3.1703 \(\int \frac{(A+B x) (d+e x)^2}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0775451, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {27, 78, 43} \[ -\frac{(d+e x)^3 (A b-a B)}{3 b (a+b x)^3 (b d-a e)}-\frac{2 B e (b d-a e)}{b^4 (a+b x)}-\frac{B (b d-a e)^2}{2 b^4 (a+b x)^2}+\frac{B e^2 \log (a+b x)}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(B*(b*d - a*e)^2)/(2*b^4*(a + b*x)^2) - (2*B*e*(b*d - a*e))/(b^4*(a + b*x)) - ((A*b - a*B)*(d + e*x)^3)/(3*b*
(b*d - a*e)*(a + b*x)^3) + (B*e^2*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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

Mathematica [A]  time = 0.0744043, size = 138, normalized size = 1.37 \[ \frac{-2 A b \left (a^2 e^2+a b e (d+3 e x)+b^2 \left (d^2+3 d e x+3 e^2 x^2\right )\right )+B \left (a^2 b e (27 e x-4 d)+11 a^3 e^2-a b^2 \left (d^2+12 d e x-18 e^2 x^2\right )-3 b^3 d x (d+4 e x)\right )+6 B e^2 (a+b x)^3 \log (a+b x)}{6 b^4 (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(11*a^3*e^2 - 3*b^3*d*x*(d + 4*e*x) + a^2*b*e*(-4*d + 27*e*x) - a*b^2*(d^2 + 12*d*e*x - 18*e^2*x^2)) - 2*A*
b*(a^2*e^2 + a*b*e*(d + 3*e*x) + b^2*(d^2 + 3*d*e*x + 3*e^2*x^2)) + 6*B*e^2*(a + b*x)^3*Log[a + b*x])/(6*b^4*(
a + b*x)^3)

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Maple [B]  time = 0.008, size = 251, normalized size = 2.5 \begin{align*} -{\frac{A{e}^{2}}{{b}^{3} \left ( bx+a \right ) }}+3\,{\frac{aB{e}^{2}}{{b}^{4} \left ( bx+a \right ) }}-2\,{\frac{Bde}{{b}^{3} \left ( bx+a \right ) }}-{\frac{A{a}^{2}{e}^{2}}{3\,{b}^{3} \left ( bx+a \right ) ^{3}}}+{\frac{2\,Aade}{3\,{b}^{2} \left ( bx+a \right ) ^{3}}}-{\frac{A{d}^{2}}{3\,b \left ( bx+a \right ) ^{3}}}+{\frac{B{a}^{3}{e}^{2}}{3\,{b}^{4} \left ( bx+a \right ) ^{3}}}-{\frac{2\,B{a}^{2}de}{3\,{b}^{3} \left ( bx+a \right ) ^{3}}}+{\frac{Ba{d}^{2}}{3\,{b}^{2} \left ( bx+a \right ) ^{3}}}+{\frac{aA{e}^{2}}{{b}^{3} \left ( bx+a \right ) ^{2}}}-{\frac{Ade}{{b}^{2} \left ( bx+a \right ) ^{2}}}-{\frac{3\,{a}^{2}B{e}^{2}}{2\,{b}^{4} \left ( bx+a \right ) ^{2}}}+2\,{\frac{aBde}{{b}^{3} \left ( bx+a \right ) ^{2}}}-{\frac{B{d}^{2}}{2\,{b}^{2} \left ( bx+a \right ) ^{2}}}+{\frac{B{e}^{2}\ln \left ( bx+a \right ) }{{b}^{4}}} \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.06898, size = 255, normalized size = 2.52 \begin{align*} -\frac{{\left (B a b^{2} + 2 \, A b^{3}\right )} d^{2} + 2 \,{\left (2 \, B a^{2} b + A a b^{2}\right )} d e -{\left (11 \, B a^{3} - 2 \, A a^{2} b\right )} e^{2} + 6 \,{\left (2 \, B b^{3} d e -{\left (3 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 3 \,{\left (B b^{3} d^{2} + 2 \,{\left (2 \, B a b^{2} + A b^{3}\right )} d e -{\left (9 \, B a^{2} b - 2 \, A a b^{2}\right )} e^{2}\right )} x}{6 \,{\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} + \frac{B e^{2} \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)^2,x, algorithm="maxima")

[Out]

-1/6*((B*a*b^2 + 2*A*b^3)*d^2 + 2*(2*B*a^2*b + A*a*b^2)*d*e - (11*B*a^3 - 2*A*a^2*b)*e^2 + 6*(2*B*b^3*d*e - (3
*B*a*b^2 - A*b^3)*e^2)*x^2 + 3*(B*b^3*d^2 + 2*(2*B*a*b^2 + A*b^3)*d*e - (9*B*a^2*b - 2*A*a*b^2)*e^2)*x)/(b^7*x
^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4) + B*e^2*log(b*x + a)/b^4

