3.1686 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^2}{(d+e x)^6} \, dx\)

Optimal. Leaf size=155 \[ -\frac{(a+b x)^5 (B d-A e)}{5 e (d+e x)^5 (b d-a e)}+\frac{4 b^3 B (b d-a e)}{e^6 (d+e x)}-\frac{3 b^2 B (b d-a e)^2}{e^6 (d+e x)^2}+\frac{4 b B (b d-a e)^3}{3 e^6 (d+e x)^3}-\frac{B (b d-a e)^4}{4 e^6 (d+e x)^4}+\frac{b^4 B \log (d+e x)}{e^6} \]

[Out]

-((B*d - A*e)*(a + b*x)^5)/(5*e*(b*d - a*e)*(d + e*x)^5) - (B*(b*d - a*e)^4)/(4*e^6*(d + e*x)^4) + (4*b*B*(b*d
- a*e)^3)/(3*e^6*(d + e*x)^3) - (3*b^2*B*(b*d - a*e)^2)/(e^6*(d + e*x)^2) + (4*b^3*B*(b*d - a*e))/(e^6*(d + e
*x)) + (b^4*B*Log[d + e*x])/e^6

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Rubi [A]  time = 0.154916, antiderivative size = 155, 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} \[ -\frac{(a+b x)^5 (B d-A e)}{5 e (d+e x)^5 (b d-a e)}+\frac{4 b^3 B (b d-a e)}{e^6 (d+e x)}-\frac{3 b^2 B (b d-a e)^2}{e^6 (d+e x)^2}+\frac{4 b B (b d-a e)^3}{3 e^6 (d+e x)^3}-\frac{B (b d-a e)^4}{4 e^6 (d+e x)^4}+\frac{b^4 B \log (d+e x)}{e^6} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/(d + e*x)^6,x]

[Out]

-((B*d - A*e)*(a + b*x)^5)/(5*e*(b*d - a*e)*(d + e*x)^5) - (B*(b*d - a*e)^4)/(4*e^6*(d + e*x)^4) + (4*b*B*(b*d
- a*e)^3)/(3*e^6*(d + e*x)^3) - (3*b^2*B*(b*d - a*e)^2)/(e^6*(d + e*x)^2) + (4*b^3*B*(b*d - a*e))/(e^6*(d + e
*x)) + (b^4*B*Log[d + e*x])/e^6

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) \left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^6} \, dx &=\int \frac{(a+b x)^4 (A+B x)}{(d+e x)^6} \, dx\\ &=-\frac{(B d-A e) (a+b x)^5}{5 e (b d-a e) (d+e x)^5}+\frac{B \int \frac{(a+b x)^4}{(d+e x)^5} \, dx}{e}\\ &=-\frac{(B d-A e) (a+b x)^5}{5 e (b d-a e) (d+e x)^5}+\frac{B \int \left (\frac{(-b d+a e)^4}{e^4 (d+e x)^5}-\frac{4 b (b d-a e)^3}{e^4 (d+e x)^4}+\frac{6 b^2 (b d-a e)^2}{e^4 (d+e x)^3}-\frac{4 b^3 (b d-a e)}{e^4 (d+e x)^2}+\frac{b^4}{e^4 (d+e x)}\right ) \, dx}{e}\\ &=-\frac{(B d-A e) (a+b x)^5}{5 e (b d-a e) (d+e x)^5}-\frac{B (b d-a e)^4}{4 e^6 (d+e x)^4}+\frac{4 b B (b d-a e)^3}{3 e^6 (d+e x)^3}-\frac{3 b^2 B (b d-a e)^2}{e^6 (d+e x)^2}+\frac{4 b^3 B (b d-a e)}{e^6 (d+e x)}+\frac{b^4 B \log (d+e x)}{e^6}\\ \end{align*}

Mathematica [B]  time = 0.160173, size = 332, normalized size = 2.14 \[ \frac{-6 a^2 b^2 e^2 \left (2 A e \left (d^2+5 d e x+10 e^2 x^2\right )+3 B \left (5 d^2 e x+d^3+10 d e^2 x^2+10 e^3 x^3\right )\right )-4 a^3 b e^3 \left (3 A e (d+5 e x)+2 B \left (d^2+5 d e x+10 e^2 x^2\right )\right )-3 a^4 e^4 (4 A e+B (d+5 e x))-12 a b^3 e \left (A e \left (5 d^2 e x+d^3+10 d e^2 x^2+10 e^3 x^3\right )+4 B \left (10 d^2 e^2 x^2+5 d^3 e x+d^4+10 d e^3 x^3+5 e^4 x^4\right )\right )+b^4 \left (B d \left (1100 d^2 e^2 x^2+625 d^3 e x+137 d^4+900 d e^3 x^3+300 e^4 x^4\right )-12 A e \left (10 d^2 e^2 x^2+5 d^3 e x+d^4+10 d e^3 x^3+5 e^4 x^4\right )\right )+60 b^4 B (d+e x)^5 \log (d+e x)}{60 e^6 (d+e x)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/(d + e*x)^6,x]

