3.505 \(\int \frac{x^{-1-\frac{n}{4}}}{b x^n+c x^{2 n}} \, dx\)

Optimal. Leaf size=252 \[ \frac{c^{5/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} x^{-n/4}+\sqrt{b} x^{-n/2}+\sqrt{c}\right )}{\sqrt{2} b^{9/4} n}-\frac{c^{5/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} x^{-n/4}+\sqrt{b} x^{-n/2}+\sqrt{c}\right )}{\sqrt{2} b^{9/4} n}+\frac{\sqrt{2} c^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x^{-n/4}}{\sqrt [4]{c}}\right )}{b^{9/4} n}-\frac{\sqrt{2} c^{5/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x^{-n/4}}{\sqrt [4]{c}}+1\right )}{b^{9/4} n}+\frac{4 c x^{-n/4}}{b^2 n}-\frac{4 x^{-5 n/4}}{5 b n} \]

[Out]

-4/(5*b*n*x^((5*n)/4)) + (4*c)/(b^2*n*x^(n/4)) + (Sqrt[2]*c^(5/4)*ArcTan[1 - (Sqrt[2]*b^(1/4))/(c^(1/4)*x^(n/4
))])/(b^(9/4)*n) - (Sqrt[2]*c^(5/4)*ArcTan[1 + (Sqrt[2]*b^(1/4))/(c^(1/4)*x^(n/4))])/(b^(9/4)*n) + (c^(5/4)*Lo
g[Sqrt[c] + Sqrt[b]/x^(n/2) - (Sqrt[2]*b^(1/4)*c^(1/4))/x^(n/4)])/(Sqrt[2]*b^(9/4)*n) - (c^(5/4)*Log[Sqrt[c] +
Sqrt[b]/x^(n/2) + (Sqrt[2]*b^(1/4)*c^(1/4))/x^(n/4)])/(Sqrt[2]*b^(9/4)*n)

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Rubi [A]  time = 0.224505, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {1584, 362, 345, 193, 321, 211, 1165, 628, 1162, 617, 204} \[ \frac{c^{5/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} x^{-n/4}+\sqrt{b} x^{-n/2}+\sqrt{c}\right )}{\sqrt{2} b^{9/4} n}-\frac{c^{5/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} x^{-n/4}+\sqrt{b} x^{-n/2}+\sqrt{c}\right )}{\sqrt{2} b^{9/4} n}+\frac{\sqrt{2} c^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x^{-n/4}}{\sqrt [4]{c}}\right )}{b^{9/4} n}-\frac{\sqrt{2} c^{5/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x^{-n/4}}{\sqrt [4]{c}}+1\right )}{b^{9/4} n}+\frac{4 c x^{-n/4}}{b^2 n}-\frac{4 x^{-5 n/4}}{5 b n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - n/4)/(b*x^n + c*x^(2*n)),x]

[Out]

-4/(5*b*n*x^((5*n)/4)) + (4*c)/(b^2*n*x^(n/4)) + (Sqrt[2]*c^(5/4)*ArcTan[1 - (Sqrt[2]*b^(1/4))/(c^(1/4)*x^(n/4
))])/(b^(9/4)*n) - (Sqrt[2]*c^(5/4)*ArcTan[1 + (Sqrt[2]*b^(1/4))/(c^(1/4)*x^(n/4))])/(b^(9/4)*n) + (c^(5/4)*Lo
g[Sqrt[c] + Sqrt[b]/x^(n/2) - (Sqrt[2]*b^(1/4)*c^(1/4))/x^(n/4)])/(Sqrt[2]*b^(9/4)*n) - (c^(5/4)*Log[Sqrt[c] +
Sqrt[b]/x^(n/2) + (Sqrt[2]*b^(1/4)*c^(1/4))/x^(n/4)])/(Sqrt[2]*b^(9/4)*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]

