\(\int (a+\frac {b}{x^2})^{5/2} x^3 \, dx\) [1907]

Optimal result

Integrand size = 15, antiderivative size = 86 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} x^3 \, dx=-\frac {15}{8} b^2 \sqrt {a+\frac {b}{x^2}}+\frac {5}{8} b \left (a+\frac {b}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (a+\frac {b}{x^2}\right )^{5/2} x^4+\frac {15}{8} \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 86, 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, 43, 52, 65, 214} \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} x^3 \, dx=\frac {15}{8} \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )-\frac {15}{8} b^2 \sqrt {a+\frac {b}{x^2}}+\frac {5}{8} b x^2 \left (a+\frac {b}{x^2}\right )^{3/2}+\frac {1}{4} x^4 \left (a+\frac {b}{x^2}\right )^{5/2} \]

[In]

Int[(a + b/x^2)^(5/2)*x^3,x]

[Out]

(-15*b^2*Sqrt[a + b/x^2])/8 + (5*b*(a + b/x^2)^(3/2)*x^2)/8 + ((a + b/x^2)^(5/2)*x^4)/4 + (15*Sqrt[a]*b^2*ArcT
anh[Sqrt[a + b/x^2]/Sqrt[a]])/8

Rule 43

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {(a+b x)^{5/2}}{x^3} \, dx,x,\frac {1}{x^2}\right )\right ) \\ & = \frac {1}{4} \left (a+\frac {b}{x^2}\right )^{5/2} x^4-\frac {1}{8} (5 b) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^2} \, dx,x,\frac {1}{x^2}\right ) \\ & = \frac {5}{8} b \left (a+\frac {b}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (a+\frac {b}{x^2}\right )^{5/2} x^4-\frac {1}{16} \left (15 b^2\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {15}{8} b^2 \sqrt {a+\frac {b}{x^2}}+\frac {5}{8} b \left (a+\frac {b}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (a+\frac {b}{x^2}\right )^{5/2} x^4-\frac {1}{16} \left (15 a b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {15}{8} b^2 \sqrt {a+\frac {b}{x^2}}+\frac {5}{8} b \left (a+\frac {b}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (a+\frac {b}{x^2}\right )^{5/2} x^4-\frac {1}{8} (15 a b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^2}}\right ) \\ & = -\frac {15}{8} b^2 \sqrt {a+\frac {b}{x^2}}+\frac {5}{8} b \left (a+\frac {b}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (a+\frac {b}{x^2}\right )^{5/2} x^4+\frac {15}{8} \sqrt {a} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} x^3 \, dx=\frac {\sqrt {a+\frac {b}{x^2}} \left (\sqrt {b+a x^2} \left (-8 b^2+9 a b x^2+2 a^2 x^4\right )-15 \sqrt {a} b^2 x \log \left (-\sqrt {a} x+\sqrt {b+a x^2}\right )\right )}{8 \sqrt {b+a x^2}} \]

[In]

Integrate[(a + b/x^2)^(5/2)*x^3,x]

[Out]

(Sqrt[a + b/x^2]*(Sqrt[b + a*x^2]*(-8*b^2 + 9*a*b*x^2 + 2*a^2*x^4) - 15*Sqrt[a]*b^2*x*Log[-(Sqrt[a]*x) + Sqrt[
b + a*x^2]]))/(8*Sqrt[b + a*x^2])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99

method	result	size
risch	\(\frac {\left (2 a^{2} x^{4}+9 a b \,x^{2}-8 b^{2}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}{8}+\frac {15 \sqrt {a}\, b^{2} \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}{8 \sqrt {a \,x^{2}+b}}\)	\(85\)
default	\(\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} x^{4} \left (8 a^{\frac {3}{2}} \left (a \,x^{2}+b \right )^{\frac {5}{2}} x^{2}+10 a^{\frac {3}{2}} \left (a \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{2}+15 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b}\, b^{2} x^{2}-8 \left (a \,x^{2}+b \right )^{\frac {7}{2}} \sqrt {a}+15 \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right ) a \,b^{3} x \right )}{8 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b \sqrt {a}}\)	\(127\)

[In]

int((a+b/x^2)^(5/2)*x^3,x,method=_RETURNVERBOSE)

[Out]

1/8*(2*a^2*x^4+9*a*b*x^2-8*b^2)*((a*x^2+b)/x^2)^(1/2)+15/8*a^(1/2)*b^2*ln(a^(1/2)*x+(a*x^2+b)^(1/2))*((a*x^2+b
)/x^2)^(1/2)*x/(a*x^2+b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.83 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} x^3 \, dx=\left [\frac {15}{16} \, \sqrt {a} b^{2} \log \left (-2 \, a x^{2} - 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) + \frac {1}{8} \, {\left (2 \, a^{2} x^{4} + 9 \, a b x^{2} - 8 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}, -\frac {15}{8} \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + \frac {1}{8} \, {\left (2 \, a^{2} x^{4} + 9 \, a b x^{2} - 8 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}\right ] \]