[Out]

(-3*a^4*e^4*(4*A*e + B*(d + 5*e*x)) - 4*a^3*b*e^3*(3*A*e*(d + 5*e*x) + 2*B*(d^2 + 5*d*e*x + 10*e^2*x^2)) - 6*a
^2*b^2*e^2*(2*A*e*(d^2 + 5*d*e*x + 10*e^2*x^2) + 3*B*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3)) - 12*a*b^3
*e*(A*e*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3) + 4*B*(d^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^3*x^3 +
5*e^4*x^4)) + b^4*(-12*A*e*(d^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^3*x^3 + 5*e^4*x^4) + B*d*(137*d^4 + 625
*d^3*e*x + 1100*d^2*e^2*x^2 + 900*d*e^3*x^3 + 300*e^4*x^4)) + 60*b^4*B*(d + e*x)^5*Log[d + e*x])/(60*e^6*(d +
e*x)^5)

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Maple [B]  time = 0.009, size = 651, normalized size = 4.2 \begin{align*} \text{result too large to display} \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)^2/(e*x+d)^6,x)

[Out]

8*b^3/e^5/(e*x+d)^2*B*d*a+3/e^3/(e*x+d)^4*A*d*a^2*b^2+2/e^3/(e*x+d)^4*B*d*a^3*b-9/2/e^4/(e*x+d)^4*B*d^2*a^2*b^
2-3/e^4/(e*x+d)^4*A*d^2*a*b^3+4/e^5/(e*x+d)^4*B*d^3*a*b^3+4/5/e^2/(e*x+d)^5*A*d*a^3*b-6/5/e^3/(e*x+d)^5*A*d^2*
a^2*b^2+4/5/e^4/(e*x+d)^5*A*d^3*a*b^3-4/5/e^3/(e*x+d)^5*B*d^2*a^3*b+6/5/e^4/(e*x+d)^5*B*d^3*a^2*b^2-4/5/e^5/(e
*x+d)^5*B*d^4*a*b^3+4*b^3/e^4/(e*x+d)^3*A*a*d+6*b^2/e^4/(e*x+d)^3*B*a^2*d-8*b^3/e^5/(e*x+d)^3*B*a*d^2+b^4*B*ln
(e*x+d)/e^6-1/e^5*b^4/(e*x+d)*A-1/5/e/(e*x+d)^5*A*a^4-1/4/e^2/(e*x+d)^4*B*a^4-2*b^3/e^4/(e*x+d)^2*A*a+2*b^4/e^
5/(e*x+d)^2*A*d-3*b^2/e^4/(e*x+d)^2*a^2*B-2*b^4/e^5/(e*x+d)^3*A*d^2-2*b^2/e^3/(e*x+d)^3*A*a^2+1/5/e^2/(e*x+d)^
5*B*d*a^4+1/5/e^6/(e*x+d)^5*B*b^4*d^5-4/e^5*b^3/(e*x+d)*a*B+5/e^6*b^4/(e*x+d)*B*d-5*b^4/e^6/(e*x+d)^2*B*d^2-1/
e^2/(e*x+d)^4*A*a^3*b+1/e^5/(e*x+d)^4*A*d^3*b^4-5/4/e^6/(e*x+d)^4*b^4*B*d^4-1/5/e^5/(e*x+d)^5*A*d^4*b^4+10/3*b
^4/e^6/(e*x+d)^3*B*d^3-4/3*b/e^3/(e*x+d)^3*B*a^3

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Maxima [B]  time = 1.12775, size = 620, normalized size = 4. \begin{align*} \frac{137 \, B b^{4} d^{5} - 12 \, A a^{4} e^{5} - 12 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{4} e - 6 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{3} e^{2} - 4 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d^{2} e^{3} - 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} d e^{4} + 60 \,{\left (5 \, B b^{4} d e^{4} -{\left (4 \, B a b^{3} + A b^{4}\right )} e^{5}\right )} x^{4} + 60 \,{\left (15 \, B b^{4} d^{2} e^{3} - 2 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d e^{4} -{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} e^{5}\right )} x^{3} + 20 \,{\left (55 \, B b^{4} d^{3} e^{2} - 6 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{2} e^{3} - 3 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d e^{4} - 2 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} e^{5}\right )} x^{2} + 5 \,{\left (125 \, B b^{4} d^{4} e - 12 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{3} e^{2} - 6 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{2} e^{3} - 4 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d e^{4} - 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} e^{5}\right )} x}{60 \,{\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} + \frac{B b^{4} \log \left (e x + d\right )}{e^{6}} \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)^2/(e*x+d)^6,x, algorithm="maxima")

