\(\int \frac {(a+b x)^2 (A+B x)}{d+e x} \, dx\) [1027]

Optimal result

Integrand size = 20, antiderivative size = 92 \[ \int \frac {(a+b x)^2 (A+B x)}{d+e x} \, dx=\frac {b (b d-a e) (B d-A e) x}{e^3}-\frac {(B d-A e) (a+b x)^2}{2 e^2}+\frac {B (a+b x)^3}{3 b e}-\frac {(b d-a e)^2 (B d-A e) \log (d+e x)}{e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 92, 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} \[ \int \frac {(a+b x)^2 (A+B x)}{d+e x} \, dx=-\frac {(b d-a e)^2 (B d-A e) \log (d+e x)}{e^4}+\frac {b x (b d-a e) (B d-A e)}{e^3}-\frac {(a+b x)^2 (B d-A e)}{2 e^2}+\frac {B (a+b x)^3}{3 b e} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^2 (A+B x)}{d+e x} \, dx=\frac {e x \left (6 a^2 B e^2+6 a b e (-2 B d+2 A e+B e x)+b^2 \left (3 A e (-2 d+e x)+B \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )\right )-6 (b d-a e)^2 (B d-A e) \log (d+e x)}{6 e^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.61

method	result	size
norman	\(\frac {\left (2 A a b \,e^{2}-A \,b^{2} d e +B \,a^{2} e^{2}-2 B a b d e +b^{2} B \,d^{2}\right ) x}{e^{3}}+\frac {b \left (A b e +2 B a e -B b d \right ) x^{2}}{2 e^{2}}+\frac {b^{2} B \,x^{3}}{3 e}+\frac {\left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right ) \ln \left (e x +d \right )}{e^{4}}\)	\(148\)
default	\(\frac {\frac {1}{3} b^{2} B \,x^{3} e^{2}+\frac {1}{2} A \,b^{2} e^{2} x^{2}+B a b \,e^{2} x^{2}-\frac {1}{2} B \,b^{2} d e \,x^{2}+2 A a b \,e^{2} x -A \,b^{2} d e x +B \,a^{2} e^{2} x -2 B a b d e x +b^{2} B \,d^{2} x}{e^{3}}+\frac {\left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right ) \ln \left (e x +d \right )}{e^{4}}\)	\(161\)
risch	\(\frac {b^{2} B \,x^{3}}{3 e}+\frac {A \,b^{2} x^{2}}{2 e}+\frac {B a b \,x^{2}}{e}-\frac {B \,b^{2} d \,x^{2}}{2 e^{2}}+\frac {2 A a b x}{e}-\frac {A \,b^{2} d x}{e^{2}}+\frac {B \,a^{2} x}{e}-\frac {2 B a b d x}{e^{2}}+\frac {b^{2} B \,d^{2} x}{e^{3}}+\frac {\ln \left (e x +d \right ) a^{2} A}{e}-\frac {2 \ln \left (e x +d \right ) A a b d}{e^{2}}+\frac {\ln \left (e x +d \right ) A \,b^{2} d^{2}}{e^{3}}-\frac {\ln \left (e x +d \right ) B \,a^{2} d}{e^{2}}+\frac {2 \ln \left (e x +d \right ) B a b \,d^{2}}{e^{3}}-\frac {\ln \left (e x +d \right ) b^{2} B \,d^{3}}{e^{4}}\)	\(197\)
parallelrisch	\(\frac {2 b^{2} B \,x^{3} e^{3}+3 A \,x^{2} b^{2} e^{3}+6 B \,x^{2} a b \,e^{3}-3 B \,x^{2} b^{2} d \,e^{2}+6 A \ln \left (e x +d \right ) a^{2} e^{3}-12 A \ln \left (e x +d \right ) a b d \,e^{2}+6 A \ln \left (e x +d \right ) b^{2} d^{2} e +12 A x a b \,e^{3}-6 A x \,b^{2} d \,e^{2}-6 B \ln \left (e x +d \right ) a^{2} d \,e^{2}+12 B \ln \left (e x +d \right ) a b \,d^{2} e -6 B \ln \left (e x +d \right ) b^{2} d^{3}+6 B x \,a^{2} e^{3}-12 B x a b d \,e^{2}+6 B x \,b^{2} d^{2} e}{6 e^{4}}\)	\(198\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.66 \[ \int \frac {(a+b x)^2 (A+B x)}{d+e x} \, dx=\frac {2 \, B b^{2} e^{3} x^{3} - 3 \, {\left (B b^{2} d e^{2} - {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 6 \, {\left (B b^{2} d^{2} e - {\left (2 \, B a b + A b^{2}\right )} d e^{2} + {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x - 6 \, {\left (B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2}\right )} \log \left (e x + d\right )}{6 \, e^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x)^2 (A+B x)}{d+e x} \, dx=\frac {B b^{2} x^{3}}{3 e} + x^{2} \left (\frac {A b^{2}}{2 e} + \frac {B a b}{e} - \frac {B b^{2} d}{2 e^{2}}\right ) + x \left (\frac {2 A a b}{e} - \frac {A b^{2} d}{e^{2}} + \frac {B a^{2}}{e} - \frac {2 B a b d}{e^{2}} + \frac {B b^{2} d^{2}}{e^{3}}\right ) - \frac {\left (- A e + B d\right ) \left (a e - b d\right )^{2} \log {\left (d + e x \right )}}{e^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.65 \[ \int \frac {(a+b x)^2 (A+B x)}{d+e x} \, dx=\frac {2 \, B b^{2} e^{2} x^{3} - 3 \, {\left (B b^{2} d e - {\left (2 \, B a b + A b^{2}\right )} e^{2}\right )} x^{2} + 6 \, {\left (B b^{2} d^{2} - {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} x}{6 \, e^{3}} - \frac {{\left (B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.80 \[ \int \frac {(a+b x)^2 (A+B x)}{d+e x} \, dx=\frac {2 \, B b^{2} e^{2} x^{3} - 3 \, B b^{2} d e x^{2} + 6 \, B a b e^{2} x^{2} + 3 \, A b^{2} e^{2} x^{2} + 6 \, B b^{2} d^{2} x - 12 \, B a b d e x - 6 \, A b^{2} d e x + 6 \, B a^{2} e^{2} x + 12 \, A a b e^{2} x}{6 \, e^{3}} - \frac {{\left (B b^{2} d^{3} - 2 \, B a b d^{2} e - A b^{2} d^{2} e + B a^{2} d e^{2} + 2 \, A a b d e^{2} - A a^{2} e^{3}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} \]

