\(\int \frac {x^5}{(a+\frac {b}{x^3})^{3/2}} \, dx\) [2037]

Optimal result

Integrand size = 15, antiderivative size = 95 \[ \int \frac {x^5}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=-\frac {5 b^2}{4 a^3 \sqrt {a+\frac {b}{x^3}}}-\frac {5 b x^3}{12 a^2 \sqrt {a+\frac {b}{x^3}}}+\frac {x^6}{6 a \sqrt {a+\frac {b}{x^3}}}+\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{4 a^{7/2}} \]

[Out]

5/4*b^2*arctanh((a+b/x^3)^(1/2)/a^(1/2))/a^(7/2)-5/4*b^2/a^3/(a+b/x^3)^(1/2)-5/12*b*x^3/a^2/(a+b/x^3)^(1/2)+1/
6*x^6/a/(a+b/x^3)^(1/2)

Rubi [A] (verified)

Time = 0.04 (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} \[ \int \frac {x^5}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{4 a^{7/2}}-\frac {5 b^2}{4 a^3 \sqrt {a+\frac {b}{x^3}}}-\frac {5 b x^3}{12 a^2 \sqrt {a+\frac {b}{x^3}}}+\frac {x^6}{6 a \sqrt {a+\frac {b}{x^3}}} \]

[In]

Int[x^5/(a + b/x^3)^(3/2),x]

[Out]

(-5*b^2)/(4*a^3*Sqrt[a + b/x^3]) - (5*b*x^3)/(12*a^2*Sqrt[a + b/x^3]) + x^6/(6*a*Sqrt[a + b/x^3]) + (5*b^2*Arc
Tanh[Sqrt[a + b/x^3]/Sqrt[a]])/(4*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}{3} \text {Subst}\left (\int \frac {1}{x^3 (a+b x)^{3/2}} \, dx,x,\frac {1}{x^3}\right )\right ) \\ & = \frac {x^6}{6 a \sqrt {a+\frac {b}{x^3}}}+\frac {(5 b) \text {Subst}\left (\int \frac {1}{x^2 (a+b x)^{3/2}} \, dx,x,\frac {1}{x^3}\right )}{12 a} \\ & = -\frac {5 b x^3}{12 a^2 \sqrt {a+\frac {b}{x^3}}}+\frac {x^6}{6 a \sqrt {a+\frac {b}{x^3}}}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\frac {1}{x^3}\right )}{8 a^2} \\ & = -\frac {5 b^2}{4 a^3 \sqrt {a+\frac {b}{x^3}}}-\frac {5 b x^3}{12 a^2 \sqrt {a+\frac {b}{x^3}}}+\frac {x^6}{6 a \sqrt {a+\frac {b}{x^3}}}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^3}\right )}{8 a^3} \\ & = -\frac {5 b^2}{4 a^3 \sqrt {a+\frac {b}{x^3}}}-\frac {5 b x^3}{12 a^2 \sqrt {a+\frac {b}{x^3}}}+\frac {x^6}{6 a \sqrt {a+\frac {b}{x^3}}}-\frac {(5 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^3}}\right )}{4 a^3} \\ & = -\frac {5 b^2}{4 a^3 \sqrt {a+\frac {b}{x^3}}}-\frac {5 b x^3}{12 a^2 \sqrt {a+\frac {b}{x^3}}}+\frac {x^6}{6 a \sqrt {a+\frac {b}{x^3}}}+\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{4 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.94 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.03 \[ \int \frac {x^5}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {\sqrt {a} x^{3/2} \left (-15 b^2-5 a b x^3+2 a^2 x^6\right )+15 b^2 \sqrt {b+a x^3} \log \left (\sqrt {a} x^{3/2}+\sqrt {b+a x^3}\right )}{12 a^{7/2} \sqrt {a+\frac {b}{x^3}} x^{3/2}} \]

[In]

Integrate[x^5/(a + b/x^3)^(3/2),x]

[Out]

