Integrand size = 22, antiderivative size = 103 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {a^2 (b c-a d) \left (a+b x^4\right )^{3/4}}{3 b^4}-\frac {a (2 b c-3 a d) \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac {(b c-3 a d) \left (a+b x^4\right )^{11/4}}{11 b^4}+\frac {d \left (a+b x^4\right )^{15/4}}{15 b^4} \] Output:
1/3*a^2*(-a*d+b*c)*(b*x^4+a)^(3/4)/b^4-1/7*a*(-3*a*d+2*b*c)*(b*x^4+a)^(7/4 )/b^4+1/11*(-3*a*d+b*c)*(b*x^4+a)^(11/4)/b^4+1/15*d*(b*x^4+a)^(15/4)/b^4
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {\left (a+b x^4\right )^{3/4} \left (160 a^2 b c-128 a^3 d-120 a b^2 c x^4+96 a^2 b d x^4+105 b^3 c x^8-84 a b^2 d x^8+77 b^3 d x^{12}\right )}{1155 b^4} \] Input:
Integrate[(x^11*(c + d*x^4))/(a + b*x^4)^(1/4),x]
Output:
((a + b*x^4)^(3/4)*(160*a^2*b*c - 128*a^3*d - 120*a*b^2*c*x^4 + 96*a^2*b*d *x^4 + 105*b^3*c*x^8 - 84*a*b^2*d*x^8 + 77*b^3*d*x^12))/(1155*b^4)
Time = 0.41 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {948, 86, 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.
\(\displaystyle \int \frac {x^{11} \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {x^8 \left (d x^4+c\right )}{\sqrt [4]{b x^4+a}}dx^4\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{4} \int \left (\frac {d \left (b x^4+a\right )^{11/4}}{b^3}+\frac {(b c-3 a d) \left (b x^4+a\right )^{7/4}}{b^3}+\frac {a (3 a d-2 b c) \left (b x^4+a\right )^{3/4}}{b^3}-\frac {a^2 (a d-b c)}{b^3 \sqrt [4]{b x^4+a}}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {4 a^2 \left (a+b x^4\right )^{3/4} (b c-a d)}{3 b^4}+\frac {4 \left (a+b x^4\right )^{11/4} (b c-3 a d)}{11 b^4}-\frac {4 a \left (a+b x^4\right )^{7/4} (2 b c-3 a d)}{7 b^4}+\frac {4 d \left (a+b x^4\right )^{15/4}}{15 b^4}\right )\) |
Input:
Int[(x^11*(c + d*x^4))/(a + b*x^4)^(1/4),x]
Output:
((4*a^2*(b*c - a*d)*(a + b*x^4)^(3/4))/(3*b^4) - (4*a*(2*b*c - 3*a*d)*(a + b*x^4)^(7/4))/(7*b^4) + (4*(b*c - 3*a*d)*(a + b*x^4)^(11/4))/(11*b^4) + ( 4*d*(a + b*x^4)^(15/4))/(15*b^4))/4
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(-\frac {128 \left (-\frac {105 x^{8} \left (\frac {11 d \,x^{4}}{15}+c \right ) b^{3}}{128}+\frac {15 \left (\frac {7 d \,x^{4}}{10}+c \right ) x^{4} a \,b^{2}}{16}-\frac {5 \left (\frac {3 d \,x^{4}}{5}+c \right ) a^{2} b}{4}+a^{3} d \right ) \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{1155 b^{4}}\) | \(68\) |
gosper | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (-77 b^{3} d \,x^{12}+84 a \,b^{2} d \,x^{8}-105 c \,b^{3} x^{8}-96 a^{2} b d \,x^{4}+120 a \,b^{2} c \,x^{4}+128 a^{3} d -160 a^{2} b c \right )}{1155 b^{4}}\) | \(77\) |
trager | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (-77 b^{3} d \,x^{12}+84 a \,b^{2} d \,x^{8}-105 c \,b^{3} x^{8}-96 a^{2} b d \,x^{4}+120 a \,b^{2} c \,x^{4}+128 a^{3} d -160 a^{2} b c \right )}{1155 b^{4}}\) | \(77\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (-77 b^{3} d \,x^{12}+84 a \,b^{2} d \,x^{8}-105 c \,b^{3} x^{8}-96 a^{2} b d \,x^{4}+120 a \,b^{2} c \,x^{4}+128 a^{3} d -160 a^{2} b c \right )}{1155 b^{4}}\) | \(77\) |
orering | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (-77 b^{3} d \,x^{12}+84 a \,b^{2} d \,x^{8}-105 c \,b^{3} x^{8}-96 a^{2} b d \,x^{4}+120 a \,b^{2} c \,x^{4}+128 a^{3} d -160 a^{2} b c \right )}{1155 b^{4}}\) | \(77\) |
Input:
int(x^11*(d*x^4+c)/(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
-128/1155*(-105/128*x^8*(11/15*d*x^4+c)*b^3+15/16*(7/10*d*x^4+c)*x^4*a*b^2 -5/4*(3/5*d*x^4+c)*a^2*b+a^3*d)*(b*x^4+a)^(3/4)/b^4
Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {{\left (77 \, b^{3} d x^{12} + 21 \, {\left (5 \, b^{3} c - 4 \, a b^{2} d\right )} x^{8} - 24 \, {\left (5 \, a b^{2} c - 4 \, a^{2} b d\right )} x^{4} + 160 \, a^{2} b c - 128 \, a^{3} d\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{1155 \, b^{4}} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
1/1155*(77*b^3*d*x^12 + 21*(5*b^3*c - 4*a*b^2*d)*x^8 - 24*(5*a*b^2*c - 4*a ^2*b*d)*x^4 + 160*a^2*b*c - 128*a^3*d)*(b*x^4 + a)^(3/4)/b^4
