3.2782 \(\int \frac{(c x)^{-1+\frac{5 n}{2}}}{\sqrt{a+b x^n}} \, dx\)

Optimal. Leaf size=137 \[ \frac{3 a^2 x^{-5 n/2} (c x)^{5 n/2} \tanh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a+b x^n}}\right )}{4 b^{5/2} c n}-\frac{3 a x^{-2 n} (c x)^{5 n/2} \sqrt{a+b x^n}}{4 b^2 c n}+\frac{x^{-n} (c x)^{5 n/2} \sqrt{a+b x^n}}{2 b c n} \]

[Out]

(-3*a*(c*x)^((5*n)/2)*Sqrt[a + b*x^n])/(4*b^2*c*n*x^(2*n)) + ((c*x)^((5*n)/2)*Sqrt[a + b*x^n])/(2*b*c*n*x^n) +
(3*a^2*(c*x)^((5*n)/2)*ArcTanh[(Sqrt[b]*x^(n/2))/Sqrt[a + b*x^n]])/(4*b^(5/2)*c*n*x^((5*n)/2))

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Rubi [A]  time = 0.0570214, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {357, 355, 288, 206} \[ \frac{3 a^2 x^{-5 n/2} (c x)^{5 n/2} \tanh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a+b x^n}}\right )}{4 b^{5/2} c n}-\frac{3 a x^{-2 n} (c x)^{5 n/2} \sqrt{a+b x^n}}{4 b^2 c n}+\frac{x^{-n} (c x)^{5 n/2} \sqrt{a+b x^n}}{2 b c n} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(-1 + (5*n)/2)/Sqrt[a + b*x^n],x]

[Out]

(-3*a*(c*x)^((5*n)/2)*Sqrt[a + b*x^n])/(4*b^2*c*n*x^(2*n)) + ((c*x)^((5*n)/2)*Sqrt[a + b*x^n])/(2*b*c*n*x^n) +
(3*a^2*(c*x)^((5*n)/2)*ArcTanh[(Sqrt[b]*x^(n/2))/Sqrt[a + b*x^n]])/(4*b^(5/2)*c*n*x^((5*n)/2))

Rule 357

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracPart[m])/x^FracP
art[m], Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IntegerQ[p + Simplify[(m + 1)/n]] && LtQ
[-1, p, 0]

Rule 355

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[p]}, Dist[(k*a^(p + Simplify[
(m + 1)/n]))/n, Subst[Int[x^(k*Simplify[(m + 1)/n] - 1)/(1 - b*x^k)^(p + Simplify[(m + 1)/n] + 1), x], x, x^(n
/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p + Simplify[(m + 1)/n]] && LtQ[-1, p, 0]

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.0907107, size = 121, normalized size = 0.88 \[ \frac{a x^{-5 n/2} (c x)^{5 n/2} \sqrt{\frac{b x^n}{a}+1} \left (3 a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a}}\right )+\sqrt{b} x^{n/2} \left (2 b x^n-3 a\right ) \sqrt{\frac{b x^n}{a}+1}\right )}{4 b^{5/2} c n \sqrt{a+b x^n}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(-1 + (5*n)/2)/Sqrt[a + b*x^n],x]

[Out]

(a*(c*x)^((5*n)/2)*Sqrt[1 + (b*x^n)/a]*(Sqrt[b]*x^(n/2)*(-3*a + 2*b*x^n)*Sqrt[1 + (b*x^n)/a] + 3*a^(3/2)*ArcSi
nh[(Sqrt[b]*x^(n/2))/Sqrt[a]]))/(4*b^(5/2)*c*n*x^((5*n)/2)*Sqrt[a + b*x^n])

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{-1+{\frac{5\,n}{2}}}{\frac{1}{\sqrt{a+b{x}^{n}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x)

[Out]

int((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{5}{2} \, n - 1}}{\sqrt{b x^{n} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

integrate((c*x)^(5/2*n - 1)/sqrt(b*x^n + a), x)

