\(\int \frac {(a+\frac {b}{x^4})^{5/2}}{x^3} \, dx\) [2077]

Optimal result

Integrand size = 15, antiderivative size = 92 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=-\frac {5 a^2 \sqrt {a+\frac {b}{x^4}}}{32 x^2}-\frac {5 a \left (a+\frac {b}{x^4}\right )^{3/2}}{48 x^2}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{12 x^2}-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^4}} x^2}\right )}{32 \sqrt {b}} \]

[Out]

-5/48*a*(a+b/x^4)^(3/2)/x^2-1/12*(a+b/x^4)^(5/2)/x^2-5/32*a^3*arctanh(b^(1/2)/x^2/(a+b/x^4)^(1/2))/b^(1/2)-5/3
2*a^2*(a+b/x^4)^(1/2)/x^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {342, 281, 201, 223, 212} \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^4}}}\right )}{32 \sqrt {b}}-\frac {5 a^2 \sqrt {a+\frac {b}{x^4}}}{32 x^2}-\frac {5 a \left (a+\frac {b}{x^4}\right )^{3/2}}{48 x^2}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{12 x^2} \]

[In]

Int[(a + b/x^4)^(5/2)/x^3,x]

[Out]

(-5*a^2*Sqrt[a + b/x^4])/(32*x^2) - (5*a*(a + b/x^4)^(3/2))/(48*x^2) - (a + b/x^4)^(5/2)/(12*x^2) - (5*a^3*Arc
Tanh[Sqrt[b]/(Sqrt[a + b/x^4]*x^2)])/(32*Sqrt[b])

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int x \left (a+b x^4\right )^{5/2} \, dx,x,\frac {1}{x}\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \left (a+b x^2\right )^{5/2} \, dx,x,\frac {1}{x^2}\right )\right ) \\ & = -\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{12 x^2}-\frac {1}{12} (5 a) \text {Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {5 a \left (a+\frac {b}{x^4}\right )^{3/2}}{48 x^2}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{12 x^2}-\frac {1}{16} \left (5 a^2\right ) \text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {5 a^2 \sqrt {a+\frac {b}{x^4}}}{32 x^2}-\frac {5 a \left (a+\frac {b}{x^4}\right )^{3/2}}{48 x^2}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{12 x^2}-\frac {1}{32} \left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {5 a^2 \sqrt {a+\frac {b}{x^4}}}{32 x^2}-\frac {5 a \left (a+\frac {b}{x^4}\right )^{3/2}}{48 x^2}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{12 x^2}-\frac {1}{32} \left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^4}} x^2}\right ) \\ & = -\frac {5 a^2 \sqrt {a+\frac {b}{x^4}}}{32 x^2}-\frac {5 a \left (a+\frac {b}{x^4}\right )^{3/2}}{48 x^2}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{12 x^2}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^4}} x^2}\right )}{32 \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\frac {\sqrt {a+\frac {b}{x^4}} \left (-8 b^2-26 a b x^4-33 a^2 x^8-\frac {15 a^3 x^{12} \text {arctanh}\left (\frac {\sqrt {b+a x^4}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {b+a x^4}}\right )}{96 x^{10}} \]

[In]

Integrate[(a + b/x^4)^(5/2)/x^3,x]

[Out]

(Sqrt[a + b/x^4]*(-8*b^2 - 26*a*b*x^4 - 33*a^2*x^8 - (15*a^3*x^12*ArcTanh[Sqrt[b + a*x^4]/Sqrt[b]])/(Sqrt[b]*S
qrt[b + a*x^4])))/(96*x^10)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05

method	result	size
risch	\(-\frac {\left (33 a^{2} x^{8}+26 a b \,x^{4}+8 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{96 x^{10}}-\frac {5 a^{3} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{4}+b}}{x^{2}}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{32 \sqrt {b}\, \sqrt {a \,x^{4}+b}}\)	\(97\)
default	\(-\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (15 a^{3} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{4}+b}}{x^{2}}\right ) x^{12}+33 a^{2} \sqrt {a \,x^{4}+b}\, x^{8} \sqrt {b}+26 b^{\frac {3}{2}} a \sqrt {a \,x^{4}+b}\, x^{4}+8 b^{\frac {5}{2}} \sqrt {a \,x^{4}+b}\right )}{96 x^{2} \left (a \,x^{4}+b \right )^{\frac {5}{2}} \sqrt {b}}\)	\(113\)

[In]

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

[Out]

