Integrand size = 15, antiderivative size = 78 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{a+b x^4}}{4 a x^4}+\frac {3 b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}} \]
-1/4*(b*x^4+a)^(1/4)/a/x^4+3/8*b*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(7/4)+3 /8*b*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(7/4)
Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{a+b x^4}}{4 a x^4}+\frac {3 b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}} \]
-1/4*(a + b*x^4)^(1/4)/(a*x^4) + (3*b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/( 8*a^(7/4)) + (3*b*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(7/4))
Time = 0.20 (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.
\(\displaystyle \int \frac {1}{x^5 \left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \left (b x^4+a\right )^{3/4}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 b \int \frac {1}{x^4 \left (b x^4+a\right )^{3/4}}dx^4}{4 a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \int \frac {1}{\frac {x^{16}}{b}-\frac {a}{b}}d\sqrt [4]{b x^4+a}}{a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}\right )}{a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \left (-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )\) |
(-((a + b*x^4)^(1/4)/(a*x^4)) - (3*(-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))))/a)/4
3.12.16.3.1 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 = 4.41 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10
method | result | size |
pseudoelliptic | \(\frac {6 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b \,x^{4}+3 \ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b \,x^{4}-4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{\frac {3}{4}}}{16 a^{\frac {7}{4}} x^{4}}\) | \(86\) |
1/16*(6*arctan((b*x^4+a)^(1/4)/a^(1/4))*b*x^4+3*ln((-(b*x^4+a)^(1/4)-a^(1/ 4))/(-(b*x^4+a)^(1/4)+a^(1/4)))*b*x^4-4*(b*x^4+a)^(1/4)*a^(3/4))/a^(7/4)/x ^4
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.46 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=\frac {3 \, a x^{4} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (3 \, a^{2} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b\right ) + 3 i \, a x^{4} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (3 i \, a^{2} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b\right ) - 3 i \, a x^{4} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-3 i \, a^{2} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b\right ) - 3 \, a x^{4} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-3 \, a^{2} \left (\frac {b^{4}}{a^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{16 \, a x^{4}} \]
1/16*(3*a*x^4*(b^4/a^7)^(1/4)*log(3*a^2*(b^4/a^7)^(1/4) + 3*(b*x^4 + a)^(1 /4)*b) + 3*I*a*x^4*(b^4/a^7)^(1/4)*log(3*I*a^2*(b^4/a^7)^(1/4) + 3*(b*x^4 + a)^(1/4)*b) - 3*I*a*x^4*(b^4/a^7)^(1/4)*log(-3*I*a^2*(b^4/a^7)^(1/4) + 3 *(b*x^4 + a)^(1/4)*b) - 3*a*x^4*(b^4/a^7)^(1/4)*log(-3*a^2*(b^4/a^7)^(1/4) + 3*(b*x^4 + a)^(1/4)*b) - 4*(b*x^4 + a)^(1/4))/(a*x^4)
Result contains complex when optimal does not.
Time = 0.82 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {3}{4}} x^{7} \Gamma \left (\frac {11}{4}\right )} \]
-gamma(7/4)*hyper((3/4, 7/4), (11/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b**( 3/4)*x**7*gamma(11/4))
Time = 0.30 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{4 \, {\left ({\left (b x^{4} + a\right )} a - a^{2}\right )}} + \frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )}}{16 \, a} \]
-1/4*(b*x^4 + a)^(1/4)*b/((b*x^4 + a)*a - a^2) + 3/16*(2*b*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(3/4) - b*log(((b*x^4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4)))/a^(3/4))/a
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (58) = 116\).
Time = 0.30 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.83 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=\frac {\frac {6 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {6 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a^{2}} + \frac {3 \, \sqrt {2} b^{2} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}} a} - \frac {8 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{a x^{4}}}{32 \, b} \]
1/32*(6*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2* (b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2 + 6*sqrt(2)*(-a)^(1/4)*b^2*arctan(-1/2* sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2 + 3*sqr t(2)*(-a)^(1/4)*b^2*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a^2 + 3*sqrt(2)*b^2*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^( 1/4) + sqrt(b*x^4 + a) + sqrt(-a))/((-a)^(3/4)*a) - 8*(b*x^4 + a)^(1/4)*b/ (a*x^4))/b
Time = 5.94 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=\frac {3\,b\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{7/4}}-\frac {{\left (b\,x^4+a\right )}^{1/4}}{4\,a\,x^4}+\frac {3\,b\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{7/4}} \]