Integrand size = 19, antiderivative size = 121 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {(b c-a d) x}{13 a b \left (a+b x^4\right )^{13/4}}+\frac {(12 b c+a d) x}{117 a^2 b \left (a+b x^4\right )^{9/4}}+\frac {8 (12 b c+a d) x}{585 a^3 b \left (a+b x^4\right )^{5/4}}+\frac {32 (12 b c+a d) x}{585 a^4 b \sqrt [4]{a+b x^4}} \] Output:
1/13*(-a*d+b*c)*x/a/b/(b*x^4+a)^(13/4)+1/117*(a*d+12*b*c)*x/a^2/b/(b*x^4+a )^(9/4)+8/585*(a*d+12*b*c)*x/a^3/b/(b*x^4+a)^(5/4)+32/585*(a*d+12*b*c)*x/a ^4/b/(b*x^4+a)^(1/4)
Time = 2.36 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {384 b^3 c x^{13}+32 a b^2 x^9 \left (39 c+d x^4\right )+52 a^2 b x^5 \left (27 c+2 d x^4\right )+117 a^3 \left (5 c x+d x^5\right )}{585 a^4 \left (a+b x^4\right )^{13/4}} \] Input:
Integrate[(c + d*x^4)/(a + b*x^4)^(17/4),x]
Output:
(384*b^3*c*x^13 + 32*a*b^2*x^9*(39*c + d*x^4) + 52*a^2*b*x^5*(27*c + 2*d*x ^4) + 117*a^3*(5*c*x + d*x^5))/(585*a^4*(a + b*x^4)^(13/4))
Time = 0.38 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {910, 749, 749, 746}
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.
\(\displaystyle \int \frac {c+d x^4}{\left (a+b x^4\right )^{17/4}} \, dx\) |
\(\Big \downarrow \) 910 |
\(\displaystyle \frac {(a d+12 b c) \int \frac {1}{\left (b x^4+a\right )^{13/4}}dx}{13 a b}+\frac {x (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {(a d+12 b c) \left (\frac {8 \int \frac {1}{\left (b x^4+a\right )^{9/4}}dx}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {(a d+12 b c) \left (\frac {8 \left (\frac {4 \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle \frac {\left (\frac {8 \left (\frac {4 x}{5 a^2 \sqrt [4]{a+b x^4}}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right ) (a d+12 b c)}{13 a b}+\frac {x (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\) |
Input:
Int[(c + d*x^4)/(a + b*x^4)^(17/4),x]
Output:
((b*c - a*d)*x)/(13*a*b*(a + b*x^4)^(13/4)) + ((12*b*c + a*d)*(x/(9*a*(a + b*x^4)^(9/4)) + (8*(x/(5*a*(a + b*x^4)^(5/4)) + (4*x)/(5*a^2*(a + b*x^4)^ (1/4))))/(9*a)))/(13*a*b)
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1) /a), x] /; FreeQ[{a, b, n, p}, x] && EqQ[1/n + p + 1, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)) Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ n + p, 0])
Time = 0.90 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.59
method | result | size |
pseudoelliptic | \(\frac {\left (\left (\frac {d \,x^{4}}{5}+c \right ) a^{3}+\frac {12 b \left (\frac {2 d \,x^{4}}{27}+c \right ) x^{4} a^{2}}{5}+\frac {32 \left (\frac {d \,x^{4}}{39}+c \right ) b^{2} x^{8} a}{15}+\frac {128 b^{3} c \,x^{12}}{195}\right ) x}{\left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{4}}\) | \(71\) |
gosper | \(\frac {x \left (32 a \,b^{2} d \,x^{12}+384 b^{3} c \,x^{12}+104 a^{2} b d \,x^{8}+1248 a \,b^{2} c \,x^{8}+117 a^{3} d \,x^{4}+1404 a^{2} b c \,x^{4}+585 a^{3} c \right )}{585 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{4}}\) | \(81\) |
trager | \(\frac {x \left (32 a \,b^{2} d \,x^{12}+384 b^{3} c \,x^{12}+104 a^{2} b d \,x^{8}+1248 a \,b^{2} c \,x^{8}+117 a^{3} d \,x^{4}+1404 a^{2} b c \,x^{4}+585 a^{3} c \right )}{585 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{4}}\) | \(81\) |
