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]

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

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Rubi [A]  time = 0.05, antiderivative size = 71, normalized size of antiderivative = 1.00, 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 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.05, size = 85, normalized size = 1.20 \[ -\frac {\sqrt {a+\frac {b}{x^4}} \left (3 a^2 x^8 \sqrt {\frac {a x^4}{b}+1} \tanh ^{-1}\left (\sqrt {\frac {a x^4}{b}+1}\right )+5 a^2 x^8+7 a b x^4+2 b^2\right )}{16 x^6 \left (a x^4+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(3/2)/x^3,x]

[Out]

-1/16*(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]]))/(x^6*(b + a*x^4))

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fricas [A]  time = 1.04, size = 160, normalized size = 2.25 \[ \left [\frac {3 \, a^{2} \sqrt {b} x^{6} \log \left (\frac {a x^{4} - 2 \, \sqrt {b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + 2 \, b}{x^{4}}\right ) - 2 \, {\left (5 \, a b x^{4} + 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{32 \, b x^{6}}, \frac {3 \, a^{2} \sqrt {-b} x^{6} \arctan \left (\frac {\sqrt {-b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{b}\right ) - {\left (5 \, a b x^{4} + 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{16 \, b x^{6}}\right ] \]

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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giac [A]  time = 0.16, size = 70, normalized size = 0.99 \[ \frac {\frac {3 \, a^{3} \arctan \left (\frac {\sqrt {a x^{4} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {5 \, {\left (a x^{4} + b\right )}^{\frac {3}{2}} a^{3} - 3 \, \sqrt {a x^{4} + b} a^{3} b}{a^{2} x^{8}}}{16 \, a} \]

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

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maple [A]  time = 0.02, size = 93, normalized size = 1.31 \[ -\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} \left (3 a^{2} x^{8} \ln \left (\frac {2 b +2 \sqrt {a \,x^{4}+b}\, \sqrt {b}}{x^{2}}\right )+5 \sqrt {a \,x^{4}+b}\, a \sqrt {b}\, x^{4}+2 \sqrt {a \,x^{4}+b}\, b^{\frac {3}{2}}\right )}{16 \left (a \,x^{4}+b \right )^{\frac {3}{2}} \sqrt {b}\, x^{2}} \]

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+(a*x^4+b)^(1/2)*b^(1/2))/x^2)*x^8+5*(a*x^4+b)^(1/2)*a*b^(1/2)*x
^4+2*b^(3/2)*(a*x^4+b)^(1/2))/(a*x^4+b)^(3/2)/b^(1/2)

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maxima [B]  time = 1.88, size = 119, normalized size = 1.68 \[ \frac {3 \, a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} x^{2} - \sqrt {b}}{\sqrt {a + \frac {b}{x^{4}}} x^{2} + \sqrt {b}}\right )}{32 \, \sqrt {b}} - \frac {5 \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{2} x^{6} - 3 \, \sqrt {a + \frac {b}{x^{4}}} a^{2} b x^{2}}{16 \, {\left ({\left (a + \frac {b}{x^{4}}\right )}^{2} x^{8} - 2 \, {\left (a + \frac {b}{x^{4}}\right )} b x^{4} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*a^2*log((sqrt(a + b/x^4)*x^2 - sqrt(b))/(sqrt(a + b/x^4)*x^2 + sqrt(b)))/sqrt(b) - 1/16*(5*(a + b/x^4)^(3
/2)*a^2*x^6 - 3*sqrt(a + b/x^4)*a^2*b*x^2)/((a + b/x^4)^2*x^8 - 2*(a + b/x^4)*b*x^4 + b^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{x^4}\right )}^{3/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^4)^(3/2)/x^3,x)

[Out]

int((a + b/x^4)^(3/2)/x^3, x)

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sympy [A]  time = 3.63, size = 75, normalized size = 1.06 \[ - \frac {5 a^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x^{4}}}}{16 x^{2}} - \frac {\sqrt {a} b \sqrt {1 + \frac {b}{a x^{4}}}}{8 x^{6}} - \frac {3 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{16 \sqrt {b}} \]

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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