[Out]

1/60*(137*B*b^4*d^5 - 12*A*a^4*e^5 - 12*(4*B*a*b^3 + A*b^4)*d^4*e - 6*(3*B*a^2*b^2 + 2*A*a*b^3)*d^3*e^2 - 4*(2
*B*a^3*b + 3*A*a^2*b^2)*d^2*e^3 - 3*(B*a^4 + 4*A*a^3*b)*d*e^4 + 60*(5*B*b^4*d*e^4 - (4*B*a*b^3 + A*b^4)*e^5)*x
^4 + 60*(15*B*b^4*d^2*e^3 - 2*(4*B*a*b^3 + A*b^4)*d*e^4 - (3*B*a^2*b^2 + 2*A*a*b^3)*e^5)*x^3 + 20*(55*B*b^4*d^
3*e^2 - 6*(4*B*a*b^3 + A*b^4)*d^2*e^3 - 3*(3*B*a^2*b^2 + 2*A*a*b^3)*d*e^4 - 2*(2*B*a^3*b + 3*A*a^2*b^2)*e^5)*x
^2 + 5*(125*B*b^4*d^4*e - 12*(4*B*a*b^3 + A*b^4)*d^3*e^2 - 6*(3*B*a^2*b^2 + 2*A*a*b^3)*d^2*e^3 - 4*(2*B*a^3*b
+ 3*A*a^2*b^2)*d*e^4 - 3*(B*a^4 + 4*A*a^3*b)*e^5)*x)/(e^11*x^5 + 5*d*e^10*x^4 + 10*d^2*e^9*x^3 + 10*d^3*e^8*x^
2 + 5*d^4*e^7*x + d^5*e^6) + B*b^4*log(e*x + d)/e^6

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Fricas [B]  time = 1.51732, size = 1104, normalized size = 7.12 \begin{align*} \frac{137 \, B b^{4} d^{5} - 12 \, A a^{4} e^{5} - 12 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{4} e - 6 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{3} e^{2} - 4 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d^{2} e^{3} - 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} d e^{4} + 60 \,{\left (5 \, B b^{4} d e^{4} -{\left (4 \, B a b^{3} + A b^{4}\right )} e^{5}\right )} x^{4} + 60 \,{\left (15 \, B b^{4} d^{2} e^{3} - 2 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d e^{4} -{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} e^{5}\right )} x^{3} + 20 \,{\left (55 \, B b^{4} d^{3} e^{2} - 6 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{2} e^{3} - 3 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d e^{4} - 2 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} e^{5}\right )} x^{2} + 5 \,{\left (125 \, B b^{4} d^{4} e - 12 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{3} e^{2} - 6 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{2} e^{3} - 4 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d e^{4} - 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} e^{5}\right )} x + 60 \,{\left (B b^{4} e^{5} x^{5} + 5 \, B b^{4} d e^{4} x^{4} + 10 \, B b^{4} d^{2} e^{3} x^{3} + 10 \, B b^{4} d^{3} e^{2} x^{2} + 5 \, B b^{4} d^{4} e x + B b^{4} d^{5}\right )} \log \left (e x + d\right )}{60 \,{\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{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)^2/(e*x+d)^6,x, algorithm="fricas")

[Out]

1/60*(137*B*b^4*d^5 - 12*A*a^4*e^5 - 12*(4*B*a*b^3 + A*b^4)*d^4*e - 6*(3*B*a^2*b^2 + 2*A*a*b^3)*d^3*e^2 - 4*(2
*B*a^3*b + 3*A*a^2*b^2)*d^2*e^3 - 3*(B*a^4 + 4*A*a^3*b)*d*e^4 + 60*(5*B*b^4*d*e^4 - (4*B*a*b^3 + A*b^4)*e^5)*x
^4 + 60*(15*B*b^4*d^2*e^3 - 2*(4*B*a*b^3 + A*b^4)*d*e^4 - (3*B*a^2*b^2 + 2*A*a*b^3)*e^5)*x^3 + 20*(55*B*b^4*d^
3*e^2 - 6*(4*B*a*b^3 + A*b^4)*d^2*e^3 - 3*(3*B*a^2*b^2 + 2*A*a*b^3)*d*e^4 - 2*(2*B*a^3*b + 3*A*a^2*b^2)*e^5)*x
^2 + 5*(125*B*b^4*d^4*e - 12*(4*B*a*b^3 + A*b^4)*d^3*e^2 - 6*(3*B*a^2*b^2 + 2*A*a*b^3)*d^2*e^3 - 4*(2*B*a^3*b
+ 3*A*a^2*b^2)*d*e^4 - 3*(B*a^4 + 4*A*a^3*b)*e^5)*x + 60*(B*b^4*e^5*x^5 + 5*B*b^4*d*e^4*x^4 + 10*B*b^4*d^2*e^3
*x^3 + 10*B*b^4*d^3*e^2*x^2 + 5*B*b^4*d^4*e*x + B*b^4*d^5)*log(e*x + d))/(e^11*x^5 + 5*d*e^10*x^4 + 10*d^2*e^9
*x^3 + 10*d^3*e^8*x^2 + 5*d^4*e^7*x + d^5*e^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**2/(e*x+d)**6,x)

