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

Optimal. Leaf size=72 \[ \frac {2 c^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^n}}\right )}{a^{3/2} (2-n)}-\frac {2 c^2 x}{a (2-n) \sqrt {a x^2+b x^n}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 2030, 2008, 206} \[ \frac {2 c^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^n}}\right )}{a^{3/2} (2-n)}-\frac {2 c^2 x}{a (2-n) \sqrt {a x^2+b x^n}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2008

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 91, normalized size = 1.26 \[ \frac {2 c^2 \left (\sqrt {a} x-\sqrt {b} x^{n/2} \sqrt {\frac {a x^{2-n}}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {a} x^{1-\frac {n}{2}}}{\sqrt {b}}\right )\right )}{a^{3/2} (n-2) \sqrt {a x^2+b x^n}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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 (co
nstant residues)

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

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(c^2*x^2/(a*x^2 + b*x^n)^(3/2), x)

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maple [F]  time = 0.70, size = 0, normalized size = 0.00 \[ \int \frac {c^{2} x^{2}}{\left (a \,x^{2}+b \,x^{n}\right )^{\frac {3}{2}}}\, dx \]

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 \[ c^{2} \int \frac {x^{2}}{{\left (a x^{2} + b x^{n}\right )}^{\frac {3}{2}}}\,{d x} \]

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(x^2/(a*x^2 + b*x^n)^(3/2), x)

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

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} \int \frac {x^{2}}{a x^{2} \sqrt {a x^{2} + b x^{n}} + b x^{n} \sqrt {a x^{2} + b x^{n}}}\, dx \]

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

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