3.1029 \(\int \frac {(a+b x)^2 (A+B x)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=106 \[ -\frac {(b d-a e) (-a B e-2 A b e+3 b B d)}{e^4 (d+e x)}+\frac {(b d-a e)^2 (B d-A e)}{2 e^4 (d+e x)^2}-\frac {b \log (d+e x) (-2 a B e-A b e+3 b B d)}{e^4}+\frac {b^2 B x}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \[ -\frac {(b d-a e) (-a B e-2 A b e+3 b B d)}{e^4 (d+e x)}+\frac {(b d-a e)^2 (B d-A e)}{2 e^4 (d+e x)^2}-\frac {b \log (d+e x) (-2 a B e-A b e+3 b B d)}{e^4}+\frac {b^2 B x}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 143, normalized size = 1.35 \[ -\frac {a^2 e^2 (A e+B (d+2 e x))+2 a b e (A e (d+2 e x)-B d (3 d+4 e x))+2 b (d+e x)^2 \log (d+e x) (-2 a B e-A b e+3 b B d)-\left (b^2 \left (A d e (3 d+4 e x)+B \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )\right )\right )}{2 e^4 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.96, size = 247, normalized size = 2.33 \[ \frac {2 \, B b^{2} e^{3} x^{3} + 4 \, B b^{2} d e^{2} x^{2} - 5 \, B b^{2} d^{3} - A a^{2} e^{3} + 3 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e - {\left (B a^{2} + 2 \, A a b\right )} d e^{2} - 2 \, {\left (2 \, B b^{2} d^{2} e - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x - 2 \, {\left (3 \, B b^{2} d^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (3 \, B b^{2} d e^{2} - {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} d^{2} e - {\left (2 \, B a b + A b^{2}\right )} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.17, size = 156, normalized size = 1.47 \[ B b^{2} x e^{\left (-3\right )} - {\left (3 \, B b^{2} d - 2 \, B a b e - A b^{2} e\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) - \frac {{\left (5 \, B b^{2} d^{3} - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e + B a^{2} d e^{2} + 2 \, A a b d e^{2} + A a^{2} e^{3} + 2 \, {\left (3 \, B b^{2} d^{2} e - 4 \, B a b d e^{2} - 2 \, A b^{2} d e^{2} + B a^{2} e^{3} + 2 \, A a b e^{3}\right )} x\right )} e^{\left (-4\right )}}{2 \, {\left (x e + d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 242, normalized size = 2.28 \[ -\frac {A \,a^{2}}{2 \left (e x +d \right )^{2} e}+\frac {A a b d}{\left (e x +d \right )^{2} e^{2}}-\frac {A \,b^{2} d^{2}}{2 \left (e x +d \right )^{2} e^{3}}+\frac {B \,a^{2} d}{2 \left (e x +d \right )^{2} e^{2}}-\frac {B a b \,d^{2}}{\left (e x +d \right )^{2} e^{3}}+\frac {B \,b^{2} d^{3}}{2 \left (e x +d \right )^{2} e^{4}}-\frac {2 A a b}{\left (e x +d \right ) e^{2}}+\frac {2 A \,b^{2} d}{\left (e x +d \right ) e^{3}}+\frac {A \,b^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {B \,a^{2}}{\left (e x +d \right ) e^{2}}+\frac {4 B a b d}{\left (e x +d \right ) e^{3}}+\frac {2 B a b \ln \left (e x +d \right )}{e^{3}}-\frac {3 B \,b^{2} d^{2}}{\left (e x +d \right ) e^{4}}-\frac {3 B \,b^{2} d \ln \left (e x +d \right )}{e^{4}}+\frac {B \,b^{2} x}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.58, size = 166, normalized size = 1.57 \[ \frac {B b^{2} x}{e^{3}} - \frac {5 \, B b^{2} d^{3} + A a^{2} e^{3} - 3 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 2 \, {\left (3 \, B b^{2} d^{2} e - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} - \frac {{\left (3 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} \log \left (e x + d\right )}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.13, size = 170, normalized size = 1.60 \[ \frac {\ln \left (d+e\,x\right )\,\left (A\,b^2\,e-3\,B\,b^2\,d+2\,B\,a\,b\,e\right )}{e^4}-\frac {x\,\left (B\,a^2\,e^2-4\,B\,a\,b\,d\,e+2\,A\,a\,b\,e^2+3\,B\,b^2\,d^2-2\,A\,b^2\,d\,e\right )+\frac {B\,a^2\,d\,e^2+A\,a^2\,e^3-6\,B\,a\,b\,d^2\,e+2\,A\,a\,b\,d\,e^2+5\,B\,b^2\,d^3-3\,A\,b^2\,d^2\,e}{2\,e}}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}+\frac {B\,b^2\,x}{e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.67, size = 187, normalized size = 1.76 \[ \frac {B b^{2} x}{e^{3}} + \frac {b \left (A b e + 2 B a e - 3 B b d\right ) \log {\left (d + e x \right )}}{e^{4}} + \frac {- A a^{2} e^{3} - 2 A a b d e^{2} + 3 A b^{2} d^{2} e - B a^{2} d e^{2} + 6 B a b d^{2} e - 5 B b^{2} d^{3} + x \left (- 4 A a b e^{3} + 4 A b^{2} d e^{2} - 2 B a^{2} e^{3} + 8 B a b d e^{2} - 6 B b^{2} d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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