(Sqrt[a]*x^(3/2)*(-15*b^2 - 5*a*b*x^3 + 2*a^2*x^6) + 15*b^2*Sqrt[b + a*x^3]*Log[Sqrt[a]*x^(3/2) + Sqrt[b + a*x
^3]])/(12*a^(7/2)*Sqrt[a + b/x^3]*x^(3/2))

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {\left (a \,x^{3}+b \right ) \left (2 a^{\frac {11}{2}} x^{8}-5 a^{\frac {9}{2}} b \,x^{5}-15 b^{2} x^{2} a^{\frac {7}{2}}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (a \,x^{3}+b \right )}}{x^{2} \sqrt {a}}\right ) a^{3} \sqrt {x \left (a \,x^{3}+b \right )}\, b^{2}\right )}{12 \left (\frac {a \,x^{3}+b}{x^{3}}\right )^{\frac {3}{2}} x^{5} a^{\frac {13}{2}}}\)	\(96\)
risch	\(\frac {\left (2 a \,x^{3}-7 b \right ) \left (a \,x^{3}+b \right )}{12 a^{3} \sqrt {\frac {a \,x^{3}+b}{x^{3}}}}+\frac {b^{2} \left (\frac {10 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (a \,x^{3}+b \right )}}{x^{2} \sqrt {a}}\right )}{\sqrt {a}}-\frac {16 x^{2}}{3 \sqrt {\left (x^{3}+\frac {b}{a}\right ) a x}}\right ) \sqrt {x \left (a \,x^{3}+b \right )}}{8 a^{3} x^{2} \sqrt {\frac {a \,x^{3}+b}{x^{3}}}}\)	\(116\)

[In]

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

[Out]

1/12/((a*x^3+b)/x^3)^(3/2)/x^5*(a*x^3+b)*(2*a^(11/2)*x^8-5*a^(9/2)*b*x^5-15*b^2*x^2*a^(7/2)+15*arctanh((x*(a*x
^3+b))^(1/2)/x^2/a^(1/2))*a^3*(x*(a*x^3+b))^(1/2)*b^2)/a^(13/2)

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.56 \[ \int \frac {x^5}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (a b^{2} x^{3} + b^{3}\right )} \sqrt {a} \log \left (-8 \, a^{2} x^{6} - 8 \, a b x^{3} - b^{2} - 4 \, {\left (2 \, a x^{6} + b x^{3}\right )} \sqrt {a} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right ) + 4 \, {\left (2 \, a^{3} x^{9} - 5 \, a^{2} b x^{6} - 15 \, a b^{2} x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{48 \, {\left (a^{5} x^{3} + a^{4} b\right )}}, -\frac {15 \, {\left (a b^{2} x^{3} + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right ) - 2 \, {\left (2 \, a^{3} x^{9} - 5 \, a^{2} b x^{6} - 15 \, a b^{2} x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{24 \, {\left (a^{5} x^{3} + a^{4} b\right )}}\right ] \]

[In]

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

[Out]

[1/48*(15*(a*b^2*x^3 + b^3)*sqrt(a)*log(-8*a^2*x^6 - 8*a*b*x^3 - b^2 - 4*(2*a*x^6 + b*x^3)*sqrt(a)*sqrt((a*x^3
+ b)/x^3)) + 4*(2*a^3*x^9 - 5*a^2*b*x^6 - 15*a*b^2*x^3)*sqrt((a*x^3 + b)/x^3))/(a^5*x^3 + a^4*b), -1/24*(15*(
a*b^2*x^3 + b^3)*sqrt(-a)*arctan(2*sqrt(-a)*x^3*sqrt((a*x^3 + b)/x^3)/(2*a*x^3 + b)) - 2*(2*a^3*x^9 - 5*a^2*b*
x^6 - 15*a*b^2*x^3)*sqrt((a*x^3 + b)/x^3))/(a^5*x^3 + a^4*b)]

Sympy [A] (verification not implemented)

