3.2069 \(\int \frac{(a+\frac{b}{x^4})^{3/2}}{x^3} \, dx\)

Optimal. Leaf size=71 \[ -\frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{x^2 \sqrt{a+\frac{b}{x^4}}}\right )}{16 \sqrt{b}}-\frac{3 a \sqrt{a+\frac{b}{x^4}}}{16 x^2}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2} \]

[Out]

(-3*a*Sqrt[a + b/x^4])/(16*x^2) - (a + b/x^4)^(3/2)/(8*x^2) - (3*a^2*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^4]*x^2)])/(
16*Sqrt[b])

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Rubi [A]  time = 0.0520998, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{x^2 \sqrt{a+\frac{b}{x^4}}}\right )}{16 \sqrt{b}}-\frac{3 a \sqrt{a+\frac{b}{x^4}}}{16 x^2}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(3/2)/x^3,x]

[Out]

(-3*a*Sqrt[a + b/x^4])/(16*x^2) - (a + b/x^4)^(3/2)/(8*x^2) - (3*a^2*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^4]*x^2)])/(
16*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 )^{3/2}}{x^3} \, dx &=-\operatorname{Subst}\left (\int x \left (a+b x^4\right )^{3/2} \, dx,x,\frac{1}{x}\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}-\frac{1}{8} (3 a) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{3 a \sqrt{a+\frac{b}{x^4}}}{16 x^2}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}-\frac{1}{16} \left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{3 a \sqrt{a+\frac{b}{x^4}}}{16 x^2}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}-\frac{1}{16} \left (3 a^2\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{3 a \sqrt{a+\frac{b}{x^4}}}{16 x^2}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}-\frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x^4}} x^2}\right )}{16 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.036271, size = 85, normalized size = 1.2 \[ -\frac{\sqrt{a+\frac{b}{x^4}} \left (3 a^2 x^8 \sqrt{\frac{a x^4}{b}+1} \tanh ^{-1}\left (\sqrt{\frac{a x^4}{b}+1}\right )+5 a^2 x^8+7 a b x^4+2 b^2\right )}{16 x^6 \left (a x^4+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(3/2)/x^3,x]

[Out]

-(Sqrt[a + b/x^4]*(2*b^2 + 7*a*b*x^4 + 5*a^2*x^8 + 3*a^2*x^8*Sqrt[1 + (a*x^4)/b]*ArcTanh[Sqrt[1 + (a*x^4)/b]])
)/(16*x^6*(b + a*x^4))

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Maple [A]  time = 0.015, size = 93, normalized size = 1.3 \begin{align*} -{\frac{1}{16\,{x}^{2}} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{{\frac{3}{2}}} \left ( 3\,{a}^{2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{4}+b}+b}{{x}^{2}}} \right ){x}^{8}+5\,a\sqrt{a{x}^{4}+b}{x}^{4}\sqrt{b}+2\,{b}^{3/2}\sqrt{a{x}^{4}+b} \right ) \left ( a{x}^{4}+b \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((a*x^4+b)/x^4)^(3/2)/x^2*(3*a^2*ln(2*(b^(1/2)*(a*x^4+b)^(1/2)+b)/x^2)*x^8+5*a*(a*x^4+b)^(1/2)*x^4*b^(1/
2)+2*b^(3/2)*(a*x^4+b)^(1/2))/(a*x^4+b)^(3/2)/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)^(3/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51737, size = 363, normalized size = 5.11 \begin{align*} \left [\frac{3 \, a^{2} \sqrt{b} x^{6} \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 (5 \, a b x^{4} + 2 \, b^{2}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{32 \, b x^{6}}, \frac{3 \, a^{2} \sqrt{-b} x^{6} \arctan \left (\frac{\sqrt{-b} x^{2} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{b}\right ) -{\left (5 \, a b x^{4} + 2 \, b^{2}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{16 \, b x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*a^2*sqrt(b)*x^6*log((a*x^4 - 2*sqrt(b)*x^2*sqrt((a*x^4 + b)/x^4) + 2*b)/x^4) - 2*(5*a*b*x^4 + 2*b^2)*
sqrt((a*x^4 + b)/x^4))/(b*x^6), 1/16*(3*a^2*sqrt(-b)*x^6*arctan(sqrt(-b)*x^2*sqrt((a*x^4 + b)/x^4)/b) - (5*a*b
*x^4 + 2*b^2)*sqrt((a*x^4 + b)/x^4))/(b*x^6)]

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Sympy [A]  time = 4.45369, size = 75, normalized size = 1.06 \begin{align*} - \frac{5 a^{\frac{3}{2}} \sqrt{1 + \frac{b}{a x^{4}}}}{16 x^{2}} - \frac{\sqrt{a} b \sqrt{1 + \frac{b}{a x^{4}}}}{8 x^{6}} - \frac{3 a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} x^{2}} \right )}}{16 \sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*a**(3/2)*sqrt(1 + b/(a*x**4))/(16*x**2) - sqrt(a)*b*sqrt(1 + b/(a*x**4))/(8*x**6) - 3*a**2*asinh(sqrt(b)/(s
qrt(a)*x**2))/(16*sqrt(b))

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Giac [A]  time = 1.122, size = 82, normalized size = 1.15 \begin{align*} \frac{1}{16} \, a^{2}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{a x^{4} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{5 \,{\left (a x^{4} + b\right )}^{\frac{3}{2}} - 3 \, \sqrt{a x^{4} + b} b}{a^{2} x^{8}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a^2*(3*arctan(sqrt(a*x^4 + b)/sqrt(-b))/sqrt(-b) - (5*(a*x^4 + b)^(3/2) - 3*sqrt(a*x^4 + b)*b)/(a^2*x^8))