3.1028 \(\int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 101, 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} \[ \frac {(b d-a e)^2 (B d-A e)}{e^4 (d+e x)}-\frac {b x (-2 a B e-A b e+2 b B d)}{e^3}+\frac {(b d-a e) \log (d+e x) (-a B e-2 A b e+3 b B d)}{e^4}+\frac {b^2 B x^2}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.09, size = 98, normalized size = 0.97 \[ \frac {\frac {2 (b d-a e)^2 (B d-A e)}{d+e x}+2 b e x (2 a B e+A b e-2 b B d)+2 (b d-a e) \log (d+e x) (-a B e-2 A b e+3 b B d)+b^2 B e^2 x^2}{2 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.81, size = 238, normalized size = 2.36 \[ \frac {B b^{2} e^{3} x^{3} + 2 \, B b^{2} d^{3} - 2 \, A a^{2} e^{3} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 2 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} - {\left (3 \, B b^{2} d e^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} - 2 \, {\left (2 \, B b^{2} d^{2} e - {\left (2 \, B a b + A b^{2}\right )} d e^{2}\right )} x + 2 \, {\left (3 \, B b^{2} d^{3} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + {\left (3 \, B b^{2} d^{2} e - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x + d e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.22, size = 227, normalized size = 2.25 \[ \frac {1}{2} \, {\left (B b^{2} - \frac {2 \, {\left (3 \, B b^{2} d e - 2 \, B a b e^{2} - A b^{2} e^{2}\right )} e^{\left (-1\right )}}{x e + d}\right )} {\left (x e + d\right )}^{2} e^{\left (-4\right )} - {\left (3 \, B b^{2} d^{2} - 4 \, B a b d e - 2 \, A b^{2} d e + B a^{2} e^{2} + 2 \, A a b e^{2}\right )} e^{\left (-4\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + {\left (\frac {B b^{2} d^{3} e^{2}}{x e + d} - \frac {2 \, B a b d^{2} e^{3}}{x e + d} - \frac {A b^{2} d^{2} e^{3}}{x e + d} + \frac {B a^{2} d e^{4}}{x e + d} + \frac {2 \, A a b d e^{4}}{x e + d} - \frac {A a^{2} e^{5}}{x e + d}\right )} e^{\left (-6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 223, normalized size = 2.21 \[ \frac {B \,b^{2} x^{2}}{2 e^{2}}-\frac {A \,a^{2}}{\left (e x +d \right ) e}+\frac {2 A a b d}{\left (e x +d \right ) e^{2}}+\frac {2 A a b \ln \left (e x +d \right )}{e^{2}}-\frac {A \,b^{2} d^{2}}{\left (e x +d \right ) e^{3}}-\frac {2 A \,b^{2} d \ln \left (e x +d \right )}{e^{3}}+\frac {A \,b^{2} x}{e^{2}}+\frac {B \,a^{2} d}{\left (e x +d \right ) e^{2}}+\frac {B \,a^{2} \ln \left (e x +d \right )}{e^{2}}-\frac {2 B a b \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {4 B a b d \ln \left (e x +d \right )}{e^{3}}+\frac {2 B a b x}{e^{2}}+\frac {B \,b^{2} d^{3}}{\left (e x +d \right ) e^{4}}+\frac {3 B \,b^{2} d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {2 B \,b^{2} d x}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 156, normalized size = 1.54 \[ \frac {B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2}}{e^{5} x + d e^{4}} + \frac {B b^{2} e x^{2} - 2 \, {\left (2 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} x}{2 \, e^{3}} + \frac {{\left (3 \, B b^{2} d^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.09, size = 165, normalized size = 1.63 \[ x\,\left (\frac {A\,b^2+2\,B\,a\,b}{e^2}-\frac {2\,B\,b^2\,d}{e^3}\right )+\frac {\ln \left (d+e\,x\right )\,\left (B\,a^2\,e^2-4\,B\,a\,b\,d\,e+2\,A\,a\,b\,e^2+3\,B\,b^2\,d^2-2\,A\,b^2\,d\,e\right )}{e^4}-\frac {-B\,a^2\,d\,e^2+A\,a^2\,e^3+2\,B\,a\,b\,d^2\,e-2\,A\,a\,b\,d\,e^2-B\,b^2\,d^3+A\,b^2\,d^2\,e}{e\,\left (x\,e^4+d\,e^3\right )}+\frac {B\,b^2\,x^2}{2\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.98, size = 151, normalized size = 1.50 \[ \frac {B b^{2} x^{2}}{2 e^{2}} + x \left (\frac {A b^{2}}{e^{2}} + \frac {2 B a b}{e^{2}} - \frac {2 B b^{2} d}{e^{3}}\right ) + \frac {- A a^{2} e^{3} + 2 A a b d e^{2} - A b^{2} d^{2} e + B a^{2} d e^{2} - 2 B a b d^{2} e + B b^{2} d^{3}}{d e^{4} + e^{5} x} + \frac {\left (a e - b d\right ) \left (2 A b e + B a e - 3 B b d\right ) \log {\left (d + e x \right )}}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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