-1/96*(33*a^2*x^8+26*a*b*x^4+8*b^2)/x^10*((a*x^4+b)/x^4)^(1/2)-5/32*a^3/b^(1/2)*ln((2*b+2*b^(1/2)*(a*x^4+b)^(1
/2))/x^2)*((a*x^4+b)/x^4)^(1/2)*x^2/(a*x^4+b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} x^{10} \log \left (\frac {a x^{4} - 2 \, \sqrt {b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + 2 \, b}{x^{4}}\right ) - 2 \, {\left (33 \, a^{2} b x^{8} + 26 \, a b^{2} x^{4} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{192 \, b x^{10}}, \frac {15 \, a^{3} \sqrt {-b} x^{10} \arctan \left (\frac {\sqrt {-b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{b}\right ) - {\left (33 \, a^{2} b x^{8} + 26 \, a b^{2} x^{4} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{96 \, b x^{10}}\right ] \]

[In]

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

[Out]

[1/192*(15*a^3*sqrt(b)*x^10*log((a*x^4 - 2*sqrt(b)*x^2*sqrt((a*x^4 + b)/x^4) + 2*b)/x^4) - 2*(33*a^2*b*x^8 + 2
6*a*b^2*x^4 + 8*b^3)*sqrt((a*x^4 + b)/x^4))/(b*x^10), 1/96*(15*a^3*sqrt(-b)*x^10*arctan(sqrt(-b)*x^2*sqrt((a*x
^4 + b)/x^4)/b) - (33*a^2*b*x^8 + 26*a*b^2*x^4 + 8*b^3)*sqrt((a*x^4 + b)/x^4))/(b*x^10)]

Sympy [A] (verification not implemented)

Time = 2.76 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=- \frac {11 a^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x^{4}}}}{32 x^{2}} - \frac {13 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x^{4}}}}{48 x^{6}} - \frac {\sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x^{4}}}}{12 x^{10}} - \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{32 \sqrt {b}} \]

[In]

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

[Out]

-11*a**(5/2)*sqrt(1 + b/(a*x**4))/(32*x**2) - 13*a**(3/2)*b*sqrt(1 + b/(a*x**4))/(48*x**6) - sqrt(a)*b**2*sqrt
(1 + b/(a*x**4))/(12*x**10) - 5*a**3*asinh(sqrt(b)/(sqrt(a)*x**2))/(32*sqrt(b))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (72) = 144\).

Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.72 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\frac {5 \, a^{3} \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} x^{2} - \sqrt {b}}{\sqrt {a + \frac {b}{x^{4}}} x^{2} + \sqrt {b}}\right )}{64 \, \sqrt {b}} - \frac {33 \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}} a^{3} x^{10} - 40 \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{3} b x^{6} + 15 \, \sqrt {a + \frac {b}{x^{4}}} a^{3} b^{2} x^{2}}{96 \, {\left ({\left (a + \frac {b}{x^{4}}\right )}^{3} x^{12} - 3 \, {\left (a + \frac {b}{x^{4}}\right )}^{2} b x^{8} + 3 \, {\left (a + \frac {b}{x^{4}}\right )} b^{2} x^{4} - b^{3}\right )}} \]

[In]

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

[Out]

5/64*a^3*log((sqrt(a + b/x^4)*x^2 - sqrt(b))/(sqrt(a + b/x^4)*x^2 + sqrt(b)))/sqrt(b) - 1/96*(33*(a + b/x^4)^(
5/2)*a^3*x^10 - 40*(a + b/x^4)^(3/2)*a^3*b*x^6 + 15*sqrt(a + b/x^4)*a^3*b^2*x^2)/((a + b/x^4)^3*x^12 - 3*(a +
b/x^4)^2*b*x^8 + 3*(a + b/x^4)*b^2*x^4 - b^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\frac {\frac {15 \, a^{4} \arctan \left (\frac {\sqrt {a x^{4} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {33 \, {\left (a x^{4} + b\right )}^{\frac {5}{2}} a^{4} - 40 \, {\left (a x^{4} + b\right )}^{\frac {3}{2}} a^{4} b + 15 \, \sqrt {a x^{4} + b} a^{4} b^{2}}{a^{3} x^{12}}}{96 \, a} \]

[In]

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

[Out]

1/96*(15*a^4*arctan(sqrt(a*x^4 + b)/sqrt(-b))/sqrt(-b) - (33*(a*x^4 + b)^(5/2)*a^4 - 40*(a*x^4 + b)^(3/2)*a^4*
b + 15*sqrt(a*x^4 + b)*a^4*b^2)/(a^3*x^12))/a

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\int \frac {{\left (a+\frac {b}{x^4}\right )}^{5/2}}{x^3} \,d x \]

[In]

int((a + b/x^4)^(5/2)/x^3,x)

[Out]

int((a + b/x^4)^(5/2)/x^3, x)