\(\int \frac {x^2}{a+\frac {b}{x}} \, dx\) [1609]

Optimal result

Integrand size = 13, antiderivative size = 44 \[ \int \frac {x^2}{a+\frac {b}{x}} \, dx=\frac {b^2 x}{a^3}-\frac {b x^2}{2 a^2}+\frac {x^3}{3 a}-\frac {b^3 \log (b+a x)}{a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {269, 45} \[ \int \frac {x^2}{a+\frac {b}{x}} \, dx=-\frac {b^3 \log (a x+b)}{a^4}+\frac {b^2 x}{a^3}-\frac {b x^2}{2 a^2}+\frac {x^3}{3 a} \]

[In]

Int[x^2/(a + b/x),x]

[Out]

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

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])

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3}{b+a x} \, dx \\ & = \int \left (\frac {b^2}{a^3}-\frac {b x}{a^2}+\frac {x^2}{a}-\frac {b^3}{a^3 (b+a x)}\right ) \, dx \\ & = \frac {b^2 x}{a^3}-\frac {b x^2}{2 a^2}+\frac {x^3}{3 a}-\frac {b^3 \log (b+a x)}{a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{a+\frac {b}{x}} \, dx=\frac {b^2 x}{a^3}-\frac {b x^2}{2 a^2}+\frac {x^3}{3 a}-\frac {b^3 \log (b+a x)}{a^4} \]

[In]

Integrate[x^2/(a + b/x),x]

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\frac {1}{3} a^{2} x^{3}-\frac {1}{2} a b \,x^{2}+b^{2} x}{a^{3}}-\frac {b^{3} \ln \left (a x +b \right )}{a^{4}}\)	\(41\)
norman	\(\frac {b^{2} x}{a^{3}}-\frac {b \,x^{2}}{2 a^{2}}+\frac {x^{3}}{3 a}-\frac {b^{3} \ln \left (a x +b \right )}{a^{4}}\)	\(41\)
risch	\(\frac {b^{2} x}{a^{3}}-\frac {b \,x^{2}}{2 a^{2}}+\frac {x^{3}}{3 a}-\frac {b^{3} \ln \left (a x +b \right )}{a^{4}}\)	\(41\)
parallelrisch	\(-\frac {-2 a^{3} x^{3}+3 a^{2} b \,x^{2}+6 b^{3} \ln \left (a x +b \right )-6 a \,b^{2} x}{6 a^{4}}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{a+\frac {b}{x}} \, dx=\frac {2 \, a^{3} x^{3} - 3 \, a^{2} b x^{2} + 6 \, a b^{2} x - 6 \, b^{3} \log \left (a x + b\right )}{6 \, a^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {x^2}{a+\frac {b}{x}} \, dx=\frac {x^{3}}{3 a} - \frac {b x^{2}}{2 a^{2}} + \frac {b^{2} x}{a^{3}} - \frac {b^{3} \log {\left (a x + b \right )}}{a^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {x^2}{a+\frac {b}{x}} \, dx=-\frac {b^{3} \log \left (a x + b\right )}{a^{4}} + \frac {2 \, a^{2} x^{3} - 3 \, a b x^{2} + 6 \, b^{2} x}{6 \, a^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {x^2}{a+\frac {b}{x}} \, dx=-\frac {b^{3} \log \left ({\left | a x + b \right |}\right )}{a^{4}} + \frac {2 \, a^{2} x^{3} - 3 \, a b x^{2} + 6 \, b^{2} x}{6 \, a^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91 \[ \int \frac {x^2}{a+\frac {b}{x}} \, dx=\frac {x^3}{3\,a}-\frac {b^3\,\ln \left (b+a\,x\right )}{a^4}-\frac {b\,x^2}{2\,a^2}+\frac {b^2\,x}{a^3} \]

[In]

int(x^2/(a + b/x),x)

[Out]

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