3.1472 \(\int \frac{(a^2+2 a b x+b^2 x^2)^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=103 \[ -\frac{b^3 x (3 b d-4 a e)}{e^4}+\frac{6 b^2 (b d-a e)^2 \log (d+e x)}{e^5}+\frac{4 b (b d-a e)^3}{e^5 (d+e x)}-\frac{(b d-a e)^4}{2 e^5 (d+e x)^2}+\frac{b^4 x^2}{2 e^3} \]

[Out]

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

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Rubi [A]  time = 0.0852951, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {27, 43} \[ -\frac{b^3 x (3 b d-4 a e)}{e^4}+\frac{6 b^2 (b d-a e)^2 \log (d+e x)}{e^5}+\frac{4 b (b d-a e)^3}{e^5 (d+e x)}-\frac{(b d-a e)^4}{2 e^5 (d+e x)^2}+\frac{b^4 x^2}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0562678, size = 167, normalized size = 1.62 \[ \frac{6 a^2 b^2 d e^2 (3 d+4 e x)-4 a^3 b e^3 (d+2 e x)-a^4 e^4+4 a b^3 e \left (-4 d^2 e x-5 d^3+4 d e^2 x^2+2 e^3 x^3\right )+12 b^2 (d+e x)^2 (b d-a e)^2 \log (d+e x)+b^4 \left (-11 d^2 e^2 x^2+2 d^3 e x+7 d^4-4 d e^3 x^3+e^4 x^4\right )}{2 e^5 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^4*e^4) - 4*a^3*b*e^3*(d + 2*e*x) + 6*a^2*b^2*d*e^2*(3*d + 4*e*x) + 4*a*b^3*e*(-5*d^3 - 4*d^2*e*x + 4*d*e^
2*x^2 + 2*e^3*x^3) + b^4*(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) + 12*b^2*(b*d - a*e)^2*(
d + e*x)^2*Log[d + e*x])/(2*e^5*(d + e*x)^2)

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Maple [B]  time = 0.049, size = 245, normalized size = 2.4 \begin{align*}{\frac{{b}^{4}{x}^{2}}{2\,{e}^{3}}}+4\,{\frac{a{b}^{3}x}{{e}^{3}}}-3\,{\frac{{b}^{4}xd}{{e}^{4}}}-{\frac{{a}^{4}}{2\,e \left ( ex+d \right ) ^{2}}}+2\,{\frac{d{a}^{3}b}{{e}^{2} \left ( ex+d \right ) ^{2}}}-3\,{\frac{{b}^{2}{d}^{2}{a}^{2}}{{e}^{3} \left ( ex+d \right ) ^{2}}}+2\,{\frac{{d}^{3}a{b}^{3}}{{e}^{4} \left ( ex+d \right ) ^{2}}}-{\frac{{b}^{4}{d}^{4}}{2\,{e}^{5} \left ( ex+d \right ) ^{2}}}+6\,{\frac{{b}^{2}\ln \left ( ex+d \right ){a}^{2}}{{e}^{3}}}-12\,{\frac{{b}^{3}\ln \left ( ex+d \right ) ad}{{e}^{4}}}+6\,{\frac{{b}^{4}\ln \left ( ex+d \right ){d}^{2}}{{e}^{5}}}-4\,{\frac{{a}^{3}b}{{e}^{2} \left ( ex+d \right ) }}+12\,{\frac{{b}^{2}{a}^{2}d}{{e}^{3} \left ( ex+d \right ) }}-12\,{\frac{{d}^{2}a{b}^{3}}{{e}^{4} \left ( ex+d \right ) }}+4\,{\frac{{b}^{4}{d}^{3}}{{e}^{5} \left ( ex+d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^4*x^2/e^3+4*b^3/e^3*a*x-3*b^4/e^4*x*d-1/2/e/(e*x+d)^2*a^4+2/e^2/(e*x+d)^2*d*a^3*b-3/e^3/(e*x+d)^2*d^2*b^
2*a^2+2/e^4/(e*x+d)^2*d^3*a*b^3-1/2/e^5/(e*x+d)^2*b^4*d^4+6*b^2/e^3*ln(e*x+d)*a^2-12*b^3/e^4*ln(e*x+d)*a*d+6*b
^4/e^5*ln(e*x+d)*d^2-4*b/e^2/(e*x+d)*a^3+12*b^2/e^3/(e*x+d)*a^2*d-12*b^3/e^4/(e*x+d)*a*d^2+4*b^4/e^5/(e*x+d)*d
^3