[In]

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

[Out]

1/6*(2*B*b^2*e^2*x^3 - 3*B*b^2*d*e*x^2 + 6*B*a*b*e^2*x^2 + 3*A*b^2*e^2*x^2 + 6*B*b^2*d^2*x - 12*B*a*b*d*e*x -
6*A*b^2*d*e*x + 6*B*a^2*e^2*x + 12*A*a*b*e^2*x)/e^3 - (B*b^2*d^3 - 2*B*a*b*d^2*e - A*b^2*d^2*e + B*a^2*d*e^2 +
2*A*a*b*d*e^2 - A*a^2*e^3)*log(abs(e*x + d))/e^4

Mupad [B] (verification not implemented)

Time = 1.38 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.73 \[ \int \frac {(a+b x)^2 (A+B x)}{d+e x} \, dx=x\,\left (\frac {B\,a^2+2\,A\,b\,a}{e}-\frac {d\,\left (\frac {A\,b^2+2\,B\,a\,b}{e}-\frac {B\,b^2\,d}{e^2}\right )}{e}\right )+x^2\,\left (\frac {A\,b^2+2\,B\,a\,b}{2\,e}-\frac {B\,b^2\,d}{2\,e^2}\right )+\frac {\ln \left (d+e\,x\right )\,\left (-B\,a^2\,d\,e^2+A\,a^2\,e^3+2\,B\,a\,b\,d^2\,e-2\,A\,a\,b\,d\,e^2-B\,b^2\,d^3+A\,b^2\,d^2\,e\right )}{e^4}+\frac {B\,b^2\,x^3}{3\,e} \]

[In]

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

[Out]

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