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

Optimal. Leaf size=206 \[ \frac{3 a^2 b^4 x^{n+1} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(n+1) \left (a b+b^2 x^n\right )^3}+\frac{3 a b^5 x^{2 n+1} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(2 n+1) \left (a b+b^2 x^n\right )^3}+\frac{b^6 x^{3 n+1} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(3 n+1) \left (a b+b^2 x^n\right )^3}+\frac{a^3 x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{\left (a+b x^n\right )^3} \]

[Out]

(a^3*x*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2))/(a + b*x^n)^3 + (3*a^2*b^4*x^(1 + n)*(a^2 + 2*a*b*x^n + b^2*x^(2
*n))^(3/2))/((1 + n)*(a*b + b^2*x^n)^3) + (3*a*b^5*x^(1 + 2*n)*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2))/((1 + 2*
n)*(a*b + b^2*x^n)^3) + (b^6*x^(1 + 3*n)*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2))/((1 + 3*n)*(a*b + b^2*x^n)^3)

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Rubi [A]  time = 0.0518595, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1343, 244} \[ \frac{3 a^2 b^4 x^{n+1} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(n+1) \left (a b+b^2 x^n\right )^3}+\frac{3 a b^5 x^{2 n+1} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(2 n+1) \left (a b+b^2 x^n\right )^3}+\frac{b^6 x^{3 n+1} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(3 n+1) \left (a b+b^2 x^n\right )^3}+\frac{a^3 x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{\left (a+b x^n\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2))/(a + b*x^n)^3 + (3*a^2*b^4*x^(1 + n)*(a^2 + 2*a*b*x^n + b^2*x^(2
*n))^(3/2))/((1 + n)*(a*b + b^2*x^n)^3) + (3*a*b^5*x^(1 + 2*n)*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2))/((1 + 2*
n)*(a*b + b^2*x^n)^3) + (b^6*x^(1 + 3*n)*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2))/((1 + 3*n)*(a*b + b^2*x^n)^3)

Rule 1343

Int[((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^p/(b + 2*c*x
^n)^(2*p), Int[(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0]

Rule 244

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n},
x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, dx &=\frac{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \int \left (2 a b+2 b^2 x^n\right )^3 \, dx}{\left (2 a b+2 b^2 x^n\right )^3}\\ &=\frac{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \int \left (8 a^3 b^3+24 a^2 b^4 x^n+24 a b^5 x^{2 n}+8 b^6 x^{3 n}\right ) \, dx}{\left (2 a b+2 b^2 x^n\right )^3}\\ &=\frac{a^3 x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{\left (a+b x^n\right )^3}+\frac{3 a^2 b^4 x^{1+n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(1+n) \left (a b+b^2 x^n\right )^3}+\frac{3 a b^5 x^{1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(1+2 n) \left (a b+b^2 x^n\right )^3}+\frac{b^6 x^{1+3 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}{(1+3 n) \left (a b+b^2 x^n\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.0755736, size = 122, normalized size = 0.59 \[ \frac{x \sqrt{\left (a+b x^n\right )^2} \left (3 a^2 b \left (6 n^2+5 n+1\right ) x^n+a^3 \left (6 n^3+11 n^2+6 n+1\right )+3 a b^2 \left (3 n^2+4 n+1\right ) x^{2 n}+b^3 \left (2 n^2+3 n+1\right ) x^{3 n}\right )}{(n+1) (2 n+1) (3 n+1) \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x^n)^2]*(a^3*(1 + 6*n + 11*n^2 + 6*n^3) + 3*a^2*b*(1 + 5*n + 6*n^2)*x^n + 3*a*b^2*(1 + 4*n + 3*
n^2)*x^(2*n) + b^3*(1 + 3*n + 2*n^2)*x^(3*n)))/((1 + n)*(1 + 2*n)*(1 + 3*n)*(a + b*x^n))

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Maple [A]  time = 0.014, size = 138, normalized size = 0.7 \begin{align*}{\frac{{a}^{3}x}{a+b{x}^{n}}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}+{\frac{{b}^{3}x \left ({x}^{n} \right ) ^{3}}{ \left ( a+b{x}^{n} \right ) \left ( 1+3\,n \right ) }\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}+3\,{\frac{\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}a{b}^{2}x \left ({x}^{n} \right ) ^{2}}{ \left ( a+b{x}^{n} \right ) \left ( 1+2\,n \right ) }}+3\,{\frac{\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}{a}^{2}bx{x}^{n}}{ \left ( a+b{x}^{n} \right ) \left ( 1+n \right ) }} \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)

[Out]

