Integrand size = 15, antiderivative size = 47 \[ \int \frac {\sqrt {a+c x^4}}{x^5} \, dx=-\frac {\sqrt {a+c x^4}}{4 x^4}-\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}} \] Output:
-1/4*(c*x^4+a)^(1/2)/x^4-1/4*c*arctanh((c*x^4+a)^(1/2)/a^(1/2))/a^(1/2)
Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {a+c x^4}}{x^5} \, dx=-\frac {\sqrt {a+c x^4}}{4 x^4}-\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}} \] Input:
Integrate[Sqrt[a + c*x^4]/x^5,x]
Output:
-1/4*Sqrt[a + c*x^4]/x^4 - (c*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/(4*Sqrt[a] )
Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {798, 51, 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.
\(\displaystyle \int \frac {\sqrt {a+c x^4}}{x^5} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {\sqrt {c x^4+a}}{x^8}dx^4\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{2} c \int \frac {1}{x^4 \sqrt {c x^4+a}}dx^4-\frac {\sqrt {a+c x^4}}{x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (\int \frac {1}{\frac {x^8}{c}-\frac {a}{c}}d\sqrt {c x^4+a}-\frac {\sqrt {a+c x^4}}{x^4}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4} \left (-\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {a+c x^4}}{x^4}\right )\) |
Input:
Int[Sqrt[a + c*x^4]/x^5,x]
Output:
(-(Sqrt[a + c*x^4]/x^4) - (c*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/Sqrt[a])/4
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] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[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 = 0.64 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(\frac {c \left (-\frac {\sqrt {c \,x^{4}+a}}{x^{4} c}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {c \,x^{4}+a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{4}\) | \(41\) |
risch | \(-\frac {\sqrt {c \,x^{4}+a}}{4 x^{4}}-\frac {c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{4}+a}}{x^{2}}\right )}{4 \sqrt {a}}\) | \(45\) |
default | \(-\frac {\left (c \,x^{4}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{4}+a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {c \sqrt {c \,x^{4}+a}}{4 a}\) | \(63\) |
elliptic | \(-\frac {\left (c \,x^{4}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{4}+a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {c \sqrt {c \,x^{4}+a}}{4 a}\) | \(63\) |
Input:
int((c*x^4+a)^(1/2)/x^5,x,method=_RETURNVERBOSE)
Output:
1/4*c*(-(c*x^4+a)^(1/2)/x^4/c-1/a^(1/2)*arctanh((c*x^4+a)^(1/2)/a^(1/2)))
Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.23 \[ \int \frac {\sqrt {a+c x^4}}{x^5} \, dx=\left [\frac {\sqrt {a} c x^{4} \log \left (\frac {c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 2 \, \sqrt {c x^{4} + a} a}{8 \, a x^{4}}, \frac {\sqrt {-a} c x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{4} + a}}\right ) - \sqrt {c x^{4} + a} a}{4 \, a x^{4}}\right ] \] Input:
integrate((c*x^4+a)^(1/2)/x^5,x, algorithm="fricas")
Output:
[1/8*(sqrt(a)*c*x^4*log((c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(a) + 2*a)/x^4) - 2 *sqrt(c*x^4 + a)*a)/(a*x^4), 1/4*(sqrt(-a)*c*x^4*arctan(sqrt(-a)/sqrt(c*x^ 4 + a)) - sqrt(c*x^4 + a)*a)/(a*x^4)]
Time = 1.00 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {a+c x^4}}{x^5} \, dx=- \frac {\sqrt {c} \sqrt {\frac {a}{c x^{4}} + 1}}{4 x^{2}} - \frac {c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x^{2}} \right )}}{4 \sqrt {a}} \] Input:
integrate((c*x**4+a)**(1/2)/x**5,x)
Output:
-sqrt(c)*sqrt(a/(c*x**4) + 1)/(4*x**2) - c*asinh(sqrt(a)/(sqrt(c)*x**2))/( 4*sqrt(a))
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {a+c x^4}}{x^5} \, dx=\frac {c \log \left (\frac {\sqrt {c x^{4} + a} - \sqrt {a}}{\sqrt {c x^{4} + a} + \sqrt {a}}\right )}{8 \, \sqrt {a}} - \frac {\sqrt {c x^{4} + a}}{4 \, x^{4}} \] Input:
integrate((c*x^4+a)^(1/2)/x^5,x, algorithm="maxima")
Output:
1/8*c*log((sqrt(c*x^4 + a) - sqrt(a))/(sqrt(c*x^4 + a) + sqrt(a)))/sqrt(a) - 1/4*sqrt(c*x^4 + a)/x^4
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {a+c x^4}}{x^5} \, dx=\frac {1}{4} \, c {\left (\frac {\arctan \left (\frac {\sqrt {c x^{4} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\sqrt {c x^{4} + a}}{c x^{4}}\right )} \] Input:
integrate((c*x^4+a)^(1/2)/x^5,x, algorithm="giac")
Output:
1/4*c*(arctan(sqrt(c*x^4 + a)/sqrt(-a))/sqrt(-a) - sqrt(c*x^4 + a)/(c*x^4) )
Time = 0.49 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {a+c x^4}}{x^5} \, dx=-\frac {\sqrt {c\,x^4+a}}{4\,x^4}-\frac {c\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^4+a}}{\sqrt {a}}\right )}{4\,\sqrt {a}} \] Input:
int((a + c*x^4)^(1/2)/x^5,x)
Output:
- (a + c*x^4)^(1/2)/(4*x^4) - (c*atanh((a + c*x^4)^(1/2)/a^(1/2)))/(4*a^(1 /2))
Time = 0.21 (sec) , antiderivative size = 289, normalized size of antiderivative = 6.15 \[ \int \frac {\sqrt {a+c x^4}}{x^5} \, dx=\frac {-\sqrt {a}\, \sqrt {c \,x^{4}+a}\, a -2 \sqrt {a}\, \sqrt {c \,x^{4}+a}\, c \,x^{4}+2 \sqrt {c}\, \sqrt {c \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}-\sqrt {a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) c \,x^{6}-2 \sqrt {c}\, \sqrt {c \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}+\sqrt {a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) c \,x^{6}-2 \sqrt {c}\, \sqrt {a}\, a \,x^{2}-2 \sqrt {c}\, \sqrt {a}\, c \,x^{6}+\mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}-\sqrt {a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) a c \,x^{4}+2 \,\mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}-\sqrt {a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) c^{2} x^{8}-\mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}+\sqrt {a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) a c \,x^{4}-2 \,\mathrm {log}\left (\frac {\sqrt {c \,x^{4}+a}+\sqrt {a}+\sqrt {c}\, x^{2}}{\sqrt {a}}\right ) c^{2} x^{8}}{4 \sqrt {a}\, x^{4} \left (2 \sqrt {c}\, \sqrt {c \,x^{4}+a}\, x^{2}+a +2 c \,x^{4}\right )} \] Input:
int((c*x^4+a)^(1/2)/x^5,x)
Output:
( - sqrt(a)*sqrt(a + c*x**4)*a - 2*sqrt(a)*sqrt(a + c*x**4)*c*x**4 + 2*sqr t(c)*sqrt(a + c*x**4)*log((sqrt(a + c*x**4) - sqrt(a) + sqrt(c)*x**2)/sqrt (a))*c*x**6 - 2*sqrt(c)*sqrt(a + c*x**4)*log((sqrt(a + c*x**4) + sqrt(a) + sqrt(c)*x**2)/sqrt(a))*c*x**6 - 2*sqrt(c)*sqrt(a)*a*x**2 - 2*sqrt(c)*sqrt (a)*c*x**6 + log((sqrt(a + c*x**4) - sqrt(a) + sqrt(c)*x**2)/sqrt(a))*a*c* x**4 + 2*log((sqrt(a + c*x**4) - sqrt(a) + sqrt(c)*x**2)/sqrt(a))*c**2*x** 8 - log((sqrt(a + c*x**4) + sqrt(a) + sqrt(c)*x**2)/sqrt(a))*a*c*x**4 - 2* log((sqrt(a + c*x**4) + sqrt(a) + sqrt(c)*x**2)/sqrt(a))*c**2*x**8)/(4*sqr t(a)*x**4*(2*sqrt(c)*sqrt(a + c*x**4)*x**2 + a + 2*c*x**4))