 \,{\left (6 \, c^{2} d^{3} e - 6 \, b c d^{2} e^{2} +{\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} 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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.88416, size = 209, normalized size = 1.51 \begin{align*} \frac{c^{2} x^{2}}{2 e^{3}} - \frac{a^{2} e^{4} + 2 a b d e^{3} - 6 a c d^{2} e^{2} - 3 b^{2} d^{2} e^{2} + 10 b c d^{3} e - 7 c^{2} d^{4} + x \left (4 a b e^{4} - 8 a c d e^{3} - 4 b^{2} d e^{3} + 12 b c d^{2} e^{2} - 8 c^{2} d^{3} e\right )}{2 d^{2} e^{5} + 4 d e^{6} x + 2 e^{7} x^{2}} + \frac{x \left (2 b c e - 3 c^{2} d\right )}{e^{4}} + \frac{\left (2 a c e^{2} + b^{2} e^{2} - 6 b c d e + 6 c^{2} d^{2}\right ) \log{\left (d + e x \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)**3,x)

[Out]

c**2*x**2/(2*e**3) - (a**2*e**4 + 2*a*b*d*e**3 - 6*a*c*d**2*e**2 - 3*b**2*d**2*e**2 + 10*b*c*d**3*e - 7*c**2*d
**4 + x*(4*a*b*e**4 - 8*a*c*d*e**3 - 4*b**2*d*e**3 + 12*b*c*d**2*e**2 - 8*c**2*d**3*e))/(2*d**2*e**5 + 4*d*e**
6*x + 2*e**7*x**2) + x*(2*b*c*e - 3*c**2*d)/e**4 + (2*a*c*e**2 + b**2*e**2 - 6*b*c*d*e + 6*c**2*d**2)*log(d +
e*x)/e**5

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Giac [A]  time = 1.13492, size = 238, normalized size = 1.72 \begin{align*}{\left (6 \, c^{2} d^{2} - 6 \, b c d e + b^{2} e^{2} + 2 \, a c e^{2}\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{2} \,{\left (c^{2} x^{2} e^{3} - 6 \, c^{2} d x e^{2} + 4 \, b c x e^{3}\right )} e^{\left (-6\right )} + \frac{{\left (7 \, c^{2} d^{4} - 10 \, b c d^{3} e + 3 \, b^{2} d^{2} e^{2} + 6 \, a c d^{2} e^{2} - 2 \, a b d e^{3} - a^{2} e^{4} + 4 \,{\left (2 \, c^{2} d^{3} e - 3 \, 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 )}}{2 \,{\left (x e + d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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