Integrand size = 15, antiderivative size = 97 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {5 b}{4 a^2 \sqrt [4]{a+b x^4}}-\frac {1}{4 a x^4 \sqrt [4]{a+b x^4}}-\frac {5 b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}}+\frac {5 b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}} \]
-5/4*b/a^2/(b*x^4+a)^(1/4)-1/4/a/x^4/(b*x^4+a)^(1/4)-5/8*b*arctan((b*x^4+a )^(1/4)/a^(1/4))/a^(9/4)+5/8*b*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(9/4)
Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=\frac {-\frac {2 \sqrt [4]{a} \left (a+5 b x^4\right )}{x^4 \sqrt [4]{a+b x^4}}-5 b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )+5 b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}} \]
((-2*a^(1/4)*(a + 5*b*x^4))/(x^4*(a + b*x^4)^(1/4)) - 5*b*ArcTan[(a + b*x^ 4)^(1/4)/a^(1/4)] + 5*b*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(9/4))
Time = 0.23 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {798, 52, 61, 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.
\(\displaystyle \int \frac {1}{x^5 \left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \left (b x^4+a\right )^{5/4}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \int \frac {1}{x^4 \left (b x^4+a\right )^{5/4}}dx^4}{4 a}-\frac {1}{a x^4 \sqrt [4]{a+b x^4}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {\int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx^4}{a}+\frac {4}{a \sqrt [4]{a+b x^4}}\right )}{4 a}-\frac {1}{a x^4 \sqrt [4]{a+b x^4}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {4 \int -\frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a b}+\frac {4}{a \sqrt [4]{a+b x^4}}\right )}{4 a}-\frac {1}{a x^4 \sqrt [4]{a+b x^4}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \int \frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a b}\right )}{4 a}-\frac {1}{a x^4 \sqrt [4]{a+b x^4}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \int \frac {x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a}\right )}{4 a}-\frac {1}{a x^4 \sqrt [4]{a+b x^4}}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {1}{2} \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}\right )}{a}\right )}{4 a}-\frac {1}{a x^4 \sqrt [4]{a+b x^4}}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}\right )}{4 a}-\frac {1}{a x^4 \sqrt [4]{a+b x^4}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-\frac {5 b \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}\right )}{4 a}-\frac {1}{a x^4 \sqrt [4]{a+b x^4}}\right )\) |
(-(1/(a*x^4*(a + b*x^4)^(1/4))) - (5*b*(4/(a*(a + b*x^4)^(1/4)) - (4*(-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*a))/4
3.12.45.3.1 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] :> 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] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
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 = 4.49 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.15
method | result | size |
pseudoelliptic | \(-\frac {5 \left (\arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b \,x^{4} \left (b \,x^{4}+a \right )^{\frac {1}{4}}-\frac {\ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b \,x^{4} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{2}+2 b \,x^{4} a^{\frac {1}{4}}+\frac {2 a^{\frac {5}{4}}}{5}\right )}{8 a^{\frac {9}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{4}}\) | \(112\) |
-5/8/a^(9/4)*(arctan((b*x^4+a)^(1/4)/a^(1/4))*b*x^4*(b*x^4+a)^(1/4)-1/2*ln ((-(b*x^4+a)^(1/4)-a^(1/4))/(-(b*x^4+a)^(1/4)+a^(1/4)))*b*x^4*(b*x^4+a)^(1 /4)+2*b*x^4*a^(1/4)+2/5*a^(5/4))/(b*x^4+a)^(1/4)/x^4
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.80 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=\frac {5 \, {\left (a^{2} b x^{8} + a^{3} x^{4}\right )} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - 5 \, {\left (i \, a^{2} b x^{8} + i \, a^{3} x^{4}\right )} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 i \, a^{7} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - 5 \, {\left (-i \, a^{2} b x^{8} - i \, a^{3} x^{4}\right )} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 i \, a^{7} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - 5 \, {\left (a^{2} b x^{8} + a^{3} x^{4}\right )} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} \left (\frac {b^{4}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - 4 \, {\left (5 \, b x^{4} + a\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{16 \, {\left (a^{2} b x^{8} + a^{3} x^{4}\right )}} \]
1/16*(5*(a^2*b*x^8 + a^3*x^4)*(b^4/a^9)^(1/4)*log(125*a^7*(b^4/a^9)^(3/4) + 125*(b*x^4 + a)^(1/4)*b^3) - 5*(I*a^2*b*x^8 + I*a^3*x^4)*(b^4/a^9)^(1/4) *log(125*I*a^7*(b^4/a^9)^(3/4) + 125*(b*x^4 + a)^(1/4)*b^3) - 5*(-I*a^2*b* x^8 - I*a^3*x^4)*(b^4/a^9)^(1/4)*log(-125*I*a^7*(b^4/a^9)^(3/4) + 125*(b*x ^4 + a)^(1/4)*b^3) - 5*(a^2*b*x^8 + a^3*x^4)*(b^4/a^9)^(1/4)*log(-125*a^7* (b^4/a^9)^(3/4) + 125*(b*x^4 + a)^(1/4)*b^3) - 4*(5*b*x^4 + a)*(b*x^4 + a) ^(3/4))/(a^2*b*x^8 + a^3*x^4)
Result contains complex when optimal does not.
Time = 0.99 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=- \frac {\Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {5}{4}} x^{9} \Gamma \left (\frac {13}{4}\right )} \]
-gamma(9/4)*hyper((5/4, 9/4), (13/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b**( 5/4)*x**9*gamma(13/4))
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {5 \, {\left (b x^{4} + a\right )} b - 4 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}\right )}} - \frac {5 \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{16 \, a^{2}} \]
-1/4*(5*(b*x^4 + a)*b - 4*a*b)/((b*x^4 + a)^(5/4)*a^2 - (b*x^4 + a)^(1/4)* a^3) - 5/16*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^2
Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (73) = 146\).
Time = 0.29 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.41 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=\frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{16 \, a^{3}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{16 \, a^{3}} + \frac {5 \, \sqrt {2} b \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{32 \, \left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{32 \, a^{3}} - \frac {5 \, {\left (b x^{4} + a\right )} b - 4 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {5}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a\right )} a^{2}} \]
5/16*sqrt(2)*(-a)^(3/4)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^ 4 + a)^(1/4))/(-a)^(1/4))/a^3 + 5/16*sqrt(2)*(-a)^(3/4)*b*arctan(-1/2*sqrt (2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/a^3 + 5/32*sqrt (2)*b*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a ))/((-a)^(1/4)*a^2) + 5/32*sqrt(2)*(-a)^(3/4)*b*log(-sqrt(2)*(b*x^4 + a)^( 1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a^3 - 1/4*(5*(b*x^4 + a)*b - 4*a*b)/(((b*x^4 + a)^(5/4) - (b*x^4 + a)^(1/4)*a)*a^2)
Time = 5.97 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=\frac {5\,b\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{9/4}}-\frac {5\,b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{9/4}}-\frac {\frac {b}{a}-\frac {5\,b\,\left (b\,x^4+a\right )}{4\,a^2}}{a\,{\left (b\,x^4+a\right )}^{1/4}-{\left (b\,x^4+a\right )}^{5/4}} \]