Integrand size = 22, antiderivative size = 66 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {(b c-a d) x^9}{13 a b \left (a+b x^4\right )^{13/4}}+\frac {(4 b c+9 a d) x^9}{117 a^2 b \left (a+b x^4\right )^{9/4}} \] Output:
1/13*(-a*d+b*c)*x^9/a/b/(b*x^4+a)^(13/4)+1/117*(9*a*d+4*b*c)*x^9/a^2/b/(b* x^4+a)^(9/4)
Time = 5.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.61 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {x^9 \left (13 a c+4 b c x^4+9 a d x^4\right )}{117 a^2 \left (a+b x^4\right )^{13/4}} \] Input:
Integrate[(x^8*(c + d*x^4))/(a + b*x^4)^(17/4),x]
Output:
(x^9*(13*a*c + 4*b*c*x^4 + 9*a*d*x^4))/(117*a^2*(a + b*x^4)^(13/4))
Time = 0.32 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {957, 796}
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.
\(\displaystyle \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx\) |
\(\Big \downarrow \) 957 |
\(\displaystyle \frac {(9 a d+4 b c) \int \frac {x^8}{\left (b x^4+a\right )^{13/4}}dx}{13 a b}+\frac {x^9 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {x^9 (9 a d+4 b c)}{117 a^2 b \left (a+b x^4\right )^{9/4}}+\frac {x^9 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
Input:
Int[(x^8*(c + d*x^4))/(a + b*x^4)^(17/4),x]
Output:
((b*c - a*d)*x^9)/(13*a*b*(a + b*x^4)^(13/4)) + ((4*b*c + 9*a*d)*x^9)/(117 *a^2*b*(a + b*x^4)^(9/4))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.56
method | result | size |
gosper | \(\frac {x^{9} \left (9 a d \,x^{4}+4 b c \,x^{4}+13 a c \right )}{117 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{2}}\) | \(37\) |
trager | \(\frac {x^{9} \left (9 a d \,x^{4}+4 b c \,x^{4}+13 a c \right )}{117 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{2}}\) | \(37\) |
pseudoelliptic | \(\frac {x^{9} \left (9 a d \,x^{4}+4 b c \,x^{4}+13 a c \right )}{117 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{2}}\) | \(37\) |
orering | \(\frac {x^{9} \left (9 a d \,x^{4}+4 b c \,x^{4}+13 a c \right )}{117 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{2}}\) | \(37\) |
Input:
int(x^8*(d*x^4+c)/(b*x^4+a)^(17/4),x,method=_RETURNVERBOSE)
Output:
1/117*x^9*(9*a*d*x^4+4*b*c*x^4+13*a*c)/(b*x^4+a)^(13/4)/a^2
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.20 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {{\left ({\left (4 \, b c + 9 \, a d\right )} x^{13} + 13 \, a c x^{9}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{117 \, {\left (a^{2} b^{4} x^{16} + 4 \, a^{3} b^{3} x^{12} + 6 \, a^{4} b^{2} x^{8} + 4 \, a^{5} b x^{4} + a^{6}\right )}} \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="fricas")
Output:
1/117*((4*b*c + 9*a*d)*x^13 + 13*a*c*x^9)*(b*x^4 + a)^(3/4)/(a^2*b^4*x^16 + 4*a^3*b^3*x^12 + 6*a^4*b^2*x^8 + 4*a^5*b*x^4 + a^6)
Timed out. \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\text {Timed out} \] Input:
integrate(x**8*(d*x**4+c)/(b*x**4+a)**(17/4),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=-\frac {{\left (9 \, b - \frac {13 \, {\left (b x^{4} + a\right )}}{x^{4}}\right )} c x^{13}}{117 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{2}} + \frac {d x^{13}}{13 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a} \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="maxima")
Output:
-1/117*(9*b - 13*(b*x^4 + a)/x^4)*c*x^13/((b*x^4 + a)^(13/4)*a^2) + 1/13*d *x^13/((b*x^4 + a)^(13/4)*a)
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(17/4), x)
Time = 3.57 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.36 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {x\,\left (\frac {a^2\,d-a\,b\,c}{117\,a\,b^3}+\frac {a\,\left (\frac {d}{9\,b^2}-\frac {13\,b^2\,c-13\,a\,b\,d}{117\,a\,b^3}\right )}{b}\right )}{{\left (b\,x^4+a\right )}^{9/4}}-\frac {x\,\left (\frac {d}{5\,b^3}+\frac {18\,a\,d-5\,b\,c}{585\,a\,b^3}\right )}{{\left (b\,x^4+a\right )}^{5/4}}+\frac {a\,x\,\left (\frac {c}{13\,b}-\frac {a\,d}{13\,b^2}\right )}{b\,{\left (b\,x^4+a\right )}^{13/4}}+\frac {x\,\left (9\,a\,d+4\,b\,c\right )}{117\,a^2\,b^3\,{\left (b\,x^4+a\right )}^{1/4}} \] Input:
int((x^8*(c + d*x^4))/(a + b*x^4)^(17/4),x)
Output:
(x*((a^2*d - a*b*c)/(117*a*b^3) + (a*(d/(9*b^2) - (13*b^2*c - 13*a*b*d)/(1 17*a*b^3)))/b))/(a + b*x^4)^(9/4) - (x*(d/(5*b^3) + (18*a*d - 5*b*c)/(585* a*b^3)))/(a + b*x^4)^(5/4) + (a*x*(c/(13*b) - (a*d)/(13*b^2)))/(b*(a + b*x ^4)^(13/4)) + (x*(9*a*d + 4*b*c))/(117*a^2*b^3*(a + b*x^4)^(1/4))
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\left (\int \frac {x^{12}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) d +\left (\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) c \] Input:
int(x^8*(d*x^4+c)/(b*x^4+a)^(17/4),x)
Output:
int(x**12/((a + b*x**4)**(1/4)*a**4 + 4*(a + b*x**4)**(1/4)*a**3*b*x**4 + 6*(a + b*x**4)**(1/4)*a**2*b**2*x**8 + 4*(a + b*x**4)**(1/4)*a*b**3*x**12 + (a + b*x**4)**(1/4)*b**4*x**16),x)*d + int(x**8/((a + b*x**4)**(1/4)*a** 4 + 4*(a + b*x**4)**(1/4)*a**3*b*x**4 + 6*(a + b*x**4)**(1/4)*a**2*b**2*x* *8 + 4*(a + b*x**4)**(1/4)*a*b**3*x**12 + (a + b*x**4)**(1/4)*b**4*x**16), x)*c