Integrand size = 15, antiderivative size = 70 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x} \, dx=\frac {1}{3} \left (a+b x^4\right )^{3/4}+\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right ) \]
1/3*(b*x^4+a)^(3/4)+1/2*a^(3/4)*arctan((b*x^4+a)^(1/4)/a^(1/4))-1/2*a^(3/4 )*arctanh((b*x^4+a)^(1/4)/a^(1/4))
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x} \, dx=\frac {1}{3} \left (a+b x^4\right )^{3/4}+\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right ) \]
(a + b*x^4)^(3/4)/3 + (a^(3/4)*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/2 - (a^( 3/4)*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/2
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.
\(\displaystyle \int \frac {\left (a+b x^4\right )^{3/4}}{x} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {\left (b x^4+a\right )^{3/4}}{x^4}dx^4\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (a \int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx^4+\frac {4}{3} \left (a+b x^4\right )^{3/4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (\frac {4 a \int -\frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{b}+\frac {4}{3} \left (a+b x^4\right )^{3/4}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a+b x^4\right )^{3/4}-\frac {4 a \int \frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{b}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a+b x^4\right )^{3/4}-4 a \int \frac {x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a+b x^4\right )^{3/4}-4 a \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {1}{2} \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}\right )\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a+b x^4\right )^{3/4}-4 a \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a+b x^4\right )^{3/4}-4 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )\right )\) |
((4*(a + b*x^4)^(3/4))/3 - 4*a*(-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))))/4
3.11.22.3.1 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 = 4.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{3}+\frac {a^{\frac {3}{4}} \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{2}-\frac {\ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) a^{\frac {3}{4}}}{4}\) | \(73\) |
1/3*(b*x^4+a)^(3/4)+1/2*a^(3/4)*arctan((b*x^4+a)^(1/4)/a^(1/4))-1/4*ln((-( b*x^4+a)^(1/4)-a^(1/4))/(-(b*x^4+a)^(1/4)+a^(1/4)))*a^(3/4)
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x} \, dx=-\frac {1}{4} \, {\left (a^{3}\right )}^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} + {\left (a^{3}\right )}^{\frac {3}{4}}\right ) + \frac {1}{4} i \, {\left (a^{3}\right )}^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} + i \, {\left (a^{3}\right )}^{\frac {3}{4}}\right ) - \frac {1}{4} i \, {\left (a^{3}\right )}^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} - i \, {\left (a^{3}\right )}^{\frac {3}{4}}\right ) + \frac {1}{4} \, {\left (a^{3}\right )}^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} - {\left (a^{3}\right )}^{\frac {3}{4}}\right ) + \frac {1}{3} \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} \]
-1/4*(a^3)^(1/4)*log((b*x^4 + a)^(1/4)*a^2 + (a^3)^(3/4)) + 1/4*I*(a^3)^(1 /4)*log((b*x^4 + a)^(1/4)*a^2 + I*(a^3)^(3/4)) - 1/4*I*(a^3)^(1/4)*log((b* x^4 + a)^(1/4)*a^2 - I*(a^3)^(3/4)) + 1/4*(a^3)^(1/4)*log((b*x^4 + a)^(1/4 )*a^2 - (a^3)^(3/4)) + 1/3*(b*x^4 + a)^(3/4)
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.63 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x} \, dx=- \frac {b^{\frac {3}{4}} x^{3} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \Gamma \left (\frac {1}{4}\right )} \]
-b**(3/4)*x**3*gamma(-3/4)*hyper((-3/4, -3/4), (1/4,), a*exp_polar(I*pi)/( b*x**4))/(4*gamma(1/4))
Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x} \, dx=\frac {1}{4} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )} + \frac {1}{3} \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} \]
1/4*a*(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)) + 1/3*(b*x^4 + a)^(3/ 4)
Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (50) = 100\).
Time = 0.29 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.64 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x} \, dx=-\frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{8} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right ) - \frac {1}{8} \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right ) + \frac {1}{3} \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} \]
-1/4*sqrt(2)*(-a)^(3/4)*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)^(3/4)*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)^(3/4) *log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a)) - 1/8*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b* x^4 + a) + sqrt(-a)) + 1/3*(b*x^4 + a)^(3/4)
Time = 5.64 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.71 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x} \, dx=\frac {a^{3/4}\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2}-\frac {a^{3/4}\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2}+\frac {{\left (b\,x^4+a\right )}^{3/4}}{3} \]