3.855 \(\int \frac {1}{x^5 (a+b x^4)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 71, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac {3 \sqrt {a+b x^4}}{4 a^2 x^4}+\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {1}{2 a x^4 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^4)^(3/2)),x]

[Out]

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

Rule 51

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

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.54 \[ -\frac {b \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {b x^4}{a}+1\right )}{2 a^2 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^4)^(3/2)),x]

[Out]

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

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fricas [A]  time = 0.97, size = 173, normalized size = 2.51 \[ \left [\frac {3 \, {\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt {a} \log \left (\frac {b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 2 \, {\left (3 \, a b x^{4} + a^{2}\right )} \sqrt {b x^{4} + a}}{8 \, {\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}, -\frac {3 \, {\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{4} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x^{4} + a^{2}\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 72, normalized size = 1.04 \[ -\frac {3 \, b \arctan \left (\frac {\sqrt {b x^{4} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{2}} - \frac {3 \, {\left (b x^{4} + a\right )} b - 2 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {3}{2}} - \sqrt {b x^{4} + a} a\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 63, normalized size = 0.91 \[ \frac {3 b \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}-\frac {3 b}{4 \sqrt {b \,x^{4}+a}\, a^{2}}-\frac {1}{4 \sqrt {b \,x^{4}+a}\, a \,x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.98, size = 86, normalized size = 1.25 \[ -\frac {3 \, {\left (b x^{4} + a\right )} b - 2 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {3}{2}} a^{2} - \sqrt {b x^{4} + a} a^{3}\right )}} - \frac {3 \, b \log \left (\frac {\sqrt {b x^{4} + a} - \sqrt {a}}{\sqrt {b x^{4} + a} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.37, size = 53, normalized size = 0.77 \[ \frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^4+a}}{\sqrt {a}}\right )}{4\,a^{5/2}}-\frac {1}{4\,a\,x^4\,\sqrt {b\,x^4+a}}-\frac {3\,b}{4\,a^2\,\sqrt {b\,x^4+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^4)^(3/2)),x)

[Out]

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

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sympy [A]  time = 6.88, size = 76, normalized size = 1.10 \[ - \frac {1}{4 a \sqrt {b} x^{6} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {3 \sqrt {b}}{4 a^{2} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**4+a)**(3/2),x)

[Out]

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

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