\(\int \frac {1}{x^5 (a+b x^4)^{3/2}} \, dx\) [855]

Optimal result

Integrand size = 15, antiderivative size = 69 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {3 b}{4 a^2 \sqrt {a+b x^4}}-\frac {1}{4 a x^4 \sqrt {a+b x^4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{5/2}} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 44, 53, 65, 214} \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{5/2}}-\frac {3 b}{4 a^2 \sqrt {a+b x^4}}-\frac {1}{4 a x^4 \sqrt {a+b x^4}} \]

[In]

Int[1/(x^5*(a + b*x^4)^(3/2)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\frac {-a-3 b x^4}{4 a^2 x^4 \sqrt {a+b x^4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 4.41 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{4}+a}}{\sqrt {a}}\right ) \sqrt {b \,x^{4}+a}\, b \,x^{4}-3 b \,x^{4} \sqrt {a}-a^{\frac {3}{2}}}{4 x^{4} a^{\frac {5}{2}} \sqrt {b \,x^{4}+a}}\)	\(62\)
default	\(-\frac {1}{4 a \,x^{4} \sqrt {b \,x^{4}+a}}-\frac {3 b}{4 a^{2} \sqrt {b \,x^{4}+a}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\)	\(63\)
risch	\(-\frac {\sqrt {b \,x^{4}+a}}{4 a^{2} x^{4}}-\frac {b}{2 a^{2} \sqrt {b \,x^{4}+a}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\)	\(63\)
elliptic	\(-\frac {1}{4 a \,x^{4} \sqrt {b \,x^{4}+a}}-\frac {3 b}{4 a^{2} \sqrt {b \,x^{4}+a}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\)	\(63\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.51 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt {a} \log \left (\frac {b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 2 \, {\left (3 \, a b x^{4} + a^{2}\right )} \sqrt {b x^{4} + a}}{8 \, {\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}, -\frac {3 \, {\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{4} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x^{4} + a^{2}\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 1.75 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=- \frac {1}{4 a \sqrt {b} x^{6} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {3 \sqrt {b}}{4 a^{2} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 a^{\frac {5}{2}}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (b x^{4} + a\right )} b - 2 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {3}{2}} a^{2} - \sqrt {b x^{4} + a} a^{3}\right )}} - \frac {3 \, b \log \left (\frac {\sqrt {b x^{4} + a} - \sqrt {a}}{\sqrt {b x^{4} + a} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {3 \, b \arctan \left (\frac {\sqrt {b x^{4} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{2}} - \frac {3 \, {\left (b x^{4} + a\right )} b - 2 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {3}{2}} - \sqrt {b x^{4} + a} a\right )} a^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 5.85 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^4+a}}{\sqrt {a}}\right )}{4\,a^{5/2}}-\frac {1}{4\,a\,x^4\,\sqrt {b\,x^4+a}}-\frac {3\,b}{4\,a^2\,\sqrt {b\,x^4+a}} \]

[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))