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Fricas [A]  time = 1.50845, size = 482, normalized size = 3.52 \begin{align*} \left [\frac{3 \, a^{2} \sqrt{b} c^{\frac{5}{2} \, n - 1} \log \left (-2 \, \sqrt{b x^{n} + a} \sqrt{b} x^{\frac{1}{2} \, n} - 2 \, b x^{n} - a\right ) + 2 \,{\left (2 \, b^{2} c^{\frac{5}{2} \, n - 1} x^{\frac{3}{2} \, n} - 3 \, a b c^{\frac{5}{2} \, n - 1} x^{\frac{1}{2} \, n}\right )} \sqrt{b x^{n} + a}}{8 \, b^{3} n}, -\frac{3 \, a^{2} \sqrt{-b} c^{\frac{5}{2} \, n - 1} \arctan \left (\frac{\sqrt{-b} x^{\frac{1}{2} \, n}}{\sqrt{b x^{n} + a}}\right ) -{\left (2 \, b^{2} c^{\frac{5}{2} \, n - 1} x^{\frac{3}{2} \, n} - 3 \, a b c^{\frac{5}{2} \, n - 1} x^{\frac{1}{2} \, n}\right )} \sqrt{b x^{n} + a}}{4 \, b^{3} n}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(3*a^2*sqrt(b)*c^(5/2*n - 1)*log(-2*sqrt(b*x^n + a)*sqrt(b)*x^(1/2*n) - 2*b*x^n - a) + 2*(2*b^2*c^(5/2*n
- 1)*x^(3/2*n) - 3*a*b*c^(5/2*n - 1)*x^(1/2*n))*sqrt(b*x^n + a))/(b^3*n), -1/4*(3*a^2*sqrt(-b)*c^(5/2*n - 1)*a
rctan(sqrt(-b)*x^(1/2*n)/sqrt(b*x^n + a)) - (2*b^2*c^(5/2*n - 1)*x^(3/2*n) - 3*a*b*c^(5/2*n - 1)*x^(1/2*n))*sq
rt(b*x^n + a))/(b^3*n)]

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Sympy [A]  time = 16.7758, size = 150, normalized size = 1.09 \begin{align*} - \frac{3 a^{\frac{3}{2}} c^{\frac{5 n}{2}} x^{\frac{n}{2}}}{4 b^{2} c n \sqrt{1 + \frac{b x^{n}}{a}}} - \frac{\sqrt{a} c^{\frac{5 n}{2}} x^{\frac{3 n}{2}}}{4 b c n \sqrt{1 + \frac{b x^{n}}{a}}} + \frac{3 a^{2} c^{\frac{5 n}{2}} \operatorname{asinh}{\left (\frac{\sqrt{b} x^{\frac{n}{2}}}{\sqrt{a}} \right )}}{4 b^{\frac{5}{2}} c n} + \frac{c^{\frac{5 n}{2}} x^{\frac{5 n}{2}}}{2 \sqrt{a} c n \sqrt{1 + \frac{b x^{n}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(-1+5/2*n)/(a+b*x**n)**(1/2),x)

[Out]

-3*a**(3/2)*c**(5*n/2)*x**(n/2)/(4*b**2*c*n*sqrt(1 + b*x**n/a)) - sqrt(a)*c**(5*n/2)*x**(3*n/2)/(4*b*c*n*sqrt(
1 + b*x**n/a)) + 3*a**2*c**(5*n/2)*asinh(sqrt(b)*x**(n/2)/sqrt(a))/(4*b**(5/2)*c*n) + c**(5*n/2)*x**(5*n/2)/(2
*sqrt(a)*c*n*sqrt(1 + b*x**n/a))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{5}{2} \, n - 1}}{\sqrt{b x^{n} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate((c*x)^(5/2*n - 1)/sqrt(b*x^n + a), x)