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

Optimal. Leaf size=98 \[ -\frac{2 a^{3/2} c^2 \tanh ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{\frac{a}{x^2}+b x^n}}\right )}{n+2}+\frac{2 c^2 x^3 \left (\frac{a}{x^2}+b x^n\right )^{3/2}}{3 (n+2)}+\frac{2 a c^2 x \sqrt{\frac{a}{x^2}+b x^n}}{n+2} \]

[Out]

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

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Rubi [A]  time = 0.171814, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {12, 2028, 2007, 2029, 206} \[ -\frac{2 a^{3/2} c^2 \tanh ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{\frac{a}{x^2}+b x^n}}\right )}{n+2}+\frac{2 c^2 x^3 \left (\frac{a}{x^2}+b x^n\right )^{3/2}}{3 (n+2)}+\frac{2 a c^2 x \sqrt{\frac{a}{x^2}+b x^n}}{n+2} \]

Antiderivative was successfully verified.

[In]

Int[c^2*x^2*(a/x^2 + b*x^n)^(3/2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2007

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(p*(n - j)), x] + Dist
[a, Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, j, n}, x] && IGtQ[p + 1/2, 0] && NeQ[n, j] && EqQ[
Simplify[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 c^2 x^2 \left (\frac{a}{x^2}+b x^n\right )^{3/2} \, dx &=c^2 \int x^2 \left (\frac{a}{x^2}+b x^n\right )^{3/2} \, dx\\ &=\frac{2 c^2 x^3 \left (\frac{a}{x^2}+b x^n\right )^{3/2}}{3 (2+n)}+\left (a c^2\right ) \int \sqrt{\frac{a}{x^2}+b x^n} \, dx\\ &=\frac{2 a c^2 x \sqrt{\frac{a}{x^2}+b x^n}}{2+n}+\frac{2 c^2 x^3 \left (\frac{a}{x^2}+b x^n\right )^{3/2}}{3 (2+n)}+\left (a^2 c^2\right ) \int \frac{1}{x^2 \sqrt{\frac{a}{x^2}+b x^n}} \, dx\\ &=\frac{2 a c^2 x \sqrt{\frac{a}{x^2}+b x^n}}{2+n}+\frac{2 c^2 x^3 \left (\frac{a}{x^2}+b x^n\right )^{3/2}}{3 (2+n)}-\frac{\left (2 a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{1}{x \sqrt{\frac{a}{x^2}+b x^n}}\right )}{2+n}\\ &=\frac{2 a c^2 x \sqrt{\frac{a}{x^2}+b x^n}}{2+n}+\frac{2 c^2 x^3 \left (\frac{a}{x^2}+b x^n\right )^{3/2}}{3 (2+n)}-\frac{2 a^{3/2} c^2 \tanh ^{-1}\left (\frac{\sqrt{a}}{x \sqrt{\frac{a}{x^2}+b x^n}}\right )}{2+n}\\ \end{align*}

Mathematica [A]  time = 0.0678041, size = 94, normalized size = 0.96 \[ \frac{2 c^2 x \sqrt{\frac{a}{x^2}+b x^n} \left (\sqrt{a+b x^{n+2}} \left (4 a+b x^{n+2}\right )-3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^{n+2}}}{\sqrt{a}}\right )\right )}{3 (n+2) \sqrt{a+b x^{n+2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.322, size = 0, normalized size = 0. \begin{align*} \int{c}^{2}{x}^{2} \left ({\frac{a}{{x}^{2}}}+b{x}^{n} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*(Integral(a*sqrt(a/x**2 + b*x**n), x) + Integral(b*x**2*x**n*sqrt(a/x**2 + b*x**n), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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