Integrand size = 22, antiderivative size = 101 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=-\frac {a^2 (b c-a d)}{3 b^4 \left (a+b x^4\right )^{3/4}}-\frac {a (2 b c-3 a d) \sqrt [4]{a+b x^4}}{b^4}+\frac {(b c-3 a d) \left (a+b x^4\right )^{5/4}}{5 b^4}+\frac {d \left (a+b x^4\right )^{9/4}}{9 b^4} \] Output:
-1/3*a^2*(-a*d+b*c)/b^4/(b*x^4+a)^(3/4)-a*(-3*a*d+2*b*c)*(b*x^4+a)^(1/4)/b ^4+1/5*(-3*a*d+b*c)*(b*x^4+a)^(5/4)/b^4+1/9*d*(b*x^4+a)^(9/4)/b^4
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.73 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {128 a^3 d-96 a^2 b \left (c-d x^4\right )-12 a b^2 x^4 \left (6 c+d x^4\right )+b^3 x^8 \left (9 c+5 d x^4\right )}{45 b^4 \left (a+b x^4\right )^{3/4}} \] Input:
Integrate[(x^11*(c + d*x^4))/(a + b*x^4)^(7/4),x]
Output:
(128*a^3*d - 96*a^2*b*(c - d*x^4) - 12*a*b^2*x^4*(6*c + d*x^4) + b^3*x^8*( 9*c + 5*d*x^4))/(45*b^4*(a + b*x^4)^(3/4))
Time = 0.41 (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.
\(\displaystyle \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {x^8 \left (d x^4+c\right )}{\left (b x^4+a\right )^{7/4}}dx^4\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{4} \int \left (-\frac {(a d-b c) a^2}{b^3 \left (b x^4+a\right )^{7/4}}+\frac {(3 a d-2 b c) a}{b^3 \left (b x^4+a\right )^{3/4}}+\frac {d \left (b x^4+a\right )^{5/4}}{b^3}+\frac {(b c-3 a d) \sqrt [4]{b x^4+a}}{b^3}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\frac {4 a^2 (b c-a d)}{3 b^4 \left (a+b x^4\right )^{3/4}}+\frac {4 \left (a+b x^4\right )^{5/4} (b c-3 a d)}{5 b^4}-\frac {4 a \sqrt [4]{a+b x^4} (2 b c-3 a d)}{b^4}+\frac {4 d \left (a+b x^4\right )^{9/4}}{9 b^4}\right )\) |
Input:
Int[(x^11*(c + d*x^4))/(a + b*x^4)^(7/4),x]
Output:
((-4*a^2*(b*c - a*d))/(3*b^4*(a + b*x^4)^(3/4)) - (4*a*(2*b*c - 3*a*d)*(a + b*x^4)^(1/4))/b^4 + (4*(b*c - 3*a*d)*(a + b*x^4)^(5/4))/(5*b^4) + (4*d*( a + b*x^4)^(9/4))/(9*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 = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {\frac {x^{8} \left (\frac {5 d \,x^{4}}{9}+c \right ) b^{3}}{5}-\frac {8 \left (\frac {d \,x^{4}}{6}+c \right ) x^{4} a \,b^{2}}{5}-\frac {32 a^{2} \left (-d \,x^{4}+c \right ) b}{15}+\frac {128 a^{3} d}{45}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{4}}\) | \(68\) |
gosper | \(\frac {5 b^{3} d \,x^{12}-12 a \,b^{2} d \,x^{8}+9 c \,b^{3} x^{8}+96 a^{2} b d \,x^{4}-72 a \,b^{2} c \,x^{4}+128 a^{3} d -96 a^{2} b c}{45 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{4}}\) | \(77\) |
trager | \(\frac {5 b^{3} d \,x^{12}-12 a \,b^{2} d \,x^{8}+9 c \,b^{3} x^{8}+96 a^{2} b d \,x^{4}-72 a \,b^{2} c \,x^{4}+128 a^{3} d -96 a^{2} b c}{45 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{4}}\) | \(77\) |
orering | \(\frac {5 b^{3} d \,x^{12}-12 a \,b^{2} d \,x^{8}+9 c \,b^{3} x^{8}+96 a^{2} b d \,x^{4}-72 a \,b^{2} c \,x^{4}+128 a^{3} d -96 a^{2} b c}{45 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{4}}\) | \(77\) |
risch | \(\frac {\left (5 d \,b^{2} x^{8}-17 a b d \,x^{4}+9 b^{2} c \,x^{4}+113 a^{2} d -81 a b c \right ) \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{45 b^{4}}+\frac {a^{2} \left (a d -c b \right )}{3 b^{4} \left (b \,x^{4}+a \right )^{\frac {3}{4}}}\) | \(79\) |
Input:
int(x^11*(d*x^4+c)/(b*x^4+a)^(7/4),x,method=_RETURNVERBOSE)
Output:
128/45/(b*x^4+a)^(3/4)*(9/128*x^8*(5/9*d*x^4+c)*b^3-9/16*(1/6*d*x^4+c)*x^4 *a*b^2-3/4*a^2*(-d*x^4+c)*b+a^3*d)/b^4
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {{\left (5 \, b^{3} d x^{12} + 3 \, {\left (3 \, b^{3} c - 4 \, a b^{2} d\right )} x^{8} - 24 \, {\left (3 \, a b^{2} c - 4 \, a^{2} b d\right )} x^{4} - 96 \, a^{2} b c + 128 \, a^{3} d\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{45 \, {\left (b^{5} x^{4} + a b^{4}\right )}} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="fricas")
Output:
1/45*(5*b^3*d*x^12 + 3*(3*b^3*c - 4*a*b^2*d)*x^8 - 24*(3*a*b^2*c - 4*a^2*b *d)*x^4 - 96*a^2*b*c + 128*a^3*d)*(b*x^4 + a)^(1/4)/(b^5*x^4 + a*b^4)
