Integrand size = 15, antiderivative size = 78 \[ \int \frac {1}{x^5 \left (b+a x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{b+a x^4}}{4 b x^4}+\frac {3 a \arctan \left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{8 b^{7/4}}+\frac {3 a \text {arctanh}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{8 b^{7/4}} \]
-1/4*(a*x^4+b)^(1/4)/b/x^4+3/8*a*arctan((a*x^4+b)^(1/4)/b^(1/4))/b^(7/4)+3 /8*a*arctanh((a*x^4+b)^(1/4)/b^(1/4))/b^(7/4)
Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (b+a x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{b+a x^4}}{4 b x^4}+\frac {3 a \arctan \left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{8 b^{7/4}}+\frac {3 a \text {arctanh}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{8 b^{7/4}} \]
-1/4*(b + a*x^4)^(1/4)/(b*x^4) + (3*a*ArcTan[(b + a*x^4)^(1/4)/b^(1/4)])/( 8*b^(7/4)) + (3*a*ArcTanh[(b + a*x^4)^(1/4)/b^(1/4)])/(8*b^(7/4))
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {798, 52, 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.
\(\displaystyle \int \frac {1}{x^5 \left (a x^4+b\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \left (a x^4+b\right )^{3/4}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 a \int \frac {1}{x^4 \left (a x^4+b\right )^{3/4}}dx^4}{4 b}-\frac {\sqrt [4]{a x^4+b}}{b x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \int \frac {1}{\frac {x^{16}}{a}-\frac {b}{a}}d\sqrt [4]{a x^4+b}}{b}-\frac {\sqrt [4]{a x^4+b}}{b x^4}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \left (-\frac {a \int \frac {1}{\sqrt {b}-x^8}d\sqrt [4]{a x^4+b}}{2 \sqrt {b}}-\frac {a \int \frac {1}{x^8+\sqrt {b}}d\sqrt [4]{a x^4+b}}{2 \sqrt {b}}\right )}{b}-\frac {\sqrt [4]{a x^4+b}}{b x^4}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \left (-\frac {a \int \frac {1}{\sqrt {b}-x^8}d\sqrt [4]{a x^4+b}}{2 \sqrt {b}}-\frac {a \arctan \left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{2 b^{3/4}}\right )}{b}-\frac {\sqrt [4]{a x^4+b}}{b x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \left (-\frac {a \arctan \left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{2 b^{3/4}}-\frac {a \text {arctanh}\left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{2 b^{3/4}}\right )}{b}-\frac {\sqrt [4]{a x^4+b}}{b x^4}\right )\) |
(-((b + a*x^4)^(1/4)/(b*x^4)) - (3*(-1/2*(a*ArcTan[(b + a*x^4)^(1/4)/b^(1/ 4)])/b^(3/4) - (a*ArcTanh[(b + a*x^4)^(1/4)/b^(1/4)])/(2*b^(3/4))))/b)/4
3.11.32.3.1 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 + 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[((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 = 1.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.05
method | result | size |
pseudoelliptic | \(\frac {3 \ln \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{\left (a \,x^{4}+b \right )^{\frac {1}{4}}-b^{\frac {1}{4}}}\right ) a \,x^{4}+6 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) a \,x^{4}-4 \left (a \,x^{4}+b \right )^{\frac {1}{4}} b^{\frac {3}{4}}}{16 b^{\frac {7}{4}} x^{4}}\) | \(82\) |
1/16/b^(7/4)*(3*ln(((a*x^4+b)^(1/4)+b^(1/4))/((a*x^4+b)^(1/4)-b^(1/4)))*a* x^4+6*arctan((a*x^4+b)^(1/4)/b^(1/4))*a*x^4-4*(a*x^4+b)^(1/4)*b^(3/4))/x^4
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.46 \[ \int \frac {1}{x^5 \left (b+a x^4\right )^{3/4}} \, dx=\frac {3 \, b x^{4} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (3 \, b^{2} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} a\right ) + 3 i \, b x^{4} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (3 i \, b^{2} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} a\right ) - 3 i \, b x^{4} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-3 i \, b^{2} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} a\right ) - 3 \, b x^{4} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-3 \, b^{2} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} a\right ) - 4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{16 \, b x^{4}} \]
1/16*(3*b*x^4*(a^4/b^7)^(1/4)*log(3*b^2*(a^4/b^7)^(1/4) + 3*(a*x^4 + b)^(1 /4)*a) + 3*I*b*x^4*(a^4/b^7)^(1/4)*log(3*I*b^2*(a^4/b^7)^(1/4) + 3*(a*x^4 + b)^(1/4)*a) - 3*I*b*x^4*(a^4/b^7)^(1/4)*log(-3*I*b^2*(a^4/b^7)^(1/4) + 3 *(a*x^4 + b)^(1/4)*a) - 3*b*x^4*(a^4/b^7)^(1/4)*log(-3*b^2*(a^4/b^7)^(1/4) + 3*(a*x^4 + b)^(1/4)*a) - 4*(a*x^4 + b)^(1/4))/(b*x^4)
Result contains complex when optimal does not.
