3.1695 \(\int \frac{(A+B x) (d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=99 \[ -\frac{(A b-a B) (b d-a e)^2}{b^4 (a+b x)}+\frac{e x (-2 a B e+A b e+2 b B d)}{b^3}+\frac{(b d-a e) \log (a+b x) (-3 a B e+2 A b e+b B d)}{b^4}+\frac{B e^2 x^2}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0980695, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {27, 77} \[ -\frac{(A b-a B) (b d-a e)^2}{b^4 (a+b x)}+\frac{e x (-2 a B e+A b e+2 b B d)}{b^3}+\frac{(b d-a e) \log (a+b x) (-3 a B e+2 A b e+b B d)}{b^4}+\frac{B e^2 x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0794667, size = 153, normalized size = 1.55 \[ \frac{-a^2 A b e^2-2 a^2 b B d e+a^3 B e^2+2 a A b^2 d e+a b^2 B d^2-A b^3 d^2}{b^4 (a+b x)}+\frac{\log (a+b x) \left (3 a^2 B e^2-2 a A b e^2-4 a b B d e+2 A b^2 d e+b^2 B d^2\right )}{b^4}+\frac{e x (-2 a B e+A b e+2 b B d)}{b^3}+\frac{B e^2 x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.009, size = 223, normalized size = 2.3 \begin{align*}{\frac{B{e}^{2}{x}^{2}}{2\,{b}^{2}}}+{\frac{A{e}^{2}x}{{b}^{2}}}-2\,{\frac{aB{e}^{2}x}{{b}^{3}}}+2\,{\frac{Bdex}{{b}^{2}}}-{\frac{A{a}^{2}{e}^{2}}{{b}^{3} \left ( bx+a \right ) }}+2\,{\frac{aAde}{{b}^{2} \left ( bx+a \right ) }}-{\frac{A{d}^{2}}{b \left ( bx+a \right ) }}+{\frac{B{a}^{3}{e}^{2}}{{b}^{4} \left ( bx+a \right ) }}-2\,{\frac{B{a}^{2}de}{{b}^{3} \left ( bx+a \right ) }}+{\frac{aB{d}^{2}}{{b}^{2} \left ( bx+a \right ) }}-2\,{\frac{\ln \left ( bx+a \right ) Aa{e}^{2}}{{b}^{3}}}+2\,{\frac{\ln \left ( bx+a \right ) Ade}{{b}^{2}}}+3\,{\frac{B\ln \left ( bx+a \right ){a}^{2}{e}^{2}}{{b}^{4}}}-4\,{\frac{B\ln \left ( bx+a \right ) dae}{{b}^{3}}}+{\frac{B\ln \left ( bx+a \right ){d}^{2}}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04048, size = 213, normalized size = 2.15 \begin{align*} \frac{{\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \,{\left (B a^{2} b - A a b^{2}\right )} d e +{\left (B a^{3} - A a^{2} b\right )} e^{2}}{b^{5} x + a b^{4}} + \frac{B b e^{2} x^{2} + 2 \,{\left (2 \, B b d e -{\left (2 \, B a - A b\right )} e^{2}\right )} x}{2 \, b^{3}} + \frac{{\left (B b^{2} d^{2} - 2 \,{\left (2 \, B a b - A b^{2}\right )} d e +{\left (3 \, B a^{2} - 2 \, A a b\right )} e^{2}\right )} \log \left (b x + a\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.53892, size = 510, normalized size = 5.15 \begin{align*} \frac{B b^{3} e^{2} x^{3} + 2 \,{\left (B a b^{2} - A b^{3}\right )} d^{2} - 4 \,{\left (B a^{2} b - A a b^{2}\right )} d e + 2 \,{\left (B a^{3} - A a^{2} b\right )} e^{2} +{\left (4 \, B b^{3} d e -{\left (3 \, B a b^{2} - 2 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \,{\left (2 \, B a b^{2} d e -{\left (2 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x + 2 \,{\left (B a b^{2} d^{2} - 2 \,{\left (2 \, B a^{2} b - A a b^{2}\right )} d e +{\left (3 \, B a^{3} - 2 \, A a^{2} b\right )} e^{2} +{\left (B b^{3} d^{2} - 2 \,{\left (2 \, B a b^{2} - A b^{3}\right )} d e +{\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} e^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \,{\left (b^{5} x + a b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.23034, size = 148, normalized size = 1.49 \begin{align*} \frac{B e^{2} x^{2}}{2 b^{2}} + \frac{- A a^{2} b e^{2} + 2 A a b^{2} d e - A b^{3} d^{2} + B a^{3} e^{2} - 2 B a^{2} b d e + B a b^{2} d^{2}}{a b^{4} + b^{5} x} - \frac{x \left (- A b e^{2} + 2 B a e^{2} - 2 B b d e\right )}{b^{3}} + \frac{\left (a e - b d\right ) \left (- 2 A b e + 3 B a e - B b d\right ) \log{\left (a + b x \right )}}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13735, size = 219, normalized size = 2.21 \begin{align*} \frac{{\left (B b^{2} d^{2} - 4 \, B a b d e + 2 \, A b^{2} d e + 3 \, B a^{2} e^{2} - 2 \, A a b e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac{B b^{2} x^{2} e^{2} + 4 \, B b^{2} d x e - 4 \, B a b x e^{2} + 2 \, A b^{2} x e^{2}}{2 \, b^{4}} + \frac{B a b^{2} d^{2} - A b^{3} d^{2} - 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + B a^{3} e^{2} - A a^{2} b e^{2}}{{\left (b x + a\right )} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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