\(\int \frac {x^2 (A+B x)}{a+b x} \, dx\) [176]

Optimal result

Integrand size = 16, antiderivative size = 66 \[ \int \frac {x^2 (A+B x)}{a+b x} \, dx=-\frac {a (A b-a B) x}{b^3}+\frac {(A b-a B) x^2}{2 b^2}+\frac {B x^3}{3 b}+\frac {a^2 (A b-a B) \log (a+b x)}{b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {x^2 (A+B x)}{a+b x} \, dx=\frac {a^2 (A b-a B) \log (a+b x)}{b^4}-\frac {a x (A b-a B)}{b^3}+\frac {x^2 (A b-a B)}{2 b^2}+\frac {B x^3}{3 b} \]

[In]

Int[(x^2*(A + B*x))/(a + b*x),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int \frac {x^2 (A+B x)}{a+b x} \, dx=\frac {b x \left (6 a^2 B-3 a b (2 A+B x)+b^2 x (3 A+2 B x)\right )+6 a^2 (A b-a B) \log (a+b x)}{6 b^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95

method	result	size
norman	\(-\frac {a \left (A b -B a \right ) x}{b^{3}}+\frac {\left (A b -B a \right ) x^{2}}{2 b^{2}}+\frac {B \,x^{3}}{3 b}+\frac {a^{2} \left (A b -B a \right ) \ln \left (b x +a \right )}{b^{4}}\)	\(63\)
default	\(-\frac {-\frac {1}{3} b^{2} B \,x^{3}-\frac {1}{2} A \,b^{2} x^{2}+\frac {1}{2} B a b \,x^{2}+a A b x -a^{2} B x}{b^{3}}+\frac {a^{2} \left (A b -B a \right ) \ln \left (b x +a \right )}{b^{4}}\)	\(67\)
risch	\(\frac {B \,x^{3}}{3 b}+\frac {A \,x^{2}}{2 b}-\frac {B a \,x^{2}}{2 b^{2}}-\frac {a A x}{b^{2}}+\frac {a^{2} B x}{b^{3}}+\frac {a^{2} \ln \left (b x +a \right ) A}{b^{3}}-\frac {a^{3} \ln \left (b x +a \right ) B}{b^{4}}\)	\(76\)
parallelrisch	\(\frac {2 b^{3} B \,x^{3}+3 A \,b^{3} x^{2}-3 B a \,b^{2} x^{2}+6 A \ln \left (b x +a \right ) a^{2} b -6 a \,b^{2} A x -6 B \ln \left (b x +a \right ) a^{3}+6 a^{2} b B x}{6 b^{4}}\)	\(76\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08 \[ \int \frac {x^2 (A+B x)}{a+b x} \, dx=\frac {2 \, B b^{3} x^{3} - 3 \, {\left (B a b^{2} - A b^{3}\right )} x^{2} + 6 \, {\left (B a^{2} b - A a b^{2}\right )} x - 6 \, {\left (B a^{3} - A a^{2} b\right )} \log \left (b x + a\right )}{6 \, b^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int \frac {x^2 (A+B x)}{a+b x} \, dx=\frac {B x^{3}}{3 b} - \frac {a^{2} \left (- A b + B a\right ) \log {\left (a + b x \right )}}{b^{4}} + x^{2} \left (\frac {A}{2 b} - \frac {B a}{2 b^{2}}\right ) + x \left (- \frac {A a}{b^{2}} + \frac {B a^{2}}{b^{3}}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06 \[ \int \frac {x^2 (A+B x)}{a+b x} \, dx=\frac {2 \, B b^{2} x^{3} - 3 \, {\left (B a b - A b^{2}\right )} x^{2} + 6 \, {\left (B a^{2} - A a b\right )} x}{6 \, b^{3}} - \frac {{\left (B a^{3} - A a^{2} b\right )} \log \left (b x + a\right )}{b^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08 \[ \int \frac {x^2 (A+B x)}{a+b x} \, dx=\frac {2 \, B b^{2} x^{3} - 3 \, B a b x^{2} + 3 \, A b^{2} x^{2} + 6 \, B a^{2} x - 6 \, A a b x}{6 \, b^{3}} - \frac {{\left (B a^{3} - A a^{2} b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.09 \[ \int \frac {x^2 (A+B x)}{a+b x} \, dx=x^2\,\left (\frac {A}{2\,b}-\frac {B\,a}{2\,b^2}\right )-\frac {\ln \left (a+b\,x\right )\,\left (B\,a^3-A\,a^2\,b\right )}{b^4}+\frac {B\,x^3}{3\,b}-\frac {a\,x\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{b} \]

[In]

int((x^2*(A + B*x))/(a + b*x),x)

[Out]

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