3.3005 \(\int (a+b (c x^n)^{\frac{1}{n}})^3 \, dx\)

Optimal. Leaf size=34 \[ \frac{x \left (c x^n\right )^{-1/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^4}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.0077302, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {254, 32} \[ \frac{x \left (c x^n\right )^{-1/n} \left (a+b \left (c x^n\right )^{\frac{1}{n}}\right )^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(c*x^n)^n^(-1))^3,x]

[Out]

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

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0197438, size = 60, normalized size = 1.76 \[ \frac{3}{2} a^2 b x \left (c x^n\right )^{\frac{1}{n}}+a^3 x+a b^2 x \left (c x^n\right )^{2/n}+\frac{1}{4} b^3 x \left (c x^n\right )^{3/n} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(c*x^n)^n^(-1))^3,x]

[Out]

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

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sqrt [n]{c{x}^{n}} \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.36354, size = 99, normalized size = 2.91 \begin{align*} \frac{1}{4} \, b^{3} c^{\frac{3}{n}} x^{4} + a b^{2} c^{\frac{2}{n}} x^{3} + \frac{3}{2} \, a^{2} b c^{\left (\frac{1}{n}\right )} x^{2} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.831868, size = 63, normalized size = 1.85 \begin{align*} a^{3} x + \frac{3 a^{2} b c^{\frac{1}{n}} x \left (x^{n}\right )^{\frac{1}{n}}}{2} + a b^{2} c^{\frac{2}{n}} x \left (x^{n}\right )^{\frac{2}{n}} + \frac{b^{3} 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((a+b*(c*x**n)**(1/n))**3,x)

[Out]

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

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Giac [A]  time = 1.1624, size = 68, normalized size = 2. \begin{align*} \frac{1}{4} \, b^{3} c^{\frac{3}{n}} x^{4} + a b^{2} c^{\frac{2}{n}} x^{3} + \frac{3}{2} \, a^{2} b c^{\left (\frac{1}{n}\right )} x^{2} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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