3.6.40 \(\int \frac {1}{x (a^2+2 a b x^n+b^2 x^{2 n})^{3/2}} \, dx\) [540]

Optimal. Leaf size=159 \[ \frac {1}{a^2 n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}+\frac {1}{2 a n \left (a+b x^n\right ) \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}+\frac {\left (a+b x^n\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {\left (a+b x^n\right ) \log \left (a+b x^n\right )}{a^3 n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1369, 272, 46} \begin {gather*} \frac {1}{a^2 n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}+\frac {1}{2 a n \left (a+b x^n\right ) \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}+\frac {\log (x) \left (a+b x^n\right )}{a^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {\left (a+b x^n\right ) \log \left (a+b x^n\right )}{a^3 n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2)),x]

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

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 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 {1}{x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x^n\right )\right ) \int \frac {1}{x \left (a b+b^2 x^n\right )^3} \, dx}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac {\left (b^2 \left (a b+b^2 x^n\right )\right ) \text {Subst}\left (\int \frac {1}{x \left (a b+b^2 x\right )^3} \, dx,x,x^n\right )}{n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac {\left (b^2 \left (a b+b^2 x^n\right )\right ) \text {Subst}\left (\int \left (\frac {1}{a^3 b^3 x}-\frac {1}{a b^2 (a+b x)^3}-\frac {1}{a^2 b^2 (a+b x)^2}-\frac {1}{a^3 b^2 (a+b x)}\right ) \, dx,x,x^n\right )}{n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac {1}{a^2 n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}+\frac {1}{2 a n \left (a+b x^n\right ) \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}+\frac {\left (a+b x^n\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {\left (a+b x^n\right ) \log \left (a+b x^n\right )}{a^3 n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 79, normalized size = 0.50 \begin {gather*} \frac {a \left (3 a+2 b x^n\right )+2 \left (a+b x^n\right )^2 \log \left (x^n\right )-2 \left (a+b x^n\right )^2 \log \left (a+b x^n\right )}{2 a^3 n \left (a+b x^n\right ) \sqrt {\left (a+b x^n\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2)),x]

[Out]

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

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Maple [A]
time = 0.02, size = 104, normalized size = 0.65

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, \ln \left (x \right )}{\left (a +b \,x^{n}\right ) a^{3}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, \left (2 b \,x^{n}+3 a \right )}{2 \left (a +b \,x^{n}\right )^{3} a^{2} n}-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, \ln \left (x^{n}+\frac {a}{b}\right )}{\left (a +b \,x^{n}\right ) a^{3} n}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 70, normalized size = 0.44 \begin {gather*} \frac {2 \, b x^{n} + 3 \, a}{2 \, {\left (a^{2} b^{2} n x^{2 \, n} + 2 \, a^{3} b n x^{n} + a^{4} n\right )}} + \frac {\log \left (x\right )}{a^{3}} - \frac {\log \left (\frac {b x^{n} + a}{b}\right )}{a^{3} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="maxima")

[Out]

1/2*(2*b*x^n + 3*a)/(a^2*b^2*n*x^(2*n) + 2*a^3*b*n*x^n + a^4*n) + log(x)/a^3 - log((b*x^n + a)/b)/(a^3*n)

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Fricas [A]
time = 0.34, size = 106, normalized size = 0.67 \begin {gather*} \frac {2 \, b^{2} n x^{2 \, n} \log \left (x\right ) + 2 \, a^{2} n \log \left (x\right ) + 3 \, a^{2} + 2 \, {\left (2 \, a b n \log \left (x\right ) + a b\right )} x^{n} - 2 \, {\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )} \log \left (b x^{n} + a\right )}{2 \, {\left (a^{3} b^{2} n x^{2 \, n} + 2 \, a^{4} b n x^{n} + a^{5} n\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="fricas")

[Out]

1/2*(2*b^2*n*x^(2*n)*log(x) + 2*a^2*n*log(x) + 3*a^2 + 2*(2*a*b*n*log(x) + a*b)*x^n - 2*(b^2*x^(2*n) + 2*a*b*x
^n + a^2)*log(b*x^n + a))/(a^3*b^2*n*x^(2*n) + 2*a^4*b*n*x^n + a^5*n)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a**2+2*a*b*x**n+b**2*x**(2*n))**(3/2),x)

[Out]

Integral(1/(x*((a + b*x**n)**2)**(3/2)), 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(1/x/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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