Integrand size = 15, antiderivative size = 70 \[ \int \frac {\left (b+a x^5\right )^{3/4}}{x} \, dx=\frac {4}{15} \left (b+a x^5\right )^{3/4}+\frac {2}{5} b^{3/4} \arctan \left (\frac {\sqrt [4]{b+a x^5}}{\sqrt [4]{b}}\right )-\frac {2}{5} b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^5}}{\sqrt [4]{b}}\right ) \]
4/15*(a*x^5+b)^(3/4)+2/5*b^(3/4)*arctan((a*x^5+b)^(1/4)/b^(1/4))-2/5*b^(3/ 4)*arctanh((a*x^5+b)^(1/4)/b^(1/4))
Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b+a x^5\right )^{3/4}}{x} \, dx=\frac {4}{15} \left (b+a x^5\right )^{3/4}+\frac {2}{5} b^{3/4} \arctan \left (\frac {\sqrt [4]{b+a x^5}}{\sqrt [4]{b}}\right )-\frac {2}{5} b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^5}}{\sqrt [4]{b}}\right ) \]
(4*(b + a*x^5)^(3/4))/15 + (2*b^(3/4)*ArcTan[(b + a*x^5)^(1/4)/b^(1/4)])/5 - (2*b^(3/4)*ArcTanh[(b + a*x^5)^(1/4)/b^(1/4)])/5
Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {798, 60, 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.
\(\displaystyle \int \frac {\left (a x^5+b\right )^{3/4}}{x} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{5} \int \frac {\left (a x^5+b\right )^{3/4}}{x^5}dx^5\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{5} \left (b \int \frac {1}{x^5 \sqrt [4]{a x^5+b}}dx^5+\frac {4}{3} \left (a x^5+b\right )^{3/4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{5} \left (\frac {4 b \int -\frac {a x^{10}}{b-x^{20}}d\sqrt [4]{a x^5+b}}{a}+\frac {4}{3} \left (a x^5+b\right )^{3/4}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \left (\frac {4}{3} \left (a x^5+b\right )^{3/4}-\frac {4 b \int \frac {a x^{10}}{b-x^{20}}d\sqrt [4]{a x^5+b}}{a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {4}{3} \left (a x^5+b\right )^{3/4}-4 b \int \frac {x^{10}}{b-x^{20}}d\sqrt [4]{a x^5+b}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{5} \left (\frac {4}{3} \left (a x^5+b\right )^{3/4}-4 b \left (\frac {1}{2} \int \frac {1}{\sqrt {b}-x^{10}}d\sqrt [4]{a x^5+b}-\frac {1}{2} \int \frac {1}{x^{10}+\sqrt {b}}d\sqrt [4]{a x^5+b}\right )\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{5} \left (\frac {4}{3} \left (a x^5+b\right )^{3/4}-4 b \left (\frac {1}{2} \int \frac {1}{\sqrt {b}-x^{10}}d\sqrt [4]{a x^5+b}-\frac {\arctan \left (\frac {\sqrt [4]{a x^5+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{5} \left (\frac {4}{3} \left (a x^5+b\right )^{3/4}-4 b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a x^5+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}-\frac {\arctan \left (\frac {\sqrt [4]{a x^5+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}\right )\right )\) |
((4*(b + a*x^5)^(3/4))/3 - 4*b*(-1/2*ArcTan[(b + a*x^5)^(1/4)/b^(1/4)]/b^( 1/4) + ArcTanh[(b + a*x^5)^(1/4)/b^(1/4)]/(2*b^(1/4))))/5
3.10.22.3.1 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/(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[(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 = 0.98 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.99
method | result | size |
pseudoelliptic | \(\frac {4 \left (a \,x^{5}+b \right )^{\frac {3}{4}}}{15}+\frac {2 b^{\frac {3}{4}} \arctan \left (\frac {\left (a \,x^{5}+b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{5}-\frac {\ln \left (\frac {\left (a \,x^{5}+b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{\left (a \,x^{5}+b \right )^{\frac {1}{4}}-b^{\frac {1}{4}}}\right ) b^{\frac {3}{4}}}{5}\) | \(69\) |
4/15*(a*x^5+b)^(3/4)+2/5*b^(3/4)*arctan((a*x^5+b)^(1/4)/b^(1/4))-1/5*ln((( a*x^5+b)^(1/4)+b^(1/4))/((a*x^5+b)^(1/4)-b^(1/4)))*b^(3/4)
