3.403 \(\int \frac{1}{(c x)^{5/2} (\frac{a}{x}+b x^n)^{3/2}} \, dx\)

Optimal. Leaf size=90 \[ \frac{2}{a c^2 (n+1) \sqrt{c x} \sqrt{\frac{a}{x}+b x^n}}-\frac{2 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{a}}{\sqrt{x} \sqrt{\frac{a}{x}+b x^n}}\right )}{a^{3/2} c^2 (n+1) \sqrt{c x}} \]

[Out]

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

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Rubi [A]  time = 0.193509, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2030, 2031, 2029, 206} \[ \frac{2}{a c^2 (n+1) \sqrt{c x} \sqrt{\frac{a}{x}+b x^n}}-\frac{2 \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{a}}{\sqrt{x} \sqrt{\frac{a}{x}+b x^n}}\right )}{a^{3/2} c^2 (n+1) \sqrt{c x}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(5/2)*(a/x + b*x^n)^(3/2)),x]

[Out]

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

Rule 2030

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] && ILtQ[p + 1/2, 0] && NeQ[n
, j] && EqQ[Simplify[m + j*p + 1], 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2031

Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracPar
t[m])/x^FracPart[m], Int[x^m*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && IntegerQ[p + 1/2]
&& NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0]

Rule 2029

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

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

Mathematica [C]  time = 0.046005, size = 55, normalized size = 0.61 \[ \frac{2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b x^{n+1}}{a}+1\right )}{a c^2 (n+1) \sqrt{c x} \sqrt{\frac{a}{x}+b x^n}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(5/2)*(a/x + b*x^n)^(3/2)),x]

[Out]

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

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Maple [F]  time = 0.328, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{-{\frac{5}{2}}} \left ({\frac{a}{x}}+b{x}^{n} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(5/2)/(a/x+b*x^n)^(3/2),x)

[Out]

int(1/(c*x)^(5/2)/(a/x+b*x^n)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{n} + \frac{a}{x}\right )}^{\frac{3}{2}} \left (c x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(a/x+b*x^n)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^n + a/x)^(3/2)*(c*x)^(5/2)), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(a/x+b*x^n)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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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(1/(c*x)**(5/2)/(a/x+b*x**n)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{n} + \frac{a}{x}\right )}^{\frac{3}{2}} \left (c x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(a/x+b*x^n)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^n + a/x)^(3/2)*(c*x)^(5/2)), x)