\(\int \frac {(c x)^{-1-\frac {n}{2}}}{a+b x^n} \, dx\) [2766]

Optimal result

Integrand size = 21, antiderivative size = 74 \[ \int \frac {(c x)^{-1-\frac {n}{2}}}{a+b x^n} \, dx=-\frac {2 (c x)^{-n/2}}{a c n}+\frac {2 \sqrt {b} x^{n/2} (c x)^{-n/2} \arctan \left (\frac {\sqrt {a} x^{-n/2}}{\sqrt {b}}\right )}{a^{3/2} c n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {354, 352, 199, 327, 211} \[ \int \frac {(c x)^{-1-\frac {n}{2}}}{a+b x^n} \, dx=\frac {2 \sqrt {b} x^{n/2} (c x)^{-n/2} \arctan \left (\frac {\sqrt {a} x^{-n/2}}{\sqrt {b}}\right )}{a^{3/2} c n}-\frac {2 (c x)^{-n/2}}{a c n} \]

[In]

Int[(c*x)^(-1 - n/2)/(a + b*x^n),x]

[Out]

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 352

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/(m + 1), Subst[Int[(a + b*x^Simplify[n/(m +
1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] &&  !IntegerQ[n]

Rule 354

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPa
rt[m]), Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] &&  !In
tegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{n/2} (c x)^{-n/2}\right ) \int \frac {x^{-1-\frac {n}{2}}}{a+b x^n} \, dx}{c} \\ & = -\frac {\left (2 x^{n/2} (c x)^{-n/2}\right ) \text {Subst}\left (\int \frac {1}{a+\frac {b}{x^2}} \, dx,x,x^{-n/2}\right )}{c n} \\ & = -\frac {\left (2 x^{n/2} (c x)^{-n/2}\right ) \text {Subst}\left (\int \frac {x^2}{b+a x^2} \, dx,x,x^{-n/2}\right )}{c n} \\ & = -\frac {2 (c x)^{-n/2}}{a c n}+\frac {\left (2 b x^{n/2} (c x)^{-n/2}\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,x^{-n/2}\right )}{a c n} \\ & = -\frac {2 (c x)^{-n/2}}{a c n}+\frac {2 \sqrt {b} x^{n/2} (c x)^{-n/2} \tan ^{-1}\left (\frac {\sqrt {a} x^{-n/2}}{\sqrt {b}}\right )}{a^{3/2} c n} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.50 \[ \int \frac {(c x)^{-1-\frac {n}{2}}}{a+b x^n} \, dx=-\frac {2 x (c x)^{-1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {b x^n}{a}\right )}{a n} \]

[In]

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

[Out]

(-2*x*(c*x)^(-1 - n/2)*Hypergeometric2F1[-1/2, 1, 1/2, -((b*x^n)/a)])/(a*n)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 225, normalized size of antiderivative = 3.04 \[ \int \frac {(c x)^{-1-\frac {n}{2}}}{a+b x^n} \, dx=\left [-\frac {2 \, x e^{\left (-\frac {1}{2} \, {\left (n + 2\right )} \log \left (c\right ) - \frac {1}{2} \, {\left (n + 2\right )} \log \left (x\right )\right )} - \sqrt {-\frac {b c^{-n - 2}}{a}} \log \left (\frac {a x^{2} e^{\left (-{\left (n + 2\right )} \log \left (c\right ) - {\left (n + 2\right )} \log \left (x\right )\right )} + 2 \, a \sqrt {-\frac {b c^{-n - 2}}{a}} x e^{\left (-\frac {1}{2} \, {\left (n + 2\right )} \log \left (c\right ) - \frac {1}{2} \, {\left (n + 2\right )} \log \left (x\right )\right )} - b c^{-n - 2}}{a x^{2} e^{\left (-{\left (n + 2\right )} \log \left (c\right ) - {\left (n + 2\right )} \log \left (x\right )\right )} + b c^{-n - 2}}\right )}{a n}, -\frac {2 \, {\left (x e^{\left (-\frac {1}{2} \, {\left (n + 2\right )} \log \left (c\right ) - \frac {1}{2} \, {\left (n + 2\right )} \log \left (x\right )\right )} + \sqrt {\frac {b c^{-n - 2}}{a}} \arctan \left (\frac {\sqrt {\frac {b c^{-n - 2}}{a}} e^{\left (\frac {1}{2} \, {\left (n + 2\right )} \log \left (c\right ) + \frac {1}{2} \, {\left (n + 2\right )} \log \left (x\right )\right )}}{x}\right )\right )}}{a n}\right ] \]

[In]

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

[Out]

[-(2*x*e^(-1/2*(n + 2)*log(c) - 1/2*(n + 2)*log(x)) - sqrt(-b*c^(-n - 2)/a)*log((a*x^2*e^(-(n + 2)*log(c) - (n
+ 2)*log(x)) + 2*a*sqrt(-b*c^(-n - 2)/a)*x*e^(-1/2*(n + 2)*log(c) - 1/2*(n + 2)*log(x)) - b*c^(-n - 2))/(a*x^
2*e^(-(n + 2)*log(c) - (n + 2)*log(x)) + b*c^(-n - 2))))/(a*n), -2*(x*e^(-1/2*(n + 2)*log(c) - 1/2*(n + 2)*log
(x)) + sqrt(b*c^(-n - 2)/a)*arctan(sqrt(b*c^(-n - 2)/a)*e^(1/2*(n + 2)*log(c) + 1/2*(n + 2)*log(x))/x))/(a*n)]

Sympy [A] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.78 \[ \int \frac {(c x)^{-1-\frac {n}{2}}}{a+b x^n} \, dx=- \frac {2 c^{- \frac {n}{2} - 1} x^{- \frac {n}{2}}}{a n} - \frac {2 \sqrt {b} c^{- \frac {n}{2} - 1} \operatorname {atan}{\left (\frac {\sqrt {b} x^{\frac {n}{2}}}{\sqrt {a}} \right )}}{a^{\frac {3}{2}} n} \]

[In]

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

[Out]

-2*c**(-n/2 - 1)/(a*n*x**(n/2)) - 2*sqrt(b)*c**(-n/2 - 1)*atan(sqrt(b)*x**(n/2)/sqrt(a))/(a**(3/2)*n)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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