\(\int (c x^n)^{\frac {1}{n}} (a+b (c x^n)^{\frac {1}{n}})^2 \, dx\) [3072]

Optimal result

Integrand size = 25, antiderivative size = 55 \[ \int \left (c x^n\right )^{\frac {1}{n}} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2 \, dx=\frac {1}{2} a^2 x \left (c x^n\right )^{\frac {1}{n}}+\frac {2}{3} a b x \left (c x^n\right )^{2/n}+\frac {1}{4} b^2 x \left (c x^n\right )^{3/n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {15, 375, 45} \[ \int \left (c x^n\right )^{\frac {1}{n}} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2 \, dx=\frac {1}{2} a^2 x \left (c x^n\right )^{\frac {1}{n}}+\frac {2}{3} a b x \left (c x^n\right )^{2/n}+\frac {1}{4} b^2 x \left (c x^n\right )^{3/n} \]

[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 45

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])

Rule 375

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)]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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] (verified)

Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \left (c x^n\right )^{\frac {1}{n}} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2 \, dx=\frac {1}{12} x \left (c x^n\right )^{\frac {1}{n}} \left (6 a^2+8 a b \left (c x^n\right )^{\frac {1}{n}}+3 b^2 \left (c x^n\right )^{2/n}\right ) \]

[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

Maple [A] (verified)

Time = 6.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(\frac {3 x^{2} \left (c \,x^{n}\right )^{\frac {3}{n}} b^{2}+8 x^{2} \left (c \,x^{n}\right )^{\frac {2}{n}} a b +6 x^{2} \left (c \,x^{n}\right )^{\frac {1}{n}} a^{2}}{12 x}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \left (c x^n\right )^{\frac {1}{n}} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2 \, dx=\frac {1}{4} \, b^{2} c^{\frac {3}{n}} x^{4} + \frac {2}{3} \, a b c^{\frac {2}{n}} x^{3} + \frac {1}{2} \, a^{2} c^{\left (\frac {1}{n}\right )} x^{2} \]

[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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \left (c x^n\right )^{\frac {1}{n}} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2 \, dx=\frac {a^{2} x \left (c x^{n}\right )^{\frac {1}{n}}}{2} + \frac {2 a b x \left (c x^{n}\right )^{\frac {2}{n}}}{3} + \frac {b^{2} x \left (c x^{n}\right )^{\frac {3}{n}}}{4} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \left (c x^n\right )^{\frac {1}{n}} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2 \, dx=\int { {\left (\left (c x^{n}\right )^{\left (\frac {1}{n}\right )} b + a\right )}^{2} \left (c x^{n}\right )^{\left (\frac {1}{n}\right )} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \left (c x^n\right )^{\frac {1}{n}} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2 \, dx=\frac {1}{4} \, b^{2} c^{\frac {3}{n}} x^{4} + \frac {2}{3} \, a b c^{\frac {2}{n}} x^{3} + \frac {1}{2} \, a^{2} c^{\left (\frac {1}{n}\right )} x^{2} \]

[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

Mupad [B] (verification not implemented)

Time = 5.36 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \left (c x^n\right )^{\frac {1}{n}} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2 \, dx=\frac {x\,{\left (c\,x^n\right )}^{1/n}\,\left (3\,b^2\,{\left (c\,x^n\right )}^{2/n}+6\,a^2+8\,a\,b\,{\left (c\,x^n\right )}^{1/n}\right )}{12} \]

[In]

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

[Out]

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