3.4.79 \(\int c^2 x^2 (\frac {a}{x^2}+b x^n)^{3/2} \, dx\) [379]

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

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 98, normalized size of antiderivative = 1.00, 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, 2053, 2032, 2054, 212} \begin {gather*} -\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 a c^2 x \sqrt {\frac {a}{x^2}+b x^n}}{n+2}+\frac {2 c^2 x^3 \left (\frac {a}{x^2}+b x^n\right )^{3/2}}{3 (n+2)} \end {gather*}

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 212

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

Rule 2032

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 2053

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 2054

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.09, size = 94, normalized size = 0.96 \begin {gather*} \frac {2 c^2 x \sqrt {\frac {a}{x^2}+b x^n} \left (\sqrt {a+b x^{2+n}} \left (4 a+b x^{2+n}\right )-3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^{2+n}}}{\sqrt {a}}\right )\right )}{3 (2+n) \sqrt {a+b x^{2+n}}} \end {gather*}

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 = 0.07, size = 0, normalized size = 0.00 \[\int c^{2} x^{2} \left (\frac {a}{x^{2}}+b \,x^{n}\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int c^2\,x^2\,{\left (b\,x^n+\frac {a}{x^2}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(c^2*x^2*(b*x^n + a/x^2)^(3/2),x)

[Out]

int(c^2*x^2*(b*x^n + a/x^2)^(3/2), x)

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