Integrand size = 22, antiderivative size = 99 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {(b c-a d) x^5}{13 a b \left (a+b x^4\right )^{13/4}}+\frac {(8 b c+5 a d) x^5}{117 a^2 b \left (a+b x^4\right )^{9/4}}+\frac {4 (8 b c+5 a d) x^5}{585 a^3 b \left (a+b x^4\right )^{5/4}} \] Output:
1/13*(-a*d+b*c)*x^5/a/b/(b*x^4+a)^(13/4)+1/117*(5*a*d+8*b*c)*x^5/a^2/b/(b* x^4+a)^(9/4)+4/585*(5*a*d+8*b*c)*x^5/a^3/b/(b*x^4+a)^(5/4)
Time = 2.96 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.63 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {x^5 \left (117 a^2 c+104 a b c x^4+65 a^2 d x^4+32 b^2 c x^8+20 a b d x^8\right )}{585 a^3 \left (a+b x^4\right )^{13/4}} \] Input:
Integrate[(x^4*(c + d*x^4))/(a + b*x^4)^(17/4),x]
Output:
(x^5*(117*a^2*c + 104*a*b*c*x^4 + 65*a^2*d*x^4 + 32*b^2*c*x^8 + 20*a*b*d*x ^8))/(585*a^3*(a + b*x^4)^(13/4))
Time = 0.37 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {957, 803, 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.
\(\displaystyle \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx\) |
\(\Big \downarrow \) 957 |
\(\displaystyle \frac {(5 a d+8 b c) \int \frac {x^4}{\left (b x^4+a\right )^{13/4}}dx}{13 a b}+\frac {x^5 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {(5 a d+8 b c) \left (\frac {4 b \int \frac {x^8}{\left (b x^4+a\right )^{13/4}}dx}{5 a}+\frac {x^5}{5 a \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^5 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {\left (\frac {4 b x^9}{45 a^2 \left (a+b x^4\right )^{9/4}}+\frac {x^5}{5 a \left (a+b x^4\right )^{9/4}}\right ) (5 a d+8 b c)}{13 a b}+\frac {x^5 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
Input:
Int[(x^4*(c + d*x^4))/(a + b*x^4)^(17/4),x]
Output:
((b*c - a*d)*x^5)/(13*a*b*(a + b*x^4)^(13/4)) + ((8*b*c + 5*a*d)*(x^5/(5*a *(a + b*x^4)^(9/4)) + (4*b*x^9)/(45*a^2*(a + b*x^4)^(9/4))))/(13*a*b)
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[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 ))) Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I LtQ[Simplify[(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 = 0.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.56
method | result | size |
pseudoelliptic | \(\frac {\left (\left (\frac {5 d \,x^{4}}{9}+c \right ) a^{2}+\frac {8 b \left (\frac {5 d \,x^{4}}{26}+c \right ) x^{4} a}{9}+\frac {32 b^{2} c \,x^{8}}{117}\right ) x^{5}}{5 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{3}}\) | \(55\) |
gosper | \(\frac {x^{5} \left (20 a b d \,x^{8}+32 b^{2} c \,x^{8}+65 a^{2} d \,x^{4}+104 a b c \,x^{4}+117 a^{2} c \right )}{585 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{3}}\) | \(59\) |
trager | \(\frac {x^{5} \left (20 a b d \,x^{8}+32 b^{2} c \,x^{8}+65 a^{2} d \,x^{4}+104 a b c \,x^{4}+117 a^{2} c \right )}{585 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{3}}\) | \(59\) |
orering | \(\frac {x^{5} \left (20 a b d \,x^{8}+32 b^{2} c \,x^{8}+65 a^{2} d \,x^{4}+104 a b c \,x^{4}+117 a^{2} c \right )}{585 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{3}}\) | \(59\) |
Input:
int(x^4*(d*x^4+c)/(b*x^4+a)^(17/4),x,method=_RETURNVERBOSE)
Output:
1/5*((5/9*d*x^4+c)*a^2+8/9*b*(5/26*d*x^4+c)*x^4*a+32/117*b^2*c*x^8)/(b*x^4 +a)^(13/4)*x^5/a^3
Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.03 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {{\left (4 \, {\left (8 \, b^{2} c + 5 \, a b d\right )} x^{13} + 13 \, {\left (8 \, a b c + 5 \, a^{2} d\right )} x^{9} + 117 \, a^{2} c x^{5}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{585 \, {\left (a^{3} b^{4} x^{16} + 4 \, a^{4} b^{3} x^{12} + 6 \, a^{5} b^{2} x^{8} + 4 \, a^{6} b x^{4} + a^{7}\right )}} \] Input:
integrate(x^4*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="fricas")
Output:
1/585*(4*(8*b^2*c + 5*a*b*d)*x^13 + 13*(8*a*b*c + 5*a^2*d)*x^9 + 117*a^2*c *x^5)*(b*x^4 + a)^(3/4)/(a^3*b^4*x^16 + 4*a^4*b^3*x^12 + 6*a^5*b^2*x^8 + 4 *a^6*b*x^4 + a^7)
Timed out. \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\text {Timed out} \] Input:
integrate(x**4*(d*x**4+c)/(b*x**4+a)**(17/4),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.87 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=-\frac {{\left (9 \, b - \frac {13 \, {\left (b x^{4} + a\right )}}{x^{4}}\right )} d x^{13}}{117 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{2}} + \frac {{\left (45 \, b^{2} - \frac {130 \, {\left (b x^{4} + a\right )} b}{x^{4}} + \frac {117 \, {\left (b x^{4} + a\right )}^{2}}{x^{8}}\right )} c x^{13}}{585 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{3}} \] Input:
integrate(x^4*(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)*d*x^13/((b*x^4 + a)^(13/4)*a^2) + 1/585* (45*b^2 - 130*(b*x^4 + a)*b/x^4 + 117*(b*x^4 + a)^2/x^8)*c*x^13/((b*x^4 + a)^(13/4)*a^3)
\[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{4}}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:
integrate(x^4*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)*x^4/(b*x^4 + a)^(17/4), x)
Time = 3.42 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {x\,\left (5\,a\,d+8\,b\,c\right )}{585\,a^2\,b^2\,{\left (b\,x^4+a\right )}^{5/4}}-\frac {x\,\left (\frac {d}{9\,b^2}+\frac {a\,d-b\,c}{117\,a\,b^2}\right )}{{\left (b\,x^4+a\right )}^{9/4}}-\frac {x\,\left (\frac {c}{13\,b}-\frac {a\,d}{13\,b^2}\right )}{{\left (b\,x^4+a\right )}^{13/4}}+\frac {x\,\left (20\,a\,d+32\,b\,c\right )}{585\,a^3\,b^2\,{\left (b\,x^4+a\right )}^{1/4}} \] Input:
int((x^4*(c + d*x^4))/(a + b*x^4)^(17/4),x)
Output:
(x*(5*a*d + 8*b*c))/(585*a^2*b^2*(a + b*x^4)^(5/4)) - (x*(d/(9*b^2) + (a*d - b*c)/(117*a*b^2)))/(a + b*x^4)^(9/4) - (x*(c/(13*b) - (a*d)/(13*b^2)))/ (a + b*x^4)^(13/4) + (x*(20*a*d + 32*b*c))/(585*a^3*b^2*(a + b*x^4)^(1/4))
\[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\left (\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) d +\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) c \] Input:
int(x^4*(d*x^4+c)/(b*x^4+a)^(17/4),x)
Output:
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)*d + int(x**4/((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