3.2648 \(\int \frac{x^{-1-\frac{5 n}{4}}}{a+b x^n} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.197332, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.526, Rules used = {362, 345, 193, 321, 211, 1165, 628, 1162, 617, 204} \[ \frac{b^{5/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x^{-n/4}+\sqrt{a} x^{-n/2}+\sqrt{b}\right )}{\sqrt{2} a^{9/4} n}-\frac{b^{5/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x^{-n/4}+\sqrt{a} x^{-n/2}+\sqrt{b}\right )}{\sqrt{2} a^{9/4} n}+\frac{\sqrt{2} b^{5/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{b}}\right )}{a^{9/4} n}-\frac{\sqrt{2} b^{5/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{b}}+1\right )}{a^{9/4} n}+\frac{4 b x^{-n/4}}{a^2 n}-\frac{4 x^{-5 n/4}}{5 a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - (5*n)/4)/(a + b*x^n),x]

[Out]

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

Mathematica [C]  time = 0.0071868, size = 34, normalized size = 0.13 \[ -\frac{4 x^{-5 n/4} \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\frac{b x^n}{a}\right )}{5 a n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - (5*n)/4)/(a + b*x^n),x]

[Out]

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

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Maple [C]  time = 0.072, size = 73, normalized size = 0.3 \begin{align*} 4\,{\frac{b}{{a}^{2}n{x}^{n/4}}}-{\frac{4}{5\,an} \left ({x}^{{\frac{n}{4}}} \right ) ^{-5}}+\sum _{{\it \_R}={\it RootOf} \left ({a}^{9}{n}^{4}{{\it \_Z}}^{4}+{b}^{5} \right ) }{\it \_R}\,\ln \left ({x}^{{\frac{n}{4}}}+{\frac{{a}^{7}{n}^{3}{{\it \_R}}^{3}}{{b}^{4}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1-5/4*n)/(a+b*x^n),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-5/4*n)/(a+b*x^n),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.45495, size = 667, normalized size = 2.65 \begin{align*} -\frac{20 \, a^{2} n \left (-\frac{b^{5}}{a^{9} n^{4}}\right )^{\frac{1}{4}} \arctan \left (-\frac{a^{7} b n^{3} x^{\frac{1}{5}} x^{-\frac{1}{4} \, n - \frac{1}{5}} \left (-\frac{b^{5}}{a^{9} n^{4}}\right )^{\frac{3}{4}} - a^{7} n^{3} x^{\frac{1}{5}} \sqrt{\frac{a^{4} n^{2} x^{\frac{3}{5}} \sqrt{-\frac{b^{5}}{a^{9} n^{4}}} + b^{2} x x^{-\frac{1}{2} \, n - \frac{2}{5}}}{x}} \left (-\frac{b^{5}}{a^{9} n^{4}}\right )^{\frac{3}{4}}}{b^{5}}\right ) + 5 \, a^{2} n \left (-\frac{b^{5}}{a^{9} n^{4}}\right )^{\frac{1}{4}} \log \left (\frac{a^{2} n x^{\frac{4}{5}} \left (-\frac{b^{5}}{a^{9} n^{4}}\right )^{\frac{1}{4}} + b x x^{-\frac{1}{4} \, n - \frac{1}{5}}}{x}\right ) - 5 \, a^{2} n \left (-\frac{b^{5}}{a^{9} n^{4}}\right )^{\frac{1}{4}} \log \left (-\frac{a^{2} n x^{\frac{4}{5}} \left (-\frac{b^{5}}{a^{9} n^{4}}\right )^{\frac{1}{4}} - b x x^{-\frac{1}{4} \, n - \frac{1}{5}}}{x}\right ) + 4 \, a x x^{-\frac{5}{4} \, n - 1} - 20 \, b x^{\frac{1}{5}} x^{-\frac{1}{4} \, n - \frac{1}{5}}}{5 \, a^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-5/4*n)/(a+b*x^n),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 5.54571, size = 309, normalized size = 1.23 \begin{align*} \frac{x^{- \frac{5 n}{4}} \Gamma \left (- \frac{5}{4}\right )}{a n \Gamma \left (- \frac{1}{4}\right )} - \frac{5 b x^{- \frac{n}{4}} \Gamma \left (- \frac{5}{4}\right )}{a^{2} n \Gamma \left (- \frac{1}{4}\right )} + \frac{5 b^{\frac{5}{4}} e^{- \frac{3 i \pi }{4}} \log{\left (1 - \frac{\sqrt [4]{b} x^{\frac{n}{4}} e^{\frac{i \pi }{4}}}{\sqrt [4]{a}} \right )} \Gamma \left (- \frac{5}{4}\right )}{4 a^{\frac{9}{4}} n \Gamma \left (- \frac{1}{4}\right )} + \frac{5 i b^{\frac{5}{4}} e^{- \frac{3 i \pi }{4}} \log{\left (1 - \frac{\sqrt [4]{b} x^{\frac{n}{4}} e^{\frac{3 i \pi }{4}}}{\sqrt [4]{a}} \right )} \Gamma \left (- \frac{5}{4}\right )}{4 a^{\frac{9}{4}} n \Gamma \left (- \frac{1}{4}\right )} - \frac{5 b^{\frac{5}{4}} e^{- \frac{3 i \pi }{4}} \log{\left (1 - \frac{\sqrt [4]{b} x^{\frac{n}{4}} e^{\frac{5 i \pi }{4}}}{\sqrt [4]{a}} \right )} \Gamma \left (- \frac{5}{4}\right )}{4 a^{\frac{9}{4}} n \Gamma \left (- \frac{1}{4}\right )} - \frac{5 i b^{\frac{5}{4}} e^{- \frac{3 i \pi }{4}} \log{\left (1 - \frac{\sqrt [4]{b} x^{\frac{n}{4}} e^{\frac{7 i \pi }{4}}}{\sqrt [4]{a}} \right )} \Gamma \left (- \frac{5}{4}\right )}{4 a^{\frac{9}{4}} n \Gamma \left (- \frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1-5/4*n)/(a+b*x**n),x)

[Out]

x**(-5*n/4)*gamma(-5/4)/(a*n*gamma(-1/4)) - 5*b*x**(-n/4)*gamma(-5/4)/(a**2*n*gamma(-1/4)) + 5*b**(5/4)*exp(-3
*I*pi/4)*log(1 - b**(1/4)*x**(n/4)*exp_polar(I*pi/4)/a**(1/4))*gamma(-5/4)/(4*a**(9/4)*n*gamma(-1/4)) + 5*I*b*
*(5/4)*exp(-3*I*pi/4)*log(1 - b**(1/4)*x**(n/4)*exp_polar(3*I*pi/4)/a**(1/4))*gamma(-5/4)/(4*a**(9/4)*n*gamma(
-1/4)) - 5*b**(5/4)*exp(-3*I*pi/4)*log(1 - b**(1/4)*x**(n/4)*exp_polar(5*I*pi/4)/a**(1/4))*gamma(-5/4)/(4*a**(
9/4)*n*gamma(-1/4)) - 5*I*b**(5/4)*exp(-3*I*pi/4)*log(1 - b**(1/4)*x**(n/4)*exp_polar(7*I*pi/4)/a**(1/4))*gamm
a(-5/4)/(4*a**(9/4)*n*gamma(-1/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{-\frac{5}{4} \, n - 1}}{b x^{n} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-5/4*n)/(a+b*x^n),x, algorithm="giac")

[Out]

integrate(x^(-5/4*n - 1)/(b*x^n + a), x)