Integrand size = 19, antiderivative size = 91 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {(b c-a d) x}{9 a b \left (a+b x^4\right )^{9/4}}+\frac {(8 b c+a d) x}{45 a^2 b \left (a+b x^4\right )^{5/4}}+\frac {4 (8 b c+a d) x}{45 a^3 b \sqrt [4]{a+b x^4}} \] Output:
1/9*(-a*d+b*c)*x/a/b/(b*x^4+a)^(9/4)+1/45*(a*d+8*b*c)*x/a^2/b/(b*x^4+a)^(5 /4)+4/45*(a*d+8*b*c)*x/a^3/b/(b*x^4+a)^(1/4)
Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {45 a^2 c x+72 a b c x^5+9 a^2 d x^5+32 b^2 c x^9+4 a b d x^9}{45 a^3 \left (a+b x^4\right )^{9/4}} \] Input:
Integrate[(c + d*x^4)/(a + b*x^4)^(13/4),x]
Output:
(45*a^2*c*x + 72*a*b*c*x^5 + 9*a^2*d*x^5 + 32*b^2*c*x^9 + 4*a*b*d*x^9)/(45 *a^3*(a + b*x^4)^(9/4))
Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {910, 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.
\(\displaystyle \int \frac {c+d x^4}{\left (a+b x^4\right )^{13/4}} \, dx\) |
\(\Big \downarrow \) 910 |
\(\displaystyle \frac {(a d+8 b c) \int \frac {1}{\left (b x^4+a\right )^{9/4}}dx}{9 a b}+\frac {x (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {(a d+8 b c) \left (\frac {4 \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{9 a b}+\frac {x (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle \frac {\left (\frac {4 x}{5 a^2 \sqrt [4]{a+b x^4}}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right ) (a d+8 b c)}{9 a b}+\frac {x (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}\) |
Input:
Int[(c + d*x^4)/(a + b*x^4)^(13/4),x]
Output:
((b*c - a*d)*x)/(9*a*b*(a + b*x^4)^(9/4)) + ((8*b*c + a*d)*(x/(5*a*(a + b* x^4)^(5/4)) + (4*x)/(5*a^2*(a + b*x^4)^(1/4))))/(9*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 = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57
method | result | size |
pseudoelliptic | \(\frac {x \left (\left (\frac {d \,x^{4}}{5}+c \right ) a^{2}+\frac {8 b \,x^{4} \left (\frac {d \,x^{4}}{18}+c \right ) a}{5}+\frac {32 b^{2} c \,x^{8}}{45}\right )}{\left (b \,x^{4}+a \right )^{\frac {9}{4}} a^{3}}\) | \(52\) |
gosper | \(\frac {x \left (4 a b d \,x^{8}+32 b^{2} c \,x^{8}+9 a^{2} d \,x^{4}+72 a b c \,x^{4}+45 a^{2} c \right )}{45 \left (b \,x^{4}+a \right )^{\frac {9}{4}} a^{3}}\) | \(57\) |
trager | \(\frac {x \left (4 a b d \,x^{8}+32 b^{2} c \,x^{8}+9 a^{2} d \,x^{4}+72 a b c \,x^{4}+45 a^{2} c \right )}{45 \left (b \,x^{4}+a \right )^{\frac {9}{4}} a^{3}}\) | \(57\) |
orering | \(\frac {x \left (4 a b d \,x^{8}+32 b^{2} c \,x^{8}+9 a^{2} d \,x^{4}+72 a b c \,x^{4}+45 a^{2} c \right )}{45 \left (b \,x^{4}+a \right )^{\frac {9}{4}} a^{3}}\) | \(57\) |
Input:
int((d*x^4+c)/(b*x^4+a)^(13/4),x,method=_RETURNVERBOSE)
Output:
1/(b*x^4+a)^(9/4)*x*((1/5*d*x^4+c)*a^2+8/5*b*x^4*(1/18*d*x^4+c)*a+32/45*b^ 2*c*x^8)/a^3
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {{\left (4 \, {\left (8 \, b^{2} c + a b d\right )} x^{9} + 9 \, {\left (8 \, a b c + a^{2} d\right )} x^{5} + 45 \, a^{2} c x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{45 \, {\left (a^{3} b^{3} x^{12} + 3 \, a^{4} b^{2} x^{8} + 3 \, a^{5} b x^{4} + a^{6}\right )}} \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="fricas")
Output:
1/45*(4*(8*b^2*c + a*b*d)*x^9 + 9*(8*a*b*c + a^2*d)*x^5 + 45*a^2*c*x)*(b*x ^4 + a)^(3/4)/(a^3*b^3*x^12 + 3*a^4*b^2*x^8 + 3*a^5*b*x^4 + a^6)
Leaf count of result is larger than twice the leaf count of optimal. 709 vs. \(2 (83) = 166\).