[In]

integrate((a+b/x^2)^(5/2)*x^3,x, algorithm="fricas")

[Out]

[15/16*sqrt(a)*b^2*log(-2*a*x^2 - 2*sqrt(a)*x^2*sqrt((a*x^2 + b)/x^2) - b) + 1/8*(2*a^2*x^4 + 9*a*b*x^2 - 8*b^
2)*sqrt((a*x^2 + b)/x^2), -15/8*sqrt(-a)*b^2*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + 1/8*(2*a
^2*x^4 + 9*a*b*x^2 - 8*b^2)*sqrt((a*x^2 + b)/x^2)]

Sympy [A] (verification not implemented)

Time = 2.55 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.36 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} x^3 \, dx=\frac {15 \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x}{\sqrt {b}} \right )}}{8} + \frac {a^{3} x^{5}}{4 \sqrt {b} \sqrt {\frac {a x^{2}}{b} + 1}} + \frac {11 a^{2} \sqrt {b} x^{3}}{8 \sqrt {\frac {a x^{2}}{b} + 1}} + \frac {a b^{\frac {3}{2}} x}{8 \sqrt {\frac {a x^{2}}{b} + 1}} - \frac {b^{\frac {5}{2}}}{x \sqrt {\frac {a x^{2}}{b} + 1}} \]

[In]

integrate((a+b/x**2)**(5/2)*x**3,x)

[Out]

15*sqrt(a)*b**2*asinh(sqrt(a)*x/sqrt(b))/8 + a**3*x**5/(4*sqrt(b)*sqrt(a*x**2/b + 1)) + 11*a**2*sqrt(b)*x**3/(
8*sqrt(a*x**2/b + 1)) + a*b**(3/2)*x/(8*sqrt(a*x**2/b + 1)) - b**(5/2)/(x*sqrt(a*x**2/b + 1))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.34 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} x^3 \, dx=-\frac {15}{16} \, \sqrt {a} b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right ) - \sqrt {a + \frac {b}{x^{2}}} b^{2} + \frac {9 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a b^{2} - 7 \, \sqrt {a + \frac {b}{x^{2}}} a^{2} b^{2}}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{2} - 2 \, {\left (a + \frac {b}{x^{2}}\right )} a + a^{2}\right )}} \]

[In]

integrate((a+b/x^2)^(5/2)*x^3,x, algorithm="maxima")

[Out]

-15/16*sqrt(a)*b^2*log((sqrt(a + b/x^2) - sqrt(a))/(sqrt(a + b/x^2) + sqrt(a))) - sqrt(a + b/x^2)*b^2 + 1/8*(9
*(a + b/x^2)^(3/2)*a*b^2 - 7*sqrt(a + b/x^2)*a^2*b^2)/((a + b/x^2)^2 - 2*(a + b/x^2)*a + a^2)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} x^3 \, dx=-\frac {15}{16} \, \sqrt {a} b^{2} \log \left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\left (x\right ) + \frac {2 \, \sqrt {a} b^{3} \mathrm {sgn}\left (x\right )}{{\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2} - b} + \frac {1}{8} \, {\left (2 \, a^{2} x^{2} \mathrm {sgn}\left (x\right ) + 9 \, a b \mathrm {sgn}\left (x\right )\right )} \sqrt {a x^{2} + b} x \]

[In]

integrate((a+b/x^2)^(5/2)*x^3,x, algorithm="giac")

[Out]

-15/16*sqrt(a)*b^2*log((sqrt(a)*x - sqrt(a*x^2 + b))^2)*sgn(x) + 2*sqrt(a)*b^3*sgn(x)/((sqrt(a)*x - sqrt(a*x^2
+ b))^2 - b) + 1/8*(2*a^2*x^2*sgn(x) + 9*a*b*sgn(x))*sqrt(a*x^2 + b)*x

Mupad [B] (verification not implemented)

Time = 6.55 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} x^3 \, dx=\frac {9\,a\,x^4\,{\left (a+\frac {b}{x^2}\right )}^{3/2}}{8}-b^2\,\sqrt {a+\frac {b}{x^2}}-\frac {7\,a^2\,x^4\,\sqrt {a+\frac {b}{x^2}}}{8}-\frac {\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x^2}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,15{}\mathrm {i}}{8} \]

[In]

int(x^3*(a + b/x^2)^(5/2),x)

[Out]

(9*a*x^4*(a + b/x^2)^(3/2))/8 - (a^(1/2)*b^2*atan(((a + b/x^2)^(1/2)*1i)/a^(1/2))*15i)/8 - b^2*(a + b/x^2)^(1/
2) - (7*a^2*x^4*(a + b/x^2)^(1/2))/8