\(\int \frac {(a+b x)^3}{x^2} \, dx\) [70]

Optimal result

Integrand size = 11, antiderivative size = 34 \[ \int \frac {(a+b x)^3}{x^2} \, dx=-\frac {a^3}{x}+3 a b^2 x+\frac {b^3 x^2}{2}+3 a^2 b \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, 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} \[ \int \frac {(a+b x)^3}{x^2} \, dx=-\frac {a^3}{x}+3 a^2 b \log (x)+3 a b^2 x+\frac {b^3 x^2}{2} \]

[In]

Int[(a + b*x)^3/x^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^3}{x^2} \, dx=-\frac {a^3}{x}+3 a b^2 x+\frac {b^3 x^2}{2}+3 a^2 b \log (x) \]

[In]

Integrate[(a + b*x)^3/x^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {a^{3}}{x}+3 a \,b^{2} x +\frac {b^{3} x^{2}}{2}+3 a^{2} b \ln \left (x \right )\)	\(33\)
risch	\(-\frac {a^{3}}{x}+3 a \,b^{2} x +\frac {b^{3} x^{2}}{2}+3 a^{2} b \ln \left (x \right )\)	\(33\)
norman	\(\frac {-a^{3}+\frac {1}{2} b^{3} x^{3}+3 a \,b^{2} x^{2}}{x}+3 a^{2} b \ln \left (x \right )\)	\(37\)
parallelrisch	\(\frac {b^{3} x^{3}+6 a^{2} b \ln \left (x \right ) x +6 a \,b^{2} x^{2}-2 a^{3}}{2 x}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^3}{x^2} \, dx=\frac {b^{3} x^{3} + 6 \, a b^{2} x^{2} + 6 \, a^{2} b x \log \left (x\right ) - 2 \, a^{3}}{2 \, x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x)^3}{x^2} \, dx=- \frac {a^{3}}{x} + 3 a^{2} b \log {\left (x \right )} + 3 a b^{2} x + \frac {b^{3} x^{2}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b x)^3}{x^2} \, dx=\frac {1}{2} \, b^{3} x^{2} + 3 \, a b^{2} x + 3 \, a^{2} b \log \left (x\right ) - \frac {a^{3}}{x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^3}{x^2} \, dx=\frac {1}{2} \, b^{3} x^{2} + 3 \, a b^{2} x + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) - \frac {a^{3}}{x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b x)^3}{x^2} \, dx=\frac {b^3\,x^2}{2}-\frac {a^3}{x}+3\,a^2\,b\,\ln \left (x\right )+3\,a\,b^2\,x \]

[In]

int((a + b*x)^3/x^2,x)

[Out]

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