\(\int \frac {x^3}{(a+b (c x^n)^{\frac {1}{n}})^2} \, dx\) [3013]

Optimal result

Integrand size = 19, antiderivative size = 114 \[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=-\frac {2 a x^4 \left (c x^n\right )^{-3/n}}{b^3}+\frac {x^4 \left (c x^n\right )^{-2/n}}{2 b^2}+\frac {a^3 x^4 \left (c x^n\right )^{-4/n}}{b^4 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}+\frac {3 a^2 x^4 \left (c x^n\right )^{-4/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {375, 45} \[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {a^3 x^4 \left (c x^n\right )^{-4/n}}{b^4 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}+\frac {3 a^2 x^4 \left (c x^n\right )^{-4/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{b^4}-\frac {2 a x^4 \left (c x^n\right )^{-3/n}}{b^3}+\frac {x^4 \left (c x^n\right )^{-2/n}}{2 b^2} \]

[In]

Int[x^3/(a + b*(c*x^n)^n^(-1))^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {x^4 \left (c x^n\right )^{-4/n} \left (-4 a b \left (c x^n\right )^{\frac {1}{n}}+b^2 \left (c x^n\right )^{2/n}+\frac {2 a^3}{a+b \left (c x^n\right )^{\frac {1}{n}}}+6 a^2 \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )\right )}{2 b^4} \]

[In]

Integrate[x^3/(a + b*(c*x^n)^n^(-1))^2,x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 4.15 (sec) , antiderivative size = 463, normalized size of antiderivative = 4.06

method	result	size
risch	\(\frac {x^{4}}{a \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 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}}+a \right )}-\frac {\left (x^{n}\right )^{-\frac {1}{n}} c^{-\frac {1}{n}} x^{4} {\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}}}{a b}+\frac {3 \left (x^{n}\right )^{-\frac {2}{n}} c^{-\frac {2}{n}} x^{4} {\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}}}{2 b^{2}}-\frac {3 a \left (x^{n}\right )^{-\frac {3}{n}} c^{-\frac {3}{n}} x^{4} {\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}}}{b^{3}}+\frac {3 a^{2} \ln \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 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}}+a \right ) \left (x^{n}\right )^{-\frac {1}{n}} c^{-\frac {1}{n}} c^{-\frac {3}{n}} \left (x^{n}\right )^{-\frac {3}{n}} x^{4} {\mathrm e}^{-\frac {2 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}}}{b^{4}}\)	\(463\)

[In]

int(x^3/(a+b*(c*x^n)^(1/n))^2,x,method=_RETURNVERBOSE)

[Out]

x^4/a/(b*(x^n)^(1/n)*c^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))
/n)+a)-1/a/((x^n)^(1/n))/(c^(1/n))*x^4*exp(-1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csg
n(I*c*x^n))/n)/b+3/2/((x^n)^(1/n))^2/(c^(1/n))^2*x^4*exp(-I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csg
n(I*c)-csgn(I*c*x^n))/n)/b^2-3*a/((x^n)^(1/n))^3/(c^(1/n))^3*x^4*exp(-3/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csg
n(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)/b^3+3*a^2*ln(b*(x^n)^(1/n)*c^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(
I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)+a)/((x^n)^(1/n))/(c^(1/n))*c^(-3/n)*(x^n)^(-3/n)*x^4/b^4*ex
p(-2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {b^{3} c^{\frac {3}{n}} x^{3} - 3 \, a b^{2} c^{\frac {2}{n}} x^{2} - 4 \, a^{2} b c^{\left (\frac {1}{n}\right )} x + 2 \, a^{3} + 6 \, {\left (a^{2} b c^{\left (\frac {1}{n}\right )} x + a^{3}\right )} \log \left (b c^{\left (\frac {1}{n}\right )} x + a\right )}{2 \, {\left (b^{5} c^{\frac {5}{n}} x + a b^{4} c^{\frac {4}{n}}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\int \frac {x^{3}}{\left (a + b \left (c x^{n}\right )^{\frac {1}{n}}\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\int { \frac {x^{3}}{{\left (\left (c x^{n}\right )^{\left (\frac {1}{n}\right )} b + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\int { \frac {x^{3}}{{\left (\left (c x^{n}\right )^{\left (\frac {1}{n}\right )} b + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\int \frac {x^3}{{\left (a+b\,{\left (c\,x^n\right )}^{1/n}\right )}^2} \,d x \]

[In]

int(x^3/(a + b*(c*x^n)^(1/n))^2,x)

[Out]

int(x^3/(a + b*(c*x^n)^(1/n))^2, x)