3.1030 \(\int \frac{(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0809283, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {78, 43} \[ -\frac{(a+b x)^3 (B d-A e)}{3 e (d+e x)^3 (b d-a e)}+\frac{2 b B (b d-a e)}{e^4 (d+e x)}-\frac{B (b d-a e)^2}{2 e^4 (d+e x)^2}+\frac{b^2 B \log (d+e x)}{e^4} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*(A + B*x))/(d + e*x)^4,x]

[Out]

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

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

Mathematica [A]  time = 0.0678053, size = 138, normalized size = 1.37 \[ \frac{-a^2 e^2 (2 A e+B (d+3 e x))-2 a b e \left (A e (d+3 e x)+2 B \left (d^2+3 d e x+3 e^2 x^2\right )\right )+b^2 \left (B d \left (11 d^2+27 d e x+18 e^2 x^2\right )-2 A e \left (d^2+3 d e x+3 e^2 x^2\right )\right )+6 b^2 B (d+e x)^3 \log (d+e x)}{6 e^4 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^2*(A + B*x))/(d + e*x)^4,x]

[Out]

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

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Maple [B]  time = 0.006, size = 251, normalized size = 2.5 \begin{align*} -{\frac{A{a}^{2}}{3\,e \left ( ex+d \right ) ^{3}}}+{\frac{2\,Adab}{3\,{e}^{2} \left ( ex+d \right ) ^{3}}}-{\frac{A{d}^{2}{b}^{2}}{3\,{e}^{3} \left ( ex+d \right ) ^{3}}}+{\frac{Bd{a}^{2}}{3\,{e}^{2} \left ( ex+d \right ) ^{3}}}-{\frac{2\,B{d}^{2}ab}{3\,{e}^{3} \left ( ex+d \right ) ^{3}}}+{\frac{{b}^{2}B{d}^{3}}{3\,{e}^{4} \left ( ex+d \right ) ^{3}}}-{\frac{Aba}{{e}^{2} \left ( ex+d \right ) ^{2}}}+{\frac{Ad{b}^{2}}{{e}^{3} \left ( ex+d \right ) ^{2}}}-{\frac{B{a}^{2}}{2\,{e}^{2} \left ( ex+d \right ) ^{2}}}+2\,{\frac{Bdab}{{e}^{3} \left ( ex+d \right ) ^{2}}}-{\frac{3\,{b}^{2}B{d}^{2}}{2\,{e}^{4} \left ( ex+d \right ) ^{2}}}-{\frac{A{b}^{2}}{{e}^{3} \left ( ex+d \right ) }}-2\,{\frac{Bba}{{e}^{3} \left ( ex+d \right ) }}+3\,{\frac{{b}^{2}Bd}{{e}^{4} \left ( ex+d \right ) }}+{\frac{B{b}^{2}\ln \left ( ex+d \right ) }{{e}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*(B*x+A)/(e*x+d)^4,x)

[Out]

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

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Maxima [A]  time = 1.16426, size = 248, normalized size = 2.46 \begin{align*} \frac{11 \, B b^{2} d^{3} - 2 \, A a^{2} e^{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} + 6 \,{\left (3 \, B b^{2} d e^{2} -{\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 3 \,{\left (9 \, 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}{6 \,{\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} + \frac{B b^{2} \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)^2*(B*x+A)/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.80428, size = 460, normalized size = 4.55 \begin{align*} \frac{11 \, B b^{2} d^{3} - 2 \, A a^{2} e^{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} + 6 \,{\left (3 \, B b^{2} d e^{2} -{\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 3 \,{\left (9 \, 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 + 6 \,{\left (B b^{2} e^{3} x^{3} + 3 \, B b^{2} d e^{2} x^{2} + 3 \, B b^{2} d^{2} e x + B b^{2} d^{3}\right )} \log \left (e x + d\right )}{6 \,{\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*(B*x+A)/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 7.45954, size = 211, normalized size = 2.09 \begin{align*} \frac{B b^{2} \log{\left (d + e x \right )}}{e^{4}} - \frac{2 A a^{2} e^{3} + 2 A a b d e^{2} + 2 A b^{2} d^{2} e + B a^{2} d e^{2} + 4 B a b d^{2} e - 11 B b^{2} d^{3} + x^{2} \left (6 A b^{2} e^{3} + 12 B a b e^{3} - 18 B b^{2} d e^{2}\right ) + x \left (6 A a b e^{3} + 6 A b^{2} d e^{2} + 3 B a^{2} e^{3} + 12 B a b d e^{2} - 27 B b^{2} d^{2} e\right )}{6 d^{3} e^{4} + 18 d^{2} e^{5} x + 18 d e^{6} x^{2} + 6 e^{7} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*(B*x+A)/(e*x+d)**4,x)

[Out]

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

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Giac [A]  time = 2.51406, size = 220, normalized size = 2.18 \begin{align*} B b^{2} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{{\left (6 \,{\left (3 \, B b^{2} d e - 2 \, B a b e^{2} - A b^{2} e^{2}\right )} x^{2} + 3 \,{\left (9 \, 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 )} x +{\left (11 \, B b^{2} d^{3} - 4 \, B a b d^{2} e - 2 \, A b^{2} d^{2} e - B a^{2} d e^{2} - 2 \, A a b d e^{2} - 2 \, A a^{2} e^{3}\right )} e^{\left (-1\right )}\right )} e^{\left (-3\right )}}{6 \,{\left (x e + d\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*(B*x+A)/(e*x+d)^4,x, algorithm="giac")

[Out]

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