Integrand size = 22, antiderivative size = 100 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {a^2 (b c-a d) \sqrt [4]{a+b x^4}}{b^4}-\frac {a (2 b c-3 a d) \left (a+b x^4\right )^{5/4}}{5 b^4}+\frac {(b c-3 a d) \left (a+b x^4\right )^{9/4}}{9 b^4}+\frac {d \left (a+b x^4\right )^{13/4}}{13 b^4} \] Output:
a^2*(-a*d+b*c)*(b*x^4+a)^(1/4)/b^4-1/5*a*(-3*a*d+2*b*c)*(b*x^4+a)^(5/4)/b^ 4+1/9*(-3*a*d+b*c)*(b*x^4+a)^(9/4)/b^4+1/13*d*(b*x^4+a)^(13/4)/b^4
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{a+b x^4} \left (416 a^2 b c-384 a^3 d-104 a b^2 c x^4+96 a^2 b d x^4+65 b^3 c x^8-60 a b^2 d x^8+45 b^3 d x^{12}\right )}{585 b^4} \] Input:
Integrate[(x^11*(c + d*x^4))/(a + b*x^4)^(3/4),x]
Output:
((a + b*x^4)^(1/4)*(416*a^2*b*c - 384*a^3*d - 104*a*b^2*c*x^4 + 96*a^2*b*d *x^4 + 65*b^3*c*x^8 - 60*a*b^2*d*x^8 + 45*b^3*d*x^12))/(585*b^4)
Time = 0.41 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {x^8 \left (d x^4+c\right )}{\left (b x^4+a\right )^{3/4}}dx^4\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{4} \int \left (\frac {d \left (b x^4+a\right )^{9/4}}{b^3}+\frac {(b c-3 a d) \left (b x^4+a\right )^{5/4}}{b^3}+\frac {a (3 a d-2 b c) \sqrt [4]{b x^4+a}}{b^3}-\frac {a^2 (a d-b c)}{b^3 \left (b x^4+a\right )^{3/4}}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {4 a^2 \sqrt [4]{a+b x^4} (b c-a d)}{b^4}+\frac {4 \left (a+b x^4\right )^{9/4} (b c-3 a d)}{9 b^4}-\frac {4 a \left (a+b x^4\right )^{5/4} (2 b c-3 a d)}{5 b^4}+\frac {4 d \left (a+b x^4\right )^{13/4}}{13 b^4}\right )\) |
Input:
Int[(x^11*(c + d*x^4))/(a + b*x^4)^(3/4),x]
Output:
((4*a^2*(b*c - a*d)*(a + b*x^4)^(1/4))/b^4 - (4*a*(2*b*c - 3*a*d)*(a + b*x ^4)^(5/4))/(5*b^4) + (4*(b*c - 3*a*d)*(a + b*x^4)^(9/4))/(9*b^4) + (4*d*(a + b*x^4)^(13/4))/(13*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 = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(-\frac {128 \left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-\frac {65 \left (\frac {9 d \,x^{4}}{13}+c \right ) x^{8} b^{3}}{384}+\frac {13 x^{4} a \left (\frac {15 d \,x^{4}}{26}+c \right ) b^{2}}{48}-\frac {13 a^{2} \left (\frac {3 d \,x^{4}}{13}+c \right ) b}{12}+a^{3} d \right )}{195 b^{4}}\) | \(68\) |
gosper | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-45 b^{3} d \,x^{12}+60 a \,b^{2} d \,x^{8}-65 c \,b^{3} x^{8}-96 a^{2} b d \,x^{4}+104 a \,b^{2} c \,x^{4}+384 a^{3} d -416 a^{2} b c \right )}{585 b^{4}}\) | \(77\) |
trager | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-45 b^{3} d \,x^{12}+60 a \,b^{2} d \,x^{8}-65 c \,b^{3} x^{8}-96 a^{2} b d \,x^{4}+104 a \,b^{2} c \,x^{4}+384 a^{3} d -416 a^{2} b c \right )}{585 b^{4}}\) | \(77\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-45 b^{3} d \,x^{12}+60 a \,b^{2} d \,x^{8}-65 c \,b^{3} x^{8}-96 a^{2} b d \,x^{4}+104 a \,b^{2} c \,x^{4}+384 a^{3} d -416 a^{2} b c \right )}{585 b^{4}}\) | \(77\) |
orering | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-45 b^{3} d \,x^{12}+60 a \,b^{2} d \,x^{8}-65 c \,b^{3} x^{8}-96 a^{2} b d \,x^{4}+104 a \,b^{2} c \,x^{4}+384 a^{3} d -416 a^{2} b c \right )}{585 b^{4}}\) | \(77\) |
Input:
int(x^11*(d*x^4+c)/(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)
Output:
-128/195*(b*x^4+a)^(1/4)*(-65/384*(9/13*d*x^4+c)*x^8*b^3+13/48*x^4*a*(15/2 6*d*x^4+c)*b^2-13/12*a^2*(3/13*d*x^4+c)*b+a^3*d)/b^4
Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.76 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {{\left (45 \, b^{3} d x^{12} + 5 \, {\left (13 \, b^{3} c - 12 \, a b^{2} d\right )} x^{8} - 8 \, {\left (13 \, a b^{2} c - 12 \, a^{2} b d\right )} x^{4} + 416 \, a^{2} b c - 384 \, a^{3} d\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{585 \, b^{4}} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