Time = 0.97 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.67 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\begin {cases} - \frac {128 a^{3} d \left (a + b x^{4}\right )^{\frac {3}{4}}}{1155 b^{4}} + \frac {32 a^{2} c \left (a + b x^{4}\right )^{\frac {3}{4}}}{231 b^{3}} + \frac {32 a^{2} d x^{4} \left (a + b x^{4}\right )^{\frac {3}{4}}}{385 b^{3}} - \frac {8 a c x^{4} \left (a + b x^{4}\right )^{\frac {3}{4}}}{77 b^{2}} - \frac {4 a d x^{8} \left (a + b x^{4}\right )^{\frac {3}{4}}}{55 b^{2}} + \frac {c x^{8} \left (a + b x^{4}\right )^{\frac {3}{4}}}{11 b} + \frac {d x^{12} \left (a + b x^{4}\right )^{\frac {3}{4}}}{15 b} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{12}}{12} + \frac {d x^{16}}{16}}{\sqrt [4]{a}} & \text {otherwise} \end {cases} \] Input:
integrate(x**11*(d*x**4+c)/(b*x**4+a)**(1/4),x)
Output:
Piecewise((-128*a**3*d*(a + b*x**4)**(3/4)/(1155*b**4) + 32*a**2*c*(a + b* x**4)**(3/4)/(231*b**3) + 32*a**2*d*x**4*(a + b*x**4)**(3/4)/(385*b**3) - 8*a*c*x**4*(a + b*x**4)**(3/4)/(77*b**2) - 4*a*d*x**8*(a + b*x**4)**(3/4)/ (55*b**2) + c*x**8*(a + b*x**4)**(3/4)/(11*b) + d*x**12*(a + b*x**4)**(3/4 )/(15*b), Ne(b, 0)), ((c*x**12/12 + d*x**16/16)/a**(1/4), True))
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.15 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {1}{1155} \, d {\left (\frac {77 \, {\left (b x^{4} + a\right )}^{\frac {15}{4}}}{b^{4}} - \frac {315 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a}{b^{4}} + \frac {495 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2}}{b^{4}} - \frac {385 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{3}}{b^{4}}\right )} + \frac {1}{231} \, c {\left (\frac {21 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}}}{b^{3}} - \frac {66 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a}{b^{3}} + \frac {77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2}}{b^{3}}\right )} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
1/1155*d*(77*(b*x^4 + a)^(15/4)/b^4 - 315*(b*x^4 + a)^(11/4)*a/b^4 + 495*( b*x^4 + a)^(7/4)*a^2/b^4 - 385*(b*x^4 + a)^(3/4)*a^3/b^4) + 1/231*c*(21*(b *x^4 + a)^(11/4)/b^3 - 66*(b*x^4 + a)^(7/4)*a/b^3 + 77*(b*x^4 + a)^(3/4)*a ^2/b^3)
Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {\frac {5 \, {\left (21 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} - 66 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a + 77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2}\right )} c}{b^{2}} + \frac {{\left (77 \, {\left (b x^{4} + a\right )}^{\frac {15}{4}} - 315 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a + 495 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2} - 385 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{3}\right )} d}{b^{3}}}{1155 \, b} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
1/1155*(5*(21*(b*x^4 + a)^(11/4) - 66*(b*x^4 + a)^(7/4)*a + 77*(b*x^4 + a) ^(3/4)*a^2)*c/b^2 + (77*(b*x^4 + a)^(15/4) - 315*(b*x^4 + a)^(11/4)*a + 49 5*(b*x^4 + a)^(7/4)*a^2 - 385*(b*x^4 + a)^(3/4)*a^3)*d/b^3)/b
Time = 3.73 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx={\left (b\,x^4+a\right )}^{3/4}\,\left (\frac {d\,x^{12}}{15\,b}-\frac {128\,a^3\,d-160\,a^2\,b\,c}{1155\,b^4}+\frac {x^8\,\left (105\,b^3\,c-84\,a\,b^2\,d\right )}{1155\,b^4}+\frac {8\,a\,x^4\,\left (4\,a\,d-5\,b\,c\right )}{385\,b^3}\right ) \] Input:
int((x^11*(c + d*x^4))/(a + b*x^4)^(1/4),x)
Output:
(a + b*x^4)^(3/4)*((d*x^12)/(15*b) - (128*a^3*d - 160*a^2*b*c)/(1155*b^4) + (x^8*(105*b^3*c - 84*a*b^2*d))/(1155*b^4) + (8*a*x^4*(4*a*d - 5*b*c))/(3 85*b^3))
\[ \int \frac {x^{11} \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\left (\int \frac {x^{15}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) d +\left (\int \frac {x^{11}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) c \] Input:
int(x^11*(d*x^4+c)/(b*x^4+a)^(1/4),x)
Output:
int(x**15/(a + b*x**4)**(1/4),x)*d + int(x**11/(a + b*x**4)**(1/4),x)*c