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.80 \[ \int \frac {\left (b+a x^5\right )^{3/4}}{x} \, dx=-\frac {1}{5} \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{5} + b\right )}^{\frac {1}{4}} b^{2} + {\left (b^{3}\right )}^{\frac {3}{4}}\right ) + \frac {1}{5} i \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{5} + b\right )}^{\frac {1}{4}} b^{2} + i \, {\left (b^{3}\right )}^{\frac {3}{4}}\right ) - \frac {1}{5} i \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{5} + b\right )}^{\frac {1}{4}} b^{2} - i \, {\left (b^{3}\right )}^{\frac {3}{4}}\right ) + \frac {1}{5} \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{5} + b\right )}^{\frac {1}{4}} b^{2} - {\left (b^{3}\right )}^{\frac {3}{4}}\right ) + \frac {4}{15} \, {\left (a x^{5} + b\right )}^{\frac {3}{4}} \]
-1/5*(b^3)^(1/4)*log((a*x^5 + b)^(1/4)*b^2 + (b^3)^(3/4)) + 1/5*I*(b^3)^(1 /4)*log((a*x^5 + b)^(1/4)*b^2 + I*(b^3)^(3/4)) - 1/5*I*(b^3)^(1/4)*log((a* x^5 + b)^(1/4)*b^2 - I*(b^3)^(3/4)) + 1/5*(b^3)^(1/4)*log((a*x^5 + b)^(1/4 )*b^2 - (b^3)^(3/4)) + 4/15*(a*x^5 + b)^(3/4)
Result contains complex when optimal does not.
Time = 0.77 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.66 \[ \int \frac {\left (b+a x^5\right )^{3/4}}{x} \, dx=- \frac {a^{\frac {3}{4}} x^{\frac {15}{4}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{5}}} \right )}}{5 \Gamma \left (\frac {1}{4}\right )} \]
-a**(3/4)*x**(15/4)*gamma(-3/4)*hyper((-3/4, -3/4), (1/4,), b*exp_polar(I* pi)/(a*x**5))/(5*gamma(1/4))
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \frac {\left (b+a x^5\right )^{3/4}}{x} \, dx=\frac {1}{5} \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{5} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (a x^{5} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{5} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}}\right )} + \frac {4}{15} \, {\left (a x^{5} + b\right )}^{\frac {3}{4}} \]
1/5*b*(2*arctan((a*x^5 + b)^(1/4)/b^(1/4))/b^(1/4) + log(((a*x^5 + b)^(1/4 ) - b^(1/4))/((a*x^5 + b)^(1/4) + b^(1/4)))/b^(1/4)) + 4/15*(a*x^5 + b)^(3 /4)
Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (50) = 100\).
Time = 0.28 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.64 \[ \int \frac {\left (b+a x^5\right )^{3/4}}{x} \, dx=-\frac {1}{5} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{5} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right ) - \frac {1}{5} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{5} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right ) + \frac {1}{10} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (a x^{5} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{5} + b} + \sqrt {-b}\right ) - \frac {1}{10} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (a x^{5} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{5} + b} + \sqrt {-b}\right ) + \frac {4}{15} \, {\left (a x^{5} + b\right )}^{\frac {3}{4}} \]
-1/5*sqrt(2)*(-b)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(a*x^5 + b)^(1/4))/(-b)^(1/4)) - 1/5*sqrt(2)*(-b)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt (2)*(-b)^(1/4) - 2*(a*x^5 + b)^(1/4))/(-b)^(1/4)) + 1/10*sqrt(2)*(-b)^(3/4 )*log(sqrt(2)*(a*x^5 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^5 + b) + sqrt(-b)) - 1/10*sqrt(2)*(-b)^(3/4)*log(-sqrt(2)*(a*x^5 + b)^(1/4)*(-b)^(1/4) + sqrt( a*x^5 + b) + sqrt(-b)) + 4/15*(a*x^5 + b)^(3/4)
Time = 5.40 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.71 \[ \int \frac {\left (b+a x^5\right )^{3/4}}{x} \, dx=\frac {2\,b^{3/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^5+b\right )}^{1/4}}{b^{1/4}}\right )}{5}-\frac {2\,b^{3/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^5+b\right )}^{1/4}}{b^{1/4}}\right )}{5}+\frac {4\,{\left (a\,x^5+b\right )}^{3/4}}{15} \]