3.503 \(\int \frac {x^{-1-\frac {n}{2}}}{b x^n+c x^{2 n}} \, dx\)

Optimal. Leaf size=68 \[ -\frac {2 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^{-n/2}}{\sqrt {c}}\right )}{b^{5/2} n}+\frac {2 c x^{-n/2}}{b^2 n}-\frac {2 x^{-3 n/2}}{3 b n} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1584, 362, 345, 193, 321, 205} \[ -\frac {2 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^{-n/2}}{\sqrt {c}}\right )}{b^{5/2} n}+\frac {2 c x^{-n/2}}{b^2 n}-\frac {2 x^{-3 n/2}}{3 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 193

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 205

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

Rule 321

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 345

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 362

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]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.50 \[ -\frac {2 x^{-3 n/2} \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {c x^n}{b}\right )}{3 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.05, size = 161, normalized size = 2.37 \[ \left [-\frac {2 \, b x^{3} x^{-\frac {3}{2} \, n - 3} - 6 \, c x x^{-\frac {1}{2} \, n - 1} - 3 \, c \sqrt {-\frac {c}{b}} \log \left (\frac {b x^{2} x^{-n - 2} - 2 \, b x x^{-\frac {1}{2} \, n - 1} \sqrt {-\frac {c}{b}} - c}{b x^{2} x^{-n - 2} + c}\right )}{3 \, b^{2} n}, -\frac {2 \, {\left (b x^{3} x^{-\frac {3}{2} \, n - 3} - 3 \, c x x^{-\frac {1}{2} \, n - 1} - 3 \, c \sqrt {\frac {c}{b}} \arctan \left (\frac {\sqrt {\frac {c}{b}}}{x x^{-\frac {1}{2} \, n - 1}}\right )\right )}}{3 \, b^{2} n}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{-\frac {1}{2} \, n - 1}}{c x^{2 \, n} + b x^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 97, normalized size = 1.43 \[ -\frac {2 x^{-\frac {3 n}{2}}}{3 b n}+\frac {2 c \,x^{-\frac {n}{2}}}{b^{2} n}-\frac {\sqrt {-b c}\, c \ln \left (x^{\frac {n}{2}}-\frac {\sqrt {-b c}}{c}\right )}{b^{3} n}+\frac {\sqrt {-b c}\, c \ln \left (x^{\frac {n}{2}}+\frac {\sqrt {-b c}}{c}\right )}{b^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} \int \frac {x^{\frac {1}{2} \, n}}{b^{2} c x x^{n} + b^{3} x}\,{d x} + \frac {2 \, {\left (3 \, c x^{n} - b\right )}}{3 \, b^{2} n x^{\frac {3}{2} \, n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^{\frac {n}{2}+1}\,\left (b\,x^n+c\,x^{2\,n}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 24.92, size = 58, normalized size = 0.85 \[ - \frac {2 x^{- \frac {3 n}{2}}}{3 b n} + \frac {2 c x^{- \frac {n}{2}}}{b^{2} n} - \frac {2 c^{2} \operatorname {atan}{\left (\frac {x^{- \frac {n}{2}}}{\sqrt {\frac {c}{b}}} \right )}}{b^{3} n \sqrt {\frac {c}{b}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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