Time = 0.79 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^5 \left (b+a x^4\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{4}} x^{7} \Gamma \left (\frac {11}{4}\right )} \]
-gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**( 3/4)*x**7*gamma(11/4))
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^5 \left (b+a x^4\right )^{3/4}} \, dx=-\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} a}{4 \, {\left ({\left (a x^{4} + b\right )} b - b^{2}\right )}} + \frac {3 \, {\left (\frac {2 \, a \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} - \frac {a \log \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{4} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}}\right )}}{16 \, b} \]
-1/4*(a*x^4 + b)^(1/4)*a/((a*x^4 + b)*b - b^2) + 3/16*(2*a*arctan((a*x^4 + b)^(1/4)/b^(1/4))/b^(3/4) - a*log(((a*x^4 + b)^(1/4) - b^(1/4))/((a*x^4 + b)^(1/4) + b^(1/4)))/b^(3/4))/b
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (58) = 116\).
Time = 0.27 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.83 \[ \int \frac {1}{x^5 \left (b+a x^4\right )^{3/4}} \, dx=\frac {\frac {6 \, \sqrt {2} a^{2} \left (-b\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{b^{2}} + \frac {6 \, \sqrt {2} a^{2} \left (-b\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{b^{2}} + \frac {3 \, \sqrt {2} a^{2} \left (-b\right )^{\frac {1}{4}} \log \left (\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right )}{b^{2}} + \frac {3 \, \sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right )}{\left (-b\right )^{\frac {3}{4}} b} - \frac {8 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} a}{b x^{4}}}{32 \, a} \]
1/32*(6*sqrt(2)*a^2*(-b)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2* (a*x^4 + b)^(1/4))/(-b)^(1/4))/b^2 + 6*sqrt(2)*a^2*(-b)^(1/4)*arctan(-1/2* sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(a*x^4 + b)^(1/4))/(-b)^(1/4))/b^2 + 3*sqr t(2)*a^2*(-b)^(1/4)*log(sqrt(2)*(a*x^4 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^4 + b) + sqrt(-b))/b^2 + 3*sqrt(2)*a^2*log(-sqrt(2)*(a*x^4 + b)^(1/4)*(-b)^( 1/4) + sqrt(a*x^4 + b) + sqrt(-b))/((-b)^(3/4)*b) - 8*(a*x^4 + b)^(1/4)*a/ (b*x^4))/a
Time = 6.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^5 \left (b+a x^4\right )^{3/4}} \, dx=\frac {3\,a\,\mathrm {atan}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}}{b^{1/4}}\right )}{8\,b^{7/4}}-\frac {{\left (a\,x^4+b\right )}^{1/4}}{4\,b\,x^4}+\frac {3\,a\,\mathrm {atanh}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}}{b^{1/4}}\right )}{8\,b^{7/4}} \]