\(\int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx\) [1028]

Optimal result

Integrand size = 20, antiderivative size = 101 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx=-\frac {b (2 b B d-A b e-2 a B e) x}{e^3}+\frac {b^2 B x^2}{2 e^2}+\frac {(b d-a e)^2 (B d-A e)}{e^4 (d+e x)}+\frac {(b d-a e) (3 b B d-2 A b e-a B e) \log (d+e x)}{e^4} \]

[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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx=\frac {(b d-a e)^2 (B d-A e)}{e^4 (d+e x)}+\frac {(b d-a e) \log (d+e x) (-a B e-2 A b e+3 b B d)}{e^4}-\frac {b x (-2 a B e-A b e+2 b B d)}{e^3}+\frac {b^2 B x^2}{2 e^2} \]

[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 78

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx=\frac {2 b e (-2 b B d+A b e+2 a B e) x+b^2 B e^2 x^2+\frac {2 (b d-a e)^2 (B d-A e)}{d+e x}+2 (b d-a e) (3 b B d-2 A b e-a B e) \log (d+e x)}{2 e^4} \]

[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)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {b \left (\frac {1}{2} B b e \,x^{2}+A b e x +2 B a e x -2 B b d x \right )}{e^{3}}-\frac {a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}}{e^{4} \left (e x +d \right )}+\frac {\left (2 A a b \,e^{2}-2 A \,b^{2} d e +B \,a^{2} e^{2}-4 B a b d e +3 b^{2} B \,d^{2}\right ) \ln \left (e x +d \right )}{e^{4}}\)	\(151\)
norman	\(\frac {\frac {\left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+2 A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+4 B a b \,d^{2} e -3 b^{2} B \,d^{3}\right ) x}{e^{3} d}+\frac {b \left (2 A b e +4 B a e -3 B b d \right ) x^{2}}{2 e^{2}}+\frac {b^{2} B \,x^{3}}{2 e}}{e x +d}+\frac {\left (2 A a b \,e^{2}-2 A \,b^{2} d e +B \,a^{2} e^{2}-4 B a b d e +3 b^{2} B \,d^{2}\right ) \ln \left (e x +d \right )}{e^{4}}\)	\(163\)
risch	\(\frac {b^{2} B \,x^{2}}{2 e^{2}}+\frac {b^{2} A x}{e^{2}}+\frac {2 b B a x}{e^{2}}-\frac {2 b^{2} B d x}{e^{3}}-\frac {a^{2} A}{e \left (e x +d \right )}+\frac {2 A a b d}{e^{2} \left (e x +d \right )}-\frac {A \,b^{2} d^{2}}{e^{3} \left (e x +d \right )}+\frac {B \,a^{2} d}{e^{2} \left (e x +d \right )}-\frac {2 B a b \,d^{2}}{e^{3} \left (e x +d \right )}+\frac {b^{2} B \,d^{3}}{e^{4} \left (e x +d \right )}+\frac {2 \ln \left (e x +d \right ) A a b}{e^{2}}-\frac {2 \ln \left (e x +d \right ) A \,b^{2} d}{e^{3}}+\frac {\ln \left (e x +d \right ) B \,a^{2}}{e^{2}}-\frac {4 \ln \left (e x +d \right ) B a b d}{e^{3}}+\frac {3 \ln \left (e x +d \right ) b^{2} B \,d^{2}}{e^{4}}\)	\(223\)
parallelrisch	\(\frac {b^{2} B \,x^{3} e^{3}+4 A \ln \left (e x +d \right ) x a b \,e^{3}-4 A \ln \left (e x +d \right ) x \,b^{2} d \,e^{2}+2 A \,x^{2} b^{2} e^{3}+2 B \ln \left (e x +d \right ) x \,a^{2} e^{3}-8 B \ln \left (e x +d \right ) x a b d \,e^{2}+6 B \ln \left (e x +d \right ) x \,b^{2} d^{2} e +4 B \,x^{2} a b \,e^{3}-3 B \,x^{2} b^{2} d \,e^{2}+4 A \ln \left (e x +d \right ) a b d \,e^{2}-4 A \ln \left (e x +d \right ) b^{2} d^{2} e +2 B \ln \left (e x +d \right ) a^{2} d \,e^{2}-8 B \ln \left (e x +d \right ) a b \,d^{2} e +6 B \ln \left (e x +d \right ) b^{2} d^{3}-2 a^{2} A \,e^{3}+4 A a b d \,e^{2}-4 A \,b^{2} d^{2} e +2 B \,a^{2} d \,e^{2}-8 B a b \,d^{2} e +6 b^{2} B \,d^{3}}{2 e^{4} \left (e x +d \right )}\)	\(275\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (99) = 198\).

