Integrand size = 16, antiderivative size = 106 \[ \int \frac {x^{19}}{\sqrt [4]{a-b x^4}} \, dx=-\frac {a^4 \left (a-b x^4\right )^{3/4}}{3 b^5}+\frac {4 a^3 \left (a-b x^4\right )^{7/4}}{7 b^5}-\frac {6 a^2 \left (a-b x^4\right )^{11/4}}{11 b^5}+\frac {4 a \left (a-b x^4\right )^{15/4}}{15 b^5}-\frac {\left (a-b x^4\right )^{19/4}}{19 b^5} \] Output:
-1/3*a^4*(-b*x^4+a)^(3/4)/b^5+4/7*a^3*(-b*x^4+a)^(7/4)/b^5-6/11*a^2*(-b*x^ 4+a)^(11/4)/b^5+4/15*a*(-b*x^4+a)^(15/4)/b^5-1/19*(-b*x^4+a)^(19/4)/b^5
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.58 \[ \int \frac {x^{19}}{\sqrt [4]{a-b x^4}} \, dx=\frac {\left (a-b x^4\right )^{3/4} \left (-2048 a^4-1536 a^3 b x^4-1344 a^2 b^2 x^8-1232 a b^3 x^{12}-1155 b^4 x^{16}\right )}{21945 b^5} \] Input:
Integrate[x^19/(a - b*x^4)^(1/4),x]
Output:
((a - b*x^4)^(3/4)*(-2048*a^4 - 1536*a^3*b*x^4 - 1344*a^2*b^2*x^8 - 1232*a *b^3*x^12 - 1155*b^4*x^16))/(21945*b^5)
Time = 0.23 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {798, 53, 2009}
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.
\(\displaystyle \int \frac {x^{19}}{\sqrt [4]{a-b x^4}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {x^{16}}{\sqrt [4]{a-b x^4}}dx^4\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{4} \int \left (\frac {a^4}{b^4 \sqrt [4]{a-b x^4}}-\frac {4 \left (a-b x^4\right )^{3/4} a^3}{b^4}+\frac {6 \left (a-b x^4\right )^{7/4} a^2}{b^4}-\frac {4 \left (a-b x^4\right )^{11/4} a}{b^4}+\frac {\left (a-b x^4\right )^{15/4}}{b^4}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\frac {4 a^4 \left (a-b x^4\right )^{3/4}}{3 b^5}+\frac {16 a^3 \left (a-b x^4\right )^{7/4}}{7 b^5}-\frac {24 a^2 \left (a-b x^4\right )^{11/4}}{11 b^5}-\frac {4 \left (a-b x^4\right )^{19/4}}{19 b^5}+\frac {16 a \left (a-b x^4\right )^{15/4}}{15 b^5}\right )\) |
Input:
Int[x^19/(a - b*x^4)^(1/4),x]
Output:
((-4*a^4*(a - b*x^4)^(3/4))/(3*b^5) + (16*a^3*(a - b*x^4)^(7/4))/(7*b^5) - (24*a^2*(a - b*x^4)^(11/4))/(11*b^5) + (16*a*(a - b*x^4)^(15/4))/(15*b^5) - (4*(a - b*x^4)^(19/4))/(19*b^5))/4
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 = 0.56 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.56
method | result | size |
gosper | \(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (1155 x^{16} b^{4}+1232 a \,b^{3} x^{12}+1344 a^{2} b^{2} x^{8}+1536 a^{3} b \,x^{4}+2048 a^{4}\right )}{21945 b^{5}}\) | \(59\) |
trager | \(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (1155 x^{16} b^{4}+1232 a \,b^{3} x^{12}+1344 a^{2} b^{2} x^{8}+1536 a^{3} b \,x^{4}+2048 a^{4}\right )}{21945 b^{5}}\) | \(59\) |
risch | \(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (1155 x^{16} b^{4}+1232 a \,b^{3} x^{12}+1344 a^{2} b^{2} x^{8}+1536 a^{3} b \,x^{4}+2048 a^{4}\right )}{21945 b^{5}}\) | \(59\) |
pseudoelliptic | \(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (1155 x^{16} b^{4}+1232 a \,b^{3} x^{12}+1344 a^{2} b^{2} x^{8}+1536 a^{3} b \,x^{4}+2048 a^{4}\right )}{21945 b^{5}}\) | \(59\) |
orering | \(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (1155 x^{16} b^{4}+1232 a \,b^{3} x^{12}+1344 a^{2} b^{2} x^{8}+1536 a^{3} b \,x^{4}+2048 a^{4}\right )}{21945 b^{5}}\) | \(59\) |
Input:
int(x^19/(-b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
-1/21945*(-b*x^4+a)^(3/4)*(1155*b^4*x^16+1232*a*b^3*x^12+1344*a^2*b^2*x^8+ 1536*a^3*b*x^4+2048*a^4)/b^5
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.55 \[ \int \frac {x^{19}}{\sqrt [4]{a-b x^4}} \, dx=-\frac {{\left (1155 \, b^{4} x^{16} + 1232 \, a b^{3} x^{12} + 1344 \, a^{2} b^{2} x^{8} + 1536 \, a^{3} b x^{4} + 2048 \, a^{4}\right )} {\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{21945 \, b^{5}} \] Input:
