Integrand size = 22, antiderivative size = 101 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {a^2 (b c-a d)}{b^4 \sqrt [4]{a+b x^4}}-\frac {a (2 b c-3 a d) \left (a+b x^4\right )^{3/4}}{3 b^4}+\frac {(b c-3 a d) \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac {d \left (a+b x^4\right )^{11/4}}{11 b^4} \] Output:
-a^2*(-a*d+b*c)/b^4/(b*x^4+a)^(1/4)-1/3*a*(-3*a*d+2*b*c)*(b*x^4+a)^(3/4)/b ^4+1/7*(-3*a*d+b*c)*(b*x^4+a)^(7/4)/b^4+1/11*d*(b*x^4+a)^(11/4)/b^4
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {384 a^3 d+3 b^3 x^8 \left (11 c+7 d x^4\right )-4 a b^2 x^4 \left (22 c+9 d x^4\right )+a^2 b \left (-352 c+96 d x^4\right )}{231 b^4 \sqrt [4]{a+b x^4}} \] Input:
Integrate[(x^11*(c + d*x^4))/(a + b*x^4)^(5/4),x]
Output:
(384*a^3*d + 3*b^3*x^8*(11*c + 7*d*x^4) - 4*a*b^2*x^4*(22*c + 9*d*x^4) + a ^2*b*(-352*c + 96*d*x^4))/(231*b^4*(a + b*x^4)^(1/4))
Time = 0.42 (sec) , antiderivative size = 105, 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.
\(\displaystyle \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {x^8 \left (d x^4+c\right )}{\left (b x^4+a\right )^{5/4}}dx^4\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{4} \int \left (-\frac {(a d-b c) a^2}{b^3 \left (b x^4+a\right )^{5/4}}+\frac {(3 a d-2 b c) a}{b^3 \sqrt [4]{b x^4+a}}+\frac {d \left (b x^4+a\right )^{7/4}}{b^3}+\frac {(b c-3 a d) \left (b x^4+a\right )^{3/4}}{b^3}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\frac {4 a^2 (b c-a d)}{b^4 \sqrt [4]{a+b x^4}}+\frac {4 \left (a+b x^4\right )^{7/4} (b c-3 a d)}{7 b^4}-\frac {4 a \left (a+b x^4\right )^{3/4} (2 b c-3 a d)}{3 b^4}+\frac {4 d \left (a+b x^4\right )^{11/4}}{11 b^4}\right )\) |
Input:
Int[(x^11*(c + d*x^4))/(a + b*x^4)^(5/4),x]
Output:
((-4*a^2*(b*c - a*d))/(b^4*(a + b*x^4)^(1/4)) - (4*a*(2*b*c - 3*a*d)*(a + b*x^4)^(3/4))/(3*b^4) + (4*(b*c - 3*a*d)*(a + b*x^4)^(7/4))/(7*b^4) + (4*d *(a + b*x^4)^(11/4))/(11*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 = 0.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {\frac {\left (\frac {7 d \,x^{4}}{11}+c \right ) x^{8} b^{3}}{7}-\frac {8 \left (\frac {9 d \,x^{4}}{22}+c \right ) x^{4} a \,b^{2}}{21}-\frac {32 \left (-\frac {3 d \,x^{4}}{11}+c \right ) a^{2} b}{21}+\frac {128 a^{3} d}{77}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4}}\) | \(68\) |
gosper | \(\frac {21 b^{3} d \,x^{12}-36 a \,b^{2} d \,x^{8}+33 c \,b^{3} x^{8}+96 a^{2} b d \,x^{4}-88 a \,b^{2} c \,x^{4}+384 a^{3} d -352 a^{2} b c}{231 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4}}\) | \(77\) |
trager | \(\frac {21 b^{3} d \,x^{12}-36 a \,b^{2} d \,x^{8}+33 c \,b^{3} x^{8}+96 a^{2} b d \,x^{4}-88 a \,b^{2} c \,x^{4}+384 a^{3} d -352 a^{2} b c}{231 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4}}\) | \(77\) |
orering | \(\frac {21 b^{3} d \,x^{12}-36 a \,b^{2} d \,x^{8}+33 c \,b^{3} x^{8}+96 a^{2} b d \,x^{4}-88 a \,b^{2} c \,x^{4}+384 a^{3} d -352 a^{2} b c}{231 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4}}\) | \(77\) |
risch | \(\frac {\left (21 d \,b^{2} x^{8}-57 a b d \,x^{4}+33 b^{2} c \,x^{4}+153 a^{2} d -121 a b c \right ) \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{231 b^{4}}+\frac {a^{2} \left (a d -c b \right )}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4}}\) | \(78\) |
Input:
int(x^11*(d*x^4+c)/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)
Output:
128/77/(b*x^4+a)^(1/4)*(11/128*(7/11*d*x^4+c)*x^8*b^3-11/48*(9/22*d*x^4+c) *x^4*a*b^2-11/12*(-3/11*d*x^4+c)*a^2*b+a^3*d)/b^4
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (21 \, b^{3} d x^{12} + 3 \, {\left (11 \, b^{3} c - 12 \, a b^{2} d\right )} x^{8} - 8 \, {\left (11 \, a b^{2} c - 12 \, a^{2} b d\right )} x^{4} - 352 \, a^{2} b c + 384 \, a^{3} d\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{231 \, {\left (b^{5} x^{4} + a b^{4}\right )}} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(5/4),x, algorithm="fricas")
Output:
