3.2067 \(\int (a+\frac {b}{x^4})^{3/2} x \, dx\)

Optimal. Leaf size=69 \[ \frac {1}{2} x^2 \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {3 b \sqrt {a+\frac {b}{x^4}}}{4 x^2}-\frac {3}{4} a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^4}}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {335, 275, 277, 195, 217, 206} \[ \frac {1}{2} x^2 \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {3 b \sqrt {a+\frac {b}{x^4}}}{4 x^2}-\frac {3}{4} a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^4}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(3/2)*x,x]

[Out]

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

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.68 \[ \frac {a x^6 \left (a+\frac {b}{x^4}\right )^{3/2} \left (a x^4+b\right ) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {a x^4}{b}+1\right )}{10 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(3/2)*x,x]

[Out]

(a*(a + b/x^4)^(3/2)*x^6*(b + a*x^4)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (a*x^4)/b])/(10*b^2)

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fricas [A]  time = 0.73, size = 143, normalized size = 2.07 \[ \left [\frac {3 \, a \sqrt {b} x^{2} \log \left (\frac {a x^{4} - 2 \, \sqrt {b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + 2 \, b}{x^{4}}\right ) + 2 \, {\left (2 \, a x^{4} - b\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{8 \, x^{2}}, \frac {3 \, a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{b}\right ) + {\left (2 \, a x^{4} - b\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{4 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 63, normalized size = 0.91 \[ \frac {\frac {3 \, a^{2} b \arctan \left (\frac {\sqrt {a x^{4} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 2 \, \sqrt {a x^{4} + b} a^{2} - \frac {\sqrt {a x^{4} + b} a b}{x^{4}}}{4 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 86, normalized size = 1.25 \[ \frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} \left (-3 a \sqrt {b}\, x^{4} \ln \left (\frac {2 b +2 \sqrt {a \,x^{4}+b}\, \sqrt {b}}{x^{2}}\right )+2 \sqrt {a \,x^{4}+b}\, a \,x^{4}-\sqrt {a \,x^{4}+b}\, b \right ) x^{2}}{4 \left (a \,x^{4}+b \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((a*x^4+b)/x^4)^(3/2)*x^2*(-3*a*b^(1/2)*ln(2*(b+(a*x^4+b)^(1/2)*b^(1/2))/x^2)*x^4+2*(a*x^4+b)^(1/2)*a*x^4-
(a*x^4+b)^(1/2)*b)/(a*x^4+b)^(3/2)

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maxima [A]  time = 1.97, size = 95, normalized size = 1.38 \[ \frac {1}{2} \, \sqrt {a + \frac {b}{x^{4}}} a x^{2} - \frac {\sqrt {a + \frac {b}{x^{4}}} a b x^{2}}{4 \, {\left ({\left (a + \frac {b}{x^{4}}\right )} x^{4} - b\right )}} + \frac {3}{8} \, a \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} x^{2} - \sqrt {b}}{\sqrt {a + \frac {b}{x^{4}}} x^{2} + \sqrt {b}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/x^4)^(3/2),x)

[Out]

int(x*(a + b/x^4)^(3/2), x)

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sympy [A]  time = 3.08, size = 95, normalized size = 1.38 \[ \frac {a^{\frac {3}{2}} x^{2}}{2 \sqrt {1 + \frac {b}{a x^{4}}}} + \frac {\sqrt {a} b}{4 x^{2} \sqrt {1 + \frac {b}{a x^{4}}}} - \frac {3 a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{4} - \frac {b^{2}}{4 \sqrt {a} x^{6} \sqrt {1 + \frac {b}{a x^{4}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*x**2/(2*sqrt(1 + b/(a*x**4))) + sqrt(a)*b/(4*x**2*sqrt(1 + b/(a*x**4))) - 3*a*sqrt(b)*asinh(sqrt(b)/(
sqrt(a)*x**2))/4 - b**2/(4*sqrt(a)*x**6*sqrt(1 + b/(a*x**4)))

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