\(\int \frac {x^{-1-\frac {3 n}{2}}}{a+b x^n} \, dx\) [2646]

Optimal result

Integrand size = 19, antiderivative size = 68 \[ \int \frac {x^{-1-\frac {3 n}{2}}}{a+b x^n} \, dx=-\frac {2 x^{-3 n/2}}{3 a n}+\frac {2 b x^{-n/2}}{a^2 n}-\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {a} x^{-n/2}}{\sqrt {b}}\right )}{a^{5/2} n} \]

[Out]

-2/3/a/n/(x^(3/2*n))+2*b/a^2/n/(x^(1/2*n))-2*b^(3/2)*arctan(a^(1/2)/(x^(1/2*n))/b^(1/2))/a^(5/2)/n

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {369, 352, 199, 327, 211} \[ \int \frac {x^{-1-\frac {3 n}{2}}}{a+b x^n} \, dx=-\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {a} x^{-n/2}}{\sqrt {b}}\right )}{a^{5/2} n}+\frac {2 b x^{-n/2}}{a^2 n}-\frac {2 x^{-3 n/2}}{3 a n} \]

[In]

Int[x^(-1 - (3*n)/2)/(a + b*x^n),x]

[Out]

-2/(3*a*n*x^((3*n)/2)) + (2*b)/(a^2*n*x^(n/2)) - (2*b^(3/2)*ArcTan[Sqrt[a]/(Sqrt[b]*x^(n/2))])/(a^(5/2)*n)

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 369

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[x^(m + 1)/(a*(m + 1)), x] - Dist[b/a, Int[x^Simplify
[m + n]/(a + b*x^n), x], x] /; FreeQ[{a, b, m, n}, x] && FractionQ[Simplify[(m + 1)/n]] && SumSimplerQ[m, n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 x^{-3 n/2}}{3 a n}-\frac {b \int \frac {x^{-1-\frac {n}{2}}}{a+b x^n} \, dx}{a} \\ & = -\frac {2 x^{-3 n/2}}{3 a n}+\frac {(2 b) \text {Subst}\left (\int \frac {1}{a+\frac {b}{x^2}} \, dx,x,x^{-n/2}\right )}{a n} \\ & = -\frac {2 x^{-3 n/2}}{3 a n}+\frac {(2 b) \text {Subst}\left (\int \frac {x^2}{b+a x^2} \, dx,x,x^{-n/2}\right )}{a n} \\ & = -\frac {2 x^{-3 n/2}}{3 a n}+\frac {2 b x^{-n/2}}{a^2 n}-\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,x^{-n/2}\right )}{a^2 n} \\ & = -\frac {2 x^{-3 n/2}}{3 a n}+\frac {2 b x^{-n/2}}{a^2 n}-\frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} x^{-n/2}}{\sqrt {b}}\right )}{a^{5/2} n} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.50 \[ \int \frac {x^{-1-\frac {3 n}{2}}}{a+b x^n} \, dx=-\frac {2 x^{-3 n/2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {b x^n}{a}\right )}{3 a n} \]

[In]

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

[Out]

(-2*Hypergeometric2F1[-3/2, 1, -1/2, -((b*x^n)/a)])/(3*a*n*x^((3*n)/2))

Maple [A] (verified)

