\(\int \frac {1}{x^3 (a+b (c x^n)^{\frac {1}{n}})} \, dx\) [3012]

   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 = 87 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )} \, dx=-\frac {1}{2 a x^2}+\frac {b \left (c x^n\right )^{\frac {1}{n}}}{a^2 x^2}+\frac {b^2 \left (c x^n\right )^{2/n} \log (x)}{a^3 x^2}-\frac {b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{a^3 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 87, 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} \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )} \, dx=\frac {b^2 \log (x) \left (c x^n\right )^{2/n}}{a^3 x^2}-\frac {b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{a^3 x^2}+\frac {b \left (c x^n\right )^{\frac {1}{n}}}{a^2 x^2}-\frac {1}{2 a x^2} \]

[In]

Int[1/(x^3*(a + b*(c*x^n)^n^(-1))),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )} \, dx=-\frac {a^2-2 a b \left (c x^n\right )^{\frac {1}{n}}-2 b^2 \left (c x^n\right )^{2/n} \log (x)+2 b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{2 a^3 x^2} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 4.35 (sec) , antiderivative size = 302, normalized size of antiderivative = 3.47

method result size
risch \(-\frac {\left (x^{n}\right )^{\frac {2}{n}} c^{\frac {2}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{n}} b^{2} \ln \left (b \left (x^{n}\right )^{\frac {1}{n}} c^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}+a \right )}{a^{3} x^{2}}+\frac {b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}}{a^{2} x^{2}}-\frac {1}{2 a \,x^{2}}+\frac {c^{\frac {2}{n}} \left (x^{n}\right )^{\frac {2}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{n}} b^{2} \ln \left (x \right )}{a^{3} x^{2}}\) \(302\)

[In]

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

[Out]

-1/a^3*((x^n)^(1/n))^2*(c^(1/n))^2/x^2*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(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^2*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)-1/2/a/x^2+1/a^3/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 = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )} \, dx=-\frac {2 \, b^{2} c^{\frac {2}{n}} x^{2} \log \left (b c^{\left (\frac {1}{n}\right )} x + a\right ) - 2 \, b^{2} c^{\frac {2}{n}} x^{2} \log \left (x\right ) - 2 \, a b c^{\left (\frac {1}{n}\right )} x + a^{2}}{2 \, a^{3} x^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(x^3*(a + b*(c*x^n)^(1/n))),x)

[Out]

int(1/(x^3*(a + b*(c*x^n)^(1/n))), x)