Integrand size = 15, antiderivative size = 97 \[ \int \frac {x^8}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {x^5}{b \sqrt [4]{a+b x^4}}+\frac {5 x \left (a+b x^4\right )^{3/4}}{4 b^2}-\frac {5 a \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}}-\frac {5 a \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}} \]
-x^5/b/(b*x^4+a)^(1/4)+5/4*x*(b*x^4+a)^(3/4)/b^2-5/8*a*arctan(b^(1/4)*x/(b *x^4+a)^(1/4))/b^(9/4)-5/8*a*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(9/4)
Time = 0.64 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {x^8}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {\frac {2 \sqrt [4]{b} x \left (5 a+b x^4\right )}{\sqrt [4]{a+b x^4}}-5 a \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-5 a \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}} \]
((2*b^(1/4)*x*(5*a + b*x^4))/(a + b*x^4)^(1/4) - 5*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] - 5*a*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(9/4))
Time = 0.22 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {817, 843, 770, 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.
\(\displaystyle \int \frac {x^8}{\left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {5 \int \frac {x^4}{\sqrt [4]{b x^4+a}}dx}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{\sqrt [4]{b x^4+a}}dx}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\) |
-(x^5/(b*(a + b*x^4)^(1/4))) + (5*((x*(a + b*x^4)^(3/4))/(4*b) - (a*(ArcTa n[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4)) + ArcTanh[(b^(1/4)*x)/(a + b* x^4)^(1/4)]/(2*b^(1/4))))/(4*b)))/b
3.12.55.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 /n]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Time = 4.55 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.14
method | result | size |
pseudoelliptic | \(\frac {4 b^{\frac {5}{4}} x^{5}+10 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a \left (b \,x^{4}+a \right )^{\frac {1}{4}}-5 \ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a \left (b \,x^{4}+a \right )^{\frac {1}{4}}+20 a x \,b^{\frac {1}{4}}}{16 b^{\frac {9}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\) | \(111\) |
1/16/b^(9/4)*(4*b^(5/4)*x^5+10*arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a*(b*x^ 4+a)^(1/4)-5*ln((-b^(1/4)*x-(b*x^4+a)^(1/4))/(b^(1/4)*x-(b*x^4+a)^(1/4)))* a*(b*x^4+a)^(1/4)+20*a*x*b^(1/4))/(b*x^4+a)^(1/4)
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.89 \[ \int \frac {x^8}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {5 \, {\left (b^{3} x^{4} + a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {125 \, {\left (b^{7} x \left (\frac {a^{4}}{b^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}\right )}}{x}\right ) - 5 \, {\left (b^{3} x^{4} + a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {125 \, {\left (b^{7} x \left (\frac {a^{4}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}\right )}}{x}\right ) + 5 \, {\left (i \, b^{3} x^{4} + i \, a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {125 \, {\left (i \, b^{7} x \left (\frac {a^{4}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}\right )}}{x}\right ) + 5 \, {\left (-i \, b^{3} x^{4} - i \, a b^{2}\right )} \left (\frac {a^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {125 \, {\left (-i \, b^{7} x \left (\frac {a^{4}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}\right )}}{x}\right ) - 4 \, {\left (b x^{5} + 5 \, a x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{16 \, {\left (b^{3} x^{4} + a b^{2}\right )}} \]
-1/16*(5*(b^3*x^4 + a*b^2)*(a^4/b^9)^(1/4)*log(125*(b^7*x*(a^4/b^9)^(3/4) + (b*x^4 + a)^(1/4)*a^3)/x) - 5*(b^3*x^4 + a*b^2)*(a^4/b^9)^(1/4)*log(-125 *(b^7*x*(a^4/b^9)^(3/4) - (b*x^4 + a)^(1/4)*a^3)/x) + 5*(I*b^3*x^4 + I*a*b ^2)*(a^4/b^9)^(1/4)*log(-125*(I*b^7*x*(a^4/b^9)^(3/4) - (b*x^4 + a)^(1/4)* a^3)/x) + 5*(-I*b^3*x^4 - I*a*b^2)*(a^4/b^9)^(1/4)*log(-125*(-I*b^7*x*(a^4 /b^9)^(3/4) - (b*x^4 + a)^(1/4)*a^3)/x) - 4*(b*x^5 + 5*a*x)*(b*x^4 + a)^(3 /4))/(b^3*x^4 + a*b^2)
Result contains complex when optimal does not.
Time = 0.90 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.38 \[ \int \frac {x^8}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {13}{4}\right )} \]
x**9*gamma(9/4)*hyper((5/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a* *(5/4)*gamma(13/4))
Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.34 \[ \int \frac {x^8}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {4 \, a b - \frac {5 \, {\left (b x^{4} + a\right )} a}{x^{4}}}{4 \, {\left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}}{x} - \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{2}}{x^{5}}\right )}} + \frac {5 \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{16 \, b^{2}} \]
1/4*(4*a*b - 5*(b*x^4 + a)*a/x^4)/((b*x^4 + a)^(1/4)*b^3/x - (b*x^4 + a)^( 5/4)*b^2/x^5) + 5/16*a*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(1 /4))/b^2
\[ \int \frac {x^8}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{8}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^8}{\left (a+b x^4\right )^{5/4}} \, dx=\int \frac {x^8}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \]