\(\int \frac {(a+b x)^3}{x^3} \, dx\) [71]

Optimal result

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

[Out]

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

Rubi [A] (verified)

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

[In]

Int[(a + b*x)^3/x^3,x]

[Out]

-1/2*a^3/x^2 - (3*a^2*b)/x + b^3*x + 3*a*b^2*Log[x]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^3}{x^3} \, dx=-\frac {a^3}{2 x^2}-\frac {3 a^2 b}{x}+b^3 x+3 a b^2 \log (x) \]

[In]

Integrate[(a + b*x)^3/x^3,x]

[Out]

-1/2*a^3/x^2 - (3*a^2*b)/x + b^3*x + 3*a*b^2*Log[x]

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {a^{3}}{2 x^{2}}-\frac {3 a^{2} b}{x}+b^{3} x +3 a \,b^{2} \ln \left (x \right )\)	\(32\)
risch	\(b^{3} x +\frac {-3 a^{2} b x -\frac {1}{2} a^{3}}{x^{2}}+3 a \,b^{2} \ln \left (x \right )\)	\(32\)
norman	\(\frac {b^{3} x^{3}-\frac {1}{2} a^{3}-3 a^{2} b x}{x^{2}}+3 a \,b^{2} \ln \left (x \right )\)	\(34\)
parallelrisch	\(\frac {6 a \,b^{2} \ln \left (x \right ) x^{2}+2 b^{3} x^{3}-6 a^{2} b x -a^{3}}{2 x^{2}}\)	\(38\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^3}{x^3} \, dx=\frac {2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} \log \left (x\right ) - 6 \, a^{2} b x - a^{3}}{2 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^3}{x^3} \, dx=3 a b^{2} \log {\left (x \right )} + b^{3} x + \frac {- a^{3} - 6 a^{2} b x}{2 x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x)^3}{x^3} \, dx=b^{3} x + 3 \, a b^{2} \log \left (x\right ) - \frac {6 \, a^{2} b x + a^{3}}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b x)^3}{x^3} \, dx=b^{3} x + 3 \, a b^{2} \log \left ({\left | x \right |}\right ) - \frac {6 \, a^{2} b x + a^{3}}{2 \, x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^3}{x^3} \, dx=b^3\,x-\frac {\frac {a^3}{2}+3\,b\,x\,a^2}{x^2}+3\,a\,b^2\,\ln \left (x\right ) \]

[In]

int((a + b*x)^3/x^3,x)

[Out]

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