Integrand size = 15, antiderivative size = 78 \[ \int \frac {1}{x^5 \sqrt [4]{a+b x^4}} \, dx=-\frac {\left (a+b x^4\right )^{3/4}}{4 a x^4}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}} \]
-1/4*(b*x^4+a)^(3/4)/a/x^4-1/8*b*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(5/4)+1 /8*b*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(5/4)
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \sqrt [4]{a+b x^4}} \, dx=-\frac {\left (a+b x^4\right )^{3/4}}{4 a x^4}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}} \]
-1/4*(a + b*x^4)^(3/4)/(a*x^4) - (b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(8* a^(5/4)) + (b*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(5/4))
Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {798, 52, 73, 25, 27, 827, 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.
\(\displaystyle \int \frac {1}{x^5 \sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \sqrt [4]{b x^4+a}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {b \int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx^4}{4 a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {\int -\frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {b \int \frac {x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} \left (\frac {b \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {1}{2} \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}\right )}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (\frac {b \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )\) |
(-((a + b*x^4)^(3/4)/(a*x^4)) + (b*(-1/2*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)] /a^(1/4) + ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)]/(2*a^(1/4))))/a)/4
3.11.86.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(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[(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]]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Time = 4.58 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.09
method | result | size |
pseudoelliptic | \(\frac {\ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b \,x^{4}-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b \,x^{4}-4 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{\frac {1}{4}}}{16 a^{\frac {5}{4}} x^{4}}\) | \(85\) |
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*ar ctan((b*x^4+a)^(1/4)/a^(1/4))*b*x^4-4*(b*x^4+a)^(3/4)*a^(1/4))/a^(5/4)/x^4
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.49 \[ \int \frac {1}{x^5 \sqrt [4]{a+b x^4}} \, dx=\frac {a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - i \, a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (i \, a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) + i \, a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (-i \, a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (-a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{16 \, a x^{4}} \]
1/16*(a*x^4*(b^4/a^5)^(1/4)*log(a^4*(b^4/a^5)^(3/4) + (b*x^4 + a)^(1/4)*b^ 3) - I*a*x^4*(b^4/a^5)^(1/4)*log(I*a^4*(b^4/a^5)^(3/4) + (b*x^4 + a)^(1/4) *b^3) + I*a*x^4*(b^4/a^5)^(1/4)*log(-I*a^4*(b^4/a^5)^(3/4) + (b*x^4 + a)^( 1/4)*b^3) - a*x^4*(b^4/a^5)^(1/4)*log(-a^4*(b^4/a^5)^(3/4) + (b*x^4 + a)^( 1/4)*b^3) - 4*(b*x^4 + a)^(3/4))/(a*x^4)
Result contains complex when optimal does not.
Time = 0.77 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^5 \sqrt [4]{a+b x^4}} \, dx=- \frac {\Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \sqrt [4]{b} x^{5} \Gamma \left (\frac {9}{4}\right )} \]
-gamma(5/4)*hyper((1/4, 5/4), (9/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b**(1 /4)*x**5*gamma(9/4))
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x^5 \sqrt [4]{a+b x^4}} \, dx=-\frac {b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{16 \, a} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} b}{4 \, {\left ({\left (b x^{4} + a\right )} a - a^{2}\right )}} \]
-1/16*b*(2*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(1/4) + log(((b*x^4 + a)^(1 /4) - a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4)))/a^(1/4))/a - 1/4*(b*x^4 + a) ^(3/4)*b/((b*x^4 + a)*a - a^2)
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (58) = 116\).
Time = 0.28 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.81 \[ \int \frac {1}{x^5 \sqrt [4]{a+b x^4}} \, dx=\frac {\frac {2 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {\sqrt {2} b^{2} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}} a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a^{2}} - \frac {8 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b}{a x^{4}}}{32 \, b} \]
1/32*(2*sqrt(2)*(-a)^(3/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2* (b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2 + 2*sqrt(2)*(-a)^(3/4)*b^2*arctan(-1/2* sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2 + sqrt( 2)*b^2*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(- a))/((-a)^(1/4)*a) + sqrt(2)*(-a)^(3/4)*b^2*log(-sqrt(2)*(b*x^4 + a)^(1/4) *(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a^2 - 8*(b*x^4 + a)^(3/4)*b/(a*x ^4))/b
Time = 5.84 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^5 \sqrt [4]{a+b x^4}} \, dx=\frac {b\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{5/4}}-\frac {b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{5/4}}-\frac {{\left (b\,x^4+a\right )}^{3/4}}{4\,a\,x^4} \]