\(\int (a+\frac {b}{x^4})^{5/2} x^3 \, dx\) [2074]

Optimal result

Integrand size = 15, antiderivative size = 80 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=-\frac {5}{4} a b \sqrt {a+\frac {b}{x^4}}-\frac {5}{12} b \left (a+\frac {b}{x^4}\right )^{3/2}+\frac {1}{4} \left (a+\frac {b}{x^4}\right )^{5/2} x^4+\frac {5}{4} a^{3/2} b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 80, 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} \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {5}{4} a^{3/2} b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )+\frac {1}{4} x^4 \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {5}{12} b \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {5}{4} a b \sqrt {a+\frac {b}{x^4}} \]

[In]

Int[(a + b/x^4)^(5/2)*x^3,x]

[Out]

(-5*a*b*Sqrt[a + b/x^4])/4 - (5*b*(a + b/x^4)^(3/2))/12 + ((a + b/x^4)^(5/2)*x^4)/4 + (5*a^(3/2)*b*ArcTanh[Sqr
t[a + b/x^4]/Sqrt[a]])/4

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.20 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {\sqrt {a+\frac {b}{x^4}} \left (\sqrt {b+a x^4} \left (-2 b^2-14 a b x^4+3 a^2 x^8\right )+15 a^{3/2} b x^6 \log \left (\sqrt {a} x^2+\sqrt {b+a x^4}\right )\right )}{12 x^4 \sqrt {b+a x^4}} \]

[In]

Integrate[(a + b/x^4)^(5/2)*x^3,x]

[Out]

(Sqrt[a + b/x^4]*(Sqrt[b + a*x^4]*(-2*b^2 - 14*a*b*x^4 + 3*a^2*x^8) + 15*a^(3/2)*b*x^6*Log[Sqrt[a]*x^2 + Sqrt[
b + a*x^4]]))/(12*x^4*Sqrt[b + a*x^4])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {\left (3 a^{2} x^{8}-14 a b \,x^{4}-2 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{12 x^{4}}+\frac {5 a^{\frac {3}{2}} b \ln \left (x^{2} \sqrt {a}+\sqrt {a \,x^{4}+b}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{4 \sqrt {a \,x^{4}+b}}\)	\(90\)
default	\(\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} x^{4} \left (3 a^{2} x^{8} \sqrt {a \,x^{4}+b}+15 a^{\frac {3}{2}} b \ln \left (x^{2} \sqrt {a}+\sqrt {a \,x^{4}+b}\right ) x^{6}-14 a b \sqrt {a \,x^{4}+b}\, x^{4}-2 b^{2} \sqrt {a \,x^{4}+b}\right )}{12 \left (a \,x^{4}+b \right )^{\frac {5}{2}}}\)	\(103\)

[In]

int((a+b/x^4)^(5/2)*x^3,x,method=_RETURNVERBOSE)

[Out]

1/12*(3*a^2*x^8-14*a*b*x^4-2*b^2)/x^4*((a*x^4+b)/x^4)^(1/2)+5/4*a^(3/2)*b*ln(x^2*a^(1/2)+(a*x^4+b)^(1/2))*((a*
x^4+b)/x^4)^(1/2)*x^2/(a*x^4+b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.12 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\left [\frac {15 \, a^{\frac {3}{2}} b x^{4} \log \left (-2 \, a x^{4} - 2 \, \sqrt {a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}} - b\right ) + 2 \, {\left (3 \, a^{2} x^{8} - 14 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{24 \, x^{4}}, -\frac {15 \, \sqrt {-a} a b x^{4} \arctan \left (\frac {\sqrt {-a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) - {\left (3 \, a^{2} x^{8} - 14 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{12 \, x^{4}}\right ] \]

[In]

integrate((a+b/x^4)^(5/2)*x^3,x, algorithm="fricas")

[Out]

