Integrand size = 22, antiderivative size = 135 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{17/4}} \, dx=-\frac {c}{3 a x^3 \left (a+b x^4\right )^{13/4}}-\frac {(16 b c-3 a d) x}{39 a^2 \left (a+b x^4\right )^{13/4}}-\frac {4 (16 b c-3 a d) x}{117 a^3 \left (a+b x^4\right )^{9/4}}-\frac {32 (16 b c-3 a d) x}{585 a^4 \left (a+b x^4\right )^{5/4}}-\frac {128 (16 b c-3 a d) x}{585 a^5 \sqrt [4]{a+b x^4}} \] Output:
-1/3*c/a/x^3/(b*x^4+a)^(13/4)-1/39*(-3*a*d+16*b*c)*x/a^2/(b*x^4+a)^(13/4)- 4/117*(-3*a*d+16*b*c)*x/a^3/(b*x^4+a)^(9/4)-32/585*(-3*a*d+16*b*c)*x/a^4/( b*x^4+a)^(5/4)-128/585*(-3*a*d+16*b*c)*x/a^5/(b*x^4+a)^(1/4)
Time = 1.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.76 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{17/4}} \, dx=\frac {-2048 b^4 c x^{16}-195 a^4 \left (c-3 d x^4\right )+1248 a^2 b^2 x^8 \left (-6 c+d x^4\right )+128 a b^3 x^{12} \left (-52 c+3 d x^4\right )+156 a^3 b x^4 \left (-20 c+9 d x^4\right )}{585 a^5 x^3 \left (a+b x^4\right )^{13/4}} \] Input:
Integrate[(c + d*x^4)/(x^4*(a + b*x^4)^(17/4)),x]
Output:
(-2048*b^4*c*x^16 - 195*a^4*(c - 3*d*x^4) + 1248*a^2*b^2*x^8*(-6*c + d*x^4 ) + 128*a*b^3*x^12*(-52*c + 3*d*x^4) + 156*a^3*b*x^4*(-20*c + 9*d*x^4))/(5 85*a^5*x^3*(a + b*x^4)^(13/4))
Time = 0.42 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {955, 749, 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.
| \(\displaystyle \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{17/4}} \, dx\) |
| \(\Big \downarrow \) 955 |
| \(\displaystyle -\frac {(16 b c-3 a d) \int \frac {1}{\left (b x^4+a\right )^{17/4}}dx}{3 a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{13/4}}\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle -\frac {(16 b c-3 a d) \left (\frac {12 \int \frac {1}{\left (b x^4+a\right )^{13/4}}dx}{13 a}+\frac {x}{13 a \left (a+b x^4\right )^{13/4}}\right )}{3 a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{13/4}}\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle -\frac {(16 b c-3 a d) \left (\frac {12 \left (\frac {8 \int \frac {1}{\left (b x^4+a\right )^{9/4}}dx}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x}{13 a \left (a+b x^4\right )^{13/4}}\right )}{3 a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{13/4}}\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle -\frac {(16 b c-3 a d) \left (\frac {12 \left (\frac {8 \left (\frac {4 \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x}{13 a \left (a+b x^4\right )^{13/4}}\right )}{3 a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{13/4}}\) |
| \(\Big \downarrow \) 746 |
| \(\displaystyle -\frac {\left (\frac {12 \left (\frac {8 \left (\frac {4 x}{5 a^2 \sqrt [4]{a+b x^4}}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x}{13 a \left (a+b x^4\right )^{13/4}}\right ) (16 b c-3 a d)}{3 a}-\frac {c}{3 a x^3 \left (a+b x^4\right )^{13/4}}\) |
Input:
Int[(c + d*x^4)/(x^4*(a + b*x^4)^(17/4)),x]
Output:
-1/3*c/(a*x^3*(a + b*x^4)^(13/4)) - ((16*b*c - 3*a*d)*(x/(13*a*(a + b*x^4) ^(13/4)) + (12*(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)))/(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 = 0.45 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.70
