3.21.90 \(\int \frac {x^3}{(a+\frac {b}{x^4})^{3/2}} \, dx\) [2090]

Optimal. Leaf size=69 \[ \frac {3 b}{4 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {x^4}{4 a \sqrt {a+\frac {b}{x^4}}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{4 a^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 44, 53, 65, 214} \begin {gather*} -\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {3 b}{4 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {x^4}{4 a \sqrt {a+\frac {b}{x^4}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b/x^4)^(3/2),x]

[Out]

(3*b)/(4*a^2*Sqrt[a + b/x^4]) + x^4/(4*a*Sqrt[a + b/x^4]) - (3*b*ArcTanh[Sqrt[a + b/x^4]/Sqrt[a]])/(4*a^(5/2))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 272

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

Rubi steps

\begin {align*} \int \frac {x^3}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx &=-\left (\frac {1}{4} \text {Subst}\left (\int \frac {1}{x^2 (a+b x)^{3/2}} \, dx,x,\frac {1}{x^4}\right )\right )\\ &=-\frac {x^4}{2 a \sqrt {a+\frac {b}{x^4}}}-\frac {3 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x^4}\right )}{4 a}\\ &=-\frac {x^4}{2 a \sqrt {a+\frac {b}{x^4}}}+\frac {3 \sqrt {a+\frac {b}{x^4}} x^4}{4 a^2}+\frac {(3 b) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^4}\right )}{8 a^2}\\ &=-\frac {x^4}{2 a \sqrt {a+\frac {b}{x^4}}}+\frac {3 \sqrt {a+\frac {b}{x^4}} x^4}{4 a^2}+\frac {3 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^4}}\right )}{4 a^2}\\ &=-\frac {x^4}{2 a \sqrt {a+\frac {b}{x^4}}}+\frac {3 \sqrt {a+\frac {b}{x^4}} x^4}{4 a^2}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{4 a^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 77, normalized size = 1.12 \begin {gather*} \frac {\sqrt {a} x^2 \left (3 b+a x^4\right )-3 b \sqrt {b+a x^4} \tanh ^{-1}\left (\frac {\sqrt {b+a x^4}}{\sqrt {a} x^2}\right )}{4 a^{5/2} \sqrt {a+\frac {b}{x^4}} x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b/x^4)^(3/2),x]

[Out]

(Sqrt[a]*x^2*(3*b + a*x^4) - 3*b*Sqrt[b + a*x^4]*ArcTanh[Sqrt[b + a*x^4]/(Sqrt[a]*x^2)])/(4*a^(5/2)*Sqrt[a + b
/x^4]*x^2)

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Maple [A]
time = 0.09, size = 80, normalized size = 1.16

method result size
default \(-\frac {\left (a \,x^{4}+b \right ) \left (-x^{6} a^{\frac {7}{2}}-3 b \,x^{2} a^{\frac {5}{2}}+3 b \ln \left (x^{2} \sqrt {a}+\sqrt {a \,x^{4}+b}\right ) a^{2} \sqrt {a \,x^{4}+b}\right )}{4 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x^{6} a^{\frac {9}{2}}}\) \(80\)
risch \(\frac {a \,x^{4}+b}{4 a^{2} \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}+\frac {\left (\frac {b \,x^{2}}{2 a^{2} \sqrt {a \,x^{4}+b}}-\frac {3 b \ln \left (x^{2} \sqrt {a}+\sqrt {a \,x^{4}+b}\right )}{4 a^{\frac {5}{2}}}\right ) \sqrt {a \,x^{4}+b}}{\sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}\) \(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a*x^4+b)*(-x^6*a^(7/2)-3*b*x^2*a^(5/2)+3*b*ln(x^2*a^(1/2)+(a*x^4+b)^(1/2))*a^2*(a*x^4+b)^(1/2))/((a*x^4+
b)/x^4)^(3/2)/x^6/a^(9/2)

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Maxima [A]
time = 0.51, size = 86, normalized size = 1.25 \begin {gather*} \frac {3 \, {\left (a + \frac {b}{x^{4}}\right )} b - 2 \, a b}{4 \, {\left ({\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{2} - \sqrt {a + \frac {b}{x^{4}}} a^{3}\right )}} + \frac {3 \, b \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{4}}} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*(a + b/x^4)*b - 2*a*b)/((a + b/x^4)^(3/2)*a^2 - sqrt(a + b/x^4)*a^3) + 3/8*b*log((sqrt(a + b/x^4) - sqr
t(a))/(sqrt(a + b/x^4) + sqrt(a)))/a^(5/2)

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Fricas [A]
time = 0.39, size = 192, normalized size = 2.78 \begin {gather*} \left [\frac {3 \, {\left (a b x^{4} + b^{2}\right )} \sqrt {a} \log \left (-2 \, a x^{4} + 2 \, \sqrt {a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}} - b\right ) + 2 \, {\left (a^{2} x^{8} + 3 \, a b x^{4}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{8 \, {\left (a^{4} x^{4} + a^{3} b\right )}}, \frac {3 \, {\left (a b x^{4} + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) + {\left (a^{2} x^{8} + 3 \, a b x^{4}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{4 \, {\left (a^{4} x^{4} + a^{3} b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.78, size = 75, normalized size = 1.09 \begin {gather*} \frac {x^{6}}{4 a \sqrt {b} \sqrt {\frac {a x^{4}}{b} + 1}} + \frac {3 \sqrt {b} x^{2}}{4 a^{2} \sqrt {\frac {a x^{4}}{b} + 1}} - \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a} x^{2}}{\sqrt {b}} \right )}}{4 a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**6/(4*a*sqrt(b)*sqrt(a*x**4/b + 1)) + 3*sqrt(b)*x**2/(4*a**2*sqrt(a*x**4/b + 1)) - 3*b*asinh(sqrt(a)*x**2/sq
rt(b))/(4*a**(5/2))

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Giac [A]
time = 0.51, size = 55, normalized size = 0.80 \begin {gather*} \frac {{\left (\frac {x^{4}}{a} + \frac {3 \, b}{a^{2}}\right )} x^{2}}{4 \, \sqrt {a x^{4} + b}} + \frac {3 \, b \log \left ({\left | -\sqrt {a} x^{2} + \sqrt {a x^{4} + b} \right |}\right )}{4 \, a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x^4/a + 3*b/a^2)*x^2/sqrt(a*x^4 + b) + 3/4*b*log(abs(-sqrt(a)*x^2 + sqrt(a*x^4 + b)))/a^(5/2)

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Mupad [B]
time = 1.76, size = 53, normalized size = 0.77 \begin {gather*} \frac {3\,b}{4\,a^2\,\sqrt {a+\frac {b}{x^4}}}+\frac {x^4}{4\,a\,\sqrt {a+\frac {b}{x^4}}}-\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{4\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b/x^4)^(3/2),x)

[Out]

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

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