3.6.19 \(\int \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x} \, dx\) [519]

Optimal. Leaf size=85 \[ \frac {b^2 x^n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{n \left (a b+b^2 x^n\right )}+\frac {a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \log (x)}{a+b x^n} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1369, 14} \begin {gather*} \frac {b^2 x^n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{n \left (a b+b^2 x^n\right )}+\frac {a \log (x) \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{a+b x^n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]/x,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 38, normalized size = 0.45 \begin {gather*} \frac {\sqrt {\left (a+b x^n\right )^2} \left (b x^n+a \log \left (x^n\right )\right )}{n \left (a+b x^n\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]/x,x]

[Out]

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

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Maple [A]
time = 0.03, size = 54, normalized size = 0.64

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a \ln \left (x \right )}{a +b \,x^{n}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b \,x^{n}}{\left (a +b \,x^{n}\right ) n}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/x,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.33, size = 13, normalized size = 0.15 \begin {gather*} a \log \left (x\right ) + \frac {b x^{n}}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/x,x, algorithm="maxima")

[Out]

a*log(x) + b*x^n/n

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Fricas [A]
time = 0.36, size = 15, normalized size = 0.18 \begin {gather*} \frac {a n \log \left (x\right ) + b x^{n}}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/x,x, algorithm="fricas")

[Out]

(a*n*log(x) + b*x^n)/n

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\left (a + b x^{n}\right )^{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2+2*a*b*x**n+b**2*x**(2*n))**(1/2)/x,x)

[Out]

Integral(sqrt((a + b*x**n)**2)/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/x,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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