\(\int \frac {(a+b x^n)^2}{x^2} \, dx\) [2463]

Optimal result

Integrand size = 13, antiderivative size = 44 \[ \int \frac {\left (a+b x^n\right )^2}{x^2} \, dx=-\frac {a^2}{x}-\frac {2 a b x^{-1+n}}{1-n}-\frac {b^2 x^{-1+2 n}}{1-2 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {276} \[ \int \frac {\left (a+b x^n\right )^2}{x^2} \, dx=-\frac {a^2}{x}-\frac {2 a b x^{n-1}}{1-n}-\frac {b^2 x^{2 n-1}}{1-2 n} \]

[In]

Int[(a + b*x^n)^2/x^2,x]

[Out]

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

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^2}{x^2}+2 a b x^{-2+n}+b^2 x^{2 (-1+n)}\right ) \, dx \\ & = -\frac {a^2}{x}-\frac {2 a b x^{-1+n}}{1-n}-\frac {b^2 x^{-1+2 n}}{1-2 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^n\right )^2}{x^2} \, dx=\frac {-a^2+\frac {2 a b x^n}{-1+n}+\frac {b^2 x^{2 n}}{-1+2 n}}{x} \]

[In]

Integrate[(a + b*x^n)^2/x^2,x]

[Out]

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

Maple [A] (verified)

Time = 3.62 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98

method	result	size
norman	\(\frac {\frac {b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{-1+2 n}-a^{2}+\frac {2 a b \,{\mathrm e}^{n \ln \left (x \right )}}{-1+n}}{x}\)	\(43\)
risch	\(-\frac {a^{2}}{x}+\frac {b^{2} x^{2 n}}{\left (-1+2 n \right ) x}+\frac {2 a b \,x^{n}}{\left (-1+n \right ) x}\)	\(44\)
parallelrisch	\(\frac {b^{2} x^{2 n} n -b^{2} x^{2 n}+4 a b \,x^{n} n -2 a^{2} n^{2}-2 a b \,x^{n}+3 a^{2} n -a^{2}}{x \left (-1+2 n \right ) \left (-1+n \right )}\)	\(72\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a+b x^n\right )^2}{x^2} \, dx=-\frac {2 \, a^{2} n^{2} - 3 \, a^{2} n + a^{2} - {\left (b^{2} n - b^{2}\right )} x^{2 \, n} - 2 \, {\left (2 \, a b n - a b\right )} x^{n}}{{\left (2 \, n^{2} - 3 \, n + 1\right )} x} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (34) = 68\).

Time = 0.24 (sec) , antiderivative size = 190, normalized size of antiderivative = 4.32 \[ \int \frac {\left (a+b x^n\right )^2}{x^2} \, dx=\begin {cases} - \frac {a^{2}}{x} - \frac {4 a b}{\sqrt {x}} + b^{2} \log {\left (x \right )} & \text {for}\: n = \frac {1}{2} \\- \frac {a^{2}}{x} + 2 a b \log {\left (x \right )} + b^{2} x & \text {for}\: n = 1 \\- \frac {2 a^{2} n^{2}}{2 n^{2} x - 3 n x + x} + \frac {3 a^{2} n}{2 n^{2} x - 3 n x + x} - \frac {a^{2}}{2 n^{2} x - 3 n x + x} + \frac {4 a b n x^{n}}{2 n^{2} x - 3 n x + x} - \frac {2 a b x^{n}}{2 n^{2} x - 3 n x + x} + \frac {b^{2} n x^{2 n}}{2 n^{2} x - 3 n x + x} - \frac {b^{2} x^{2 n}}{2 n^{2} x - 3 n x + x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a**2/x - 4*a*b/sqrt(x) + b**2*log(x), Eq(n, 1/2)), (-a**2/x + 2*a*b*log(x) + b**2*x, Eq(n, 1)), (-
2*a**2*n**2/(2*n**2*x - 3*n*x + x) + 3*a**2*n/(2*n**2*x - 3*n*x + x) - a**2/(2*n**2*x - 3*n*x + x) + 4*a*b*n*x
**n/(2*n**2*x - 3*n*x + x) - 2*a*b*x**n/(2*n**2*x - 3*n*x + x) + b**2*n*x**(2*n)/(2*n**2*x - 3*n*x + x) - b**2
*x**(2*n)/(2*n**2*x - 3*n*x + x), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^n\right )^2}{x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(n-2>0)', see `assume?` for mor
e details)Is

Giac [F]

\[ \int \frac {\left (a+b x^n\right )^2}{x^2} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{2}}{x^{2}} \,d x } \]

[In]

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

[Out]

integrate((b*x^n + a)^2/x^2, x)

Mupad [B] (verification not implemented)

Time = 5.67 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^n\right )^2}{x^2} \, dx=\frac {b^2\,x^{2\,n}}{x\,\left (2\,n-1\right )}-\frac {a^2}{x}+\frac {2\,a\,b\,x^n}{x\,\left (n-1\right )} \]

[In]

int((a + b*x^n)^2/x^2,x)

[Out]

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