\(\int \frac {x^2}{(a+b x)^3} \, dx\) [185]

Optimal result

Integrand size = 11, antiderivative size = 41 \[ \int \frac {x^2}{(a+b x)^3} \, dx=-\frac {a^2}{2 b^3 (a+b x)^2}+\frac {2 a}{b^3 (a+b x)}+\frac {\log (a+b x)}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, 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} \[ \int \frac {x^2}{(a+b x)^3} \, dx=-\frac {a^2}{2 b^3 (a+b x)^2}+\frac {2 a}{b^3 (a+b x)}+\frac {\log (a+b x)}{b^3} \]

[In]

Int[x^2/(a + b*x)^3,x]

[Out]

-1/2*a^2/(b^3*(a + b*x)^2) + (2*a)/(b^3*(a + b*x)) + 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 {a^2}{b^2 (a+b x)^3}-\frac {2 a}{b^2 (a+b x)^2}+\frac {1}{b^2 (a+b x)}\right ) \, dx \\ & = -\frac {a^2}{2 b^3 (a+b x)^2}+\frac {2 a}{b^3 (a+b x)}+\frac {\log (a+b x)}{b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {x^2}{(a+b x)^3} \, dx=\frac {\frac {a (3 a+4 b x)}{(a+b x)^2}+2 \log (a+b x)}{2 b^3} \]

[In]

Integrate[x^2/(a + b*x)^3,x]

[Out]

((a*(3*a + 4*b*x))/(a + b*x)^2 + 2*Log[a + b*x])/(2*b^3)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88

method	result	size
norman	\(\frac {\frac {3 a^{2}}{2 b^{3}}+\frac {2 a x}{b^{2}}}{\left (b x +a \right )^{2}}+\frac {\ln \left (b x +a \right )}{b^{3}}\)	\(36\)
risch	\(\frac {\frac {3 a^{2}}{2 b^{3}}+\frac {2 a x}{b^{2}}}{\left (b x +a \right )^{2}}+\frac {\ln \left (b x +a \right )}{b^{3}}\)	\(36\)
default	\(-\frac {a^{2}}{2 b^{3} \left (b x +a \right )^{2}}+\frac {2 a}{b^{3} \left (b x +a \right )}+\frac {\ln \left (b x +a \right )}{b^{3}}\)	\(40\)
parallelrisch	\(\frac {2 b^{2} \ln \left (b x +a \right ) x^{2}+4 \ln \left (b x +a \right ) x a b +2 a^{2} \ln \left (b x +a \right )+4 a b x +3 a^{2}}{2 b^{3} \left (b x +a \right )^{2}}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.49 \[ \int \frac {x^2}{(a+b x)^3} \, dx=\frac {4 \, a b x + 3 \, a^{2} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

[In]

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

[Out]

1/2*(4*a*b*x + 3*a^2 + 2*(b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {x^2}{(a+b x)^3} \, dx=\frac {3 a^{2} + 4 a b x}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {\log {\left (a + b x \right )}}{b^{3}} \]

[In]

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

[Out]

(3*a**2 + 4*a*b*x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + log(a + b*x)/b**3

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.17 \[ \int \frac {x^2}{(a+b x)^3} \, dx=\frac {4 \, a b x + 3 \, a^{2}}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {\log \left (b x + a\right )}{b^{3}} \]

[In]

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

[Out]

1/2*(4*a*b*x + 3*a^2)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) + log(b*x + a)/b^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int \frac {x^2}{(a+b x)^3} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{b^{3}} + \frac {4 \, a x + \frac {3 \, a^{2}}{b}}{2 \, {\left (b x + a\right )}^{2} b^{2}} \]

[In]

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

[Out]

log(abs(b*x + a))/b^3 + 1/2*(4*a*x + 3*a^2/b)/((b*x + a)^2*b^2)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {x^2}{(a+b x)^3} \, dx=\frac {\ln \left (a+b\,x\right )}{b^3}+\frac {\frac {3\,a^2}{2\,b^3}+\frac {2\,a\,x}{b^2}}{a^2+2\,a\,b\,x+b^2\,x^2} \]

[In]

int(x^2/(a + b*x)^3,x)

[Out]

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