\(\int \frac {x^2}{(a+b x)^2} \, dx\) [33]

Optimal result

Integrand size = 11, antiderivative size = 33 \[ \int \frac {x^2}{(a+b x)^2} \, dx=\frac {x}{b^2}-\frac {a^2}{b^3 (a+b x)}-\frac {2 a \log (a+b x)}{b^3} \]

[Out]

x/b^2-a^2/b^3/(b*x+a)-2*a/b^3*ln(b*x+a)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {x^2}{(a+b x)^2} \, dx=-\frac {a^2}{b^3 (a+b x)}-\frac {2 a \log (a+b x)}{b^3}+\frac {x}{b^2} \]

[In]

Int[x^2/(a + b*x)^2,x]

[Out]

x/b^2 - a^2/(b^3*(a + b*x)) - (2*a*Log[a + b*x])/b^3

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{b^2}+\frac {a^2}{b^2 (a+b x)^2}-\frac {2 a}{b^2 (a+b x)}\right ) \, dx \\ & = \frac {x}{b^2}-\frac {a^2}{b^3 (a+b x)}-\frac {2 a \log (a+b x)}{b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{(a+b x)^2} \, dx=\frac {b x-\frac {a^2}{a+b x}-2 a \log (a+b x)}{b^3} \]

[In]

Integrate[x^2/(a + b*x)^2,x]

[Out]

(b*x - a^2/(a + b*x) - 2*a*Log[a + b*x])/b^3

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {x}{b^{2}}-\frac {a^{2}}{b^{3} \left (b x +a \right )}-\frac {2 a \ln \left (b x +a \right )}{b^{3}}\)	\(34\)
risch	\(\frac {x}{b^{2}}-\frac {a^{2}}{b^{3} \left (b x +a \right )}-\frac {2 a \ln \left (b x +a \right )}{b^{3}}\)	\(34\)
norman	\(\frac {\frac {x^{2}}{b}-\frac {2 a^{2}}{b^{3}}}{b x +a}-\frac {2 a \ln \left (b x +a \right )}{b^{3}}\)	\(38\)
parallelrisch	\(-\frac {2 \ln \left (b x +a \right ) x a b -b^{2} x^{2}+2 a^{2} \ln \left (b x +a \right )+2 a^{2}}{b^{3} \left (b x +a \right )}\)	\(49\)

[In]

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

[Out]

x/b^2-a^2/b^3/(b*x+a)-2*a/b^3*ln(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {x^2}{(a+b x)^2} \, dx=\frac {b^{2} x^{2} + a b x - a^{2} - 2 \, {\left (a b x + a^{2}\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {x^2}{(a+b x)^2} \, dx=- \frac {a^{2}}{a b^{3} + b^{4} x} - \frac {2 a \log {\left (a + b x \right )}}{b^{3}} + \frac {x}{b^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {x^2}{(a+b x)^2} \, dx=-\frac {a^{2}}{b^{4} x + a b^{3}} + \frac {x}{b^{2}} - \frac {2 \, a \log \left (b x + a\right )}{b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \frac {x^2}{(a+b x)^2} \, dx=\frac {2 \, a \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{3}} + \frac {b x + a}{b^{3}} - \frac {a^{2}}{{\left (b x + a\right )} b^{3}} \]

[In]

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

[Out]

2*a*log(abs(b*x + a)/((b*x + a)^2*abs(b)))/b^3 + (b*x + a)/b^3 - a^2/((b*x + a)*b^3)

Mupad [B] (verification not implemented)

Time = 18.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {x^2}{(a+b x)^2} \, dx=\frac {x}{b^2}-\frac {a^2}{x\,b^4+a\,b^3}-\frac {2\,a\,\ln \left (a+b\,x\right )}{b^3} \]

[In]

int(x^2/(a + b*x)^2,x)

[Out]

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