3.2077 \(\int \frac{(a+\frac{b}{x^4})^{5/2}}{x^3} \, dx\)

Optimal. Leaf size=92 \[ -\frac{5 a^2 \sqrt{a+\frac{b}{x^4}}}{32 x^2}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{x^2 \sqrt{a+\frac{b}{x^4}}}\right )}{32 \sqrt{b}}-\frac{5 a \left (a+\frac{b}{x^4}\right )^{3/2}}{48 x^2}-\frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{12 x^2} \]

[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])

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Rubi [A]  time = 0.0671092, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {335, 275, 195, 217, 206} \[ -\frac{5 a^2 \sqrt{a+\frac{b}{x^4}}}{32 x^2}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{x^2 \sqrt{a+\frac{b}{x^4}}}\right )}{32 \sqrt{b}}-\frac{5 a \left (a+\frac{b}{x^4}\right )^{3/2}}{48 x^2}-\frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{12 x^2} \]

Antiderivative was successfully verified.

[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 335

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]

Rule 275

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 195

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 217

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 206

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

Rubi steps

\begin{align*} \int \frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{x^3} \, dx &=-\operatorname{Subst}\left (\int x \left (a+b x^4\right )^{5/2} \, dx,x,\frac{1}{x}\right )\\ &=-\left (\frac{1}{2} \operatorname{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) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.0423634, size = 96, normalized size = 1.04 \[ -\frac{\sqrt{a+\frac{b}{x^4}} \left (59 a^2 b x^8+15 a^3 x^{12} \sqrt{\frac{a x^4}{b}+1} \tanh ^{-1}\left (\sqrt{\frac{a x^4}{b}+1}\right )+33 a^3 x^{12}+34 a b^2 x^4+8 b^3\right )}{96 x^{10} \left (a x^4+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(5/2)/x^3,x]

[Out]

-(Sqrt[a + b/x^4]*(8*b^3 + 34*a*b^2*x^4 + 59*a^2*b*x^8 + 33*a^3*x^12 + 15*a^3*x^12*Sqrt[1 + (a*x^4)/b]*ArcTanh
[Sqrt[1 + (a*x^4)/b]]))/(96*x^10*(b + a*x^4))

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Maple [A]  time = 0.018, size = 113, normalized size = 1.2 \begin{align*} -{\frac{1}{96\,{x}^{2}} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{{\frac{5}{2}}} \left ( 15\,{a}^{3}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{4}+b}+b}{{x}^{2}}} \right ){x}^{12}+33\,{a}^{2}\sqrt{a{x}^{4}+b}\sqrt{b}{x}^{8}+26\,{b}^{3/2}a\sqrt{a{x}^{4}+b}{x}^{4}+8\,{b}^{5/2}\sqrt{a{x}^{4}+b} \right ) \left ( a{x}^{4}+b \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*((a*x^4+b)/x^4)^(5/2)/x^2*(15*a^3*ln(2*(b^(1/2)*(a*x^4+b)^(1/2)+b)/x^2)*x^12+33*a^2*(a*x^4+b)^(1/2)*b^(1
/2)*x^8+26*b^(3/2)*a*(a*x^4+b)^(1/2)*x^4+8*b^(5/2)*(a*x^4+b)^(1/2))/(a*x^4+b)^(5/2)/b^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53293, size = 421, normalized size = 4.58 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [A]  time = 5.36968, size = 102, normalized size = 1.11 \begin{align*} - \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.14566, size = 101, normalized size = 1.1 \begin{align*} \frac{1}{96} \, a^{3}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{a x^{4} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{33 \,{\left (a x^{4} + b\right )}^{\frac{5}{2}} - 40 \,{\left (a x^{4} + b\right )}^{\frac{3}{2}} b + 15 \, \sqrt{a x^{4} + b} b^{2}}{a^{3} x^{12}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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