Time = 3.80 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.43

method	result	size
risch	\(\frac {2 b \,x^{-\frac {n}{2}}}{a^{2} n}-\frac {2 x^{-\frac {3 n}{2}}}{3 a n}+\frac {\sqrt {-a b}\, b \ln \left (x^{\frac {n}{2}}+\frac {\sqrt {-a b}}{b}\right )}{a^{3} n}-\frac {\sqrt {-a b}\, b \ln \left (x^{\frac {n}{2}}-\frac {\sqrt {-a b}}{b}\right )}{a^{3} n}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 251, normalized size of antiderivative = 3.69 \[ \int \frac {x^{-1-\frac {3 n}{2}}}{a+b x^n} \, dx=\left [-\frac {2 \, a x x^{-\frac {3}{2} \, n - 1} - 3 \, b \sqrt {-\frac {b}{a}} \log \left (-\frac {2 \, a^{3} x^{\frac {5}{3}} x^{-\frac {5}{2} \, n - \frac {5}{3}} \sqrt {-\frac {b}{a}} - a^{3} x^{2} x^{-3 \, n - 2} - 2 \, a^{2} b x x^{-\frac {3}{2} \, n - 1} \sqrt {-\frac {b}{a}} + 2 \, a^{2} b x^{\frac {4}{3}} x^{-2 \, n - \frac {4}{3}} + 2 \, a b^{2} x^{\frac {1}{3}} x^{-\frac {1}{2} \, n - \frac {1}{3}} \sqrt {-\frac {b}{a}} - 2 \, a b^{2} x^{\frac {2}{3}} x^{-n - \frac {2}{3}} + b^{3}}{a^{3} x^{2} x^{-3 \, n - 2} + b^{3}}\right ) - 6 \, b x^{\frac {1}{3}} x^{-\frac {1}{2} \, n - \frac {1}{3}}}{3 \, a^{2} n}, -\frac {2 \, {\left (a x x^{-\frac {3}{2} \, n - 1} - 3 \, b \sqrt {\frac {b}{a}} \arctan \left (\frac {\sqrt {\frac {b}{a}}}{x^{\frac {1}{3}} x^{-\frac {1}{2} \, n - \frac {1}{3}}}\right ) - 3 \, b x^{\frac {1}{3}} x^{-\frac {1}{2} \, n - \frac {1}{3}}\right )}}{3 \, a^{2} n}\right ] \]

[In]

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

[Out]

[-1/3*(2*a*x*x^(-3/2*n - 1) - 3*b*sqrt(-b/a)*log(-(2*a^3*x^(5/3)*x^(-5/2*n - 5/3)*sqrt(-b/a) - a^3*x^2*x^(-3*n
- 2) - 2*a^2*b*x*x^(-3/2*n - 1)*sqrt(-b/a) + 2*a^2*b*x^(4/3)*x^(-2*n - 4/3) + 2*a*b^2*x^(1/3)*x^(-1/2*n - 1/3
)*sqrt(-b/a) - 2*a*b^2*x^(2/3)*x^(-n - 2/3) + b^3)/(a^3*x^2*x^(-3*n - 2) + b^3)) - 6*b*x^(1/3)*x^(-1/2*n - 1/3
))/(a^2*n), -2/3*(a*x*x^(-3/2*n - 1) - 3*b*sqrt(b/a)*arctan(sqrt(b/a)/(x^(1/3)*x^(-1/2*n - 1/3))) - 3*b*x^(1/3
)*x^(-1/2*n - 1/3))/(a^2*n)]

Sympy [A] (verification not implemented)

Time = 1.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \frac {x^{-1-\frac {3 n}{2}}}{a+b x^n} \, dx=- \frac {2 x^{- \frac {3 n}{2}}}{3 a n} + \frac {2 b x^{- \frac {n}{2}}}{a^{2} n} + \frac {2 b^{\frac {3}{2}} \operatorname {atan}{\left (\frac {\sqrt {b} x^{\frac {n}{2}}}{\sqrt {a}} \right )}}{a^{\frac {5}{2}} n} \]

[In]

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

[Out]

-2/(3*a*n*x**(3*n/2)) + 2*b/(a**2*n*x**(n/2)) + 2*b**(3/2)*atan(sqrt(b)*x**(n/2)/sqrt(a))/(a**(5/2)*n)

Maxima [F]

\[ \int \frac {x^{-1-\frac {3 n}{2}}}{a+b x^n} \, dx=\int { \frac {x^{-\frac {3}{2} \, n - 1}}{b x^{n} + a} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x^{-1-\frac {3 n}{2}}}{a+b x^n} \, dx=\int { \frac {x^{-\frac {3}{2} \, n - 1}}{b x^{n} + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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