3.526 \(\int \frac{(a^2+2 a b x^n+b^2 x^{2 n})^{3/2}}{x} \, dx\)

Optimal. Leaf size=196 \[ \frac{3 a^2 b^2 x^n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{n \left (a b+b^2 x^n\right )}+\frac{3 a b^3 x^{2 n} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a b+b^2 x^n\right )}+\frac{b^4 x^{3 n} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{3 n \left (a b+b^2 x^n\right )}+\frac{a^3 \log (x) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{a+b x^n} \]

[Out]

(3*a^2*b^2*x^n*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/(n*(a*b + b^2*x^n)) + (3*a*b^3*x^(2*n)*Sqrt[a^2 + 2*a*b*x^
n + b^2*x^(2*n)])/(2*n*(a*b + b^2*x^n)) + (b^4*x^(3*n)*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/(3*n*(a*b + b^2*x^
n)) + (a^3*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]*Log[x])/(a + b*x^n)

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Rubi [A]  time = 0.0516225, antiderivative size = 196, normalized size of antiderivative = 1., 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 = {1355, 266, 43} \[ \frac{3 a^2 b^2 x^n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{n \left (a b+b^2 x^n\right )}+\frac{3 a b^3 x^{2 n} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a b+b^2 x^n\right )}+\frac{b^4 x^{3 n} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{3 n \left (a b+b^2 x^n\right )}+\frac{a^3 \log (x) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{a+b x^n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*b^2*x^n*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/(n*(a*b + b^2*x^n)) + (3*a*b^3*x^(2*n)*Sqrt[a^2 + 2*a*b*x^
n + b^2*x^(2*n)])/(2*n*(a*b + b^2*x^n)) + (b^4*x^(3*n)*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/(3*n*(a*b + b^2*x^
n)) + (a^3*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]*Log[x])/(a + b*x^n)

Rule 1355

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]

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^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{x} \, dx &=\frac{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}} \int \frac{\left (a b+b^2 x^n\right )^3}{x} \, dx}{b^2 \left (a b+b^2 x^n\right )}\\ &=\frac{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}} \operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^3}{x} \, dx,x,x^n\right )}{b^2 n \left (a b+b^2 x^n\right )}\\ &=\frac{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}} \operatorname{Subst}\left (\int \left (3 a^2 b^4+\frac{a^3 b^3}{x}+3 a b^5 x+b^6 x^2\right ) \, dx,x,x^n\right )}{b^2 n \left (a b+b^2 x^n\right )}\\ &=\frac{3 a^2 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{3 a b^3 x^{2 n} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a b+b^2 x^n\right )}+\frac{b^4 x^{3 n} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{3 n \left (a b+b^2 x^n\right )}+\frac{a^3 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}} \log (x)}{a+b x^n}\\ \end{align*}

Mathematica [A]  time = 0.0385719, size = 68, normalized size = 0.35 \[ \frac{\left (\left (a+b x^n\right )^2\right )^{3/2} \left (3 a^2 b x^n+a^3 n \log (x)+\frac{3}{2} a b^2 x^{2 n}+\frac{1}{3} b^3 x^{3 n}\right )}{n \left (a+b x^n\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 127, normalized size = 0.7 \begin{align*}{\frac{{a}^{3}\ln \left ( x \right ) }{a+b{x}^{n}}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}+{\frac{{b}^{3} \left ({x}^{n} \right ) ^{3}}{ \left ( 3\,a+3\,b{x}^{n} \right ) n}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}+{\frac{3\,a{b}^{2} \left ({x}^{n} \right ) ^{2}}{ \left ( 2\,a+2\,b{x}^{n} \right ) n}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}+3\,{\frac{\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}{a}^{2}b{x}^{n}}{ \left ( a+b{x}^{n} \right ) n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.983491, size = 58, normalized size = 0.3 \begin{align*} a^{3} \log \left (x\right ) + \frac{2 \, b^{3} x^{3 \, n} + 9 \, a b^{2} x^{2 \, n} + 18 \, a^{2} b x^{n}}{6 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*log(x) + 1/6*(2*b^3*x^(3*n) + 9*a*b^2*x^(2*n) + 18*a^2*b*x^n)/n

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Fricas [A]  time = 1.56861, size = 99, normalized size = 0.51 \begin{align*} \frac{6 \, a^{3} n \log \left (x\right ) + 2 \, b^{3} x^{3 \, n} + 9 \, a b^{2} x^{2 \, n} + 18 \, a^{2} b x^{n}}{6 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*a^3*n*log(x) + 2*b^3*x^(3*n) + 9*a*b^2*x^(2*n) + 18*a^2*b*x^n)/n

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac{3}{2}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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