3.2532 \(\int \frac{(a+b x^n)^2}{x} \, dx\)

Optimal. Leaf size=32 \[ a^2 \log (x)+\frac{2 a b x^n}{n}+\frac{b^2 x^{2 n}}{2 n} \]

[Out]

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

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Rubi [A]  time = 0.0126149, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ a^2 \log (x)+\frac{2 a b x^n}{n}+\frac{b^2 x^{2 n}}{2 n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^2/x,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

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*} \int \frac{\left (a+b x^n\right )^2}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^2}{x} \, dx,x,x^n\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 a b+\frac{a^2}{x}+b^2 x\right ) \, dx,x,x^n\right )}{n}\\ &=\frac{2 a b x^n}{n}+\frac{b^2 x^{2 n}}{2 n}+a^2 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0159887, size = 27, normalized size = 0.84 \[ a^2 \log (x)+\frac{b x^n \left (4 a+b x^n\right )}{2 n} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^n)^2/x,x]

[Out]

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

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Maple [A]  time = 0., size = 36, normalized size = 1.1 \begin{align*}{\frac{ \left ({x}^{n} \right ) ^{2}{b}^{2}}{2\,n}}+2\,{\frac{ab{x}^{n}}{n}}+{\frac{{a}^{2}\ln \left ({x}^{n} \right ) }{n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.962723, size = 46, normalized size = 1.44 \begin{align*} \frac{a^{2} \log \left (x^{n}\right )}{n} + \frac{b^{2} x^{2 \, n} + 4 \, a b x^{n}}{2 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(x^n)/n + 1/2*(b^2*x^(2*n) + 4*a*b*x^n)/n

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Fricas [A]  time = 1.02017, size = 68, normalized size = 2.12 \begin{align*} \frac{2 \, a^{2} n \log \left (x\right ) + b^{2} x^{2 \, n} + 4 \, a b x^{n}}{2 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^2*n*log(x) + b^2*x^(2*n) + 4*a*b*x^n)/n

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Sympy [A]  time = 0.354659, size = 36, normalized size = 1.12 \begin{align*} \begin{cases} a^{2} \log{\left (x \right )} + \frac{2 a b x^{n}}{n} + \frac{b^{2} x^{2 n}}{2 n} & \text{for}\: n \neq 0 \\\left (a + b\right )^{2} \log{\left (x \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*log(x) + 2*a*b*x**n/n + b**2*x**(2*n)/(2*n), Ne(n, 0)), ((a + b)**2*log(x), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{n} + a\right )}^{2}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^n + a)^2/x, x)