3.31.71 \(\int (c x^n)^{\frac {1}{n}} (a+b (c x^n)^{\frac {1}{n}})^3 \, dx\) [3071]

Optimal. Leaf size=70 \[ -\frac {a x \left (c x^n\right )^{-1/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^4}{4 b^2}+\frac {x \left (c x^n\right )^{-1/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5}{5 b^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {15, 375, 45} \begin {gather*} \frac {x \left (c x^n\right )^{-1/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5}{5 b^2}-\frac {a x \left (c x^n\right )^{-1/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^4}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x^n)^n^(-1)*(a + b*(c*x^n)^n^(-1))^3,x]

[Out]

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

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

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Mathematica [A]
time = 0.09, size = 48, normalized size = 0.69 \begin {gather*} -\frac {x \left (c x^n\right )^{-1/n} \left (a-4 b \left (c x^n\right )^{\frac {1}{n}}\right ) \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^4}{20 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*x^n)^n^(-1)*(a + b*(c*x^n)^n^(-1))^3,x]

[Out]

-1/20*(x*(a - 4*b*(c*x^n)^n^(-1))*(a + b*(c*x^n)^n^(-1))^4)/(b^2*(c*x^n)^n^(-1))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (c \,x^{n}\right )^{\frac {1}{n}} \left (a +b \left (c \,x^{n}\right )^{\frac {1}{n}}\right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.40, size = 60, normalized size = 0.86 \begin {gather*} \frac {1}{5} \, b^{3} c^{\frac {4}{n}} x^{5} + \frac {3}{4} \, a b^{2} c^{\frac {3}{n}} x^{4} + a^{2} b c^{\frac {2}{n}} x^{3} + \frac {1}{2} \, a^{3} c^{\left (\frac {1}{n}\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.31, size = 63, normalized size = 0.90 \begin {gather*} \frac {a^{3} x \left (c x^{n}\right )^{\frac {1}{n}}}{2} + a^{2} b x \left (c x^{n}\right )^{\frac {2}{n}} + \frac {3 a b^{2} x \left (c x^{n}\right )^{\frac {3}{n}}}{4} + \frac {b^{3} x \left (c x^{n}\right )^{\frac {4}{n}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.09, size = 60, normalized size = 0.86 \begin {gather*} \frac {1}{5} \, b^{3} c^{\frac {4}{n}} x^{5} + \frac {3}{4} \, a b^{2} c^{\frac {3}{n}} x^{4} + a^{2} b c^{\frac {2}{n}} x^{3} + \frac {1}{2} \, a^{3} c^{\left (\frac {1}{n}\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.26, size = 68, normalized size = 0.97 \begin {gather*} \frac {b^3\,x\,{\left (c\,x^n\right )}^{4/n}}{5}+\frac {a^3\,x\,{\left (c\,x^n\right )}^{1/n}}{2}+a^2\,b\,x\,{\left (c\,x^n\right )}^{2/n}+\frac {3\,a\,b^2\,x\,{\left (c\,x^n\right )}^{3/n}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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