Integrand size = 15, antiderivative size = 83 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x} \, dx=a \sqrt [4]{a+b x^4}+\frac {1}{5} \left (a+b x^4\right )^{5/4}-\frac {1}{2} a^{5/4} \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {1}{2} a^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right ) \]
a*(b*x^4+a)^(1/4)+1/5*(b*x^4+a)^(5/4)-1/2*a^(5/4)*arctan((b*x^4+a)^(1/4)/a ^(1/4))-1/2*a^(5/4)*arctanh((b*x^4+a)^(1/4)/a^(1/4))
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x} \, dx=\frac {1}{5} \sqrt [4]{a+b x^4} \left (6 a+b x^4\right )-\frac {1}{2} a^{5/4} \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {1}{2} a^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right ) \]
((a + b*x^4)^(1/4)*(6*a + b*x^4))/5 - (a^(5/4)*ArcTan[(a + b*x^4)^(1/4)/a^ (1/4)])/2 - (a^(5/4)*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/2
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {798, 60, 60, 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.
\(\displaystyle \int \frac {\left (a+b x^4\right )^{5/4}}{x} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {\left (b x^4+a\right )^{5/4}}{x^4}dx^4\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (a \int \frac {\sqrt [4]{b x^4+a}}{x^4}dx^4+\frac {4}{5} \left (a+b x^4\right )^{5/4}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (a \left (a \int \frac {1}{x^4 \left (b x^4+a\right )^{3/4}}dx^4+4 \sqrt [4]{a+b x^4}\right )+\frac {4}{5} \left (a+b x^4\right )^{5/4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (a \left (\frac {4 a \int \frac {1}{\frac {x^{16}}{b}-\frac {a}{b}}d\sqrt [4]{b x^4+a}}{b}+4 \sqrt [4]{a+b x^4}\right )+\frac {4}{5} \left (a+b x^4\right )^{5/4}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{4} \left (a \left (\frac {4 a \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}\right )}{b}+4 \sqrt [4]{a+b x^4}\right )+\frac {4}{5} \left (a+b x^4\right )^{5/4}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (a \left (\frac {4 a \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{b}+4 \sqrt [4]{a+b x^4}\right )+\frac {4}{5} \left (a+b x^4\right )^{5/4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (a \left (\frac {4 a \left (-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{b}+4 \sqrt [4]{a+b x^4}\right )+\frac {4}{5} \left (a+b x^4\right )^{5/4}\right )\) |
((4*(a + b*x^4)^(5/4))/5 + a*(4*(a + b*x^4)^(1/4) + (4*a*(-1/2*(b*ArcTan[( a + b*x^4)^(1/4)/a^(1/4)])/a^(3/4) - (b*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)] )/(2*a^(3/4))))/b))/4
3.11.52.3.1 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 + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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[((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 = 4.41 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99
method | result | size |
pseudoelliptic | \(\frac {\left (-\ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )\right ) a^{\frac {5}{4}}}{4}+\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (b \,x^{4}+6 a \right )}{5}\) | \(82\) |
1/4*(-ln((-(b*x^4+a)^(1/4)-a^(1/4))/(-(b*x^4+a)^(1/4)+a^(1/4)))-2*arctan(( b*x^4+a)^(1/4)/a^(1/4)))*a^(5/4)+1/5*(b*x^4+a)^(1/4)*(b*x^4+6*a)
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x} \, dx=\frac {1}{5} \, {\left (b x^{4} + 6 \, a\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}} - \frac {1}{4} \, {\left (a^{5}\right )}^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a + {\left (a^{5}\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} i \, {\left (a^{5}\right )}^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a + i \, {\left (a^{5}\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} i \, {\left (a^{5}\right )}^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a - i \, {\left (a^{5}\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, {\left (a^{5}\right )}^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a - {\left (a^{5}\right )}^{\frac {1}{4}}\right ) \]
1/5*(b*x^4 + 6*a)*(b*x^4 + a)^(1/4) - 1/4*(a^5)^(1/4)*log((b*x^4 + a)^(1/4 )*a + (a^5)^(1/4)) - 1/4*I*(a^5)^(1/4)*log((b*x^4 + a)^(1/4)*a + I*(a^5)^( 1/4)) + 1/4*I*(a^5)^(1/4)*log((b*x^4 + a)^(1/4)*a - I*(a^5)^(1/4)) + 1/4*( a^5)^(1/4)*log((b*x^4 + a)^(1/4)*a - (a^5)^(1/4))
Result contains complex when optimal does not.
Time = 0.95 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x} \, dx=- \frac {b^{\frac {5}{4}} x^{5} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {5}{4} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \Gamma \left (- \frac {1}{4}\right )} \]
-b**(5/4)*x**5*gamma(-5/4)*hyper((-5/4, -5/4), (-1/4,), a*exp_polar(I*pi)/ (b*x**4))/(4*gamma(-1/4))
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x} \, dx=-\frac {1}{2} \, a^{\frac {5}{4}} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) + \frac {1}{4} \, a^{\frac {5}{4}} \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right ) + \frac {1}{5} \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a \]
-1/2*a^(5/4)*arctan((b*x^4 + a)^(1/4)/a^(1/4)) + 1/4*a^(5/4)*log(((b*x^4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4))) + 1/5*(b*x^4 + a)^(5/4 ) + (b*x^4 + a)^(1/4)*a
Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (61) = 122\).
Time = 0.29 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.41 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x} \, dx=-\frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{8} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right ) + \frac {1}{8} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right ) + \frac {1}{5} \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a \]
-1/4*sqrt(2)*(-a)^(1/4)*a*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^ 4 + a)^(1/4))/(-a)^(1/4)) - 1/4*sqrt(2)*(-a)^(1/4)*a*arctan(-1/2*sqrt(2)*( sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4)) - 1/8*sqrt(2)*(-a)^( 1/4)*a*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(- a)) + 1/8*sqrt(2)*(-a)^(1/4)*a*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a)) + 1/5*(b*x^4 + a)^(5/4) + (b*x^4 + a)^(1/4)*a
Time = 5.71 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x} \, dx=a\,{\left (b\,x^4+a\right )}^{1/4}-\frac {a^{5/4}\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2}+\frac {{\left (b\,x^4+a\right )}^{5/4}}{5}+\frac {a^{5/4}\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,1{}\mathrm {i}}{2} \]