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Maxima [A]  time = 1.11434, size = 258, normalized size = 2.5 \begin{align*} \frac{7 \, b^{4} d^{4} - 20 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - a^{4} e^{4} + 8 \,{\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x}{2 \,{\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} + \frac{b^{4} e x^{2} - 2 \,{\left (3 \, b^{4} d - 4 \, a b^{3} e\right )} x}{2 \, e^{4}} + \frac{6 \,{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} 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((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^3,x, algorithm="maxima")

[Out]

1/2*(7*b^4*d^4 - 20*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 - a^4*e^4 + 8*(b^4*d^3*e - 3*a*b^3*d^2*e^
2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)/(e^7*x^2 + 2*d*e^6*x + d^2*e^5) + 1/2*(b^4*e*x^2 - 2*(3*b^4*d - 4*a*b^3*e)
*x)/e^4 + 6*(b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*log(e*x + d)/e^5

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Fricas [B]  time = 2.03556, size = 586, normalized size = 5.69 \begin{align*} \frac{b^{4} e^{4} x^{4} + 7 \, b^{4} d^{4} - 20 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - a^{4} e^{4} - 4 \,{\left (b^{4} d e^{3} - 2 \, a b^{3} e^{4}\right )} x^{3} -{\left (11 \, b^{4} d^{2} e^{2} - 16 \, a b^{3} d e^{3}\right )} x^{2} + 2 \,{\left (b^{4} d^{3} e - 8 \, a b^{3} d^{2} e^{2} + 12 \, a^{2} b^{2} d e^{3} - 4 \, a^{3} b e^{4}\right )} x + 12 \,{\left (b^{4} d^{4} - 2 \, a b^{3} d^{3} e + a^{2} b^{2} d^{2} e^{2} +{\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \,{\left (b^{4} d^{3} e - 2 \, a b^{3} d^{2} e^{2} + a^{2} b^{2} 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((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/2*(b^4*e^4*x^4 + 7*b^4*d^4 - 20*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 - a^4*e^4 - 4*(b^4*d*e^3 -
2*a*b^3*e^4)*x^3 - (11*b^4*d^2*e^2 - 16*a*b^3*d*e^3)*x^2 + 2*(b^4*d^3*e - 8*a*b^3*d^2*e^2 + 12*a^2*b^2*d*e^3 -
4*a^3*b*e^4)*x + 12*(b^4*d^4 - 2*a*b^3*d^3*e + a^2*b^2*d^2*e^2 + (b^4*d^2*e^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4)*
x^2 + 2*(b^4*d^3*e - 2*a*b^3*d^2*e^2 + a^2*b^2*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 = 1.70429, size = 184, normalized size = 1.79 \begin{align*} \frac{b^{4} x^{2}}{2 e^{3}} + \frac{6 b^{2} \left (a e - b d\right )^{2} \log{\left (d + e x \right )}}{e^{5}} - \frac{a^{4} e^{4} + 4 a^{3} b d e^{3} - 18 a^{2} b^{2} d^{2} e^{2} + 20 a b^{3} d^{3} e - 7 b^{4} d^{4} + x \left (8 a^{3} b e^{4} - 24 a^{2} b^{2} d e^{3} + 24 a b^{3} d^{2} e^{2} - 8 b^{4} d^{3} e\right )}{2 d^{2} e^{5} + 4 d e^{6} x + 2 e^{7} x^{2}} + \frac{x \left (4 a b^{3} e - 3 b^{4} d\right )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**4*x**2/(2*e**3) + 6*b**2*(a*e - b*d)**2*log(d + e*x)/e**5 - (a**4*e**4 + 4*a**3*b*d*e**3 - 18*a**2*b**2*d**
2*e**2 + 20*a*b**3*d**3*e - 7*b**4*d**4 + x*(8*a**3*b*e**4 - 24*a**2*b**2*d*e**3 + 24*a*b**3*d**2*e**2 - 8*b**
4*d**3*e))/(2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) + x*(4*a*b**3*e - 3*b**4*d)/e**4

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Giac [A]  time = 1.13883, size = 236, normalized size = 2.29 \begin{align*} 6 \,{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{2} \,{\left (b^{4} x^{2} e^{3} - 6 \, b^{4} d x e^{2} + 8 \, a b^{3} x e^{3}\right )} e^{\left (-6\right )} + \frac{{\left (7 \, b^{4} d^{4} - 20 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - a^{4} e^{4} + 8 \,{\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} 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((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^3,x, algorithm="giac")

[Out]

6*(b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*e^(-5)*log(abs(x*e + d)) + 1/2*(b^4*x^2*e^3 - 6*b^4*d*x*e^2 + 8*a*b^3*
x*e^3)*e^(-6) + 1/2*(7*b^4*d^4 - 20*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 - a^4*e^4 + 8*(b^4*d^3*e
- 3*a*b^3*d^2*e^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)*e^(-5)/(x*e + d)^2