\(\int \frac {1}{x^3 (a+b (c x^n)^{\frac {1}{n}})^2} \, dx\) [3019]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 19, antiderivative size = 125 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=-\frac {1}{2 a^2 x^2}+\frac {2 b \left (c x^n\right )^{\frac {1}{n}}}{a^3 x^2}+\frac {b^2 \left (c x^n\right )^{2/n}}{a^3 x^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}+\frac {3 b^2 \left (c x^n\right )^{2/n} \log (x)}{a^4 x^2}-\frac {3 b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{a^4 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 125, 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, 46} \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {3 b^2 \log (x) \left (c x^n\right )^{2/n}}{a^4 x^2}-\frac {3 b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{a^4 x^2}+\frac {b^2 \left (c x^n\right )^{2/n}}{a^3 x^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}+\frac {2 b \left (c x^n\right )^{\frac {1}{n}}}{a^3 x^2}-\frac {1}{2 a^2 x^2} \]

[In]

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

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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 )^{2/n} \text {Subst}\left (\int \frac {1}{x^3 (a+b x)^2} \, dx,x,\left (c x^n\right )^{\frac {1}{n}}\right )}{x^2} \\ & = \frac {\left (c x^n\right )^{2/n} \text {Subst}\left (\int \left (\frac {1}{a^2 x^3}-\frac {2 b}{a^3 x^2}+\frac {3 b^2}{a^4 x}-\frac {b^3}{a^3 (a+b x)^2}-\frac {3 b^3}{a^4 (a+b x)}\right ) \, dx,x,\left (c x^n\right )^{\frac {1}{n}}\right )}{x^2} \\ & = -\frac {1}{2 a^2 x^2}+\frac {2 b \left (c x^n\right )^{\frac {1}{n}}}{a^3 x^2}+\frac {b^2 \left (c x^n\right )^{2/n}}{a^3 x^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}+\frac {3 b^2 \left (c x^n\right )^{2/n} \log (x)}{a^4 x^2}-\frac {3 b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{a^4 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {\left (c x^n\right )^{2/n} \left (a \left (-a \left (c x^n\right )^{-2/n}+4 b \left (c x^n\right )^{-1/n}+\frac {2 b^2}{a+b \left (c x^n\right )^{\frac {1}{n}}}\right )+6 b^2 \log (x)-6 b^2 \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )\right )}{2 a^4 x^2} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 4.56 (sec) , antiderivative size = 379, normalized size of antiderivative = 3.03

method result size
risch \(\frac {1}{a \,x^{2} \left (b \left (x^{n}\right )^{\frac {1}{n}} c^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}+a \right )}-\frac {3 \ln \left (b \left (x^{n}\right )^{\frac {1}{n}} c^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}+a \right ) b^{2} c^{\frac {2}{n}} \left (x^{n}\right )^{\frac {2}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{n}}}{a^{4} x^{2}}+\frac {3 b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}}{a^{3} x^{2}}-\frac {3}{2 a^{2} x^{2}}+\frac {3 c^{\frac {2}{n}} \left (x^{n}\right )^{\frac {2}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{n}} b^{2} \ln \left (x \right )}{a^{4} x^{2}}\) \(379\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {6 \, b^{3} c^{\frac {3}{n}} x^{3} \log \left (x\right ) + 3 \, a^{2} b c^{\left (\frac {1}{n}\right )} x - a^{3} + 6 \, {\left (a b^{2} x^{2} \log \left (x\right ) + a b^{2} x^{2}\right )} c^{\frac {2}{n}} - 6 \, {\left (b^{3} c^{\frac {3}{n}} x^{3} + a b^{2} c^{\frac {2}{n}} x^{2}\right )} \log \left (b c^{\left (\frac {1}{n}\right )} x + a\right )}{2 \, {\left (a^{4} b c^{\left (\frac {1}{n}\right )} x^{3} + a^{5} x^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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