3.2066 \(\int (a+\frac {b}{x^4})^{3/2} x^3 \, dx\)

Optimal. Leaf size=63 \[ \frac {1}{4} x^4 \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {3}{4} b \sqrt {a+\frac {b}{x^4}}+\frac {3}{4} \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 63, 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 = {266, 47, 50, 63, 208} \[ \frac {1}{4} x^4 \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {3}{4} b \sqrt {a+\frac {b}{x^4}}+\frac {3}{4} \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(3/2)*x^3,x]

[Out]

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

Rule 47

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

Rule 50

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

Rule 63

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 208

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

Rule 266

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

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Mathematica [C]  time = 0.02, size = 49, normalized size = 0.78 \[ -\frac {b \sqrt {a+\frac {b}{x^4}} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {a x^4}{b}\right )}{2 \sqrt {\frac {a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(3/2)*x^3,x]

[Out]

-1/2*(b*Sqrt[a + b/x^4]*Hypergeometric2F1[-3/2, -1/2, 1/2, -((a*x^4)/b)])/Sqrt[1 + (a*x^4)/b]

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fricas [A]  time = 0.93, size = 129, normalized size = 2.05 \[ \left [\frac {3}{8} \, \sqrt {a} b \log \left (-2 \, a x^{4} - 2 \, \sqrt {a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}} - b\right ) + \frac {1}{4} \, {\left (a x^{4} - 2 \, b\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}, -\frac {3}{4} \, \sqrt {-a} b \arctan \left (\frac {\sqrt {-a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) + \frac {1}{4} \, {\left (a x^{4} - 2 \, b\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 78, normalized size = 1.24 \[ \frac {1}{4} \, \sqrt {a x^{4} + b} a x^{2} - \frac {3}{8} \, \sqrt {a} b \log \left ({\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2}\right ) + \frac {\sqrt {a} b^{2}}{{\left (\sqrt {a} x^{2} - \sqrt {a x^{4} + b}\right )}^{2} - b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 82, normalized size = 1.30 \[ \frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} \left (\sqrt {a \,x^{4}+b}\, a \,x^{4}+3 \sqrt {a}\, b \,x^{2} \ln \left (\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}+b}\right )-2 \sqrt {a \,x^{4}+b}\, b \right ) x^{4}}{4 \left (a \,x^{4}+b \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.02, size = 66, normalized size = 1.05 \[ \frac {1}{4} \, \sqrt {a + \frac {b}{x^{4}}} a x^{4} - \frac {3}{8} \, \sqrt {a} b \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{4}}} + \sqrt {a}}\right ) - \frac {1}{2} \, \sqrt {a + \frac {b}{x^{4}}} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.74, size = 48, normalized size = 0.76 \[ \frac {a\,x^4\,\sqrt {a+\frac {b}{x^4}}}{4}-\frac {b\,\sqrt {a+\frac {b}{x^4}}}{2}+\frac {3\,\sqrt {a}\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/x^4)^(3/2),x)

[Out]

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

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sympy [A]  time = 3.28, size = 95, normalized size = 1.51 \[ \frac {3 \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a} x^{2}}{\sqrt {b}} \right )}}{4} + \frac {a^{2} x^{6}}{4 \sqrt {b} \sqrt {\frac {a x^{4}}{b} + 1}} - \frac {a \sqrt {b} x^{2}}{4 \sqrt {\frac {a x^{4}}{b} + 1}} - \frac {b^{\frac {3}{2}}}{2 x^{2} \sqrt {\frac {a x^{4}}{b} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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