\(\int x^2 (\frac {c}{a+b x^2})^{3/2} \, dx\) [243]

Optimal result

Integrand size = 19, antiderivative size = 77 \[ \int x^2 \left (\frac {c}{a+b x^2}\right )^{3/2} \, dx=-\frac {c x \sqrt {\frac {c}{a+b x^2}}}{b}+\frac {\sqrt {a} c \sqrt {\frac {c}{a+b x^2}} \sqrt {1+\frac {b x^2}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1973, 294, 221} \[ \int x^2 \left (\frac {c}{a+b x^2}\right )^{3/2} \, dx=\frac {\sqrt {a} c \sqrt {\frac {b x^2}{a}+1} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \sqrt {\frac {c}{a+b x^2}}}{b^{3/2}}-\frac {c x \sqrt {\frac {c}{a+b x^2}}}{b} \]

[In]

Int[x^2*(c/(a + b*x^2))^(3/2),x]

[Out]

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

Rule 221

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

Rule 294

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 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c \sqrt {\frac {c}{a+b x^2}} \sqrt {1+\frac {b x^2}{a}}\right ) \int \frac {x^2}{\left (1+\frac {b x^2}{a}\right )^{3/2}} \, dx}{a} \\ & = -\frac {c x \sqrt {\frac {c}{a+b x^2}}}{b}+\frac {\left (c \sqrt {\frac {c}{a+b x^2}} \sqrt {1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt {1+\frac {b x^2}{a}}} \, dx}{b} \\ & = -\frac {c x \sqrt {\frac {c}{a+b x^2}}}{b}+\frac {\sqrt {a} c \sqrt {\frac {c}{a+b x^2}} \sqrt {1+\frac {b x^2}{a}} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int x^2 \left (\frac {c}{a+b x^2}\right )^{3/2} \, dx=-\frac {c \sqrt {\frac {c}{a+b x^2}} \left (\sqrt {b} x+2 \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a}-\sqrt {a+b x^2}}\right )\right )}{b^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {\left (\frac {c}{b \,x^{2}+a}\right )^{\frac {3}{2}} \left (b \,x^{2}+a \right ) \left (x \,b^{\frac {3}{2}}-\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) b \sqrt {b \,x^{2}+a}\right )}{b^{\frac {5}{2}}}\)	\(60\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.83 \[ \int x^2 \left (\frac {c}{a+b x^2}\right )^{3/2} \, dx=\left [-\frac {2 \, c x \sqrt {\frac {c}{b x^{2} + a}} - c \sqrt {\frac {c}{b}} \log \left (-2 \, b c x^{2} - a c - 2 \, {\left (b^{2} x^{3} + a b x\right )} \sqrt {\frac {c}{b x^{2} + a}} \sqrt {\frac {c}{b}}\right )}{2 \, b}, -\frac {c x \sqrt {\frac {c}{b x^{2} + a}} + c \sqrt {-\frac {c}{b}} \arctan \left (\frac {b x \sqrt {\frac {c}{b x^{2} + a}} \sqrt {-\frac {c}{b}}}{c}\right )}{b}\right ] \]

[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]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int x^2 \left (\frac {c}{a+b x^2}\right )^{3/2} \, dx=-{\left (\frac {c x \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {b c x^{2} + a c} b} + \frac {c \log \left ({\left | -\sqrt {b c} x + \sqrt {b c x^{2} + a c} \right |}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {b c} b}\right )} c \]

[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

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (\frac {c}{a+b x^2}\right )^{3/2} \, dx=\int x^2\,{\left (\frac {c}{b\,x^2+a}\right )}^{3/2} \,d x \]

[In]

int(x^2*(c/(a + b*x^2))^(3/2),x)

[Out]

int(x^2*(c/(a + b*x^2))^(3/2), x)