\(\int \frac {x^3}{(a+\frac {b}{x^2})^{3/2}} \, dx\) [1930]

Optimal result

Integrand size = 15, antiderivative size = 95 \[ \int \frac {x^3}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=-\frac {15 b^2}{8 a^3 \sqrt {a+\frac {b}{x^2}}}-\frac {5 b x^2}{8 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {x^4}{4 a \sqrt {a+\frac {b}{x^2}}}+\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8 a^{7/2}} \]

[Out]

15/8*b^2*arctanh((a+b/x^2)^(1/2)/a^(1/2))/a^(7/2)-15/8*b^2/a^3/(a+b/x^2)^(1/2)-5/8*b*x^2/a^2/(a+b/x^2)^(1/2)+1
/4*x^4/a/(a+b/x^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 44, 53, 65, 214} \[ \int \frac {x^3}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8 a^{7/2}}-\frac {15 b^2}{8 a^3 \sqrt {a+\frac {b}{x^2}}}-\frac {5 b x^2}{8 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {x^4}{4 a \sqrt {a+\frac {b}{x^2}}} \]

[In]

Int[x^3/(a + b/x^2)^(3/2),x]

[Out]

(-15*b^2)/(8*a^3*Sqrt[a + b/x^2]) - (5*b*x^2)/(8*a^2*Sqrt[a + b/x^2]) + x^4/(4*a*Sqrt[a + b/x^2]) + (15*b^2*Ar
cTanh[Sqrt[a + b/x^2]/Sqrt[a]])/(8*a^(7/2))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^3 (a+b x)^{3/2}} \, dx,x,\frac {1}{x^2}\right )\right ) \\ & = \frac {x^4}{4 a \sqrt {a+\frac {b}{x^2}}}+\frac {(5 b) \text {Subst}\left (\int \frac {1}{x^2 (a+b x)^{3/2}} \, dx,x,\frac {1}{x^2}\right )}{8 a} \\ & = -\frac {5 b x^2}{8 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {x^4}{4 a \sqrt {a+\frac {b}{x^2}}}-\frac {\left (15 b^2\right ) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\frac {1}{x^2}\right )}{16 a^2} \\ & = -\frac {15 b^2}{8 a^3 \sqrt {a+\frac {b}{x^2}}}-\frac {5 b x^2}{8 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {x^4}{4 a \sqrt {a+\frac {b}{x^2}}}-\frac {\left (15 b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^2}\right )}{16 a^3} \\ & = -\frac {15 b^2}{8 a^3 \sqrt {a+\frac {b}{x^2}}}-\frac {5 b x^2}{8 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {x^4}{4 a \sqrt {a+\frac {b}{x^2}}}-\frac {(15 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^2}}\right )}{8 a^3} \\ & = -\frac {15 b^2}{8 a^3 \sqrt {a+\frac {b}{x^2}}}-\frac {5 b x^2}{8 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {x^4}{4 a \sqrt {a+\frac {b}{x^2}}}+\frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {\sqrt {a} x \left (-15 b^2-5 a b x^2+2 a^2 x^4\right )+30 b^2 \sqrt {b+a x^2} \text {arctanh}\left (\frac {\sqrt {a} x}{-\sqrt {b}+\sqrt {b+a x^2}}\right )}{8 a^{7/2} \sqrt {a+\frac {b}{x^2}} x} \]

[In]

Integrate[x^3/(a + b/x^2)^(3/2),x]

[Out]

(Sqrt[a]*x*(-15*b^2 - 5*a*b*x^2 + 2*a^2*x^4) + 30*b^2*Sqrt[b + a*x^2]*ArcTanh[(Sqrt[a]*x)/(-Sqrt[b] + Sqrt[b +
a*x^2])])/(8*a^(7/2)*Sqrt[a + b/x^2]*x)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92

method	result	size
default	\(\frac {\left (a \,x^{2}+b \right ) \left (2 x^{5} a^{\frac {7}{2}}-5 a^{\frac {5}{2}} b \,x^{3}-15 a^{\frac {3}{2}} b^{2} x +15 \sqrt {a \,x^{2}+b}\, \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right ) a \,b^{2}\right )}{8 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} x^{3} a^{\frac {9}{2}}}\)	\(87\)
risch	\(\frac {\left (2 a \,x^{2}-7 b \right ) \left (a \,x^{2}+b \right )}{8 a^{3} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}+\frac {\left (-\frac {b^{2} x}{a^{3} \sqrt {a \,x^{2}+b}}+\frac {15 b^{2} \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right )}{8 a^{\frac {7}{2}}}\right ) \sqrt {a \,x^{2}+b}}{\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\)	\(106\)

[In]

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

[Out]

