3.2769 \(\int \frac{(c x)^{-1-\frac{3 n}{2}}}{a+b x^n} \, dx\)

Optimal. Leaf size=100 \[ -\frac{2 b^{3/2} x^{3 n/2} (c x)^{-3 n/2} \tan ^{-1}\left (\frac{\sqrt{a} x^{-n/2}}{\sqrt{b}}\right )}{a^{5/2} c n}+\frac{2 b x^n (c x)^{-3 n/2}}{a^2 c n}-\frac{2 (c x)^{-3 n/2}}{3 a c n} \]

[Out]

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

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Rubi [A]  time = 0.0522623, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {363, 362, 345, 193, 321, 205} \[ -\frac{2 b^{3/2} x^{3 n/2} (c x)^{-3 n/2} \tan ^{-1}\left (\frac{\sqrt{a} x^{-n/2}}{\sqrt{b}}\right )}{a^{5/2} c n}+\frac{2 b x^n (c x)^{-3 n/2}}{a^2 c n}-\frac{2 (c x)^{-3 n/2}}{3 a c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 363

Int[((c_)*(x_))^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracPart[m])/x^FracPart[m
], Int[x^m/(a + b*x^n), x], x] /; FreeQ[{a, b, c, m, n}, x] && FractionQ[Simplify[(m + 1)/n]] && (SumSimplerQ[
m, n] || SumSimplerQ[m, -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 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]

Rubi steps

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

Mathematica [C]  time = 0.0108574, size = 39, normalized size = 0.39 \[ -\frac{2 x (c x)^{-\frac{3 n}{2}-1} \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{b x^n}{a}\right )}{3 a n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.067, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{a+b{x}^{n}} \left ( cx \right ) ^{-1-{\frac{3\,n}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-1-3/2*n)/(a+b*x^n),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.17901, size = 1373, normalized size = 13.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-1-3/2*n)/(a+b*x^n),x, algorithm="fricas")

[Out]

[1/3*(3*b*c^(-n - 2/3)*sqrt(-b*c^(-n - 2/3)/a)*log(-(2*a^2*b*c^(-n - 2/3)*x^(4/3)*e^(-2/3*(3*n + 2)*log(c) - 2
/3*(3*n + 2)*log(x)) - a^3*x^2*e^(-(3*n + 2)*log(c) - (3*n + 2)*log(x)) - 2*a*b^2*c^(-2*n - 4/3)*x^(2/3)*e^(-1
/3*(3*n + 2)*log(c) - 1/3*(3*n + 2)*log(x)) + b^3*c^(-3*n - 2) - 2*(a^2*b*c^(-n - 2/3)*x*e^(-1/2*(3*n + 2)*log
(c) - 1/2*(3*n + 2)*log(x)) - a^3*x^(5/3)*e^(-5/6*(3*n + 2)*log(c) - 5/6*(3*n + 2)*log(x)) - a*b^2*c^(-2*n - 4
/3)*x^(1/3)*e^(-1/6*(3*n + 2)*log(c) - 1/6*(3*n + 2)*log(x)))*sqrt(-b*c^(-n - 2/3)/a))/(a^3*x^2*e^(-(3*n + 2)*
log(c) - (3*n + 2)*log(x)) + b^3*c^(-3*n - 2))) + 6*b*c^(-n - 2/3)*x^(1/3)*e^(-1/6*(3*n + 2)*log(c) - 1/6*(3*n
+ 2)*log(x)) - 2*a*x*e^(-1/2*(3*n + 2)*log(c) - 1/2*(3*n + 2)*log(x)))/(a^2*n), 2/3*(3*b*c^(-n - 2/3)*sqrt(b*
c^(-n - 2/3)/a)*arctan(sqrt(b*c^(-n - 2/3)/a)*e^(1/6*(3*n + 2)*log(c) + 1/6*(3*n + 2)*log(x))/x^(1/3)) + 3*b*c
^(-n - 2/3)*x^(1/3)*e^(-1/6*(3*n + 2)*log(c) - 1/6*(3*n + 2)*log(x)) - a*x*e^(-1/2*(3*n + 2)*log(c) - 1/2*(3*n
+ 2)*log(x)))/(a^2*n)]

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Sympy [A]  time = 5.31654, size = 82, normalized size = 0.82 \begin{align*} - \frac{2 c^{- \frac{3 n}{2}} x^{- \frac{3 n}{2}}}{3 a c n} + \frac{2 b c^{- \frac{3 n}{2}} x^{- \frac{n}{2}}}{a^{2} c n} + \frac{2 b^{\frac{3}{2}} c^{- \frac{3 n}{2}} \operatorname{atan}{\left (\frac{\sqrt{b} x^{\frac{n}{2}}}{\sqrt{a}} \right )}}{a^{\frac{5}{2}} c n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c**(-3*n/2)*x**(-3*n/2)/(3*a*c*n) + 2*b*c**(-3*n/2)*x**(-n/2)/(a**2*c*n) + 2*b**(3/2)*c**(-3*n/2)*atan(sqrt
(b)*x**(n/2)/sqrt(a))/(a**(5/2)*c*n)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{-\frac{3}{2} \, n - 1}}{b x^{n} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-1-3/2*n)/(a+b*x^n),x, algorithm="giac")

[Out]

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