1/231*(21*b^3*d*x^12 + 3*(11*b^3*c - 12*a*b^2*d)*x^8 - 8*(11*a*b^2*c - 12* a^2*b*d)*x^4 - 352*a^2*b*c + 384*a^3*d)*(b*x^4 + a)^(3/4)/(b^5*x^4 + a*b^4 )
Time = 1.20 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.70 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\begin {cases} \frac {128 a^{3} d}{77 b^{4} \sqrt [4]{a + b x^{4}}} - \frac {32 a^{2} c}{21 b^{3} \sqrt [4]{a + b x^{4}}} + \frac {32 a^{2} d x^{4}}{77 b^{3} \sqrt [4]{a + b x^{4}}} - \frac {8 a c x^{4}}{21 b^{2} \sqrt [4]{a + b x^{4}}} - \frac {12 a d x^{8}}{77 b^{2} \sqrt [4]{a + b x^{4}}} + \frac {c x^{8}}{7 b \sqrt [4]{a + b x^{4}}} + \frac {d x^{12}}{11 b \sqrt [4]{a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{12}}{12} + \frac {d x^{16}}{16}}{a^{\frac {5}{4}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**11*(d*x**4+c)/(b*x**4+a)**(5/4),x)
Output:
Piecewise((128*a**3*d/(77*b**4*(a + b*x**4)**(1/4)) - 32*a**2*c/(21*b**3*( a + b*x**4)**(1/4)) + 32*a**2*d*x**4/(77*b**3*(a + b*x**4)**(1/4)) - 8*a*c *x**4/(21*b**2*(a + b*x**4)**(1/4)) - 12*a*d*x**8/(77*b**2*(a + b*x**4)**( 1/4)) + c*x**8/(7*b*(a + b*x**4)**(1/4)) + d*x**12/(11*b*(a + b*x**4)**(1/ 4)), Ne(b, 0)), ((c*x**12/12 + d*x**16/16)/a**(5/4), True))
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.17 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {1}{77} \, d {\left (\frac {7 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}}}{b^{4}} - \frac {33 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a}{b^{4}} + \frac {77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2}}{b^{4}} + \frac {77 \, a^{3}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{4}}\right )} + \frac {1}{21} \, c {\left (\frac {3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}}}{b^{3}} - \frac {14 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a}{b^{3}} - \frac {21 \, a^{2}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}}\right )} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(5/4),x, algorithm="maxima")
Output:
1/77*d*(7*(b*x^4 + a)^(11/4)/b^4 - 33*(b*x^4 + a)^(7/4)*a/b^4 + 77*(b*x^4 + a)^(3/4)*a^2/b^4 + 77*a^3/((b*x^4 + a)^(1/4)*b^4)) + 1/21*c*(3*(b*x^4 + a)^(7/4)/b^3 - 14*(b*x^4 + a)^(3/4)*a/b^3 - 21*a^2/((b*x^4 + a)^(1/4)*b^3) )
Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {a^{2} b c - a^{3} d}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{4}} + \frac {33 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{41} c - 154 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{41} c + 21 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} b^{40} d - 99 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a b^{40} d + 231 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2} b^{40} d}{231 \, b^{44}} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(5/4),x, algorithm="giac")
Output:
-(a^2*b*c - a^3*d)/((b*x^4 + a)^(1/4)*b^4) + 1/231*(33*(b*x^4 + a)^(7/4)*b ^41*c - 154*(b*x^4 + a)^(3/4)*a*b^41*c + 21*(b*x^4 + a)^(11/4)*b^40*d - 99 *(b*x^4 + a)^(7/4)*a*b^40*d + 231*(b*x^4 + a)^(3/4)*a^2*b^40*d)/b^44
Time = 3.70 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {\frac {d\,{\left (b\,x^4+a\right )}^3}{11}+a^3\,d-\frac {3\,a\,d\,{\left (b\,x^4+a\right )}^2}{7}+\frac {b\,c\,{\left (b\,x^4+a\right )}^2}{7}+a^2\,d\,\left (b\,x^4+a\right )-a^2\,b\,c-\frac {2\,a\,b\,c\,\left (b\,x^4+a\right )}{3}}{b^4\,{\left (b\,x^4+a\right )}^{1/4}} \] Input:
int((x^11*(c + d*x^4))/(a + b*x^4)^(5/4),x)
Output:
((d*(a + b*x^4)^3)/11 + a^3*d - (3*a*d*(a + b*x^4)^2)/7 + (b*c*(a + b*x^4) ^2)/7 + a^2*d*(a + b*x^4) - a^2*b*c - (2*a*b*c*(a + b*x^4))/3)/(b^4*(a + b *x^4)^(1/4))
\[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {x^{15}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {x^{11}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) c \] Input:
int(x^11*(d*x^4+c)/(b*x^4+a)^(5/4),x)
Output:
int(x**15/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*d + int( x**11/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*c