Integrand size = 22, antiderivative size = 79 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{9/4}} \, dx=-\frac {c}{3 a x^3 \left (a+b x^4\right )^{5/4}}-\frac {(8 b c-3 a d) x}{15 a^2 \left (a+b x^4\right )^{5/4}}-\frac {4 (8 b c-3 a d) x}{15 a^3 \sqrt [4]{a+b x^4}} \] Output:
-1/3*c/a/x^3/(b*x^4+a)^(5/4)-1/15*(-3*a*d+8*b*c)*x/a^2/(b*x^4+a)^(5/4)-4/1 5*(-3*a*d+8*b*c)*x/a^3/(b*x^4+a)^(1/4)
Time = 0.55 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{9/4}} \, dx=\frac {-5 a^2 c-40 a b c x^4+15 a^2 d x^4-32 b^2 c x^8+12 a b d x^8}{15 a^3 x^3 \left (a+b x^4\right )^{5/4}} \] Input:
Integrate[(c + d*x^4)/(x^4*(a + b*x^4)^(9/4)),x]
Output:
(-5*a^2*c - 40*a*b*c*x^4 + 15*a^2*d*x^4 - 32*b^2*c*x^8 + 12*a*b*d*x^8)/(15 *a^3*x^3*(a + b*x^4)^(5/4))
Time = 0.33 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {955, 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.
\(\displaystyle \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{9/4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(8 b c-3 a d) \int \frac {1}{\left (b x^4+a\right )^{9/4}}dx}{3 a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{5/4}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle -\frac {(8 b c-3 a d) \left (\frac {4 \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{3 a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{5/4}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle -\frac {\left (\frac {4 x}{5 a^2 \sqrt [4]{a+b x^4}}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right ) (8 b c-3 a d)}{3 a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{5/4}}\) |
Input:
Int[(c + d*x^4)/(x^4*(a + b*x^4)^(9/4)),x]
Output:
-1/3*c/(a*x^3*(a + b*x^4)^(5/4)) - ((8*b*c - 3*a*d)*(x/(5*a*(a + b*x^4)^(5 /4)) + (4*x)/(5*a^2*(a + b*x^4)^(1/4))))/(3*a)
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {\left (15 d \,x^{4}-5 c \right ) a^{2}-40 \left (-\frac {3 d \,x^{4}}{10}+c \right ) b \,x^{4} a -32 b^{2} c \,x^{8}}{15 \left (b \,x^{4}+a \right )^{\frac {5}{4}} x^{3} a^{3}}\) | \(57\) |
gosper | \(-\frac {-12 a b d \,x^{8}+32 b^{2} c \,x^{8}-15 a^{2} d \,x^{4}+40 a b c \,x^{4}+5 a^{2} c}{15 x^{3} \left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{3}}\) | \(59\) |
trager | \(-\frac {-12 a b d \,x^{8}+32 b^{2} c \,x^{8}-15 a^{2} d \,x^{4}+40 a b c \,x^{4}+5 a^{2} c}{15 x^{3} \left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{3}}\) | \(59\) |
orering | \(-\frac {-12 a b d \,x^{8}+32 b^{2} c \,x^{8}-15 a^{2} d \,x^{4}+40 a b c \,x^{4}+5 a^{2} c}{15 x^{3} \left (b \,x^{4}+a \right )^{\frac {5}{4}} a^{3}}\) | \(59\) |
risch | \(-\frac {c \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{3 a^{3} x^{3}}+\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} x \left (4 a b d \,x^{4}-9 b^{2} c \,x^{4}+5 a^{2} d -10 a b c \right )}{5 a^{3} \left (x^{8} b^{2}+2 a \,x^{4} b +a^{2}\right )}\) | \(84\) |
Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(9/4),x,method=_RETURNVERBOSE)
Output:
1/15*((15*d*x^4-5*c)*a^2-40*(-3/10*d*x^4+c)*b*x^4*a-32*b^2*c*x^8)/(b*x^4+a )^(5/4)/x^3/a^3
Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{9/4}} \, dx=-\frac {{\left (4 \, {\left (8 \, b^{2} c - 3 \, a b d\right )} x^{8} + 5 \, {\left (8 \, a b c - 3 \, a^{2} d\right )} x^{4} + 5 \, a^{2} c\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{15 \, {\left (a^{3} b^{2} x^{11} + 2 \, a^{4} b x^{7} + a^{5} x^{3}\right )}} \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(9/4),x, algorithm="fricas")
Output:
-1/15*(4*(8*b^2*c - 3*a*b*d)*x^8 + 5*(8*a*b*c - 3*a^2*d)*x^4 + 5*a^2*c)*(b *x^4 + a)^(3/4)/(a^3*b^2*x^11 + 2*a^4*b*x^7 + a^5*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (75) = 150\).