orering | \(\frac {x \left (32 a \,b^{2} d \,x^{12}+384 b^{3} c \,x^{12}+104 a^{2} b d \,x^{8}+1248 a \,b^{2} c \,x^{8}+117 a^{3} d \,x^{4}+1404 a^{2} b c \,x^{4}+585 a^{3} c \right )}{585 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{4}}\) | \(81\) |
Input:
int((d*x^4+c)/(b*x^4+a)^(17/4),x,method=_RETURNVERBOSE)
Output:
((1/5*d*x^4+c)*a^3+12/5*b*(2/27*d*x^4+c)*x^4*a^2+32/15*(1/39*d*x^4+c)*b^2* x^8*a+128/195*b^3*c*x^12)/(b*x^4+a)^(13/4)*x/a^4
Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {{\left (32 \, {\left (12 \, b^{3} c + a b^{2} d\right )} x^{13} + 104 \, {\left (12 \, a b^{2} c + a^{2} b d\right )} x^{9} + 117 \, {\left (12 \, a^{2} b c + a^{3} d\right )} x^{5} + 585 \, a^{3} c x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{585 \, {\left (a^{4} b^{4} x^{16} + 4 \, a^{5} b^{3} x^{12} + 6 \, a^{6} b^{2} x^{8} + 4 \, a^{7} b x^{4} + a^{8}\right )}} \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="fricas")
Output:
1/585*(32*(12*b^3*c + a*b^2*d)*x^13 + 104*(12*a*b^2*c + a^2*b*d)*x^9 + 117 *(12*a^2*b*c + a^3*d)*x^5 + 585*a^3*c*x)*(b*x^4 + a)^(3/4)/(a^4*b^4*x^16 + 4*a^5*b^3*x^12 + 6*a^6*b^2*x^8 + 4*a^7*b*x^4 + a^8)
Timed out. \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{17/4}} \, dx=\text {Timed out} \] Input:
integrate((d*x**4+c)/(b*x**4+a)**(17/4),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {{\left (45 \, b^{2} - \frac {130 \, {\left (b x^{4} + a\right )} b}{x^{4}} + \frac {117 \, {\left (b x^{4} + a\right )}^{2}}{x^{8}}\right )} d x^{13}}{585 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{3}} - \frac {{\left (15 \, b^{3} - \frac {65 \, {\left (b x^{4} + a\right )} b^{2}}{x^{4}} + \frac {117 \, {\left (b x^{4} + a\right )}^{2} b}{x^{8}} - \frac {195 \, {\left (b x^{4} + a\right )}^{3}}{x^{12}}\right )} c x^{13}}{195 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{4}} \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="maxima")
Output:
1/585*(45*b^2 - 130*(b*x^4 + a)*b/x^4 + 117*(b*x^4 + a)^2/x^8)*d*x^13/((b* x^4 + a)^(13/4)*a^3) - 1/195*(15*b^3 - 65*(b*x^4 + a)*b^2/x^4 + 117*(b*x^4 + a)^2*b/x^8 - 195*(b*x^4 + a)^3/x^12)*c*x^13/((b*x^4 + a)^(13/4)*a^4)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(17/4), x)
Time = 0.68 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.87 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {x\,\left (\frac {c}{13\,a}-\frac {d}{13\,b}\right )}{{\left (b\,x^4+a\right )}^{13/4}}+\frac {x\,\left (a\,d+12\,b\,c\right )}{117\,a^2\,b\,{\left (b\,x^4+a\right )}^{9/4}}+\frac {x\,\left (8\,a\,d+96\,b\,c\right )}{585\,a^3\,b\,{\left (b\,x^4+a\right )}^{5/4}}+\frac {x\,\left (32\,a\,d+384\,b\,c\right )}{585\,a^4\,b\,{\left (b\,x^4+a\right )}^{1/4}} \] Input:
int((c + d*x^4)/(a + b*x^4)^(17/4),x)
Output:
(x*(c/(13*a) - d/(13*b)))/(a + b*x^4)^(13/4) + (x*(a*d + 12*b*c))/(117*a^2 *b*(a + b*x^4)^(9/4)) + (x*(8*a*d + 96*b*c))/(585*a^3*b*(a + b*x^4)^(5/4)) + (x*(32*a*d + 384*b*c))/(585*a^4*b*(a + b*x^4)^(1/4))
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{17/4}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) c \] Input:
int((d*x^4+c)/(b*x^4+a)^(17/4),x)
Output:
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)*d + int(1/((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