Time = 0.23 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.36 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx=\frac {B b^{2} e^{3} x^{3} + 2 \, B b^{2} d^{3} - 2 \, A a^{2} e^{3} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 2 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} - {\left (3 \, B b^{2} d e^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} - 2 \, {\left (2 \, B b^{2} d^{2} e - {\left (2 \, B a b + A b^{2}\right )} d e^{2}\right )} x + 2 \, {\left (3 \, B b^{2} d^{3} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + {\left (3 \, B b^{2} d^{2} e - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x + d e^{4}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx=\frac {B b^{2} x^{2}}{2 e^{2}} + x \left (\frac {A b^{2}}{e^{2}} + \frac {2 B a b}{e^{2}} - \frac {2 B b^{2} d}{e^{3}}\right ) + \frac {- A a^{2} e^{3} + 2 A a b d e^{2} - A b^{2} d^{2} e + B a^{2} d e^{2} - 2 B a b d^{2} e + B b^{2} d^{3}}{d e^{4} + e^{5} x} + \frac {\left (a e - b d\right ) \left (2 A b e + B a e - 3 B b d\right ) \log {\left (d + e x \right )}}{e^{4}} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.54 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx=\frac {B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2}}{e^{5} x + d e^{4}} + \frac {B b^{2} e x^{2} - 2 \, {\left (2 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} x}{2 \, e^{3}} + \frac {{\left (3 \, B b^{2} d^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (99) = 198\).

Time = 0.29 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.28 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx=\frac {{\left (B b^{2} - \frac {2 \, {\left (3 \, B b^{2} d e - 2 \, B a b e^{2} - A b^{2} e^{2}\right )}}{{\left (e x + d\right )} e}\right )} {\left (e x + d\right )}^{2}}{2 \, e^{4}} - \frac {{\left (3 \, B b^{2} d^{2} - 4 \, B a b d e - 2 \, A b^{2} d e + B a^{2} e^{2} + 2 \, A a b e^{2}\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{4}} + \frac {\frac {B b^{2} d^{3} e^{2}}{e x + d} - \frac {2 \, B a b d^{2} e^{3}}{e x + d} - \frac {A b^{2} d^{2} e^{3}}{e x + d} + \frac {B a^{2} d e^{4}}{e x + d} + \frac {2 \, A a b d e^{4}}{e x + d} - \frac {A a^{2} e^{5}}{e x + d}}{e^{6}} \]

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

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^2} \, dx=x\,\left (\frac {A\,b^2+2\,B\,a\,b}{e^2}-\frac {2\,B\,b^2\,d}{e^3}\right )+\frac {\ln \left (d+e\,x\right )\,\left (B\,a^2\,e^2-4\,B\,a\,b\,d\,e+2\,A\,a\,b\,e^2+3\,B\,b^2\,d^2-2\,A\,b^2\,d\,e\right )}{e^4}-\frac {-B\,a^2\,d\,e^2+A\,a^2\,e^3+2\,B\,a\,b\,d^2\,e-2\,A\,a\,b\,d\,e^2-B\,b^2\,d^3+A\,b^2\,d^2\,e}{e\,\left (x\,e^4+d\,e^3\right )}+\frac {B\,b^2\,x^2}{2\,e^2} \]

[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)