Time = 39.94 (sec) , antiderivative size = 360, normalized size of antiderivative = 4.56 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{9/4}} \, dx=c \left (\frac {5 a^{2} b^{\frac {19}{4}} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{64 a^{5} b^{4} \Gamma \left (\frac {9}{4}\right ) + 128 a^{4} b^{5} x^{4} \Gamma \left (\frac {9}{4}\right ) + 64 a^{3} b^{6} x^{8} \Gamma \left (\frac {9}{4}\right )} + \frac {40 a b^{\frac {23}{4}} x^{4} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{64 a^{5} b^{4} \Gamma \left (\frac {9}{4}\right ) + 128 a^{4} b^{5} x^{4} \Gamma \left (\frac {9}{4}\right ) + 64 a^{3} b^{6} x^{8} \Gamma \left (\frac {9}{4}\right )} + \frac {32 b^{\frac {27}{4}} x^{8} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{64 a^{5} b^{4} \Gamma \left (\frac {9}{4}\right ) + 128 a^{4} b^{5} x^{4} \Gamma \left (\frac {9}{4}\right ) + 64 a^{3} b^{6} x^{8} \Gamma \left (\frac {9}{4}\right )}\right ) + d \left (\frac {5 a x \Gamma \left (\frac {1}{4}\right )}{16 a^{\frac {13}{4}} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right ) + 16 a^{\frac {9}{4}} b x^{4} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right )} + \frac {4 b x^{5} \Gamma \left (\frac {1}{4}\right )}{16 a^{\frac {13}{4}} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right ) + 16 a^{\frac {9}{4}} b x^{4} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right )}\right ) \] Input:
integrate((d*x**4+c)/x**4/(b*x**4+a)**(9/4),x)
Output:
c*(5*a**2*b**(19/4)*(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(64*a**5*b**4*gamm a(9/4) + 128*a**4*b**5*x**4*gamma(9/4) + 64*a**3*b**6*x**8*gamma(9/4)) + 4 0*a*b**(23/4)*x**4*(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(64*a**5*b**4*gamma (9/4) + 128*a**4*b**5*x**4*gamma(9/4) + 64*a**3*b**6*x**8*gamma(9/4)) + 32 *b**(27/4)*x**8*(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(64*a**5*b**4*gamma(9/ 4) + 128*a**4*b**5*x**4*gamma(9/4) + 64*a**3*b**6*x**8*gamma(9/4))) + d*(5 *a*x*gamma(1/4)/(16*a**(13/4)*(1 + b*x**4/a)**(1/4)*gamma(9/4) + 16*a**(9/ 4)*b*x**4*(1 + b*x**4/a)**(1/4)*gamma(9/4)) + 4*b*x**5*gamma(1/4)/(16*a**( 13/4)*(1 + b*x**4/a)**(1/4)*gamma(9/4) + 16*a**(9/4)*b*x**4*(1 + b*x**4/a) **(1/4)*gamma(9/4)))
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.11 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{9/4}} \, dx=-\frac {{\left (b - \frac {5 \, {\left (b x^{4} + a\right )}}{x^{4}}\right )} d x^{5}}{5 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2}} + \frac {1}{15} \, {\left (\frac {3 \, {\left (b^{2} - \frac {10 \, {\left (b x^{4} + a\right )} b}{x^{4}}\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{3}} - \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{a^{3} x^{3}}\right )} c \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(9/4),x, algorithm="maxima")
Output:
-1/5*(b - 5*(b*x^4 + a)/x^4)*d*x^5/((b*x^4 + a)^(5/4)*a^2) + 1/15*(3*(b^2 - 10*(b*x^4 + a)*b/x^4)*x^5/((b*x^4 + a)^(5/4)*a^3) - 5*(b*x^4 + a)^(3/4)/ (a^3*x^3))*c
\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{9/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {9}{4}} x^{4}} \,d x } \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(9/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(9/4)*x^4), x)
Time = 3.52 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{9/4}} \, dx=\frac {3\,a^2\,c-32\,c\,{\left (b\,x^4+a\right )}^2+3\,a^2\,d\,x^4+24\,a\,c\,\left (b\,x^4+a\right )+12\,a\,d\,x^4\,\left (b\,x^4+a\right )}{15\,a^3\,x^3\,{\left (b\,x^4+a\right )}^{5/4}} \] Input:
int((c + d*x^4)/(x^4*(a + b*x^4)^(9/4)),x)
Output:
(3*a^2*c - 32*c*(a + b*x^4)^2 + 3*a^2*d*x^4 + 24*a*c*(a + b*x^4) + 12*a*d* x^4*(a + b*x^4))/(15*a^3*x^3*(a + b*x^4)^(5/4))
\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{9/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} x^{4}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{12}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{8}}d x \right ) d \] Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(9/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*a**2*x**4 + 2*(a + b*x**4)**(1/4)*a*b*x**8 + (a + b*x**4)**(1/4)*b**2*x**12),x)*c + int(1/((a + b*x**4)**(1/4)*a**2 + 2*( a + b*x**4)**(1/4)*a*b*x**4 + (a + b*x**4)**(1/4)*b**2*x**8),x)*d