3.509 \(\int x^{-1+2 n} (a^2+2 a b x^n+b^2 x^{2 n})^{3/2} \, dx\)

Optimal. Leaf size=112 \[ \frac{\left (a+b x^n\right )^5 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{5 n \left (a b^2+b^3 x^n\right )}-\frac{a \left (a+b x^n\right )^4 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{4 n \left (a b^2+b^3 x^n\right )} \]

[Out]

-(a*(a + b*x^n)^4*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/(4*n*(a*b^2 + b^3*x^n)) + ((a + b*x^n)^5*Sqrt[a^2 + 2*a
*b*x^n + b^2*x^(2*n)])/(5*n*(a*b^2 + b^3*x^n))

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Rubi [A]  time = 0.0396355, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {1355, 266, 43} \[ \frac{\left (a+b x^n\right )^5 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{5 n \left (a b^2+b^3 x^n\right )}-\frac{a \left (a+b x^n\right )^4 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{4 n \left (a b^2+b^3 x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(a + b*x^n)^4*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/(4*n*(a*b^2 + b^3*x^n)) + ((a + b*x^n)^5*Sqrt[a^2 + 2*a
*b*x^n + b^2*x^(2*n)])/(5*n*(a*b^2 + b^3*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 x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx &=\frac{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}} \int x^{-1+2 n} \left (a b+b^2 x^n\right )^3 \, 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 x \left (a b+b^2 x\right )^3 \, 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 (-\frac{a \left (a b+b^2 x\right )^3}{b}+\frac{\left (a b+b^2 x\right )^4}{b^2}\right ) \, dx,x,x^n\right )}{b^2 n \left (a b+b^2 x^n\right )}\\ &=-\frac{a \left (a+b x^n\right )^4 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{4 n \left (a b^2+b^3 x^n\right )}+\frac{\left (a+b x^n\right )^5 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{5 n \left (a b^2+b^3 x^n\right )}\\ \end{align*}

Mathematica [A]  time = 0.0379362, size = 40, normalized size = 0.36 \[ -\frac{\left (a-4 b x^n\right ) \left (a+b x^n\right )^3 \sqrt{\left (a+b x^n\right )^2}}{20 b^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - 4*b*x^n)*(a + b*x^n)^3*Sqrt[(a + b*x^n)^2])/(20*b^2*n)

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Maple [A]  time = 0.023, size = 135, normalized size = 1.2 \begin{align*}{\frac{{b}^{3} \left ({x}^{n} \right ) ^{5}}{ \left ( 5\,a+5\,b{x}^{n} \right ) n}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}+{\frac{3\,a{b}^{2} \left ({x}^{n} \right ) ^{4}}{ \left ( 4\,a+4\,b{x}^{n} \right ) n}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}+{\frac{{a}^{2}b \left ({x}^{n} \right ) ^{3}}{ \left ( a+b{x}^{n} \right ) n}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}+{\frac{{a}^{3} \left ({x}^{n} \right ) ^{2}}{ \left ( 2\,a+2\,b{x}^{n} \right ) n}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.986123, size = 65, normalized size = 0.58 \begin{align*} \frac{4 \, b^{3} x^{5 \, n} + 15 \, a b^{2} x^{4 \, n} + 20 \, a^{2} b x^{3 \, n} + 10 \, a^{3} x^{2 \, n}}{20 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(4*b^3*x^(5*n) + 15*a*b^2*x^(4*n) + 20*a^2*b*x^(3*n) + 10*a^3*x^(2*n))/n

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Fricas [A]  time = 1.59965, size = 107, normalized size = 0.96 \begin{align*} \frac{4 \, b^{3} x^{5 \, n} + 15 \, a b^{2} x^{4 \, n} + 20 \, a^{2} b x^{3 \, n} + 10 \, a^{3} x^{2 \, n}}{20 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(4*b^3*x^(5*n) + 15*a*b^2*x^(4*n) + 20*a^2*b*x^(3*n) + 10*a^3*x^(2*n))/n

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac{3}{2}} x^{2 \, n - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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