Time = 3.71 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.16 \[ \int \frac {x^5}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {x^{\frac {15}{2}}}{6 a \sqrt {b} \sqrt {\frac {a x^{3}}{b} + 1}} - \frac {5 \sqrt {b} x^{\frac {9}{2}}}{12 a^{2} \sqrt {\frac {a x^{3}}{b} + 1}} - \frac {5 b^{\frac {3}{2}} x^{\frac {3}{2}}}{4 a^{3} \sqrt {\frac {a x^{3}}{b} + 1}} + \frac {5 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x^{\frac {3}{2}}}{\sqrt {b}} \right )}}{4 a^{\frac {7}{2}}} \]

[In]

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

[Out]

x**(15/2)/(6*a*sqrt(b)*sqrt(a*x**3/b + 1)) - 5*sqrt(b)*x**(9/2)/(12*a**2*sqrt(a*x**3/b + 1)) - 5*b**(3/2)*x**(
3/2)/(4*a**3*sqrt(a*x**3/b + 1)) + 5*b**2*asinh(sqrt(a)*x**(3/2)/sqrt(b))/(4*a**(7/2))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.28 \[ \int \frac {x^5}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=-\frac {15 \, {\left (a + \frac {b}{x^{3}}\right )}^{2} b^{2} - 25 \, {\left (a + \frac {b}{x^{3}}\right )} a b^{2} + 8 \, a^{2} b^{2}}{12 \, {\left ({\left (a + \frac {b}{x^{3}}\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} a^{4} + \sqrt {a + \frac {b}{x^{3}}} a^{5}\right )}} - \frac {5 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{3}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}}\right )}{8 \, a^{\frac {7}{2}}} \]

[In]

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

[Out]

-1/12*(15*(a + b/x^3)^2*b^2 - 25*(a + b/x^3)*a*b^2 + 8*a^2*b^2)/((a + b/x^3)^(5/2)*a^3 - 2*(a + b/x^3)^(3/2)*a
^4 + sqrt(a + b/x^3)*a^5) - 5/8*b^2*log((sqrt(a + b/x^3) - sqrt(a))/(sqrt(a + b/x^3) + sqrt(a)))/a^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.80 \[ \int \frac {x^5}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {1}{12} \, \sqrt {a x^{4} + b x} x {\left (\frac {2 \, x^{3}}{a^{2}} - \frac {7 \, b}{a^{3}}\right )} - \frac {5 \, b^{2} \arctan \left (\frac {\sqrt {a + \frac {b}{x^{3}}}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} - \frac {2 \, b^{2}}{3 \, \sqrt {a + \frac {b}{x^{3}}} a^{3}} \]

[In]

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

[Out]

1/12*sqrt(a*x^4 + b*x)*x*(2*x^3/a^2 - 7*b/a^3) - 5/4*b^2*arctan(sqrt(a + b/x^3)/sqrt(-a))/(sqrt(-a)*a^3) - 2/3
*b^2/(sqrt(a + b/x^3)*a^3)

Mupad [B] (verification not implemented)

Time = 6.65 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01 \[ \int \frac {x^5}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx=\frac {x^6\,\sqrt {a+\frac {b}{x^3}}}{6\,a^2}-\frac {2\,b^2}{3\,a^3\,\sqrt {a+\frac {b}{x^3}}}+\frac {5\,b^2\,\ln \left (x^6\,\left (\sqrt {a+\frac {b}{x^3}}-\sqrt {a}\right )\,{\left (\sqrt {a+\frac {b}{x^3}}+\sqrt {a}\right )}^3\right )}{8\,a^{7/2}}-\frac {7\,b\,x^3\,\sqrt {a+\frac {b}{x^3}}}{12\,a^3} \]

[In]

int(x^5/(a + b/x^3)^(3/2),x)

[Out]

(x^6*(a + b/x^3)^(1/2))/(6*a^2) - (2*b^2)/(3*a^3*(a + b/x^3)^(1/2)) + (5*b^2*log(x^6*((a + b/x^3)^(1/2) - a^(1
/2))*((a + b/x^3)^(1/2) + a^(1/2))^3))/(8*a^(7/2)) - (7*b*x^3*(a + b/x^3)^(1/2))/(12*a^3)