3.2645 \(\int \frac{x^{-1-\frac{n}{4}}}{a+b x^n} \, dx\)

Optimal. Leaf size=234 \[ -\frac{\sqrt [4]{b} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x^{-n/4}+\sqrt{a} x^{-n/2}+\sqrt{b}\right )}{\sqrt{2} a^{5/4} n}+\frac{\sqrt [4]{b} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x^{-n/4}+\sqrt{a} x^{-n/2}+\sqrt{b}\right )}{\sqrt{2} a^{5/4} n}-\frac{\sqrt{2} \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{b}}\right )}{a^{5/4} n}+\frac{\sqrt{2} \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{b}}+1\right )}{a^{5/4} n}-\frac{4 x^{-n/4}}{a n} \]

[Out]

-4/(a*n*x^(n/4)) - (Sqrt[2]*b^(1/4)*ArcTan[1 - (Sqrt[2]*a^(1/4))/(b^(1/4)*x^(n/4))])/(a^(5/4)*n) + (Sqrt[2]*b^
(1/4)*ArcTan[1 + (Sqrt[2]*a^(1/4))/(b^(1/4)*x^(n/4))])/(a^(5/4)*n) - (b^(1/4)*Log[Sqrt[b] + Sqrt[a]/x^(n/2) -
(Sqrt[2]*a^(1/4)*b^(1/4))/x^(n/4)])/(Sqrt[2]*a^(5/4)*n) + (b^(1/4)*Log[Sqrt[b] + Sqrt[a]/x^(n/2) + (Sqrt[2]*a^
(1/4)*b^(1/4))/x^(n/4)])/(Sqrt[2]*a^(5/4)*n)

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Rubi [A]  time = 0.184002, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {345, 193, 321, 211, 1165, 628, 1162, 617, 204} \[ -\frac{\sqrt [4]{b} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x^{-n/4}+\sqrt{a} x^{-n/2}+\sqrt{b}\right )}{\sqrt{2} a^{5/4} n}+\frac{\sqrt [4]{b} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x^{-n/4}+\sqrt{a} x^{-n/2}+\sqrt{b}\right )}{\sqrt{2} a^{5/4} n}-\frac{\sqrt{2} \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{b}}\right )}{a^{5/4} n}+\frac{\sqrt{2} \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{b}}+1\right )}{a^{5/4} n}-\frac{4 x^{-n/4}}{a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - n/4)/(a + b*x^n),x]

[Out]

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

Mathematica [C]  time = 0.006573, size = 32, normalized size = 0.14 \[ -\frac{4 x^{-n/4} \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\frac{b x^n}{a}\right )}{a n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - n/4)/(a + b*x^n),x]

[Out]

(-4*Hypergeometric2F1[-1/4, 1, 3/4, -((b*x^n)/a)])/(a*n*x^(n/4))

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Maple [C]  time = 0.051, size = 56, normalized size = 0.2 \begin{align*} -4\,{\frac{1}{an{x}^{n/4}}}+\sum _{{\it \_R}={\it RootOf} \left ({a}^{5}{n}^{4}{{\it \_Z}}^{4}+b \right ) }{\it \_R}\,\ln \left ({x}^{{\frac{n}{4}}}-{\frac{{a}^{4}{n}^{3}{{\it \_R}}^{3}}{b}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -b \int \frac{x^{\frac{3}{4} \, n}}{a b x x^{n} + a^{2} x}\,{d x} - \frac{4}{a n x^{\frac{1}{4} \, n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*integrate(x^(3/4*n)/(a*b*x*x^n + a^2*x), x) - 4/(a*n*x^(1/4*n))

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Fricas [A]  time = 1.14022, size = 500, normalized size = 2.14 \begin{align*} \frac{4 \, a n \left (-\frac{b}{a^{5} n^{4}}\right )^{\frac{1}{4}} \arctan \left (-\frac{a^{4} n^{3} x x^{-\frac{1}{4} \, n - 1} \left (-\frac{b}{a^{5} n^{4}}\right )^{\frac{3}{4}} - a^{4} n^{3} x \sqrt{\frac{a^{2} n^{2} \sqrt{-\frac{b}{a^{5} n^{4}}} + x^{2} x^{-\frac{1}{2} \, n - 2}}{x^{2}}} \left (-\frac{b}{a^{5} n^{4}}\right )^{\frac{3}{4}}}{b}\right ) + a n \left (-\frac{b}{a^{5} n^{4}}\right )^{\frac{1}{4}} \log \left (\frac{a n \left (-\frac{b}{a^{5} n^{4}}\right )^{\frac{1}{4}} + x x^{-\frac{1}{4} \, n - 1}}{x}\right ) - a n \left (-\frac{b}{a^{5} n^{4}}\right )^{\frac{1}{4}} \log \left (-\frac{a n \left (-\frac{b}{a^{5} n^{4}}\right )^{\frac{1}{4}} - x x^{-\frac{1}{4} \, n - 1}}{x}\right ) - 4 \, x x^{-\frac{1}{4} \, n - 1}}{a n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a*n*(-b/(a^5*n^4))^(1/4)*arctan(-(a^4*n^3*x*x^(-1/4*n - 1)*(-b/(a^5*n^4))^(3/4) - a^4*n^3*x*sqrt((a^2*n^2*s
qrt(-b/(a^5*n^4)) + x^2*x^(-1/2*n - 2))/x^2)*(-b/(a^5*n^4))^(3/4))/b) + a*n*(-b/(a^5*n^4))^(1/4)*log((a*n*(-b/
(a^5*n^4))^(1/4) + x*x^(-1/4*n - 1))/x) - a*n*(-b/(a^5*n^4))^(1/4)*log(-(a*n*(-b/(a^5*n^4))^(1/4) - x*x^(-1/4*
n - 1))/x) - 4*x*x^(-1/4*n - 1))/(a*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)/(a+b*x**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}}{b x^{n} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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