Time = 1.82 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.70 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\begin {cases} \frac {128 a^{3} d}{45 b^{4} \left (a + b x^{4}\right )^{\frac {3}{4}}} - \frac {32 a^{2} c}{15 b^{3} \left (a + b x^{4}\right )^{\frac {3}{4}}} + \frac {32 a^{2} d x^{4}}{15 b^{3} \left (a + b x^{4}\right )^{\frac {3}{4}}} - \frac {8 a c x^{4}}{5 b^{2} \left (a + b x^{4}\right )^{\frac {3}{4}}} - \frac {4 a d x^{8}}{15 b^{2} \left (a + b x^{4}\right )^{\frac {3}{4}}} + \frac {c x^{8}}{5 b \left (a + b x^{4}\right )^{\frac {3}{4}}} + \frac {d x^{12}}{9 b \left (a + b x^{4}\right )^{\frac {3}{4}}} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{12}}{12} + \frac {d x^{16}}{16}}{a^{\frac {7}{4}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**11*(d*x**4+c)/(b*x**4+a)**(7/4),x)
Output:
Piecewise((128*a**3*d/(45*b**4*(a + b*x**4)**(3/4)) - 32*a**2*c/(15*b**3*( a + b*x**4)**(3/4)) + 32*a**2*d*x**4/(15*b**3*(a + b*x**4)**(3/4)) - 8*a*c *x**4/(5*b**2*(a + b*x**4)**(3/4)) - 4*a*d*x**8/(15*b**2*(a + b*x**4)**(3/ 4)) + c*x**8/(5*b*(a + b*x**4)**(3/4)) + d*x**12/(9*b*(a + b*x**4)**(3/4)) , Ne(b, 0)), ((c*x**12/12 + d*x**16/16)/a**(7/4), True))
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.17 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {1}{45} \, d {\left (\frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}}}{b^{4}} - \frac {27 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a}{b^{4}} + \frac {135 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}}{b^{4}} + \frac {15 \, a^{3}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{4}}\right )} + \frac {1}{15} \, c {\left (\frac {3 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}}}{b^{3}} - \frac {30 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{b^{3}} - \frac {5 \, a^{2}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{3}}\right )} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="maxima")
Output:
1/45*d*(5*(b*x^4 + a)^(9/4)/b^4 - 27*(b*x^4 + a)^(5/4)*a/b^4 + 135*(b*x^4 + a)^(1/4)*a^2/b^4 + 15*a^3/((b*x^4 + a)^(3/4)*b^4)) + 1/15*c*(3*(b*x^4 + a)^(5/4)/b^3 - 30*(b*x^4 + a)^(1/4)*a/b^3 - 5*a^2/((b*x^4 + a)^(3/4)*b^3))
Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=-\frac {a^{2} b c - a^{3} d}{3 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{4}} + \frac {9 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{33} c - 90 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b^{33} c + 5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} b^{32} d - 27 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a b^{32} d + 135 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b^{32} d}{45 \, b^{36}} \] Input:
integrate(x^11*(d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="giac")
Output:
-1/3*(a^2*b*c - a^3*d)/((b*x^4 + a)^(3/4)*b^4) + 1/45*(9*(b*x^4 + a)^(5/4) *b^33*c - 90*(b*x^4 + a)^(1/4)*a*b^33*c + 5*(b*x^4 + a)^(9/4)*b^32*d - 27* (b*x^4 + a)^(5/4)*a*b^32*d + 135*(b*x^4 + a)^(1/4)*a^2*b^32*d)/b^36
Time = 3.58 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {5\,d\,{\left (b\,x^4+a\right )}^3+15\,a^3\,d-27\,a\,d\,{\left (b\,x^4+a\right )}^2+9\,b\,c\,{\left (b\,x^4+a\right )}^2+135\,a^2\,d\,\left (b\,x^4+a\right )-15\,a^2\,b\,c-90\,a\,b\,c\,\left (b\,x^4+a\right )}{45\,b^4\,{\left (b\,x^4+a\right )}^{3/4}} \] Input:
int((x^11*(c + d*x^4))/(a + b*x^4)^(7/4),x)
Output:
(5*d*(a + b*x^4)^3 + 15*a^3*d - 27*a*d*(a + b*x^4)^2 + 9*b*c*(a + b*x^4)^2 + 135*a^2*d*(a + b*x^4) - 15*a^2*b*c - 90*a*b*c*(a + b*x^4))/(45*b^4*(a + b*x^4)^(3/4))
\[ \int \frac {x^{11} \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/4}} \, dx=\left (\int \frac {x^{15}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {x^{11}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) c \] Input:
int(x^11*(d*x^4+c)/(b*x^4+a)^(7/4),x)
Output:
int(x**15/((a + b*x**4)**(3/4)*a + (a + b*x**4)**(3/4)*b*x**4),x)*d + int( x**11/((a + b*x**4)**(3/4)*a + (a + b*x**4)**(3/4)*b*x**4),x)*c