Time = 86.38 (sec) , antiderivative size = 709, normalized size of antiderivative = 7.79 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{13/4}} \, dx =\text {Too large to display} \] Input:
integrate((d*x**4+c)/(b*x**4+a)**(13/4),x)
Output:
c*(45*a**5*x*gamma(1/4)/(64*a**(33/4)*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 192*a**(29/4)*b*x**4*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 192*a**(25/4)*b** 2*x**8*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 64*a**(21/4)*b**3*x**12*(1 + b* x**4/a)**(1/4)*gamma(13/4)) + 117*a**4*b*x**5*gamma(1/4)/(64*a**(33/4)*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 192*a**(29/4)*b*x**4*(1 + b*x**4/a)**(1/4 )*gamma(13/4) + 192*a**(25/4)*b**2*x**8*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 64*a**(21/4)*b**3*x**12*(1 + b*x**4/a)**(1/4)*gamma(13/4)) + 104*a**3*b* *2*x**9*gamma(1/4)/(64*a**(33/4)*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 192*a **(29/4)*b*x**4*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 192*a**(25/4)*b**2*x** 8*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 64*a**(21/4)*b**3*x**12*(1 + b*x**4/ a)**(1/4)*gamma(13/4)) + 32*a**2*b**3*x**13*gamma(1/4)/(64*a**(33/4)*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 192*a**(29/4)*b*x**4*(1 + b*x**4/a)**(1/4)* gamma(13/4) + 192*a**(25/4)*b**2*x**8*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 64*a**(21/4)*b**3*x**12*(1 + b*x**4/a)**(1/4)*gamma(13/4))) + d*(9*a*x**5* gamma(5/4)/(16*a**(17/4)*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 32*a**(13/4)* b*x**4*(1 + b*x**4/a)**(1/4)*gamma(13/4) + 16*a**(9/4)*b**2*x**8*(1 + b*x* *4/a)**(1/4)*gamma(13/4)) + 4*b*x**9*gamma(5/4)/(16*a**(17/4)*(1 + b*x**4/ a)**(1/4)*gamma(13/4) + 32*a**(13/4)*b*x**4*(1 + b*x**4/a)**(1/4)*gamma(13 /4) + 16*a**(9/4)*b**2*x**8*(1 + b*x**4/a)**(1/4)*gamma(13/4)))
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{13/4}} \, dx=-\frac {{\left (5 \, b - \frac {9 \, {\left (b x^{4} + a\right )}}{x^{4}}\right )} d x^{9}}{45 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{2}} + \frac {{\left (5 \, b^{2} - \frac {18 \, {\left (b x^{4} + a\right )} b}{x^{4}} + \frac {45 \, {\left (b x^{4} + a\right )}^{2}}{x^{8}}\right )} c x^{9}}{45 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{3}} \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="maxima")
Output:
-1/45*(5*b - 9*(b*x^4 + a)/x^4)*d*x^9/((b*x^4 + a)^(9/4)*a^2) + 1/45*(5*b^ 2 - 18*(b*x^4 + a)*b/x^4 + 45*(b*x^4 + a)^2/x^8)*c*x^9/((b*x^4 + a)^(9/4)* a^3)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {13}{4}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(13/4), x)
Time = 3.63 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {4\,a\,d\,x\,{\left (b\,x^4+a\right )}^2-5\,a^3\,d\,x+32\,b\,c\,x\,{\left (b\,x^4+a\right )}^2+a^2\,d\,x\,\left (b\,x^4+a\right )+5\,a^2\,b\,c\,x+8\,a\,b\,c\,x\,\left (b\,x^4+a\right )}{45\,a^3\,b\,{\left (b\,x^4+a\right )}^{9/4}} \] Input:
int((c + d*x^4)/(a + b*x^4)^(13/4),x)
Output:
(4*a*d*x*(a + b*x^4)^2 - 5*a^3*d*x + 32*b*c*x*(a + b*x^4)^2 + a^2*d*x*(a + b*x^4) + 5*a^2*b*c*x + 8*a*b*c*x*(a + b*x^4))/(45*a^3*b*(a + b*x^4)^(9/4) )
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{13/4}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) c \] Input:
int((d*x^4+c)/(b*x^4+a)^(13/4),x)
Output:
int(x**4/((a + b*x**4)**(1/4)*a**3 + 3*(a + b*x**4)**(1/4)*a**2*b*x**4 + 3 *(a + b*x**4)**(1/4)*a*b**2*x**8 + (a + b*x**4)**(1/4)*b**3*x**12),x)*d + int(1/((a + b*x**4)**(1/4)*a**3 + 3*(a + b*x**4)**(1/4)*a**2*b*x**4 + 3*(a + b*x**4)**(1/4)*a*b**2*x**8 + (a + b*x**4)**(1/4)*b**3*x**12),x)*c