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

Optimal. Leaf size=77 \[ \frac {\sqrt {a} c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {c}{a+b x^2}} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}-\frac {c x \sqrt {\frac {c}{a+b x^2}}}{b} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 75, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {6720, 288, 217, 206} \[ \frac {c \sqrt {a+b x^2} \sqrt {\frac {c}{a+b x^2}} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}}-\frac {c x \sqrt {\frac {c}{a+b x^2}}}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(c/(a + b*x^2))^(3/2),x]

[Out]

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 89, normalized size = 1.16 \[ \frac {\sqrt {a} \sqrt {\frac {b x^2}{a}+1} \left (\frac {c}{a+b x^2}\right )^{3/2} \left (\left (a+b x^2\right ) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )-\sqrt {a} \sqrt {b} x \sqrt {\frac {b x^2}{a}+1}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 141, normalized size = 1.83 \[ \left [-\frac {2 \, c x \sqrt {\frac {c}{b x^{2} + a}} - c \sqrt {\frac {c}{b}} \log \left (-2 \, b c x^{2} - a c - 2 \, {\left (b^{2} x^{3} + a b x\right )} \sqrt {\frac {c}{b x^{2} + a}} \sqrt {\frac {c}{b}}\right )}{2 \, b}, -\frac {c x \sqrt {\frac {c}{b x^{2} + a}} + c \sqrt {-\frac {c}{b}} \arctan \left (\frac {b x \sqrt {\frac {c}{b x^{2} + a}} \sqrt {-\frac {c}{b}}}{c}\right )}{b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*c*x*sqrt(c/(b*x^2 + a)) - c*sqrt(c/b)*log(-2*b*c*x^2 - a*c - 2*(b^2*x^3 + a*b*x)*sqrt(c/(b*x^2 + a))*
sqrt(c/b)))/b, -(c*x*sqrt(c/(b*x^2 + a)) + c*sqrt(-c/b)*arctan(b*x*sqrt(c/(b*x^2 + a))*sqrt(-c/b)/c))/b]

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giac [A]  time = 0.46, size = 71, normalized size = 0.92 \[ -{\left (\frac {c x \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {b c x^{2} + a c} b} + \frac {c \log \left ({\left | -\sqrt {b c} x + \sqrt {b c x^{2} + a c} \right |}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {b c} b}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*x*sgn(b*x^2 + a)/(sqrt(b*c*x^2 + a*c)*b) + c*log(abs(-sqrt(b*c)*x + sqrt(b*c*x^2 + a*c)))*sgn(b*x^2 + a)/(
sqrt(b*c)*b))*c

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maple [A]  time = 0.01, size = 60, normalized size = 0.78 \[ -\frac {\left (\frac {c}{b \,x^{2}+a}\right )^{\frac {3}{2}} \left (b \,x^{2}+a \right ) \left (b^{\frac {3}{2}} x -\sqrt {b \,x^{2}+a}\, b \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )\right )}{b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c/(b*x^2+a))^(3/2)*(b*x^2+a)*(x*b^(3/2)-ln(b^(1/2)*x+(b*x^2+a)^(1/2))*b*(b*x^2+a)^(1/2))/b^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c/(a + b*x^2))^(3/2),x)

[Out]

int(x^2*(c/(a + b*x^2))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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