3.1666 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=101 \[ \frac{(b d-a e)^2 (B d-A e)}{e^4 (d+e x)}-\frac{b x (-2 a B e-A b e+2 b B d)}{e^3}+\frac{(b d-a e) \log (d+e x) (-a B e-2 A b e+3 b B d)}{e^4}+\frac{b^2 B x^2}{2 e^2} \]

[Out]

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

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Rubi [A]  time = 0.102721, antiderivative size = 101, 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} \[ \frac{(b d-a e)^2 (B d-A e)}{e^4 (d+e x)}-\frac{b x (-2 a B e-A b e+2 b B d)}{e^3}+\frac{(b d-a e) \log (d+e x) (-a B e-2 A b e+3 b B d)}{e^4}+\frac{b^2 B x^2}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.087892, size = 98, normalized size = 0.97 \[ \frac{\frac{2 (b d-a e)^2 (B d-A e)}{d+e x}+2 b e x (2 a B e+A b e-2 b B d)+2 (b d-a e) \log (d+e x) (-a B e-2 A b e+3 b B d)+b^2 B e^2 x^2}{2 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.009, size = 223, normalized size = 2.2 \begin{align*}{\frac{{b}^{2}B{x}^{2}}{2\,{e}^{2}}}+{\frac{A{b}^{2}x}{{e}^{2}}}+2\,{\frac{abBx}{{e}^{2}}}-2\,{\frac{B{b}^{2}dx}{{e}^{3}}}+2\,{\frac{\ln \left ( ex+d \right ) Aab}{{e}^{2}}}-2\,{\frac{d\ln \left ( ex+d \right ) A{b}^{2}}{{e}^{3}}}+{\frac{\ln \left ( ex+d \right ) B{a}^{2}}{{e}^{2}}}-4\,{\frac{\ln \left ( ex+d \right ) Babd}{{e}^{3}}}+3\,{\frac{{d}^{2}\ln \left ( ex+d \right ) B{b}^{2}}{{e}^{4}}}-{\frac{A{a}^{2}}{e \left ( ex+d \right ) }}+2\,{\frac{Adab}{{e}^{2} \left ( ex+d \right ) }}-{\frac{A{d}^{2}{b}^{2}}{{e}^{3} \left ( ex+d \right ) }}+{\frac{Bd{a}^{2}}{{e}^{2} \left ( ex+d \right ) }}-2\,{\frac{abB{d}^{2}}{{e}^{3} \left ( ex+d \right ) }}+{\frac{B{b}^{2}{d}^{3}}{{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)^2,x)

[Out]

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

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Maxima [A]  time = 1.12651, size = 211, normalized size = 2.09 \begin{align*} \frac{B b^{2} d^{3} - A a^{2} e^{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}}{e^{5} x + d e^{4}} + \frac{B b^{2} e x^{2} - 2 \,{\left (2 \, B b^{2} d -{\left (2 \, B a b + A b^{2}\right )} e\right )} x}{2 \, e^{3}} + \frac{{\left (3 \, B b^{2} d^{2} - 2 \,{\left (2 \, B a b + A b^{2}\right )} d e +{\left (B a^{2} + 2 \, A a b\right )} e^{2}\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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.52781, size = 505, normalized size = 5. \begin{align*} \frac{B b^{2} e^{3} x^{3} + 2 \, B b^{2} d^{3} - 2 \, A a^{2} e^{3} - 2 \,{\left (2 \, B a b + A b^{2}\right )} d^{2} e + 2 \,{\left (B a^{2} + 2 \, A a b\right )} d e^{2} -{\left (3 \, B b^{2} d e^{2} - 2 \,{\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} - 2 \,{\left (2 \, B b^{2} d^{2} e -{\left (2 \, B a b + A b^{2}\right )} d e^{2}\right )} x + 2 \,{\left (3 \, B b^{2} d^{3} - 2 \,{\left (2 \, B a b + A b^{2}\right )} d^{2} e +{\left (B a^{2} + 2 \, A a b\right )} d e^{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\right )} \log \left (e x + d\right )}{2 \,{\left (e^{5} x + d 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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.29226, size = 148, normalized size = 1.47 \begin{align*} \frac{B b^{2} x^{2}}{2 e^{2}} + \frac{- A a^{2} e^{3} + 2 A a b d e^{2} - A b^{2} d^{2} e + B a^{2} d e^{2} - 2 B a b d^{2} e + B b^{2} d^{3}}{d e^{4} + e^{5} x} + \frac{x \left (A b^{2} e + 2 B a b e - 2 B b^{2} d\right )}{e^{3}} + \frac{\left (a e - b d\right ) \left (2 A b e + B a e - 3 B b d\right ) \log{\left (d + e x \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)**2,x)

[Out]

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

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Giac [B]  time = 1.14866, size = 306, normalized size = 3.03 \begin{align*} \frac{1}{2} \,{\left (B b^{2} - \frac{2 \,{\left (3 \, B b^{2} d e - 2 \, B a b e^{2} - A b^{2} e^{2}\right )} e^{\left (-1\right )}}{x e + d}\right )}{\left (x e + d\right )}^{2} e^{\left (-4\right )} -{\left (3 \, B b^{2} d^{2} - 4 \, B a b d e - 2 \, A b^{2} d e + B a^{2} e^{2} + 2 \, A a b e^{2}\right )} e^{\left (-4\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) +{\left (\frac{B b^{2} d^{3} e^{2}}{x e + d} - \frac{2 \, B a b d^{2} e^{3}}{x e + d} - \frac{A b^{2} d^{2} e^{3}}{x e + d} + \frac{B a^{2} d e^{4}}{x e + d} + \frac{2 \, A a b d e^{4}}{x e + d} - \frac{A a^{2} e^{5}}{x e + d}\right )} e^{\left (-6\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)^2,x, algorithm="giac")

[Out]

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