3.1670 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)}{(d+e x)^6} \, dx\)

Optimal. Leaf size=120 \[ \frac {b (-2 a B e-A b e+3 b B d)}{3 e^4 (d+e x)^3}-\frac {(b d-a e) (-a B e-2 A b e+3 b B d)}{4 e^4 (d+e x)^4}+\frac {(b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^5}-\frac {b^2 B}{2 e^4 (d+e x)^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 77} \[ \frac {b (-2 a B e-A b e+3 b B d)}{3 e^4 (d+e x)^3}-\frac {(b d-a e) (-a B e-2 A b e+3 b B d)}{4 e^4 (d+e x)^4}+\frac {(b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^5}-\frac {b^2 B}{2 e^4 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.06, size = 129, normalized size = 1.08 \[ -\frac {3 a^2 e^2 (4 A e+B (d+5 e x))+2 a b e \left (3 A e (d+5 e x)+2 B \left (d^2+5 d e x+10 e^2 x^2\right )\right )+b^2 \left (2 A e \left (d^2+5 d e x+10 e^2 x^2\right )+3 B \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )\right )}{60 e^4 (d+e x)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(3*a^2*e^2*(4*A*e + B*(d + 5*e*x)) + 2*a*b*e*(3*A*e*(d + 5*e*x) + 2*B*(d^2 + 5*d*e*x + 10*e^2*x^2)) + b^
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)))/(e^4*(d + e*x)^5)

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fricas [A]  time = 0.68, size = 203, normalized size = 1.69 \[ -\frac {30 \, B b^{2} e^{3} x^{3} + 3 \, B b^{2} d^{3} + 12 \, A a^{2} e^{3} + 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 3 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 10 \, {\left (3 \, B b^{2} d e^{2} + 2 \, {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 5 \, {\left (3 \, B b^{2} d^{2} e + 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{60 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \]

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)/(e*x+d)^6,x, algorithm="fricas")

[Out]

-1/60*(30*B*b^2*e^3*x^3 + 3*B*b^2*d^3 + 12*A*a^2*e^3 + 2*(2*B*a*b + A*b^2)*d^2*e + 3*(B*a^2 + 2*A*a*b)*d*e^2 +
10*(3*B*b^2*d*e^2 + 2*(2*B*a*b + A*b^2)*e^3)*x^2 + 5*(3*B*b^2*d^2*e + 2*(2*B*a*b + A*b^2)*d*e^2 + 3*(B*a^2 +
2*A*a*b)*e^3)*x)/(e^9*x^5 + 5*d*e^8*x^4 + 10*d^2*e^7*x^3 + 10*d^3*e^6*x^2 + 5*d^4*e^5*x + d^5*e^4)

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giac [A]  time = 0.15, size = 160, normalized size = 1.33 \[ -\frac {{\left (30 \, B b^{2} x^{3} e^{3} + 30 \, B b^{2} d x^{2} e^{2} + 15 \, B b^{2} d^{2} x e + 3 \, B b^{2} d^{3} + 40 \, B a b x^{2} e^{3} + 20 \, A b^{2} x^{2} e^{3} + 20 \, B a b d x e^{2} + 10 \, A b^{2} d x e^{2} + 4 \, B a b d^{2} e + 2 \, A b^{2} d^{2} e + 15 \, B a^{2} x e^{3} + 30 \, A a b x e^{3} + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} + 12 \, A a^{2} e^{3}\right )} e^{\left (-4\right )}}{60 \, {\left (x e + d\right )}^{5}} \]

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)/(e*x+d)^6,x, algorithm="giac")

[Out]

-1/60*(30*B*b^2*x^3*e^3 + 30*B*b^2*d*x^2*e^2 + 15*B*b^2*d^2*x*e + 3*B*b^2*d^3 + 40*B*a*b*x^2*e^3 + 20*A*b^2*x^
2*e^3 + 20*B*a*b*d*x*e^2 + 10*A*b^2*d*x*e^2 + 4*B*a*b*d^2*e + 2*A*b^2*d^2*e + 15*B*a^2*x*e^3 + 30*A*a*b*x*e^3
+ 3*B*a^2*d*e^2 + 6*A*a*b*d*e^2 + 12*A*a^2*e^3)*e^(-4)/(x*e + d)^5

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maple [A]  time = 0.05, size = 166, normalized size = 1.38 \[ -\frac {B \,b^{2}}{2 \left (e x +d \right )^{2} e^{4}}-\frac {\left (A b e +2 a B e -3 B b d \right ) b}{3 \left (e x +d \right )^{3} e^{4}}-\frac {A \,a^{2} e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B d \,a^{2} e^{2}+2 B \,d^{2} a b e -b^{2} B \,d^{3}}{5 \left (e x +d \right )^{5} e^{4}}-\frac {2 A a b \,e^{2}-2 A \,b^{2} d e +B \,a^{2} e^{2}-4 B d a b e +3 b^{2} B \,d^{2}}{4 \left (e x +d \right )^{4} e^{4}} \]

