3.1667 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=106 \[ -\frac{(b d-a e) (-a B e-2 A b e+3 b B d)}{e^4 (d+e x)}+\frac{(b d-a e)^2 (B d-A e)}{2 e^4 (d+e x)^2}-\frac{b \log (d+e x) (-2 a B e-A b e+3 b B d)}{e^4}+\frac{b^2 B x}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.0915613, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {27, 77} \[ -\frac{(b d-a e) (-a B e-2 A b e+3 b B d)}{e^4 (d+e x)}+\frac{(b d-a e)^2 (B d-A e)}{2 e^4 (d+e x)^2}-\frac{b \log (d+e x) (-2 a B e-A b e+3 b B d)}{e^4}+\frac{b^2 B x}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0812665, size = 143, normalized size = 1.35 \[ -\frac{a^2 e^2 (A e+B (d+2 e x))+2 a b e (A e (d+2 e x)-B d (3 d+4 e x))+2 b (d+e x)^2 \log (d+e x) (-2 a B e-A b e+3 b B d)+b^2 \left (-\left (A d e (3 d+4 e x)+B \left (-4 d^2 e x-5 d^3+4 d e^2 x^2+2 e^3 x^3\right )\right )\right )}{2 e^4 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.007, size = 242, normalized size = 2.3 \begin{align*}{\frac{{b}^{2}Bx}{{e}^{3}}}-{\frac{A{a}^{2}}{2\,e \left ( ex+d \right ) ^{2}}}+{\frac{Adab}{{e}^{2} \left ( ex+d \right ) ^{2}}}-{\frac{A{d}^{2}{b}^{2}}{2\,{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{Bd{a}^{2}}{2\,{e}^{2} \left ( ex+d \right ) ^{2}}}-{\frac{B{d}^{2}ab}{{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{B{b}^{2}{d}^{3}}{2\,{e}^{4} \left ( ex+d \right ) ^{2}}}+{\frac{{b}^{2}\ln \left ( ex+d \right ) A}{{e}^{3}}}+2\,{\frac{b\ln \left ( ex+d \right ) aB}{{e}^{3}}}-3\,{\frac{{b}^{2}\ln \left ( ex+d \right ) Bd}{{e}^{4}}}-2\,{\frac{Aab}{{e}^{2} \left ( ex+d \right ) }}+2\,{\frac{Ad{b}^{2}}{{e}^{3} \left ( ex+d \right ) }}-{\frac{B{a}^{2}}{{e}^{2} \left ( ex+d \right ) }}+4\,{\frac{abBd}{{e}^{3} \left ( ex+d \right ) }}-3\,{\frac{B{b}^{2}{d}^{2}}{{e}^{4} \left ( ex+d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*B*x/e^3-1/2/e/(e*x+d)^2*A*a^2+1/e^2/(e*x+d)^2*A*d*a*b-1/2/e^3/(e*x+d)^2*A*d^2*b^2+1/2/e^2/(e*x+d)^2*B*d*a^
2-1/e^3/(e*x+d)^2*B*d^2*a*b+1/2/e^4/(e*x+d)^2*B*b^2*d^3+b^2/e^3*ln(e*x+d)*A+2*b/e^3*ln(e*x+d)*a*B-3*b^2/e^4*ln
(e*x+d)*B*d-2/e^2/(e*x+d)*A*a*b+2/e^3/(e*x+d)*A*b^2*d-1/e^2/(e*x+d)*a^2*B+4/e^3/(e*x+d)*B*a*b*d-3/e^4/(e*x+d)*
B*b^2*d^2

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Maxima [A]  time = 1.22933, size = 224, normalized size = 2.11 \begin{align*} \frac{B b^{2} x}{e^{3}} - \frac{5 \, B b^{2} d^{3} + A a^{2} e^{3} - 3 \,{\left (2 \, B a b + A b^{2}\right )} d^{2} e +{\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 2 \,{\left (3 \, B b^{2} d^{2} e - 2 \,{\left (2 \, B a b + A b^{2}\right )} d e^{2} +{\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{2 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} - \frac{{\left (3 \, B b^{2} d -{\left (2 \, B a b + A b^{2}\right )} e\right )} \log \left (e x + d\right )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b^2*x/e^3 - 1/2*(5*B*b^2*d^3 + A*a^2*e^3 - 3*(2*B*a*b + A*b^2)*d^2*e + (B*a^2 + 2*A*a*b)*d*e^2 + 2*(3*B*b^2*
d^2*e - 2*(2*B*a*b + A*b^2)*d*e^2 + (B*a^2 + 2*A*a*b)*e^3)*x)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) - (3*B*b^2*d - (
2*B*a*b + A*b^2)*e)*log(e*x + d)/e^4