1/8*(a*x^2+b)*(2*x^5*a^(7/2)-5*a^(5/2)*b*x^3-15*a^(3/2)*b^2*x+15*(a*x^2+b)^(1/2)*ln(a^(1/2)*x+(a*x^2+b)^(1/2))
*a*b^2)/((a*x^2+b)/x^2)^(3/2)/x^3/a^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.33 \[ \int \frac {x^3}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (a b^{2} x^{2} + b^{3}\right )} \sqrt {a} \log \left (-2 \, a x^{2} - 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) + 2 \, {\left (2 \, a^{3} x^{6} - 5 \, a^{2} b x^{4} - 15 \, a b^{2} x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{16 \, {\left (a^{5} x^{2} + a^{4} b\right )}}, -\frac {15 \, {\left (a b^{2} x^{2} + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) - {\left (2 \, a^{3} x^{6} - 5 \, a^{2} b x^{4} - 15 \, a b^{2} x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{8 \, {\left (a^{5} x^{2} + a^{4} b\right )}}\right ] \]

[In]

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

[Out]

[1/16*(15*(a*b^2*x^2 + b^3)*sqrt(a)*log(-2*a*x^2 - 2*sqrt(a)*x^2*sqrt((a*x^2 + b)/x^2) - b) + 2*(2*a^3*x^6 - 5
*a^2*b*x^4 - 15*a*b^2*x^2)*sqrt((a*x^2 + b)/x^2))/(a^5*x^2 + a^4*b), -1/8*(15*(a*b^2*x^2 + b^3)*sqrt(-a)*arcta
n(sqrt(-a)*x^2*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) - (2*a^3*x^6 - 5*a^2*b*x^4 - 15*a*b^2*x^2)*sqrt((a*x^2 + b)/
x^2))/(a^5*x^2 + a^4*b)]

Sympy [A] (verification not implemented)

Time = 3.60 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.05 \[ \int \frac {x^3}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {x^{5}}{4 a \sqrt {b} \sqrt {\frac {a x^{2}}{b} + 1}} - \frac {5 \sqrt {b} x^{3}}{8 a^{2} \sqrt {\frac {a x^{2}}{b} + 1}} - \frac {15 b^{\frac {3}{2}} x}{8 a^{3} \sqrt {\frac {a x^{2}}{b} + 1}} + \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x}{\sqrt {b}} \right )}}{8 a^{\frac {7}{2}}} \]

[In]

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

[Out]

x**5/(4*a*sqrt(b)*sqrt(a*x**2/b + 1)) - 5*sqrt(b)*x**3/(8*a**2*sqrt(a*x**2/b + 1)) - 15*b**(3/2)*x/(8*a**3*sqr
t(a*x**2/b + 1)) + 15*b**2*asinh(sqrt(a)*x/sqrt(b))/(8*a**(7/2))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.28 \[ \int \frac {x^3}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=-\frac {15 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} b^{2} - 25 \, {\left (a + \frac {b}{x^{2}}\right )} a b^{2} + 8 \, a^{2} b^{2}}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{4} + \sqrt {a + \frac {b}{x^{2}}} a^{5}\right )}} - \frac {15 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right )}{16 \, a^{\frac {7}{2}}} \]

[In]

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

[Out]

-1/8*(15*(a + b/x^2)^2*b^2 - 25*(a + b/x^2)*a*b^2 + 8*a^2*b^2)/((a + b/x^2)^(5/2)*a^3 - 2*(a + b/x^2)^(3/2)*a^
4 + sqrt(a + b/x^2)*a^5) - 15/16*b^2*log((sqrt(a + b/x^2) - sqrt(a))/(sqrt(a + b/x^2) + sqrt(a)))/a^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01 \[ \int \frac {x^3}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {2 \, x^{2}}{a \mathrm {sgn}\left (x\right )} - \frac {5 \, b}{a^{2} \mathrm {sgn}\left (x\right )}\right )} - \frac {15 \, b^{2}}{a^{3} \mathrm {sgn}\left (x\right )}\right )} x}{8 \, \sqrt {a x^{2} + b}} + \frac {15 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, a^{\frac {7}{2}}} - \frac {15 \, b^{2} \log \left ({\left | -\sqrt {a} x + \sqrt {a x^{2} + b} \right |}\right )}{8 \, a^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

1/8*(x^2*(2*x^2/(a*sgn(x)) - 5*b/(a^2*sgn(x))) - 15*b^2/(a^3*sgn(x)))*x/sqrt(a*x^2 + b) + 15/16*b^2*log(abs(b)
)*sgn(x)/a^(7/2) - 15/8*b^2*log(abs(-sqrt(a)*x + sqrt(a*x^2 + b)))/(a^(7/2)*sgn(x))

Mupad [B] (verification not implemented)

Time = 6.59 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.79 \[ \int \frac {x^3}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {15\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8\,a^{7/2}}-\frac {15\,b^2}{8\,a^3\,\sqrt {a+\frac {b}{x^2}}}+\frac {x^4}{4\,a\,\sqrt {a+\frac {b}{x^2}}}-\frac {5\,b\,x^2}{8\,a^2\,\sqrt {a+\frac {b}{x^2}}} \]

[In]

int(x^3/(a + b/x^2)^(3/2),x)

[Out]

(15*b^2*atanh((a + b/x^2)^(1/2)/a^(1/2)))/(8*a^(7/2)) - (15*b^2)/(8*a^3*(a + b/x^2)^(1/2)) + x^4/(4*a*(a + b/x
^2)^(1/2)) - (5*b*x^2)/(8*a^2*(a + b/x^2)^(1/2))