Integrand size = 15, antiderivative size = 104 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=-\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8}+\frac {5 b \left (a+b x^4\right )^{3/4}}{32 a^2 x^4}+\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}} \]
-1/8*(b*x^4+a)^(3/4)/a/x^8+5/32*b*(b*x^4+a)^(3/4)/a^2/x^4+5/64*b^2*arctan( (b*x^4+a)^(1/4)/a^(1/4))/a^(9/4)-5/64*b^2*arctanh((b*x^4+a)^(1/4)/a^(1/4)) /a^(9/4)
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {\left (a+b x^4\right )^{3/4} \left (-4 a+5 b x^4\right )}{32 a^2 x^8}+\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}} \]
((a + b*x^4)^(3/4)*(-4*a + 5*b*x^4))/(32*a^2*x^8) + (5*b^2*ArcTan[(a + b*x ^4)^(1/4)/a^(1/4)])/(64*a^(9/4)) - (5*b^2*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4 )])/(64*a^(9/4))
Time = 0.22 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {798, 52, 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.
\(\displaystyle \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^{12} \sqrt [4]{b x^4+a}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \int \frac {1}{x^8 \sqrt [4]{b x^4+a}}dx^4}{8 a}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (-\frac {b \int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx^4}{4 a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )}{8 a}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (-\frac {\int -\frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )}{8 a}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {\int \frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )}{8 a}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {b \int \frac {x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )}{8 a}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {b \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {1}{2} \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}\right )}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )}{8 a}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {b \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )}{8 a}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^4}\right )}{8 a}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^8}\right )\) |
(-1/2*(a + b*x^4)^(3/4)/(a*x^8) - (5*b*(-((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))/(8*a))/4
3.11.87.3.1 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 = 4.34 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(\frac {10 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b^{2} x^{8}-5 \ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b^{2} x^{8}+20 b \,x^{4} a^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {3}{4}}-16 a^{\frac {5}{4}} \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{128 a^{\frac {9}{4}} x^{8}}\) | \(108\) |
1/128/a^(9/4)*(10*arctan((b*x^4+a)^(1/4)/a^(1/4))*b^2*x^8-5*ln((-(b*x^4+a) ^(1/4)-a^(1/4))/(-(b*x^4+a)^(1/4)+a^(1/4)))*b^2*x^8+20*b*x^4*a^(1/4)*(b*x^ 4+a)^(3/4)-16*a^(5/4)*(b*x^4+a)^(3/4))/x^8
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.10 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=-\frac {5 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 5 i \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 i \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) + 5 i \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 i \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 5 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 4 \, {\left (5 \, b x^{4} - 4 \, a\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, a^{2} x^{8}} \]
-1/128*(5*a^2*x^8*(b^8/a^9)^(1/4)*log(125*a^7*(b^8/a^9)^(3/4) + 125*(b*x^4 + a)^(1/4)*b^6) - 5*I*a^2*x^8*(b^8/a^9)^(1/4)*log(125*I*a^7*(b^8/a^9)^(3/ 4) + 125*(b*x^4 + a)^(1/4)*b^6) + 5*I*a^2*x^8*(b^8/a^9)^(1/4)*log(-125*I*a ^7*(b^8/a^9)^(3/4) + 125*(b*x^4 + a)^(1/4)*b^6) - 5*a^2*x^8*(b^8/a^9)^(1/4 )*log(-125*a^7*(b^8/a^9)^(3/4) + 125*(b*x^4 + a)^(1/4)*b^6) - 4*(5*b*x^4 - 4*a)*(b*x^4 + a)^(3/4))/(a^2*x^8)
Result contains complex when optimal does not.
Time = 1.63 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=- \frac {\Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \sqrt [4]{b} x^{9} \Gamma \left (\frac {13}{4}\right )} \]
-gamma(9/4)*hyper((1/4, 9/4), (13/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b**( 1/4)*x**9*gamma(13/4))
Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {5 \, b^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{128 \, a^{2}} + \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{2} - 9 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{4} + a\right )} a^{3} + a^{4}\right )}} \]
5/128*b^2*(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^2 + 1/32*(5*(b* x^4 + a)^(7/4)*b^2 - 9*(b*x^4 + a)^(3/4)*a*b^2)/((b*x^4 + a)^2*a^2 - 2*(b* x^4 + a)*a^3 + a^4)
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (80) = 160\).
Time = 0.28 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.35 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {\frac {10 \, \sqrt {2} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {10 \, \sqrt {2} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{3} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a^{3}} + \frac {5 \, \sqrt {2} b^{3} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {8 \, {\left (5 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{3} - 9 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{3}\right )}}{a^{2} b^{2} x^{8}}}{256 \, b} \]
1/256*(10*sqrt(2)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/((-a)^(1/4)*a^2) + 10*sqrt(2)*b^3*arctan(-1/2*sqrt(2 )*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/((-a)^(1/4)*a^2) + 5*sqrt(2)*(-a)^(3/4)*b^3*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt (b*x^4 + a) + sqrt(-a))/a^3 + 5*sqrt(2)*b^3*log(-sqrt(2)*(b*x^4 + a)^(1/4) *(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/((-a)^(1/4)*a^2) + 8*(5*(b*x^4 + a)^(7/4)*b^3 - 9*(b*x^4 + a)^(3/4)*a*b^3)/(a^2*b^2*x^8))/b
Time = 5.95 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {5\,b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{9/4}}-\frac {9\,{\left (b\,x^4+a\right )}^{3/4}}{32\,a\,x^8}+\frac {5\,{\left (b\,x^4+a\right )}^{7/4}}{32\,a^2\,x^8}+\frac {b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,5{}\mathrm {i}}{64\,a^{9/4}} \]