\(\int \frac {x^2}{a+b x} \, dx\) [30]

Optimal result

Integrand size = 11, antiderivative size = 31 \[ \int \frac {x^2}{a+b x} \, dx=-\frac {a x}{b^2}+\frac {x^2}{2 b}+\frac {a^2 \log (a+b x)}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, 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} \[ \int \frac {x^2}{a+b x} \, dx=\frac {a^2 \log (a+b x)}{b^3}-\frac {a x}{b^2}+\frac {x^2}{2 b} \]

[In]

Int[x^2/(a + b*x),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{a+b x} \, dx=-\frac {a x}{b^2}+\frac {x^2}{2 b}+\frac {a^2 \log (a+b x)}{b^3} \]

[In]

Integrate[x^2/(a + b*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {-\frac {1}{2} x^{2} b +a x}{b^{2}}+\frac {a^{2} \ln \left (b x +a \right )}{b^{3}}\)	\(30\)
norman	\(\frac {x^{2}}{2 b}-\frac {a x}{b^{2}}+\frac {a^{2} \ln \left (b x +a \right )}{b^{3}}\)	\(30\)
risch	\(\frac {x^{2}}{2 b}-\frac {a x}{b^{2}}+\frac {a^{2} \ln \left (b x +a \right )}{b^{3}}\)	\(30\)
parallelrisch	\(\frac {b^{2} x^{2}+2 a^{2} \ln \left (b x +a \right )-2 b a x}{2 b^{3}}\)	\(30\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {x^2}{a+b x} \, dx=\frac {b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} \log \left (b x + a\right )}{2 \, b^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {x^2}{a+b x} \, dx=\frac {a^{2} \log {\left (a + b x \right )}}{b^{3}} - \frac {a x}{b^{2}} + \frac {x^{2}}{2 b} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {x^2}{a+b x} \, dx=\frac {a^{2} \log \left (b x + a\right )}{b^{3}} + \frac {b x^{2} - 2 \, a x}{2 \, b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {x^2}{a+b x} \, dx=\frac {a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{3}} + \frac {b x^{2} - 2 \, a x}{2 \, b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {x^2}{a+b x} \, dx=\frac {2\,a^2\,\ln \left (a+b\,x\right )+b^2\,x^2-2\,a\,b\,x}{2\,b^3} \]

[In]

int(x^2/(a + b*x),x)

[Out]

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