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 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 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 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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^{-1-\frac{n}{4}}}{b x^n+c x^{2 n}} \, dx &=\int \frac{x^{-1-\frac{5 n}{4}}}{b+c x^n} \, dx\\ &=-\frac{4 x^{-5 n/4}}{5 b n}-\frac{c \int \frac{x^{-1-\frac{n}{4}}}{b+c x^n} \, dx}{b}\\ &=-\frac{4 x^{-5 n/4}}{5 b n}+\frac{(4 c) \operatorname{Subst}\left (\int \frac{1}{b+\frac{c}{x^4}} \, dx,x,x^{-n/4}\right )}{b n}\\ &=-\frac{4 x^{-5 n/4}}{5 b n}+\frac{(4 c) \operatorname{Subst}\left (\int \frac{x^4}{c+b x^4} \, dx,x,x^{-n/4}\right )}{b n}\\ &=-\frac{4 x^{-5 n/4}}{5 b n}+\frac{4 c x^{-n/4}}{b^2 n}-\frac{\left (4 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{c+b x^4} \, dx,x,x^{-n/4}\right )}{b^2 n}\\ &=-\frac{4 x^{-5 n/4}}{5 b n}+\frac{4 c x^{-n/4}}{b^2 n}-\frac{\left (2 c^{3/2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c}-\sqrt{b} x^2}{c+b x^4} \, dx,x,x^{-n/4}\right )}{b^2 n}-\frac{\left (2 c^{3/2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c}+\sqrt{b} x^2}{c+b x^4} \, dx,x,x^{-n/4}\right )}{b^2 n}\\ &=-\frac{4 x^{-5 n/4}}{5 b n}+\frac{4 c x^{-n/4}}{b^2 n}+\frac{c^{5/4} \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{c}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{c}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{b}}-x^2} \, dx,x,x^{-n/4}\right )}{\sqrt{2} b^{9/4} n}+\frac{c^{5/4} \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{c}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{c}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{b}}-x^2} \, dx,x,x^{-n/4}\right )}{\sqrt{2} b^{9/4} n}-\frac{c^{3/2} \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{c}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{b}}+x^2} \, dx,x,x^{-n/4}\right )}{b^{5/2} n}-\frac{c^{3/2} \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{c}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{b}}+x^2} \, dx,x,x^{-n/4}\right )}{b^{5/2} n}\\ &=-\frac{4 x^{-5 n/4}}{5 b n}+\frac{4 c x^{-n/4}}{b^2 n}+\frac{c^{5/4} \log \left (\sqrt{c}+\sqrt{b} x^{-n/2}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} x^{-n/4}\right )}{\sqrt{2} b^{9/4} n}-\frac{c^{5/4} \log \left (\sqrt{c}+\sqrt{b} x^{-n/2}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} x^{-n/4}\right )}{\sqrt{2} b^{9/4} n}-\frac{\left (\sqrt{2} c^{5/4}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} x^{-n/4}}{\sqrt [4]{c}}\right )}{b^{9/4} n}+\frac{\left (\sqrt{2} c^{5/4}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} x^{-n/4}}{\sqrt [4]{c}}\right )}{b^{9/4} n}\\ &=-\frac{4 x^{-5 n/4}}{5 b n}+\frac{4 c x^{-n/4}}{b^2 n}+\frac{\sqrt{2} c^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x^{-n/4}}{\sqrt [4]{c}}\right )}{b^{9/4} n}-\frac{\sqrt{2} c^{5/4} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} x^{-n/4}}{\sqrt [4]{c}}\right )}{b^{9/4} n}+\frac{c^{5/4} \log \left (\sqrt{c}+\sqrt{b} x^{-n/2}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} x^{-n/4}\right )}{\sqrt{2} b^{9/4} n}-\frac{c^{5/4} \log \left (\sqrt{c}+\sqrt{b} x^{-n/2}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} x^{-n/4}\right )}{\sqrt{2} b^{9/4} n}\\ \end{align*}

Mathematica [C]  time = 0.0081725, size = 34, normalized size = 0.13 \[ -\frac{4 x^{-5 n/4} \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\frac{c x^n}{b}\right )}{5 b n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - n/4)/(b*x^n + c*x^(2*n)),x]

[Out]

(-4*Hypergeometric2F1[-5/4, 1, -1/4, -((c*x^n)/b)])/(5*b*n*x^((5*n)/4))

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Maple [C]  time = 0.075, size = 73, normalized size = 0.3 \begin{align*} 4\,{\frac{c}{{b}^{2}n{x}^{n/4}}}-{\frac{4}{5\,bn} \left ({x}^{{\frac{n}{4}}} \right ) ^{-5}}+\sum _{{\it \_R}={\it RootOf} \left ({b}^{9}{n}^{4}{{\it \_Z}}^{4}+{c}^{5} \right ) }{\it \_R}\,\ln \left ({x}^{{\frac{n}{4}}}+{\frac{{b}^{7}{n}^{3}{{\it \_R}}^{3}}{{c}^{4}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1-1/4*n)/(b*x^n+c*x^(2*n)),x)