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Fricas [B]  time = 1.52689, size = 467, normalized size = 4.62 \begin{align*} -\frac{{\left (B a b^{2} + 2 \, A b^{3}\right )} d^{2} + 2 \,{\left (2 \, B a^{2} b + A a b^{2}\right )} d e -{\left (11 \, B a^{3} - 2 \, A a^{2} b\right )} e^{2} + 6 \,{\left (2 \, B b^{3} d e -{\left (3 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 3 \,{\left (B b^{3} d^{2} + 2 \,{\left (2 \, B a b^{2} + A b^{3}\right )} d e -{\left (9 \, B a^{2} b - 2 \, A a b^{2}\right )} e^{2}\right )} x - 6 \,{\left (B b^{3} e^{2} x^{3} + 3 \, B a b^{2} e^{2} x^{2} + 3 \, B a^{2} b e^{2} x + B a^{3} e^{2}\right )} \log \left (b x + a\right )}{6 \,{\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} 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)^2,x, algorithm="fricas")

[Out]

-1/6*((B*a*b^2 + 2*A*b^3)*d^2 + 2*(2*B*a^2*b + A*a*b^2)*d*e - (11*B*a^3 - 2*A*a^2*b)*e^2 + 6*(2*B*b^3*d*e - (3
*B*a*b^2 - A*b^3)*e^2)*x^2 + 3*(B*b^3*d^2 + 2*(2*B*a*b^2 + A*b^3)*d*e - (9*B*a^2*b - 2*A*a*b^2)*e^2)*x - 6*(B*
b^3*e^2*x^3 + 3*B*a*b^2*e^2*x^2 + 3*B*a^2*b*e^2*x + B*a^3*e^2)*log(b*x + a))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^
5*x + a^3*b^4)

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Sympy [B]  time = 7.15048, size = 211, normalized size = 2.09 \begin{align*} \frac{B e^{2} \log{\left (a + b x \right )}}{b^{4}} + \frac{- 2 A a^{2} b e^{2} - 2 A a b^{2} d e - 2 A b^{3} d^{2} + 11 B a^{3} e^{2} - 4 B a^{2} b d e - B a b^{2} d^{2} + x^{2} \left (- 6 A b^{3} e^{2} + 18 B a b^{2} e^{2} - 12 B b^{3} d e\right ) + x \left (- 6 A a b^{2} e^{2} - 6 A b^{3} d e + 27 B a^{2} b e^{2} - 12 B a b^{2} d e - 3 B b^{3} d^{2}\right )}{6 a^{3} b^{4} + 18 a^{2} b^{5} x + 18 a b^{6} x^{2} + 6 b^{7} x^{3}} \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)**2,x)

[Out]

B*e**2*log(a + b*x)/b**4 + (-2*A*a**2*b*e**2 - 2*A*a*b**2*d*e - 2*A*b**3*d**2 + 11*B*a**3*e**2 - 4*B*a**2*b*d*
e - B*a*b**2*d**2 + x**2*(-6*A*b**3*e**2 + 18*B*a*b**2*e**2 - 12*B*b**3*d*e) + x*(-6*A*a*b**2*e**2 - 6*A*b**3*
d*e + 27*B*a**2*b*e**2 - 12*B*a*b**2*d*e - 3*B*b**3*d**2))/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*
b**7*x**3)

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Giac [A]  time = 1.16575, size = 217, normalized size = 2.15 \begin{align*} \frac{B e^{2} \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac{6 \,{\left (2 \, B b^{2} d e - 3 \, B a b e^{2} + A b^{2} e^{2}\right )} x^{2} + 3 \,{\left (B b^{2} d^{2} + 4 \, B a b d e + 2 \, A b^{2} d e - 9 \, B a^{2} e^{2} + 2 \, A a b e^{2}\right )} x + \frac{B a b^{2} d^{2} + 2 \, A b^{3} d^{2} + 4 \, B a^{2} b d e + 2 \, A a b^{2} d e - 11 \, B a^{3} e^{2} + 2 \, A a^{2} b e^{2}}{b}}{6 \,{\left (b x + a\right )}^{3} b^{3}} \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)^2,x, algorithm="giac")

[Out]

B*e^2*log(abs(b*x + a))/b^4 - 1/6*(6*(2*B*b^2*d*e - 3*B*a*b*e^2 + A*b^2*e^2)*x^2 + 3*(B*b^2*d^2 + 4*B*a*b*d*e
+ 2*A*b^2*d*e - 9*B*a^2*e^2 + 2*A*a*b*e^2)*x + (B*a*b^2*d^2 + 2*A*b^3*d^2 + 4*B*a^2*b*d*e + 2*A*a*b^2*d*e - 11
*B*a^3*e^2 + 2*A*a^2*b*e^2)/b)/((b*x + a)^3*b^3)