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

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Maxima [A]  time = 0.988257, size = 136, normalized size = 0.66 \begin{align*} \frac{{\left (2 \, n^{2} + 3 \, n + 1\right )} b^{3} x x^{3 \, n} + 3 \,{\left (3 \, n^{2} + 4 \, n + 1\right )} a b^{2} x x^{2 \, n} + 3 \,{\left (6 \, n^{2} + 5 \, n + 1\right )} a^{2} b x x^{n} +{\left (6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1\right )} a^{3} x}{6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1} \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, algorithm="maxima")

[Out]

((2*n^2 + 3*n + 1)*b^3*x*x^(3*n) + 3*(3*n^2 + 4*n + 1)*a*b^2*x*x^(2*n) + 3*(6*n^2 + 5*n + 1)*a^2*b*x*x^n + (6*
n^3 + 11*n^2 + 6*n + 1)*a^3*x)/(6*n^3 + 11*n^2 + 6*n + 1)

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Fricas [A]  time = 1.56252, size = 277, normalized size = 1.34 \begin{align*} \frac{{\left (2 \, b^{3} n^{2} + 3 \, b^{3} n + b^{3}\right )} x x^{3 \, n} + 3 \,{\left (3 \, a b^{2} n^{2} + 4 \, a b^{2} n + a b^{2}\right )} x x^{2 \, n} + 3 \,{\left (6 \, a^{2} b n^{2} + 5 \, a^{2} b n + a^{2} b\right )} x x^{n} +{\left (6 \, a^{3} n^{3} + 11 \, a^{3} n^{2} + 6 \, a^{3} n + a^{3}\right )} x}{6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1} \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, algorithm="fricas")

[Out]

((2*b^3*n^2 + 3*b^3*n + b^3)*x*x^(3*n) + 3*(3*a*b^2*n^2 + 4*a*b^2*n + a*b^2)*x*x^(2*n) + 3*(6*a^2*b*n^2 + 5*a^
2*b*n + a^2*b)*x*x^n + (6*a^3*n^3 + 11*a^3*n^2 + 6*a^3*n + a^3)*x)/(6*n^3 + 11*n^2 + 6*n + 1)

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

[Out]

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

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Giac [A]  time = 1.15787, size = 355, normalized size = 1.72 \begin{align*} \frac{6 \, a^{3} n^{3} x \mathrm{sgn}\left (b x^{n} + a\right ) + 2 \, b^{3} n^{2} x x^{3 \, n} \mathrm{sgn}\left (b x^{n} + a\right ) + 9 \, a b^{2} n^{2} x x^{2 \, n} \mathrm{sgn}\left (b x^{n} + a\right ) + 18 \, a^{2} b n^{2} x x^{n} \mathrm{sgn}\left (b x^{n} + a\right ) + 11 \, a^{3} n^{2} x \mathrm{sgn}\left (b x^{n} + a\right ) + 3 \, b^{3} n x x^{3 \, n} \mathrm{sgn}\left (b x^{n} + a\right ) + 12 \, a b^{2} n x x^{2 \, n} \mathrm{sgn}\left (b x^{n} + a\right ) + 15 \, a^{2} b n x x^{n} \mathrm{sgn}\left (b x^{n} + a\right ) + 6 \, a^{3} n x \mathrm{sgn}\left (b x^{n} + a\right ) + b^{3} x x^{3 \, n} \mathrm{sgn}\left (b x^{n} + a\right ) + 3 \, a b^{2} x x^{2 \, n} \mathrm{sgn}\left (b x^{n} + a\right ) + 3 \, a^{2} b x x^{n} \mathrm{sgn}\left (b x^{n} + a\right ) + a^{3} x \mathrm{sgn}\left (b x^{n} + a\right )}{6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1} \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, algorithm="giac")

[Out]

(6*a^3*n^3*x*sgn(b*x^n + a) + 2*b^3*n^2*x*x^(3*n)*sgn(b*x^n + a) + 9*a*b^2*n^2*x*x^(2*n)*sgn(b*x^n + a) + 18*a
^2*b*n^2*x*x^n*sgn(b*x^n + a) + 11*a^3*n^2*x*sgn(b*x^n + a) + 3*b^3*n*x*x^(3*n)*sgn(b*x^n + a) + 12*a*b^2*n*x*
x^(2*n)*sgn(b*x^n + a) + 15*a^2*b*n*x*x^n*sgn(b*x^n + a) + 6*a^3*n*x*sgn(b*x^n + a) + b^3*x*x^(3*n)*sgn(b*x^n
+ a) + 3*a*b^2*x*x^(2*n)*sgn(b*x^n + a) + 3*a^2*b*x*x^n*sgn(b*x^n + a) + a^3*x*sgn(b*x^n + a))/(6*n^3 + 11*n^2
+ 6*n + 1)