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

Optimal. Leaf size=50 \[ \frac {2 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {b} x^{-n/2}}{\sqrt {c}}\right )}{b^{3/2} n}-\frac {2 x^{-n/2}}{b n} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1584, 345, 193, 321, 205} \[ \frac {2 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {b} x^{-n/2}}{\sqrt {c}}\right )}{b^{3/2} n}-\frac {2 x^{-n/2}}{b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.01, size = 32, normalized size = 0.64 \[ -\frac {2 x^{-n/2} \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {c x^n}{b}\right )}{b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.84, size = 151, normalized size = 3.02 \[ \left [\frac {x x^{\frac {1}{2} \, n - 1} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} x^{n - 2} - 2 \, b x x^{\frac {1}{2} \, n - 1} \sqrt {-\frac {c}{b}} - b}{c x^{2} x^{n - 2} + b}\right ) - 2}{b n x x^{\frac {1}{2} \, n - 1}}, \frac {2 \, {\left (x x^{\frac {1}{2} \, n - 1} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c x x^{\frac {1}{2} \, n - 1}}\right ) - 1\right )}}{b n x x^{\frac {1}{2} \, n - 1}}\right ] \]

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]

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

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giac [A]  time = 0.38, size = 38, normalized size = 0.76 \[ -\frac {2 \, {\left (\frac {c \arctan \left (\frac {c \sqrt {x^{n}}}{\sqrt {b c}}\right )}{\sqrt {b c} b} + \frac {1}{b \sqrt {x^{n}}}\right )}}{n} \]

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="giac")

[Out]

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

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maple [A]  time = 0.07, size = 79, normalized size = 1.58 \[ -\frac {2 x^{-\frac {n}{2}}}{b n}+\frac {\sqrt {-b c}\, \ln \left (x^{\frac {n}{2}}-\frac {\sqrt {-b c}}{c}\right )}{b^{2} n}-\frac {\sqrt {-b c}\, \ln \left (x^{\frac {n}{2}}+\frac {\sqrt {-b c}}{c}\right )}{b^{2} n} \]

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/b/n/(x^(1/2*n))+1/b^2*(-b*c)^(1/2)/n*ln(x^(1/2*n)-(-b*c)^(1/2)/c)-1/b^2*(-b*c)^(1/2)/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 \[ -c \int \frac {x^{\frac {1}{2} \, n}}{b c x x^{n} + b^{2} x}\,{d x} - \frac {2}{b n x^{\frac {1}{2} \, n}} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^{\frac {n}{2}-1}}{b\,x^n+c\,x^{2\,n}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 13.00, size = 36, normalized size = 0.72 \[ - \frac {2 \operatorname {atan}{\left (\frac {x^{\frac {n}{2}}}{\sqrt {\frac {b}{c}}} \right )}}{b n \sqrt {\frac {b}{c}}} - \frac {2 x^{- \frac {n}{2}}}{b n} \]

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)

[Out]

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

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