1/585*(45*b^3*d*x^12 + 5*(13*b^3*c - 12*a*b^2*d)*x^8 - 8*(13*a*b^2*c - 12* a^2*b*d)*x^4 + 416*a^2*b*c - 384*a^3*d)*(b*x^4 + a)^(1/4)/b^4
Time = 1.96 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.72 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\begin {cases} - \frac {128 a^{3} d \sqrt [4]{a + b x^{4}}}{195 b^{4}} + \frac {32 a^{2} c \sqrt [4]{a + b x^{4}}}{45 b^{3}} + \frac {32 a^{2} d x^{4} \sqrt [4]{a + b x^{4}}}{195 b^{3}} - \frac {8 a c x^{4} \sqrt [4]{a + b x^{4}}}{45 b^{2}} - \frac {4 a d x^{8} \sqrt [4]{a + b x^{4}}}{39 b^{2}} + \frac {c x^{8} \sqrt [4]{a + b x^{4}}}{9 b} + \frac {d x^{12} \sqrt [4]{a + b x^{4}}}{13 b} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{12}}{12} + \frac {d x^{16}}{16}}{a^{\frac {3}{4}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**11*(d*x**4+c)/(b*x**4+a)**(3/4),x)
Output:
Piecewise((-128*a**3*d*(a + b*x**4)**(1/4)/(195*b**4) + 32*a**2*c*(a + b*x **4)**(1/4)/(45*b**3) + 32*a**2*d*x**4*(a + b*x**4)**(1/4)/(195*b**3) - 8* a*c*x**4*(a + b*x**4)**(1/4)/(45*b**2) - 4*a*d*x**8*(a + b*x**4)**(1/4)/(3 9*b**2) + c*x**8*(a + b*x**4)**(1/4)/(9*b) + d*x**12*(a + b*x**4)**(1/4)/( 13*b), Ne(b, 0)), ((c*x**12/12 + d*x**16/16)/a**(3/4), True))
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {1}{195} \, d {\left (\frac {15 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}}}{b^{4}} - \frac {65 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a}{b^{4}} + \frac {117 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2}}{b^{4}} - \frac {195 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{b^{4}}\right )} + \frac {1}{45} \, c {\left (\frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}}}{b^{3}} - \frac {18 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a}{b^{3}} + \frac {45 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}}{b^{3}}\right )} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
1/195*d*(15*(b*x^4 + a)^(13/4)/b^4 - 65*(b*x^4 + a)^(9/4)*a/b^4 + 117*(b*x ^4 + a)^(5/4)*a^2/b^4 - 195*(b*x^4 + a)^(1/4)*a^3/b^4) + 1/45*c*(5*(b*x^4 + a)^(9/4)/b^3 - 18*(b*x^4 + a)^(5/4)*a/b^3 + 45*(b*x^4 + a)^(1/4)*a^2/b^3 )
Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (a^{2} b c - a^{3} d\right )}}{b^{4}} + \frac {65 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} b c - 234 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a b c + 45 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} d - 195 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a d + 351 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2} d}{585 \, b^{4}} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(3/4),x, algorithm="giac")
Output:
(b*x^4 + a)^(1/4)*(a^2*b*c - a^3*d)/b^4 + 1/585*(65*(b*x^4 + a)^(9/4)*b*c - 234*(b*x^4 + a)^(5/4)*a*b*c + 45*(b*x^4 + a)^(13/4)*d - 195*(b*x^4 + a)^ (9/4)*a*d + 351*(b*x^4 + a)^(5/4)*a^2*d)/b^4
Time = 3.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx={\left (b\,x^4+a\right )}^{1/4}\,\left (\frac {d\,x^{12}}{13\,b}-\frac {384\,a^3\,d-416\,a^2\,b\,c}{585\,b^4}+\frac {x^8\,\left (65\,b^3\,c-60\,a\,b^2\,d\right )}{585\,b^4}+\frac {8\,a\,x^4\,\left (12\,a\,d-13\,b\,c\right )}{585\,b^3}\right ) \] Input:
int((x^11*(c + d*x^4))/(a + b*x^4)^(3/4),x)
Output:
(a + b*x^4)^(1/4)*((d*x^12)/(13*b) - (384*a^3*d - 416*a^2*b*c)/(585*b^4) + (x^8*(65*b^3*c - 60*a*b^2*d))/(585*b^4) + (8*a*x^4*(12*a*d - 13*b*c))/(58 5*b^3))
\[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/4}} \, dx=\left (\int \frac {x^{15}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) d +\left (\int \frac {x^{11}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) c \] Input:
int(x^11*(d*x^4+c)/(b*x^4+a)^(3/4),x)
Output:
int(x**15/(a + b*x**4)**(3/4),x)*d + int(x**11/(a + b*x**4)**(3/4),x)*c