3.3029 \(\int (a+b (c x^n)^{2/n})^2 \, dx\)

Optimal. Leaf size=43 \[ a^2 x+\frac{2}{3} a b x \left (c x^n\right )^{2/n}+\frac{1}{5} b^2 x \left (c x^n\right )^{4/n} \]

[Out]

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

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Rubi [A]  time = 0.0171888, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {254, 194} \[ a^2 x+\frac{2}{3} a b x \left (c x^n\right )^{2/n}+\frac{1}{5} b^2 x \left (c x^n\right )^{4/n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(c*x^n)^(2/n))^2,x]

[Out]

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

Rule 254

Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_.), x_Symbol] :> Dist[x/(c*x^q)^(1/q), Subst[Int[(a + b*x^(n*q))
^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, n, p, q}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

Rule 194

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

Rubi steps

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

Mathematica [A]  time = 0.0123648, size = 43, normalized size = 1. \[ a^2 x+\frac{2}{3} a b x \left (c x^n\right )^{2/n}+\frac{1}{5} b^2 x \left (c x^n\right )^{4/n} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(c*x^n)^(2/n))^2,x]

[Out]

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

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Maple [F]  time = 0.161, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( c{x}^{n} \right ) ^{2\,{n}^{-1}} \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b^{2} c^{\frac{4}{n}} \int{\left (x^{n}\right )}^{\frac{4}{n}}\,{d x} + 2 \, a b c^{\frac{2}{n}} \int{\left (x^{n}\right )}^{\frac{2}{n}}\,{d x} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*c^(4/n)*integrate((x^n)^(4/n), x) + 2*a*b*c^(2/n)*integrate((x^n)^(2/n), x) + a^2*x

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Fricas [A]  time = 1.3387, size = 69, normalized size = 1.6 \begin{align*} \frac{1}{5} \, b^{2} c^{\frac{4}{n}} x^{5} + \frac{2}{3} \, a b c^{\frac{2}{n}} x^{3} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.418959, size = 42, normalized size = 0.98 \begin{align*} a^{2} x + \frac{2 a b c^{\frac{2}{n}} x \left (x^{n}\right )^{\frac{2}{n}}}{3} + \frac{b^{2} c^{\frac{4}{n}} x \left (x^{n}\right )^{\frac{4}{n}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11505, size = 47, normalized size = 1.09 \begin{align*} \frac{1}{5} \, b^{2} c^{\frac{4}{n}} x^{5} + \frac{2}{3} \, a b c^{\frac{2}{n}} x^{3} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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