[Out]

Timed out

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Giac [B]  time = 1.15009, size = 560, normalized size = 3.61 \begin{align*} B b^{4} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{{\left (60 \,{\left (5 \, B b^{4} d e^{3} - 4 \, B a b^{3} e^{4} - A b^{4} e^{4}\right )} x^{4} + 60 \,{\left (15 \, B b^{4} d^{2} e^{2} - 8 \, B a b^{3} d e^{3} - 2 \, A b^{4} d e^{3} - 3 \, B a^{2} b^{2} e^{4} - 2 \, A a b^{3} e^{4}\right )} x^{3} + 20 \,{\left (55 \, B b^{4} d^{3} e - 24 \, B a b^{3} d^{2} e^{2} - 6 \, A b^{4} d^{2} e^{2} - 9 \, B a^{2} b^{2} d e^{3} - 6 \, A a b^{3} d e^{3} - 4 \, B a^{3} b e^{4} - 6 \, A a^{2} b^{2} e^{4}\right )} x^{2} + 5 \,{\left (125 \, B b^{4} d^{4} - 48 \, B a b^{3} d^{3} e - 12 \, A b^{4} d^{3} e - 18 \, B a^{2} b^{2} d^{2} e^{2} - 12 \, A a b^{3} d^{2} e^{2} - 8 \, B a^{3} b d e^{3} - 12 \, A a^{2} b^{2} d e^{3} - 3 \, B a^{4} e^{4} - 12 \, A a^{3} b e^{4}\right )} x +{\left (137 \, B b^{4} d^{5} - 48 \, B a b^{3} d^{4} e - 12 \, A b^{4} d^{4} e - 18 \, B a^{2} b^{2} d^{3} e^{2} - 12 \, A a b^{3} d^{3} e^{2} - 8 \, B a^{3} b d^{2} e^{3} - 12 \, A a^{2} b^{2} d^{2} e^{3} - 3 \, B a^{4} d e^{4} - 12 \, A a^{3} b d e^{4} - 12 \, A a^{4} e^{5}\right )} e^{\left (-1\right )}\right )} e^{\left (-5\right )}}{60 \,{\left (x e + d\right )}^{5}} \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)^2/(e*x+d)^6,x, algorithm="giac")

[Out]

B*b^4*e^(-6)*log(abs(x*e + d)) + 1/60*(60*(5*B*b^4*d*e^3 - 4*B*a*b^3*e^4 - A*b^4*e^4)*x^4 + 60*(15*B*b^4*d^2*e
^2 - 8*B*a*b^3*d*e^3 - 2*A*b^4*d*e^3 - 3*B*a^2*b^2*e^4 - 2*A*a*b^3*e^4)*x^3 + 20*(55*B*b^4*d^3*e - 24*B*a*b^3*
d^2*e^2 - 6*A*b^4*d^2*e^2 - 9*B*a^2*b^2*d*e^3 - 6*A*a*b^3*d*e^3 - 4*B*a^3*b*e^4 - 6*A*a^2*b^2*e^4)*x^2 + 5*(12
5*B*b^4*d^4 - 48*B*a*b^3*d^3*e - 12*A*b^4*d^3*e - 18*B*a^2*b^2*d^2*e^2 - 12*A*a*b^3*d^2*e^2 - 8*B*a^3*b*d*e^3
- 12*A*a^2*b^2*d*e^3 - 3*B*a^4*e^4 - 12*A*a^3*b*e^4)*x + (137*B*b^4*d^5 - 48*B*a*b^3*d^4*e - 12*A*b^4*d^4*e -
18*B*a^2*b^2*d^3*e^2 - 12*A*a*b^3*d^3*e^2 - 8*B*a^3*b*d^2*e^3 - 12*A*a^2*b^2*d^2*e^3 - 3*B*a^4*d*e^4 - 12*A*a^
3*b*d*e^4 - 12*A*a^4*e^5)*e^(-1))*e^(-5)/(x*e + d)^5