\(\int \frac {x^3}{(a+b x)^2} \, dx\) [171]

Optimal result

Integrand size = 11, antiderivative size = 46 \[ \int \frac {x^3}{(a+b x)^2} \, dx=-\frac {2 a x}{b^3}+\frac {x^2}{2 b^2}+\frac {a^3}{b^4 (a+b x)}+\frac {3 a^2 \log (a+b x)}{b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, 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} \[ \int \frac {x^3}{(a+b x)^2} \, dx=\frac {a^3}{b^4 (a+b x)}+\frac {3 a^2 \log (a+b x)}{b^4}-\frac {2 a x}{b^3}+\frac {x^2}{2 b^2} \]

[In]

Int[x^3/(a + b*x)^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \frac {x^3}{(a+b x)^2} \, dx=\frac {-4 a b x+b^2 x^2+\frac {2 a^3}{a+b x}+6 a^2 \log (a+b x)}{2 b^4} \]

[In]

Integrate[x^3/(a + b*x)^2,x]

[Out]

(-4*a*b*x + b^2*x^2 + (2*a^3)/(a + b*x) + 6*a^2*Log[a + b*x])/(2*b^4)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {2 a x}{b^{3}}+\frac {x^{2}}{2 b^{2}}+\frac {a^{3}}{b^{4} \left (b x +a \right )}+\frac {3 a^{2} \ln \left (b x +a \right )}{b^{4}}\)	\(45\)
default	\(-\frac {-\frac {1}{2} b \,x^{2}+2 a x}{b^{3}}+\frac {3 a^{2} \ln \left (b x +a \right )}{b^{4}}+\frac {a^{3}}{b^{4} \left (b x +a \right )}\)	\(46\)
norman	\(\frac {\frac {3 a^{3}}{b^{4}}+\frac {x^{3}}{2 b}-\frac {3 a \,x^{2}}{2 b^{2}}}{b x +a}+\frac {3 a^{2} \ln \left (b x +a \right )}{b^{4}}\)	\(50\)
parallelrisch	\(\frac {b^{3} x^{3}+6 \ln \left (b x +a \right ) x \,a^{2} b -3 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )+6 a^{3}}{2 b^{4} \left (b x +a \right )}\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.35 \[ \int \frac {x^3}{(a+b x)^2} \, dx=\frac {b^{3} x^{3} - 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 2 \, a^{3} + 6 \, {\left (a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x + a b^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int \frac {x^3}{(a+b x)^2} \, dx=\frac {a^{3}}{a b^{4} + b^{5} x} + \frac {3 a^{2} \log {\left (a + b x \right )}}{b^{4}} - \frac {2 a x}{b^{3}} + \frac {x^{2}}{2 b^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{(a+b x)^2} \, dx=\frac {a^{3}}{b^{5} x + a b^{4}} + \frac {3 \, a^{2} \log \left (b x + a\right )}{b^{4}} + \frac {b x^{2} - 4 \, a x}{2 \, b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.43 \[ \int \frac {x^3}{(a+b x)^2} \, dx=-\frac {{\left (b x + a\right )}^{2} {\left (\frac {6 \, a}{b x + a} - 1\right )}}{2 \, b^{4}} - \frac {3 \, a^{2} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{4}} + \frac {a^{3}}{{\left (b x + a\right )} b^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09 \[ \int \frac {x^3}{(a+b x)^2} \, dx=\frac {x^2}{2\,b^2}+\frac {3\,a^2\,\ln \left (a+b\,x\right )}{b^4}+\frac {a^3}{b\,\left (x\,b^4+a\,b^3\right )}-\frac {2\,a\,x}{b^3} \]

[In]

int(x^3/(a + b*x)^2,x)

[Out]

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