Integrand size = 22, antiderivative size = 107 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{13/4}} \, dx=-\frac {c}{3 a x^3 \left (a+b x^4\right )^{9/4}}-\frac {(4 b c-a d) x}{9 a^2 \left (a+b x^4\right )^{9/4}}-\frac {8 (4 b c-a d) x}{45 a^3 \left (a+b x^4\right )^{5/4}}-\frac {32 (4 b c-a d) x}{45 a^4 \sqrt [4]{a+b x^4}} \] Output:
-1/3*c/a/x^3/(b*x^4+a)^(9/4)-1/9*(-a*d+4*b*c)*x/a^2/(b*x^4+a)^(9/4)-8/45*( -a*d+4*b*c)*x/a^3/(b*x^4+a)^(5/4)-32/45*(-a*d+4*b*c)*x/a^4/(b*x^4+a)^(1/4)
Time = 0.79 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.76 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{13/4}} \, dx=\frac {-128 b^3 c x^{12}-15 a^3 \left (c-3 d x^4\right )+32 a b^2 x^8 \left (-9 c+d x^4\right )+36 a^2 b x^4 \left (-5 c+2 d x^4\right )}{45 a^4 x^3 \left (a+b x^4\right )^{9/4}} \] Input:
Integrate[(c + d*x^4)/(x^4*(a + b*x^4)^(13/4)),x]
Output:
(-128*b^3*c*x^12 - 15*a^3*(c - 3*d*x^4) + 32*a*b^2*x^8*(-9*c + d*x^4) + 36 *a^2*b*x^4*(-5*c + 2*d*x^4))/(45*a^4*x^3*(a + b*x^4)^(9/4))
Time = 0.37 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {955, 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.
\(\displaystyle \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{13/4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(4 b c-a d) \int \frac {1}{\left (b x^4+a\right )^{13/4}}dx}{a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{9/4}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle -\frac {(4 b c-a d) \left (\frac {8 \int \frac {1}{\left (b x^4+a\right )^{9/4}}dx}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{9/4}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle -\frac {(4 b c-a d) \left (\frac {8 \left (\frac {4 \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{9/4}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle -\frac {\left (\frac {8 \left (\frac {4 x}{5 a^2 \sqrt [4]{a+b x^4}}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right ) (4 b c-a d)}{a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{9/4}}\) |
Input:
Int[(c + d*x^4)/(x^4*(a + b*x^4)^(13/4)),x]
Output:
-1/3*c/(a*x^3*(a + b*x^4)^(9/4)) - ((4*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)))/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 = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.71
method | result | size |
pseudoelliptic | \(\frac {\left (45 d \,x^{4}-15 c \right ) a^{3}-180 \left (-\frac {2 d \,x^{4}}{5}+c \right ) b \,x^{4} a^{2}-288 b^{2} x^{8} \left (-\frac {d \,x^{4}}{9}+c \right ) a -128 b^{3} c \,x^{12}}{45 \left (b \,x^{4}+a \right )^{\frac {9}{4}} x^{3} a^{4}}\) | \(76\) |
gosper | \(-\frac {-32 a \,b^{2} d \,x^{12}+128 b^{3} c \,x^{12}-72 a^{2} b d \,x^{8}+288 a \,b^{2} c \,x^{8}-45 a^{3} d \,x^{4}+180 a^{2} b c \,x^{4}+15 c \,a^{3}}{45 x^{3} \left (b \,x^{4}+a \right )^{\frac {9}{4}} a^{4}}\) | \(83\) |
trager | \(-\frac {-32 a \,b^{2} d \,x^{12}+128 b^{3} c \,x^{12}-72 a^{2} b d \,x^{8}+288 a \,b^{2} c \,x^{8}-45 a^{3} d \,x^{4}+180 a^{2} b c \,x^{4}+15 c \,a^{3}}{45 x^{3} \left (b \,x^{4}+a \right )^{\frac {9}{4}} a^{4}}\) | \(83\) |
orering | \(-\frac {-32 a \,b^{2} d \,x^{12}+128 b^{3} c \,x^{12}-72 a^{2} b d \,x^{8}+288 a \,b^{2} c \,x^{8}-45 a^{3} d \,x^{4}+180 a^{2} b c \,x^{4}+15 c \,a^{3}}{45 x^{3} \left (b \,x^{4}+a \right )^{\frac {9}{4}} a^{4}}\) | \(83\) |
risch | \(-\frac {c \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{3 a^{4} x^{3}}+\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} x \left (32 a \,b^{2} d \,x^{8}-113 c \,b^{3} x^{8}+72 a^{2} b d \,x^{4}-243 a \,b^{2} c \,x^{4}+45 a^{3} d -135 a^{2} b c \right )}{45 a^{4} \left (b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}\right )}\) | \(119\) |
Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(13/4),x,method=_RETURNVERBOSE)
Output:
1/45*((45*d*x^4-15*c)*a^3-180*(-2/5*d*x^4+c)*b*x^4*a^2-288*b^2*x^8*(-1/9*d *x^4+c)*a-128*b^3*c*x^12)/(b*x^4+a)^(9/4)/x^3/a^4
Time = 0.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.08 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{13/4}} \, dx=-\frac {{\left (32 \, {\left (4 \, b^{3} c - a b^{2} d\right )} x^{12} + 72 \, {\left (4 \, a b^{2} c - a^{2} b d\right )} x^{8} + 45 \, {\left (4 \, a^{2} b c - a^{3} d\right )} x^{4} + 15 \, a^{3} c\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{45 \, {\left (a^{4} b^{3} x^{15} + 3 \, a^{5} b^{2} x^{11} + 3 \, a^{6} b x^{7} + a^{7} x^{3}\right )}} \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(13/4),x, algorithm="fricas")
Output:
-1/45*(32*(4*b^3*c - a*b^2*d)*x^12 + 72*(4*a*b^2*c - a^2*b*d)*x^8 + 45*(4* a^2*b*c - a^3*d)*x^4 + 15*a^3*c)*(b*x^4 + a)^(3/4)/(a^4*b^3*x^15 + 3*a^5*b ^2*x^11 + 3*a^6*b*x^7 + a^7*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 898 vs. \(2 (104) = 208\).
