3.3072 \(\int (c x^n)^{\frac{1}{n}} (a+b (c x^n)^{\frac{1}{n}})^2 \, dx\)

Optimal. Leaf size=55 \[ \frac{1}{2} a^2 x \left (c x^n\right )^{\frac{1}{n}}+\frac{2}{3} a b x \left (c x^n\right )^{2/n}+\frac{1}{4} b^2 x \left (c x^n\right )^{3/n} \]

[Out]

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

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Rubi [A]  time = 0.0252137, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {15, 368, 43} \[ \frac{1}{2} a^2 x \left (c x^n\right )^{\frac{1}{n}}+\frac{2}{3} a b x \left (c x^n\right )^{2/n}+\frac{1}{4} b^2 x \left (c x^n\right )^{3/n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 368

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

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

Mathematica [A]  time = 0.0178952, size = 49, normalized size = 0.89 \[ \frac{1}{12} x \left (c x^n\right )^{\frac{1}{n}} \left (6 a^2+8 a b \left (c x^n\right )^{\frac{1}{n}}+3 b^2 \left (c x^n\right )^{2/n}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*x^n)^n^(-1)*(6*a^2 + 8*a*b*(c*x^n)^n^(-1) + 3*b^2*(c*x^n)^(2/n)))/12

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Maple [F]  time = 0.232, size = 0, normalized size = 0. \begin{align*} \int \sqrt [n]{c{x}^{n}} \left ( a+b\sqrt [n]{c{x}^{n}} \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.64861, size = 88, normalized size = 1.6 \begin{align*} \frac{1}{4} \, b^{2} c^{\frac{3}{n}} x^{4} + \frac{2}{3} \, a b c^{\frac{2}{n}} x^{3} + \frac{1}{2} \, a^{2} c^{\left (\frac{1}{n}\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.755513, size = 56, normalized size = 1.02 \begin{align*} \frac{a^{2} c^{\frac{1}{n}} x \left (x^{n}\right )^{\frac{1}{n}}}{2} + \frac{2 a b c^{\frac{2}{n}} x \left (x^{n}\right )^{\frac{2}{n}}}{3} + \frac{b^{2} c^{\frac{3}{n}} x \left (x^{n}\right )^{\frac{3}{n}}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2783, size = 58, normalized size = 1.05 \begin{align*} \frac{1}{4} \, b^{2} c^{\frac{3}{n}} x^{4} + \frac{2}{3} \, a b c^{\frac{2}{n}} x^{3} + \frac{1}{2} \, a^{2} c^{\left (\frac{1}{n}\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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