3.1136 \(\int \frac{(A+B x) (d+e x)^2}{(a+b x)^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0933971, antiderivative size = 103, 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 = {77} \[ -\frac{(b d-a e) (-3 a B e+2 A b e+b B d)}{b^4 (a+b x)}-\frac{(A b-a B) (b d-a e)^2}{2 b^4 (a+b x)^2}+\frac{e \log (a+b x) (-3 a B e+A b e+2 b B d)}{b^4}+\frac{B e^2 x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0747482, size = 140, normalized size = 1.36 \[ \frac{B \left (2 a^2 b e (3 d-2 e x)-5 a^3 e^2+a b^2 \left (-d^2+8 d e x+4 e^2 x^2\right )+2 b^3 x \left (e^2 x^2-d^2\right )\right )+2 e (a+b x)^2 \log (a+b x) (-3 a B e+A b e+2 b B d)-A b (b d-a e) (3 a e+b (d+4 e x))}{2 b^4 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b*(b*d - a*e)*(3*a*e + b*(d + 4*e*x))) + B*(-5*a^3*e^2 + 2*a^2*b*e*(3*d - 2*e*x) + 2*b^3*x*(-d^2 + e^2*x^
2) + a*b^2*(-d^2 + 8*d*e*x + 4*e^2*x^2)) + 2*e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^2*Log[a + b*x])/(2*b^4*(a
+ b*x)^2)

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Maple [B]  time = 0.008, size = 242, normalized size = 2.4 \begin{align*}{\frac{B{e}^{2}x}{{b}^{3}}}+{\frac{{e}^{2}\ln \left ( bx+a \right ) A}{{b}^{3}}}-3\,{\frac{{e}^{2}\ln \left ( bx+a \right ) Ba}{{b}^{4}}}+2\,{\frac{e\ln \left ( bx+a \right ) Bd}{{b}^{3}}}+2\,{\frac{aA{e}^{2}}{{b}^{3} \left ( bx+a \right ) }}-2\,{\frac{Ade}{{b}^{2} \left ( bx+a \right ) }}-3\,{\frac{B{a}^{2}{e}^{2}}{{b}^{4} \left ( bx+a \right ) }}+4\,{\frac{Bade}{{b}^{3} \left ( bx+a \right ) }}-{\frac{B{d}^{2}}{{b}^{2} \left ( bx+a \right ) }}-{\frac{{a}^{2}A{e}^{2}}{2\,{b}^{3} \left ( bx+a \right ) ^{2}}}+{\frac{aAde}{{b}^{2} \left ( bx+a \right ) ^{2}}}-{\frac{A{d}^{2}}{2\,b \left ( bx+a \right ) ^{2}}}+{\frac{B{a}^{3}{e}^{2}}{2\,{b}^{4} \left ( bx+a \right ) ^{2}}}-{\frac{B{a}^{2}de}{{b}^{3} \left ( bx+a \right ) ^{2}}}+{\frac{Ba{d}^{2}}{2\,{b}^{2} \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.21851, size = 230, normalized size = 2.23 \begin{align*} \frac{B e^{2} x}{b^{3}} - \frac{{\left (B a b^{2} + A b^{3}\right )} d^{2} - 2 \,{\left (3 \, B a^{2} b - A a b^{2}\right )} d e +{\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + 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}{2 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac{{\left (2 \, B b d e -{\left (3 \, B 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*x+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.53833, size = 521, normalized size = 5.06 \begin{align*} \frac{2 \, B b^{3} e^{2} x^{3} + 4 \, B a b^{2} e^{2} x^{2} -{\left (B a b^{2} + A b^{3}\right )} d^{2} + 2 \,{\left (3 \, B a^{2} b - A a b^{2}\right )} d e -{\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} - 2 \,{\left (B b^{3} d^{2} - 2 \,{\left (2 \, B a b^{2} - A b^{3}\right )} d e + 2 \,{\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x + 2 \,{\left (2 \, B a^{2} b d e -{\left (3 \, B a^{3} - A a^{2} b\right )} e^{2} +{\left (2 \, B b^{3} d e -{\left (3 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \,{\left (2 \, B a b^{2} d e -{\left (3 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} 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*x+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.85596, size = 187, normalized size = 1.82 \begin{align*} \frac{B e^{2} x}{b^{3}} - \frac{- 3 A a^{2} b e^{2} + 2 A a b^{2} d e + A b^{3} d^{2} + 5 B a^{3} e^{2} - 6 B a^{2} b d e + B a b^{2} d^{2} + x \left (- 4 A a b^{2} e^{2} + 4 A b^{3} d e + 6 B a^{2} b e^{2} - 8 B a b^{2} d e + 2 B b^{3} d^{2}\right )}{2 a^{2} b^{4} + 4 a b^{5} x + 2 b^{6} x^{2}} - \frac{e \left (- A b e + 3 B a e - 2 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*x+a)**3,x)

[Out]

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

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Giac [A]  time = 1.39483, size = 209, normalized size = 2.03 \begin{align*} \frac{B x e^{2}}{b^{3}} + \frac{{\left (2 \, B b d e - 3 \, B a e^{2} + A b e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac{B a b^{2} d^{2} + A b^{3} d^{2} - 6 \, B a^{2} b d e + 2 \, A a b^{2} d e + 5 \, B a^{3} e^{2} - 3 \, A a^{2} b e^{2} + 2 \,{\left (B b^{3} d^{2} - 4 \, B a b^{2} d e + 2 \, A b^{3} d e + 3 \, B a^{2} b e^{2} - 2 \, A a b^{2} e^{2}\right )} x}{2 \,{\left (b x + a\right )}^{2} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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