[1/24*(15*a^(3/2)*b*x^4*log(-2*a*x^4 - 2*sqrt(a)*x^4*sqrt((a*x^4 + b)/x^4) - b) + 2*(3*a^2*x^8 - 14*a*b*x^4 -
2*b^2)*sqrt((a*x^4 + b)/x^4))/x^4, -1/12*(15*sqrt(-a)*a*b*x^4*arctan(sqrt(-a)*x^4*sqrt((a*x^4 + b)/x^4)/(a*x^4
+ b)) - (3*a^2*x^8 - 14*a*b*x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^4]

Sympy [A] (verification not implemented)

Time = 2.47 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.40 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {a^{\frac {5}{2}} x^{4} \sqrt {1 + \frac {b}{a x^{4}}}}{4} - \frac {7 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x^{4}}}}{6} - \frac {5 a^{\frac {3}{2}} b \log {\left (\frac {b}{a x^{4}} \right )}}{8} + \frac {5 a^{\frac {3}{2}} b \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{4} - \frac {\sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x^{4}}}}{6 x^{4}} \]

[In]

integrate((a+b/x**4)**(5/2)*x**3,x)

[Out]

a**(5/2)*x**4*sqrt(1 + b/(a*x**4))/4 - 7*a**(3/2)*b*sqrt(1 + b/(a*x**4))/6 - 5*a**(3/2)*b*log(b/(a*x**4))/8 +
5*a**(3/2)*b*log(sqrt(1 + b/(a*x**4)) + 1)/4 - sqrt(a)*b**2*sqrt(1 + b/(a*x**4))/(6*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.01 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {1}{4} \, \sqrt {a + \frac {b}{x^{4}}} a^{2} x^{4} - \frac {5}{8} \, a^{\frac {3}{2}} b \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{4}}} + \sqrt {a}}\right ) - \frac {1}{6} \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} b - \sqrt {a + \frac {b}{x^{4}}} a b \]

[In]

integrate((a+b/x^4)^(5/2)*x^3,x, algorithm="maxima")

[Out]

1/4*sqrt(a + b/x^4)*a^2*x^4 - 5/8*a^(3/2)*b*log((sqrt(a + b/x^4) - sqrt(a))/(sqrt(a + b/x^4) + sqrt(a))) - 1/6
*(a + b/x^4)^(3/2)*b - sqrt(a + b/x^4)*a*b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (60) = 120\).

Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.78 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {1}{4} \, \sqrt {a x^{4} + b} a^{2} x^{2} - \frac {5}{8} \, a^{\frac {3}{2}} b \log \left ({\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2}\right ) + \frac {9 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{4} a^{\frac {3}{2}} b^{2} - 12 \, {\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2} a^{\frac {3}{2}} b^{3} + 7 \, a^{\frac {3}{2}} b^{4}}{3 \, {\left ({\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2} - b\right )}^{3}} \]

[In]

integrate((a+b/x^4)^(5/2)*x^3,x, algorithm="giac")

[Out]

1/4*sqrt(a*x^4 + b)*a^2*x^2 - 5/8*a^(3/2)*b*log((sqrt(a)*x^2 - sqrt(a*x^4 + b))^2) + 1/3*(9*(sqrt(a)*x^2 - sqr
t(a*x^4 + b))^4*a^(3/2)*b^2 - 12*(sqrt(a)*x^2 - sqrt(a*x^4 + b))^2*a^(3/2)*b^3 + 7*a^(3/2)*b^4)/((sqrt(a)*x^2
- sqrt(a*x^4 + b))^2 - b)^3

Mupad [B] (verification not implemented)

Time = 6.99 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^3 \, dx=\frac {a^2\,x^4\,\sqrt {a+\frac {b}{x^4}}}{4}-\frac {b\,{\left (a+\frac {b}{x^4}\right )}^{3/2}}{6}-a\,b\,\sqrt {a+\frac {b}{x^4}}-\frac {a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x^4}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{4} \]

[In]

int(x^3*(a + b/x^4)^(5/2),x)

[Out]

(a^2*x^4*(a + b/x^4)^(1/2))/4 - (b*(a + b/x^4)^(3/2))/6 - (a^(3/2)*b*atan(((a + b/x^4)^(1/2)*1i)/a^(1/2))*5i)/
4 - a*b*(a + b/x^4)^(1/2)