[Out]

4*c/b^2/n/(x^(1/4*n))-4/5/b/n/(x^(1/4*n))^5+sum(_R*ln(x^(1/4*n)+b^7*n^3/c^4*_R^3),_R=RootOf(_Z^4*b^9*n^4+c^5))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} c^{2} \int \frac{x^{\frac{3}{4} \, n}}{b^{2} c x x^{n} + b^{3} x}\,{d x} + \frac{4 \,{\left (5 \, c x^{n} - b\right )}}{5 \, b^{2} n x^{\frac{5}{4} \, n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-1/4*n)/(b*x^n+c*x^(2*n)),x, algorithm="maxima")

[Out]

c^2*integrate(x^(3/4*n)/(b^2*c*x*x^n + b^3*x), x) + 4/5*(5*c*x^n - b)/(b^2*n*x^(5/4*n))

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Fricas [A]  time = 1.72154, size = 605, normalized size = 2.4 \begin{align*} -\frac{4 \, b x^{5} x^{-\frac{5}{4} \, n - 5} + 20 \, b^{2} n \left (-\frac{c^{5}}{b^{9} n^{4}}\right )^{\frac{1}{4}} \arctan \left (-\frac{b^{7} c n^{3} x x^{-\frac{1}{4} \, n - 1} \left (-\frac{c^{5}}{b^{9} n^{4}}\right )^{\frac{3}{4}} - b^{7} n^{3} x \sqrt{\frac{b^{4} n^{2} \sqrt{-\frac{c^{5}}{b^{9} n^{4}}} + c^{2} x^{2} x^{-\frac{1}{2} \, n - 2}}{x^{2}}} \left (-\frac{c^{5}}{b^{9} n^{4}}\right )^{\frac{3}{4}}}{c^{5}}\right ) + 5 \, b^{2} n \left (-\frac{c^{5}}{b^{9} n^{4}}\right )^{\frac{1}{4}} \log \left (\frac{b^{2} n \left (-\frac{c^{5}}{b^{9} n^{4}}\right )^{\frac{1}{4}} + c x x^{-\frac{1}{4} \, n - 1}}{x}\right ) - 5 \, b^{2} n \left (-\frac{c^{5}}{b^{9} n^{4}}\right )^{\frac{1}{4}} \log \left (-\frac{b^{2} n \left (-\frac{c^{5}}{b^{9} n^{4}}\right )^{\frac{1}{4}} - c x x^{-\frac{1}{4} \, n - 1}}{x}\right ) - 20 \, c x x^{-\frac{1}{4} \, n - 1}}{5 \, b^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-1/4*n)/(b*x^n+c*x^(2*n)),x, algorithm="fricas")

[Out]

-1/5*(4*b*x^5*x^(-5/4*n - 5) + 20*b^2*n*(-c^5/(b^9*n^4))^(1/4)*arctan(-(b^7*c*n^3*x*x^(-1/4*n - 1)*(-c^5/(b^9*
n^4))^(3/4) - b^7*n^3*x*sqrt((b^4*n^2*sqrt(-c^5/(b^9*n^4)) + c^2*x^2*x^(-1/2*n - 2))/x^2)*(-c^5/(b^9*n^4))^(3/
4))/c^5) + 5*b^2*n*(-c^5/(b^9*n^4))^(1/4)*log((b^2*n*(-c^5/(b^9*n^4))^(1/4) + c*x*x^(-1/4*n - 1))/x) - 5*b^2*n
*(-c^5/(b^9*n^4))^(1/4)*log(-(b^2*n*(-c^5/(b^9*n^4))^(1/4) - c*x*x^(-1/4*n - 1))/x) - 20*c*x*x^(-1/4*n - 1))/(
b^2*n)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1-1/4*n)/(b*x**n+c*x**(2*n)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{-\frac{1}{4} \, n - 1}}{c x^{2 \, n} + b x^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-1/4*n)/(b*x^n+c*x^(2*n)),x, algorithm="giac")

[Out]

integrate(x^(-1/4*n - 1)/(c*x^(2*n) + b*x^n), x)