Integrand size = 15, antiderivative size = 75 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {\sqrt [4]{a+b x^4}}{4 x^4}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}} \]
-1/4*(b*x^4+a)^(1/4)/x^4-1/8*b*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)-1/8 *b*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)
Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {\sqrt [4]{a+b x^4}}{4 x^4}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{3/4}} \]
-1/4*(a + b*x^4)^(1/4)/x^4 - (b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(3 /4)) - (b*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(3/4))
Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {798, 51, 73, 756, 216, 219}
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.
\(\displaystyle \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {\sqrt [4]{b x^4+a}}{x^8}dx^4\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{4} b \int \frac {1}{x^4 \left (b x^4+a\right )^{3/4}}dx^4-\frac {\sqrt [4]{a+b x^4}}{x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (\int \frac {1}{\frac {x^{16}}{b}-\frac {a}{b}}d\sqrt [4]{b x^4+a}-\frac {\sqrt [4]{a+b x^4}}{x^4}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{4} \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {\sqrt [4]{a+b x^4}}{x^4}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{x^4}\right )\) |
(-((a + b*x^4)^(1/4)/x^4) - (b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(2*a^(3/ 4)) - (b*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(2*a^(3/4)))/4
3.10.94.3.1 Defintions of rubi rules used
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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
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 = 4.46 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13
method | result | size |
pseudoelliptic | \(-\frac {\ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b \,x^{4}+2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b \,x^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{\frac {3}{4}}}{16 x^{4} a^{\frac {3}{4}}}\) | \(85\) |
-1/16*(ln((-(b*x^4+a)^(1/4)-a^(1/4))/(-(b*x^4+a)^(1/4)+a^(1/4)))*b*x^4+2*a rctan((b*x^4+a)^(1/4)/a^(1/4))*b*x^4+4*(b*x^4+a)^(1/4)*a^(3/4))/x^4/a^(3/4 )
Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.28 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {\left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b + a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) + i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b + i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) - i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b - i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) - \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x^{4} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} b - a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}}\right ) + 4 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{16 \, x^{4}} \]
-1/16*((b^4/a^3)^(1/4)*x^4*log((b*x^4 + a)^(1/4)*b + a*(b^4/a^3)^(1/4)) + I*(b^4/a^3)^(1/4)*x^4*log((b*x^4 + a)^(1/4)*b + I*a*(b^4/a^3)^(1/4)) - I*( b^4/a^3)^(1/4)*x^4*log((b*x^4 + a)^(1/4)*b - I*a*(b^4/a^3)^(1/4)) - (b^4/a ^3)^(1/4)*x^4*log((b*x^4 + a)^(1/4)*b - a*(b^4/a^3)^(1/4)) + 4*(b*x^4 + a) ^(1/4))/x^4
Result contains complex when optimal does not.
Time = 0.73 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=- \frac {\sqrt [4]{b} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \]
-b**(1/4)*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), a*exp_polar(I*pi)/(b*x**4) )/(4*x**3*gamma(7/4))
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {b \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{8 \, a^{\frac {3}{4}}} + \frac {b \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{16 \, a^{\frac {3}{4}}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{4 \, x^{4}} \]
-1/8*b*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(3/4) + 1/16*b*log(((b*x^4 + a) ^(1/4) - a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4)))/a^(3/4) - 1/4*(b*x^4 + a) ^(1/4)/x^4
Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (55) = 110\).
Time = 0.29 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.76 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=\frac {\frac {2 \, \sqrt {2} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} b^{2} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a} - \frac {8 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{x^{4}}}{32 \, b} \]
1/32*(2*sqrt(2)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a) ^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 2*sqrt(2)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2 )*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + sqrt(2)*b^2*l og(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/(-a) ^(3/4) + sqrt(2)*(-a)^(1/4)*b^2*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a - 8*(b*x^4 + a)^(1/4)*b/x^4)/b
Time = 5.88 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^5} \, dx=-\frac {{\left (b\,x^4+a\right )}^{1/4}}{4\,x^4}-\frac {b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{3/4}}-\frac {b\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{3/4}} \]