\(\int \frac {(c x^n)^{\frac {1}{n}}}{(a+b (c x^n)^{\frac {1}{n}})^5} \, dx\) [3078]

Optimal result

Integrand size = 25, antiderivative size = 70 \[ \int \frac {\left (c x^n\right )^{\frac {1}{n}}}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5} \, dx=\frac {a x \left (c x^n\right )^{-1/n}}{4 b^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^4}-\frac {x \left (c x^n\right )^{-1/n}}{3 b^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, 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} \[ \int \frac {\left (c x^n\right )^{\frac {1}{n}}}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5} \, dx=\frac {a x \left (c x^n\right )^{-1/n}}{4 b^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^4}-\frac {x \left (c x^n\right )^{-1/n}}{3 b^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^3} \]

[In]

Int[(c*x^n)^n^(-1)/(a + b*(c*x^n)^n^(-1))^5,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.69 \[ \int \frac {\left (c x^n\right )^{\frac {1}{n}}}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5} \, dx=-\frac {x \left (c x^n\right )^{-1/n} \left (a+4 b \left (c x^n\right )^{\frac {1}{n}}\right )}{12 b^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^4} \]

[In]

Integrate[(c*x^n)^n^(-1)/(a + b*(c*x^n)^n^(-1))^5,x]

[Out]

-1/12*(x*(a + 4*b*(c*x^n)^n^(-1)))/(b^2*(c*x^n)^n^(-1)*(a + b*(c*x^n)^n^(-1))^4)

Maple [A] (verified)

Time = 7.52 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.27

method	result	size
parallelrisch	\(\frac {b^{6} \left (c \,x^{n}\right )^{\frac {3}{n}} x^{2} a +4 b^{5} \left (c \,x^{n}\right )^{\frac {2}{n}} x^{2} a^{2}+6 x^{2} \left (c \,x^{n}\right )^{\frac {1}{n}} a^{3} b^{4}}{12 a^{4} b^{4} x {\left (a +b \left (c \,x^{n}\right )^{\frac {1}{n}}\right )}^{4}}\)	\(89\)
risch	\(\frac {x \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} \left (c^{\frac {2}{n}} \left (x^{n}\right )^{\frac {2}{n}} b^{2} {\mathrm e}^{\frac {3 i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}+4 \left (x^{n}\right )^{\frac {1}{n}} c^{\frac {1}{n}} a b \,{\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{n}}+6 a^{2} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}\right )}{12 a^{3} {\left (b \left (x^{n}\right )^{\frac {1}{n}} c^{\frac {1}{n}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i c \right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}+a \right )}^{4}}\)	\(279\)

[In]

int((c*x^n)^(1/n)/(a+b*(c*x^n)^(1/n))^5,x,method=_RETURNVERBOSE)

[Out]

1/12*(b^6*((c*x^n)^(1/n))^3*x^2*a+4*b^5*((c*x^n)^(1/n))^2*x^2*a^2+6*x^2*(c*x^n)^(1/n)*a^3*b^4)/a^4/b^4/x/(a+b*
(c*x^n)^(1/n))^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.31 \[ \int \frac {\left (c x^n\right )^{\frac {1}{n}}}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5} \, dx=-\frac {4 \, b c^{\left (\frac {1}{n}\right )} x + a}{12 \, {\left (b^{6} c^{\frac {5}{n}} x^{4} + 4 \, a b^{5} c^{\frac {4}{n}} x^{3} + 6 \, a^{2} b^{4} c^{\frac {3}{n}} x^{2} + 4 \, a^{3} b^{3} c^{\frac {2}{n}} x + a^{4} b^{2} c^{\left (\frac {1}{n}\right )}\right )}} \]

[In]

integrate((c*x^n)^(1/n)/(a+b*(c*x^n)^(1/n))^5,x, algorithm="fricas")

[Out]

-1/12*(4*b*c^(1/n)*x + a)/(b^6*c^(5/n)*x^4 + 4*a*b^5*c^(4/n)*x^3 + 6*a^2*b^4*c^(3/n)*x^2 + 4*a^3*b^3*c^(2/n)*x
+ a^4*b^2*c^(1/n))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (58) = 116\).

