Integrand size = 15, antiderivative size = 50 \[ \int \frac {1}{x^5 \sqrt {a+b x^4}} \, dx=-\frac {\sqrt {a+b x^4}}{4 a x^4}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{3/2}} \] Output:
-1/4*(b*x^4+a)^(1/2)/a/x^4+1/4*b*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(3/2)
Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \sqrt {a+b x^4}} \, dx=-\frac {\sqrt {a+b x^4}}{4 a x^4}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{3/2}} \] Input:
Integrate[1/(x^5*Sqrt[a + b*x^4]),x]
Output:
-1/4*Sqrt[a + b*x^4]/(a*x^4) + (b*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(4*a^( 3/2))
Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {798, 52, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^5 \sqrt {a+b x^4}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \sqrt {b x^4+a}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {b \int \frac {1}{x^4 \sqrt {b x^4+a}}dx^4}{2 a}-\frac {\sqrt {a+b x^4}}{a x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {1}{\frac {x^8}{b}-\frac {a}{b}}d\sqrt {b x^4+a}}{a}-\frac {\sqrt {a+b x^4}}{a x^4}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4} \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^4}}{a x^4}\right )\) |
Input:
Int[1/(x^5*Sqrt[a + b*x^4]),x]
Output:
(-(Sqrt[a + b*x^4]/(a*x^4)) + (b*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/a^(3/2) )/4
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Time = 0.67 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86
method | result | size |
pseudoelliptic | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{4}+a}}{\sqrt {a}}\right ) b \,x^{4}-\sqrt {a}\, \sqrt {b \,x^{4}+a}}{4 a^{\frac {3}{2}} x^{4}}\) | \(43\) |
default | \(-\frac {\sqrt {b \,x^{4}+a}}{4 a \,x^{4}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\) | \(48\) |
risch | \(-\frac {\sqrt {b \,x^{4}+a}}{4 a \,x^{4}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\) | \(48\) |
elliptic | \(-\frac {\sqrt {b \,x^{4}+a}}{4 a \,x^{4}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\) | \(48\) |
Input:
int(1/x^5/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/4*(arctanh((b*x^4+a)^(1/2)/a^(1/2))*b*x^4-a^(1/2)*(b*x^4+a)^(1/2))/a^(3/ 2)/x^4
Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.08 \[ \int \frac {1}{x^5 \sqrt {a+b x^4}} \, dx=\left [\frac {\sqrt {a} b x^{4} \log \left (\frac {b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 2 \, \sqrt {b x^{4} + a} a}{8 \, a^{2} x^{4}}, -\frac {\sqrt {-a} b x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{4} + a}}\right ) + \sqrt {b x^{4} + a} a}{4 \, a^{2} x^{4}}\right ] \] Input:
integrate(1/x^5/(b*x^4+a)^(1/2),x, algorithm="fricas")
Output:
[1/8*(sqrt(a)*b*x^4*log((b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(a) + 2*a)/x^4) - 2 *sqrt(b*x^4 + a)*a)/(a^2*x^4), -1/4*(sqrt(-a)*b*x^4*arctan(sqrt(-a)/sqrt(b *x^4 + a)) + sqrt(b*x^4 + a)*a)/(a^2*x^4)]
Time = 1.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^5 \sqrt {a+b x^4}} \, dx=- \frac {\sqrt {b} \sqrt {\frac {a}{b x^{4}} + 1}}{4 a x^{2}} + \frac {b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 a^{\frac {3}{2}}} \] Input:
integrate(1/x**5/(b*x**4+a)**(1/2),x)
Output:
-sqrt(b)*sqrt(a/(b*x**4) + 1)/(4*a*x**2) + b*asinh(sqrt(a)/(sqrt(b)*x**2)) /(4*a**(3/2))
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x^5 \sqrt {a+b x^4}} \, dx=-\frac {\sqrt {b x^{4} + a} b}{4 \, {\left ({\left (b x^{4} + a\right )} a - a^{2}\right )}} - \frac {b \log \left (\frac {\sqrt {b x^{4} + a} - \sqrt {a}}{\sqrt {b x^{4} + a} + \sqrt {a}}\right )}{8 \, a^{\frac {3}{2}}} \] Input:
integrate(1/x^5/(b*x^4+a)^(1/2),x, algorithm="maxima")
Output:
-1/4*sqrt(b*x^4 + a)*b/((b*x^4 + a)*a - a^2) - 1/8*b*log((sqrt(b*x^4 + a) - sqrt(a))/(sqrt(b*x^4 + a) + sqrt(a)))/a^(3/2)
Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^5 \sqrt {a+b x^4}} \, dx=-\frac {1}{4} \, b {\left (\frac {\arctan \left (\frac {\sqrt {b x^{4} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {b x^{4} + a}}{a b x^{4}}\right )} \] Input:
integrate(1/x^5/(b*x^4+a)^(1/2),x, algorithm="giac")
Output:
-1/4*b*(arctan(sqrt(b*x^4 + a)/sqrt(-a))/(sqrt(-a)*a) + sqrt(b*x^4 + a)/(a *b*x^4))
Time = 0.47 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^5 \sqrt {a+b x^4}} \, dx=\frac {b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^4+a}}{\sqrt {a}}\right )}{4\,a^{3/2}}-\frac {\sqrt {b\,x^4+a}}{4\,a\,x^4} \] Input:
int(1/(x^5*(a + b*x^4)^(1/2)),x)
Output:
(b*atanh((a + b*x^4)^(1/2)/a^(1/2)))/(4*a^(3/2)) - (a + b*x^4)^(1/2)/(4*a* x^4)
Time = 0.20 (sec) , antiderivative size = 292, normalized size of antiderivative = 5.84 \[ \int \frac {1}{x^5 \sqrt {a+b x^4}} \, dx=\frac {-\sqrt {a}\, \sqrt {b \,x^{4}+a}\, a -2 \sqrt {a}\, \sqrt {b \,x^{4}+a}\, b \,x^{4}-2 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}-\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) b \,x^{6}+2 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) b \,x^{6}-2 \sqrt {b}\, \sqrt {a}\, a \,x^{2}-2 \sqrt {b}\, \sqrt {a}\, b \,x^{6}-\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}-\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a b \,x^{4}-2 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}-\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) b^{2} x^{8}+\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a b \,x^{4}+2 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) b^{2} x^{8}}{4 \sqrt {a}\, a \,x^{4} \left (2 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, x^{2}+a +2 b \,x^{4}\right )} \] Input:
int(1/x^5/(b*x^4+a)^(1/2),x)
Output:
( - sqrt(a)*sqrt(a + b*x**4)*a - 2*sqrt(a)*sqrt(a + b*x**4)*b*x**4 - 2*sqr t(b)*sqrt(a + b*x**4)*log((sqrt(a + b*x**4) - sqrt(a) + sqrt(b)*x**2)/sqrt (a))*b*x**6 + 2*sqrt(b)*sqrt(a + b*x**4)*log((sqrt(a + b*x**4) + sqrt(a) + sqrt(b)*x**2)/sqrt(a))*b*x**6 - 2*sqrt(b)*sqrt(a)*a*x**2 - 2*sqrt(b)*sqrt (a)*b*x**6 - log((sqrt(a + b*x**4) - sqrt(a) + sqrt(b)*x**2)/sqrt(a))*a*b* x**4 - 2*log((sqrt(a + b*x**4) - sqrt(a) + sqrt(b)*x**2)/sqrt(a))*b**2*x** 8 + log((sqrt(a + b*x**4) + sqrt(a) + sqrt(b)*x**2)/sqrt(a))*a*b*x**4 + 2* log((sqrt(a + b*x**4) + sqrt(a) + sqrt(b)*x**2)/sqrt(a))*b**2*x**8)/(4*sqr t(a)*a*x**4*(2*sqrt(b)*sqrt(a + b*x**4)*x**2 + a + 2*b*x**4))