integrate(x^19/(-b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
-1/21945*(1155*b^4*x^16 + 1232*a*b^3*x^12 + 1344*a^2*b^2*x^8 + 1536*a^3*b* x^4 + 2048*a^4)*(-b*x^4 + a)^(3/4)/b^5
Time = 0.86 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.10 \[ \int \frac {x^{19}}{\sqrt [4]{a-b x^4}} \, dx=\begin {cases} - \frac {2048 a^{4} \left (a - b x^{4}\right )^{\frac {3}{4}}}{21945 b^{5}} - \frac {512 a^{3} x^{4} \left (a - b x^{4}\right )^{\frac {3}{4}}}{7315 b^{4}} - \frac {64 a^{2} x^{8} \left (a - b x^{4}\right )^{\frac {3}{4}}}{1045 b^{3}} - \frac {16 a x^{12} \left (a - b x^{4}\right )^{\frac {3}{4}}}{285 b^{2}} - \frac {x^{16} \left (a - b x^{4}\right )^{\frac {3}{4}}}{19 b} & \text {for}\: b \neq 0 \\\frac {x^{20}}{20 \sqrt [4]{a}} & \text {otherwise} \end {cases} \] Input:
integrate(x**19/(-b*x**4+a)**(1/4),x)
Output:
Piecewise((-2048*a**4*(a - b*x**4)**(3/4)/(21945*b**5) - 512*a**3*x**4*(a - b*x**4)**(3/4)/(7315*b**4) - 64*a**2*x**8*(a - b*x**4)**(3/4)/(1045*b**3 ) - 16*a*x**12*(a - b*x**4)**(3/4)/(285*b**2) - x**16*(a - b*x**4)**(3/4)/ (19*b), Ne(b, 0)), (x**20/(20*a**(1/4)), True))
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.81 \[ \int \frac {x^{19}}{\sqrt [4]{a-b x^4}} \, dx=-\frac {{\left (-b x^{4} + a\right )}^{\frac {19}{4}}}{19 \, b^{5}} + \frac {4 \, {\left (-b x^{4} + a\right )}^{\frac {15}{4}} a}{15 \, b^{5}} - \frac {6 \, {\left (-b x^{4} + a\right )}^{\frac {11}{4}} a^{2}}{11 \, b^{5}} + \frac {4 \, {\left (-b x^{4} + a\right )}^{\frac {7}{4}} a^{3}}{7 \, b^{5}} - \frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}} a^{4}}{3 \, b^{5}} \] Input:
integrate(x^19/(-b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
-1/19*(-b*x^4 + a)^(19/4)/b^5 + 4/15*(-b*x^4 + a)^(15/4)*a/b^5 - 6/11*(-b* x^4 + a)^(11/4)*a^2/b^5 + 4/7*(-b*x^4 + a)^(7/4)*a^3/b^5 - 1/3*(-b*x^4 + a )^(3/4)*a^4/b^5
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.03 \[ \int \frac {x^{19}}{\sqrt [4]{a-b x^4}} \, dx=-\frac {1155 \, {\left (b x^{4} - a\right )}^{4} {\left (-b x^{4} + a\right )}^{\frac {3}{4}} + 5852 \, {\left (b x^{4} - a\right )}^{3} {\left (-b x^{4} + a\right )}^{\frac {3}{4}} a + 11970 \, {\left (b x^{4} - a\right )}^{2} {\left (-b x^{4} + a\right )}^{\frac {3}{4}} a^{2} - 12540 \, {\left (-b x^{4} + a\right )}^{\frac {7}{4}} a^{3} + 7315 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}} a^{4}}{21945 \, b^{5}} \] Input:
integrate(x^19/(-b*x^4+a)^(1/4),x, algorithm="giac")
Output:
-1/21945*(1155*(b*x^4 - a)^4*(-b*x^4 + a)^(3/4) + 5852*(b*x^4 - a)^3*(-b*x ^4 + a)^(3/4)*a + 11970*(b*x^4 - a)^2*(-b*x^4 + a)^(3/4)*a^2 - 12540*(-b*x ^4 + a)^(7/4)*a^3 + 7315*(-b*x^4 + a)^(3/4)*a^4)/b^5
Time = 0.40 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.57 \[ \int \frac {x^{19}}{\sqrt [4]{a-b x^4}} \, dx=-{\left (a-b\,x^4\right )}^{3/4}\,\left (\frac {2048\,a^4}{21945\,b^5}+\frac {x^{16}}{19\,b}+\frac {16\,a\,x^{12}}{285\,b^2}+\frac {512\,a^3\,x^4}{7315\,b^4}+\frac {64\,a^2\,x^8}{1045\,b^3}\right ) \] Input:
int(x^19/(a - b*x^4)^(1/4),x)
Output:
-(a - b*x^4)^(3/4)*((2048*a^4)/(21945*b^5) + x^16/(19*b) + (16*a*x^12)/(28 5*b^2) + (512*a^3*x^4)/(7315*b^4) + (64*a^2*x^8)/(1045*b^3))
\[ \int \frac {x^{19}}{\sqrt [4]{a-b x^4}} \, dx=\int \frac {x^{19}}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}d x \] Input:
int(x^19/(-b*x^4+a)^(1/4),x)
Output:
int(x**19/(a - b*x**4)**(1/4),x)