Time = 8.62 (sec) , antiderivative size = 306, normalized size of antiderivative = 4.37 \[ \int \frac {\left (c x^n\right )^{\frac {1}{n}}}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5} \, dx=\begin {cases} \tilde {\infty } x \left (c x^{n}\right )^{- \frac {4}{n}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {x \left (c x^{n}\right )^{- \frac {4}{n}}}{3 b^{5}} & \text {for}\: a = 0 \\\tilde {\infty } x \left (c x^{n}\right )^{\frac {1}{n}} & \text {for}\: b = - a \left (c x^{n}\right )^{- \frac {1}{n}} \\\frac {6 a^{2} x \left (c x^{n}\right )^{\frac {1}{n}}}{12 a^{7} + 48 a^{6} b \left (c x^{n}\right )^{\frac {1}{n}} + 72 a^{5} b^{2} \left (c x^{n}\right )^{\frac {2}{n}} + 48 a^{4} b^{3} \left (c x^{n}\right )^{\frac {3}{n}} + 12 a^{3} b^{4} \left (c x^{n}\right )^{\frac {4}{n}}} + \frac {4 a b x \left (c x^{n}\right )^{\frac {2}{n}}}{12 a^{7} + 48 a^{6} b \left (c x^{n}\right )^{\frac {1}{n}} + 72 a^{5} b^{2} \left (c x^{n}\right )^{\frac {2}{n}} + 48 a^{4} b^{3} \left (c x^{n}\right )^{\frac {3}{n}} + 12 a^{3} b^{4} \left (c x^{n}\right )^{\frac {4}{n}}} + \frac {b^{2} x \left (c x^{n}\right )^{\frac {3}{n}}}{12 a^{7} + 48 a^{6} b \left (c x^{n}\right )^{\frac {1}{n}} + 72 a^{5} b^{2} \left (c x^{n}\right )^{\frac {2}{n}} + 48 a^{4} b^{3} \left (c x^{n}\right )^{\frac {3}{n}} + 12 a^{3} b^{4} \left (c x^{n}\right )^{\frac {4}{n}}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x**n)**(1/n)/(a+b*(c*x**n)**(1/n))**5,x)

[Out]

Piecewise((zoo*x/(c*x**n)**(4/n), Eq(a, 0) & Eq(b, 0)), (-x/(3*b**5*(c*x**n)**(4/n)), Eq(a, 0)), (zoo*x*(c*x**
n)**(1/n), Eq(b, -a/(c*x**n)**(1/n))), (6*a**2*x*(c*x**n)**(1/n)/(12*a**7 + 48*a**6*b*(c*x**n)**(1/n) + 72*a**
5*b**2*(c*x**n)**(2/n) + 48*a**4*b**3*(c*x**n)**(3/n) + 12*a**3*b**4*(c*x**n)**(4/n)) + 4*a*b*x*(c*x**n)**(2/n
)/(12*a**7 + 48*a**6*b*(c*x**n)**(1/n) + 72*a**5*b**2*(c*x**n)**(2/n) + 48*a**4*b**3*(c*x**n)**(3/n) + 12*a**3
*b**4*(c*x**n)**(4/n)) + b**2*x*(c*x**n)**(3/n)/(12*a**7 + 48*a**6*b*(c*x**n)**(1/n) + 72*a**5*b**2*(c*x**n)**
(2/n) + 48*a**4*b**3*(c*x**n)**(3/n) + 12*a**3*b**4*(c*x**n)**(4/n)), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (66) = 132\).

Time = 0.25 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.26 \[ \int \frac {\left (c x^n\right )^{\frac {1}{n}}}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5} \, dx=\frac {b^{2} c^{\frac {3}{n}} x {\left (x^{n}\right )}^{\frac {3}{n}} + 4 \, a b c^{\frac {2}{n}} x {\left (x^{n}\right )}^{\frac {2}{n}} + 6 \, a^{2} c^{\left (\frac {1}{n}\right )} x {\left (x^{n}\right )}^{\left (\frac {1}{n}\right )}}{12 \, {\left (a^{3} b^{4} c^{\frac {4}{n}} {\left (x^{n}\right )}^{\frac {4}{n}} + 4 \, a^{4} b^{3} c^{\frac {3}{n}} {\left (x^{n}\right )}^{\frac {3}{n}} + 6 \, a^{5} b^{2} c^{\frac {2}{n}} {\left (x^{n}\right )}^{\frac {2}{n}} + 4 \, a^{6} b c^{\left (\frac {1}{n}\right )} {\left (x^{n}\right )}^{\left (\frac {1}{n}\right )} + a^{7}\right )}} \]

[In]

integrate((c*x^n)^(1/n)/(a+b*(c*x^n)^(1/n))^5,x, algorithm="maxima")

[Out]

1/12*(b^2*c^(3/n)*x*(x^n)^(3/n) + 4*a*b*c^(2/n)*x*(x^n)^(2/n) + 6*a^2*c^(1/n)*x*(x^n)^(1/n))/(a^3*b^4*c^(4/n)*
(x^n)^(4/n) + 4*a^4*b^3*c^(3/n)*(x^n)^(3/n) + 6*a^5*b^2*c^(2/n)*(x^n)^(2/n) + 4*a^6*b*c^(1/n)*(x^n)^(1/n) + a^
7)

Giac [F]

\[ \int \frac {\left (c x^n\right )^{\frac {1}{n}}}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5} \, dx=\int { \frac {\left (c x^{n}\right )^{\left (\frac {1}{n}\right )}}{{\left (\left (c x^{n}\right )^{\left (\frac {1}{n}\right )} b + a\right )}^{5}} \,d x } \]

[In]

integrate((c*x^n)^(1/n)/(a+b*(c*x^n)^(1/n))^5,x, algorithm="giac")

[Out]

integrate((c*x^n)^(1/n)/((c*x^n)^(1/n)*b + a)^5, x)

Mupad [B] (verification not implemented)

Time = 5.35 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.97 \[ \int \frac {\left (c x^n\right )^{\frac {1}{n}}}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^5} \, dx=\frac {x}{12\,a\,b\,\left (b^3\,{\left (c\,x^n\right )}^{3/n}+a^3+3\,a\,b^2\,{\left (c\,x^n\right )}^{2/n}+3\,a^2\,b\,{\left (c\,x^n\right )}^{1/n}\right )}-\frac {x}{4\,b\,\left (b^4\,{\left (c\,x^n\right )}^{4/n}+a^4+4\,a\,b^3\,{\left (c\,x^n\right )}^{3/n}+6\,a^2\,b^2\,{\left (c\,x^n\right )}^{2/n}+4\,a^3\,b\,{\left (c\,x^n\right )}^{1/n}\right )}+\frac {x}{12\,a^3\,b\,\left (a+b\,{\left (c\,x^n\right )}^{1/n}\right )}+\frac {x}{12\,a^2\,b\,\left (b^2\,{\left (c\,x^n\right )}^{2/n}+a^2+2\,a\,b\,{\left (c\,x^n\right )}^{1/n}\right )} \]

[In]

int((c*x^n)^(1/n)/(a + b*(c*x^n)^(1/n))^5,x)

[Out]

x/(12*a*b*(b^3*(c*x^n)^(3/n) + a^3 + 3*a*b^2*(c*x^n)^(2/n) + 3*a^2*b*(c*x^n)^(1/n))) - x/(4*b*(b^4*(c*x^n)^(4/
n) + a^4 + 4*a*b^3*(c*x^n)^(3/n) + 6*a^2*b^2*(c*x^n)^(2/n) + 4*a^3*b*(c*x^n)^(1/n))) + x/(12*a^3*b*(a + b*(c*x
^n)^(1/n))) + x/(12*a^2*b*(b^2*(c*x^n)^(2/n) + a^2 + 2*a*b*(c*x^n)^(1/n)))