3.378 \(\int \sqrt{c x} (\frac{a}{x}+b x^n)^{3/2} \, dx\)

Optimal. Leaf size=117 \[ -\frac{2 a^{3/2} c \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{a}}{\sqrt{x} \sqrt{\frac{a}{x}+b x^n}}\right )}{(n+1) \sqrt{c x}}+\frac{2 a \sqrt{c x} \sqrt{\frac{a}{x}+b x^n}}{n+1}+\frac{2 (c x)^{3/2} \left (\frac{a}{x}+b x^n\right )^{3/2}}{3 c (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.230101, antiderivative size = 117, 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 = {2028, 2031, 2029, 206} \[ -\frac{2 a^{3/2} c \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{a}}{\sqrt{x} \sqrt{\frac{a}{x}+b x^n}}\right )}{(n+1) \sqrt{c x}}+\frac{2 a \sqrt{c x} \sqrt{\frac{a}{x}+b x^n}}{n+1}+\frac{2 (c x)^{3/2} \left (\frac{a}{x}+b x^n\right )^{3/2}}{3 c (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]*(a/x + b*x^n)^(3/2),x]

[Out]

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

Rule 2028

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*p*(n - j)), x] + Dist[a/c^j, Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c,
j, m, n}, x] && IGtQ[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 \sqrt{c x} \left (\frac{a}{x}+b x^n\right )^{3/2} \, dx &=\frac{2 (c x)^{3/2} \left (\frac{a}{x}+b x^n\right )^{3/2}}{3 c (1+n)}+(a c) \int \frac{\sqrt{\frac{a}{x}+b x^n}}{\sqrt{c x}} \, dx\\ &=\frac{2 a \sqrt{c x} \sqrt{\frac{a}{x}+b x^n}}{1+n}+\frac{2 (c x)^{3/2} \left (\frac{a}{x}+b x^n\right )^{3/2}}{3 c (1+n)}+\left (a^2 c^2\right ) \int \frac{1}{(c x)^{3/2} \sqrt{\frac{a}{x}+b x^n}} \, dx\\ &=\frac{2 a \sqrt{c x} \sqrt{\frac{a}{x}+b x^n}}{1+n}+\frac{2 (c x)^{3/2} \left (\frac{a}{x}+b x^n\right )^{3/2}}{3 c (1+n)}+\frac{\left (a^2 c \sqrt{x}\right ) \int \frac{1}{x^{3/2} \sqrt{\frac{a}{x}+b x^n}} \, dx}{\sqrt{c x}}\\ &=\frac{2 a \sqrt{c x} \sqrt{\frac{a}{x}+b x^n}}{1+n}+\frac{2 (c x)^{3/2} \left (\frac{a}{x}+b x^n\right )^{3/2}}{3 c (1+n)}-\frac{\left (2 a^2 c \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 )}{(1+n) \sqrt{c x}}\\ &=\frac{2 a \sqrt{c x} \sqrt{\frac{a}{x}+b x^n}}{1+n}+\frac{2 (c x)^{3/2} \left (\frac{a}{x}+b x^n\right )^{3/2}}{3 c (1+n)}-\frac{2 a^{3/2} c \sqrt{x} \tanh ^{-1}\left (\frac{\sqrt{a}}{\sqrt{x} \sqrt{\frac{a}{x}+b x^n}}\right )}{(1+n) \sqrt{c x}}\\ \end{align*}

Mathematica [A]  time = 0.0714926, size = 97, normalized size = 0.83 \[ \frac{2 \sqrt{c x} \sqrt{\frac{a}{x}+b x^n} \left (\sqrt{a+b x^{n+1}} \left (4 a+b x^{n+1}\right )-3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^{n+1}}}{\sqrt{a}}\right )\right )}{3 (n+1) \sqrt{a+b x^{n+1}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.323, size = 0, normalized size = 0. \begin{align*} \int \sqrt{cx} \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((c*x)^(1/2)*(a/x+b*x^n)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^n + a/x)^(3/2)*sqrt(c*x), 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((c*x)^(1/2)*(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((c*x)**(1/2)*(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{\left (b x^{n} + \frac{a}{x}\right )}^{\frac{3}{2}} \sqrt{c x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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