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Fricas [B]  time = 1.73675, size = 513, normalized size = 4.84 \begin{align*} \frac{2 \, B b^{2} e^{3} x^{3} + 4 \, B b^{2} d e^{2} x^{2} - 5 \, B b^{2} d^{3} - A a^{2} e^{3} + 3 \,{\left (2 \, B a b + A b^{2}\right )} d^{2} e -{\left (B a^{2} + 2 \, A a b\right )} d e^{2} - 2 \,{\left (2 \, B b^{2} d^{2} e - 2 \,{\left (2 \, B a b + A b^{2}\right )} d e^{2} +{\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x - 2 \,{\left (3 \, B b^{2} d^{3} -{\left (2 \, B a b + A b^{2}\right )} d^{2} e +{\left (3 \, B b^{2} d e^{2} -{\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 2 \,{\left (3 \, B b^{2} d^{2} e -{\left (2 \, B a b + A b^{2}\right )} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*B*b^2*e^3*x^3 + 4*B*b^2*d*e^2*x^2 - 5*B*b^2*d^3 - A*a^2*e^3 + 3*(2*B*a*b + A*b^2)*d^2*e - (B*a^2 + 2*A*
a*b)*d*e^2 - 2*(2*B*b^2*d^2*e - 2*(2*B*a*b + A*b^2)*d*e^2 + (B*a^2 + 2*A*a*b)*e^3)*x - 2*(3*B*b^2*d^3 - (2*B*a
*b + A*b^2)*d^2*e + (3*B*b^2*d*e^2 - (2*B*a*b + A*b^2)*e^3)*x^2 + 2*(3*B*b^2*d^2*e - (2*B*a*b + A*b^2)*d*e^2)*
x)*log(e*x + d))/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)

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Sympy [A]  time = 3.23614, size = 187, normalized size = 1.76 \begin{align*} \frac{B b^{2} x}{e^{3}} + \frac{b \left (A b e + 2 B a e - 3 B b d\right ) \log{\left (d + e x \right )}}{e^{4}} - \frac{A a^{2} e^{3} + 2 A a b d e^{2} - 3 A b^{2} d^{2} e + B a^{2} d e^{2} - 6 B a b d^{2} e + 5 B b^{2} d^{3} + x \left (4 A a b e^{3} - 4 A b^{2} d e^{2} + 2 B a^{2} e^{3} - 8 B a b d e^{2} + 6 B b^{2} d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b**2*x/e**3 + b*(A*b*e + 2*B*a*e - 3*B*b*d)*log(d + e*x)/e**4 - (A*a**2*e**3 + 2*A*a*b*d*e**2 - 3*A*b**2*d**
2*e + B*a**2*d*e**2 - 6*B*a*b*d**2*e + 5*B*b**2*d**3 + x*(4*A*a*b*e**3 - 4*A*b**2*d*e**2 + 2*B*a**2*e**3 - 8*B
*a*b*d*e**2 + 6*B*b**2*d**2*e))/(2*d**2*e**4 + 4*d*e**5*x + 2*e**6*x**2)

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Giac [A]  time = 1.15101, size = 211, normalized size = 1.99 \begin{align*} B b^{2} x e^{\left (-3\right )} -{\left (3 \, B b^{2} d - 2 \, B a b e - A b^{2} e\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) - \frac{{\left (5 \, B b^{2} d^{3} - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e + B a^{2} d e^{2} + 2 \, A a b d e^{2} + A a^{2} e^{3} + 2 \,{\left (3 \, B b^{2} d^{2} e - 4 \, B a b d e^{2} - 2 \, A b^{2} d e^{2} + B a^{2} e^{3} + 2 \, A a b e^{3}\right )} x\right )} e^{\left (-4\right )}}{2 \,{\left (x e + d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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