Time = 165.83 (sec) , antiderivative size = 898, normalized size of antiderivative = 8.39 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{13/4}} \, dx=\text {Too large to display} \] Input:
integrate((d*x**4+c)/x**4/(b*x**4+a)**(13/4),x)
Output:
c*(45*a**3*b**(39/4)*(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(256*a**7*b**9*ga mma(13/4) + 768*a**6*b**10*x**4*gamma(13/4) + 768*a**5*b**11*x**8*gamma(13 /4) + 256*a**4*b**12*x**12*gamma(13/4)) + 540*a**2*b**(43/4)*x**4*(a/(b*x* *4) + 1)**(3/4)*gamma(-3/4)/(256*a**7*b**9*gamma(13/4) + 768*a**6*b**10*x* *4*gamma(13/4) + 768*a**5*b**11*x**8*gamma(13/4) + 256*a**4*b**12*x**12*ga mma(13/4)) + 864*a*b**(47/4)*x**8*(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(256 *a**7*b**9*gamma(13/4) + 768*a**6*b**10*x**4*gamma(13/4) + 768*a**5*b**11* x**8*gamma(13/4) + 256*a**4*b**12*x**12*gamma(13/4)) + 384*b**(51/4)*x**12 *(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(256*a**7*b**9*gamma(13/4) + 768*a**6 *b**10*x**4*gamma(13/4) + 768*a**5*b**11*x**8*gamma(13/4) + 256*a**4*b**12 *x**12*gamma(13/4))) + d*(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**(2 1/4)*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)*gam ma(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/...
Time = 0.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.17 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{13/4}} \, dx=\frac {{\left (5 \, b^{2} - \frac {18 \, {\left (b x^{4} + a\right )} b}{x^{4}} + \frac {45 \, {\left (b x^{4} + a\right )}^{2}}{x^{8}}\right )} d x^{9}}{45 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{3}} - \frac {1}{45} \, {\left (\frac {{\left (5 \, b^{3} - \frac {27 \, {\left (b x^{4} + a\right )} b^{2}}{x^{4}} + \frac {135 \, {\left (b x^{4} + a\right )}^{2} b}{x^{8}}\right )} x^{9}}{{\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{4}} + \frac {15 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{a^{4} x^{3}}\right )} c \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(13/4),x, algorithm="maxima")
Output:
1/45*(5*b^2 - 18*(b*x^4 + a)*b/x^4 + 45*(b*x^4 + a)^2/x^8)*d*x^9/((b*x^4 + a)^(9/4)*a^3) - 1/45*((5*b^3 - 27*(b*x^4 + a)*b^2/x^4 + 135*(b*x^4 + a)^2 *b/x^8)*x^9/((b*x^4 + a)^(9/4)*a^4) + 15*(b*x^4 + a)^(3/4)/(a^4*x^3))*c
\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} x^{4}} \,d x } \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(13/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(13/4)*x^4), x)
Time = 3.85 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.77 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{13/4}} \, dx=-\frac {-45\,d\,a^3\,x^4+15\,c\,a^3-72\,d\,a^2\,b\,x^8+180\,c\,a^2\,b\,x^4-32\,d\,a\,b^2\,x^{12}+288\,c\,a\,b^2\,x^8+128\,c\,b^3\,x^{12}}{45\,a^4\,x^3\,{\left (b\,x^4+a\right )}^{9/4}} \] Input:
int((c + d*x^4)/(x^4*(a + b*x^4)^(13/4)),x)
Output:
-(15*a^3*c - 45*a^3*d*x^4 + 128*b^3*c*x^12 + 180*a^2*b*c*x^4 + 288*a*b^2*c *x^8 - 72*a^2*b*d*x^8 - 32*a*b^2*d*x^12)/(45*a^4*x^3*(a + b*x^4)^(9/4))
\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{13/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{8}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{16}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) d \] Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(13/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*a**3*x**4 + 3*(a + b*x**4)**(1/4)*a**2*b*x**8 + 3*(a + b*x**4)**(1/4)*a*b**2*x**12 + (a + b*x**4)**(1/4)*b**3*x**16),x)*c + 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)*d