\(\int \frac {x^2 (A+B x)}{(a+b x)^3} \, dx\) [195]

Optimal result

Integrand size = 16, antiderivative size = 71 \[ \int \frac {x^2 (A+B x)}{(a+b x)^3} \, dx=\frac {B x}{b^3}-\frac {a^2 (A b-a B)}{2 b^4 (a+b x)^2}+\frac {a (2 A b-3 a B)}{b^4 (a+b x)}+\frac {(A b-3 a B) \log (a+b x)}{b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, 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} \[ \int \frac {x^2 (A+B x)}{(a+b x)^3} \, dx=-\frac {a^2 (A b-a B)}{2 b^4 (a+b x)^2}+\frac {a (2 A b-3 a B)}{b^4 (a+b x)}+\frac {(A b-3 a B) \log (a+b x)}{b^4}+\frac {B x}{b^3} \]

[In]

Int[(x^2*(A + B*x))/(a + b*x)^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06 \[ \int \frac {x^2 (A+B x)}{(a+b x)^3} \, dx=\frac {B x}{b^3}+\frac {-a^2 A b+a^3 B}{2 b^4 (a+b x)^2}+\frac {2 a A b-3 a^2 B}{b^4 (a+b x)}+\frac {(A b-3 a B) \log (a+b x)}{b^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97

method	result	size
norman	\(\frac {\frac {B \,x^{3}}{b}+\frac {a^{2} \left (3 A b -9 B a \right )}{2 b^{4}}+\frac {2 a \left (A b -3 B a \right ) x}{b^{3}}}{\left (b x +a \right )^{2}}+\frac {\left (A b -3 B a \right ) \ln \left (b x +a \right )}{b^{4}}\)	\(69\)
default	\(\frac {B x}{b^{3}}-\frac {a^{2} \left (A b -B a \right )}{2 b^{4} \left (b x +a \right )^{2}}+\frac {a \left (2 A b -3 B a \right )}{b^{4} \left (b x +a \right )}+\frac {\left (A b -3 B a \right ) \ln \left (b x +a \right )}{b^{4}}\)	\(70\)
risch	\(\frac {B x}{b^{3}}+\frac {\left (2 a b A -3 a^{2} B \right ) x +\frac {a^{2} \left (3 A b -5 B a \right )}{2 b}}{b^{3} \left (b x +a \right )^{2}}+\frac {\ln \left (b x +a \right ) A}{b^{3}}-\frac {3 \ln \left (b x +a \right ) B a}{b^{4}}\)	\(75\)
parallelrisch	\(\frac {2 A \ln \left (b x +a \right ) x^{2} b^{3}-6 B \ln \left (b x +a \right ) x^{2} a \,b^{2}+2 b^{3} B \,x^{3}+4 A \ln \left (b x +a \right ) x a \,b^{2}-12 B \ln \left (b x +a \right ) x \,a^{2} b +2 A \ln \left (b x +a \right ) a^{2} b +4 a \,b^{2} A x -6 B \ln \left (b x +a \right ) a^{3}-12 a^{2} b B x +3 a^{2} b A -9 a^{3} B}{2 b^{4} \left (b x +a \right )^{2}}\)	\(136\)

[In]

int(x^2*(B*x+A)/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

(B*x^3/b+1/2*a^2*(3*A*b-9*B*a)/b^4+2*a*(A*b-3*B*a)/b^3*x)/(b*x+a)^2+(A*b-3*B*a)*ln(b*x+a)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.89 \[ \int \frac {x^2 (A+B x)}{(a+b x)^3} \, dx=\frac {2 \, B b^{3} x^{3} + 4 \, B a b^{2} x^{2} - 5 \, B a^{3} + 3 \, A a^{2} b - 4 \, {\left (B a^{2} b - A a b^{2}\right )} x - 2 \, {\left (3 \, B a^{3} - A a^{2} b + {\left (3 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b - A a b^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.17 \[ \int \frac {x^2 (A+B x)}{(a+b x)^3} \, dx=\frac {B x}{b^{3}} + \frac {3 A a^{2} b - 5 B a^{3} + x \left (4 A a b^{2} - 6 B a^{2} b\right )}{2 a^{2} b^{4} + 4 a b^{5} x + 2 b^{6} x^{2}} - \frac {\left (- A b + 3 B a\right ) \log {\left (a + b x \right )}}{b^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20 \[ \int \frac {x^2 (A+B x)}{(a+b x)^3} \, dx=-\frac {5 \, B a^{3} - 3 \, A a^{2} b + 2 \, {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} x}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {B x}{b^{3}} - \frac {{\left (3 \, B a - A b\right )} \log \left (b x + a\right )}{b^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 (A+B x)}{(a+b x)^3} \, dx=\frac {B x}{b^{3}} - \frac {{\left (3 \, B a - A b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac {5 \, B a^{3} - 3 \, A a^{2} b + 2 \, {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20 \[ \int \frac {x^2 (A+B x)}{(a+b x)^3} \, dx=\frac {B\,x}{b^3}-\frac {\frac {5\,B\,a^3-3\,A\,a^2\,b}{2\,b}+x\,\left (3\,B\,a^2-2\,A\,a\,b\right )}{a^2\,b^3+2\,a\,b^4\,x+b^5\,x^2}+\frac {\ln \left (a+b\,x\right )\,\left (A\,b-3\,B\,a\right )}{b^4} \]

[In]

int((x^2*(A + B*x))/(a + b*x)^3,x)

[Out]

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