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)/(e*x+d)^6,x)

[Out]

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

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maxima [A]  time = 0.61, size = 203, normalized size = 1.69 \[ -\frac {30 \, B b^{2} e^{3} x^{3} + 3 \, B b^{2} d^{3} + 12 \, A a^{2} e^{3} + 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 3 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 10 \, {\left (3 \, B b^{2} d e^{2} + 2 \, {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 5 \, {\left (3 \, B b^{2} d^{2} e + 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{60 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \]

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)/(e*x+d)^6,x, algorithm="maxima")

[Out]

-1/60*(30*B*b^2*e^3*x^3 + 3*B*b^2*d^3 + 12*A*a^2*e^3 + 2*(2*B*a*b + A*b^2)*d^2*e + 3*(B*a^2 + 2*A*a*b)*d*e^2 +
10*(3*B*b^2*d*e^2 + 2*(2*B*a*b + A*b^2)*e^3)*x^2 + 5*(3*B*b^2*d^2*e + 2*(2*B*a*b + A*b^2)*d*e^2 + 3*(B*a^2 +
2*A*a*b)*e^3)*x)/(e^9*x^5 + 5*d*e^8*x^4 + 10*d^2*e^7*x^3 + 10*d^3*e^6*x^2 + 5*d^4*e^5*x + d^5*e^4)

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mupad [B]  time = 2.23, size = 201, normalized size = 1.68 \[ -\frac {\frac {3\,B\,a^2\,d\,e^2+12\,A\,a^2\,e^3+4\,B\,a\,b\,d^2\,e+6\,A\,a\,b\,d\,e^2+3\,B\,b^2\,d^3+2\,A\,b^2\,d^2\,e}{60\,e^4}+\frac {x\,\left (3\,B\,a^2\,e^2+4\,B\,a\,b\,d\,e+6\,A\,a\,b\,e^2+3\,B\,b^2\,d^2+2\,A\,b^2\,d\,e\right )}{12\,e^3}+\frac {b\,x^2\,\left (2\,A\,b\,e+4\,B\,a\,e+3\,B\,b\,d\right )}{6\,e^2}+\frac {B\,b^2\,x^3}{2\,e}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((12*A*a^2*e^3 + 3*B*b^2*d^3 + 2*A*b^2*d^2*e + 3*B*a^2*d*e^2 + 6*A*a*b*d*e^2 + 4*B*a*b*d^2*e)/(60*e^4) + (x*(
3*B*a^2*e^2 + 3*B*b^2*d^2 + 6*A*a*b*e^2 + 2*A*b^2*d*e + 4*B*a*b*d*e))/(12*e^3) + (b*x^2*(2*A*b*e + 4*B*a*e + 3
*B*b*d))/(6*e^2) + (B*b^2*x^3)/(2*e))/(d^5 + e^5*x^5 + 5*d*e^4*x^4 + 10*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 5*d^4*e
*x)

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sympy [B]  time = 21.11, size = 238, normalized size = 1.98 \[ \frac {- 12 A a^{2} e^{3} - 6 A a b d e^{2} - 2 A b^{2} d^{2} e - 3 B a^{2} d e^{2} - 4 B a b d^{2} e - 3 B b^{2} d^{3} - 30 B b^{2} e^{3} x^{3} + x^{2} \left (- 20 A b^{2} e^{3} - 40 B a b e^{3} - 30 B b^{2} d e^{2}\right ) + x \left (- 30 A a b e^{3} - 10 A b^{2} d e^{2} - 15 B a^{2} e^{3} - 20 B a b d e^{2} - 15 B b^{2} d^{2} e\right )}{60 d^{5} e^{4} + 300 d^{4} e^{5} x + 600 d^{3} e^{6} x^{2} + 600 d^{2} e^{7} x^{3} + 300 d e^{8} x^{4} + 60 e^{9} x^{5}} \]

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)/(e*x+d)**6,x)

[Out]

(-12*A*a**2*e**3 - 6*A*a*b*d*e**2 - 2*A*b**2*d**2*e - 3*B*a**2*d*e**2 - 4*B*a*b*d**2*e - 3*B*b**2*d**3 - 30*B*
b**2*e**3*x**3 + x**2*(-20*A*b**2*e**3 - 40*B*a*b*e**3 - 30*B*b**2*d*e**2) + x*(-30*A*a*b*e**3 - 10*A*b**2*d*e
**2 - 15*B*a**2*e**3 - 20*B*a*b*d*e**2 - 15*B*b**2*d**2*e))/(60*d**5*e**4 + 300*d**4*e**5*x + 600*d**3*e**6*x*
*2 + 600*d**2*e**7*x**3 + 300*d*e**8*x**4 + 60*e**9*x**5)

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