| method | result | size |
| pseudoelliptic | \(\frac {\left (585 d \,x^{4}-195 c \right ) a^{4}-3120 \left (-\frac {9 d \,x^{4}}{20}+c \right ) b \,x^{4} a^{3}-7488 \left (-\frac {d \,x^{4}}{6}+c \right ) b^{2} x^{8} a^{2}-6656 \left (-\frac {3 d \,x^{4}}{52}+c \right ) b^{3} x^{12} a -2048 b^{4} c \,x^{16}}{585 \left (b \,x^{4}+a \right )^{\frac {13}{4}} x^{3} a^{5}}\) | \(95\) |
| gosper | \(-\frac {-384 a \,b^{3} d \,x^{16}+2048 b^{4} c \,x^{16}-1248 a^{2} b^{2} d \,x^{12}+6656 a \,b^{3} c \,x^{12}-1404 a^{3} b d \,x^{8}+7488 a^{2} b^{2} c \,x^{8}-585 a^{4} d \,x^{4}+3120 a^{3} b c \,x^{4}+195 c \,a^{4}}{585 x^{3} \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{5}}\) | \(107\) |
| trager | \(-\frac {-384 a \,b^{3} d \,x^{16}+2048 b^{4} c \,x^{16}-1248 a^{2} b^{2} d \,x^{12}+6656 a \,b^{3} c \,x^{12}-1404 a^{3} b d \,x^{8}+7488 a^{2} b^{2} c \,x^{8}-585 a^{4} d \,x^{4}+3120 a^{3} b c \,x^{4}+195 c \,a^{4}}{585 x^{3} \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{5}}\) | \(107\) |
| orering | \(-\frac {-384 a \,b^{3} d \,x^{16}+2048 b^{4} c \,x^{16}-1248 a^{2} b^{2} d \,x^{12}+6656 a \,b^{3} c \,x^{12}-1404 a^{3} b d \,x^{8}+7488 a^{2} b^{2} c \,x^{8}-585 a^{4} d \,x^{4}+3120 a^{3} b c \,x^{4}+195 c \,a^{4}}{585 x^{3} \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{5}}\) | \(107\) |
| risch | \(-\frac {c \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{3 a^{5} x^{3}}+\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} x \left (384 a \,b^{3} d \,x^{12}-1853 c \,b^{4} x^{12}+1248 a^{2} b^{2} d \,x^{8}-5876 a \,b^{3} c \,x^{8}+1404 a^{3} b d \,x^{4}-6318 a^{2} b^{2} c \,x^{4}+585 a^{4} d -2340 a^{3} b c \right )}{585 a^{5} \left (b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} x^{8} b^{2}+4 x^{4} a^{3} b +a^{4}\right )}\) | \(154\) |
Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(17/4),x,method=_RETURNVERBOSE)
Output:
1/585*((585*d*x^4-195*c)*a^4-3120*(-9/20*d*x^4+c)*b*x^4*a^3-7488*(-1/6*d*x ^4+c)*b^2*x^8*a^2-6656*(-3/52*d*x^4+c)*b^3*x^12*a-2048*b^4*c*x^16)/(b*x^4+ a)^(13/4)/x^3/a^5
Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.12 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{17/4}} \, dx=-\frac {{\left (128 \, {\left (16 \, b^{4} c - 3 \, a b^{3} d\right )} x^{16} + 416 \, {\left (16 \, a b^{3} c - 3 \, a^{2} b^{2} d\right )} x^{12} + 468 \, {\left (16 \, a^{2} b^{2} c - 3 \, a^{3} b d\right )} x^{8} + 195 \, a^{4} c + 195 \, {\left (16 \, a^{3} b c - 3 \, a^{4} d\right )} x^{4}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{585 \, {\left (a^{5} b^{4} x^{19} + 4 \, a^{6} b^{3} x^{15} + 6 \, a^{7} b^{2} x^{11} + 4 \, a^{8} b x^{7} + a^{9} x^{3}\right )}} \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(17/4),x, algorithm="fricas")
Output:
-1/585*(128*(16*b^4*c - 3*a*b^3*d)*x^16 + 416*(16*a*b^3*c - 3*a^2*b^2*d)*x ^12 + 468*(16*a^2*b^2*c - 3*a^3*b*d)*x^8 + 195*a^4*c + 195*(16*a^3*b*c - 3 *a^4*d)*x^4)*(b*x^4 + a)^(3/4)/(a^5*b^4*x^19 + 4*a^6*b^3*x^15 + 6*a^7*b^2* x^11 + 4*a^8*b*x^7 + a^9*x^3)
Timed out. \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{17/4}} \, dx=\text {Timed out} \] Input:
integrate((d*x**4+c)/x**4/(b*x**4+a)**(17/4),x)
Output:
Timed out
Time = 0.06 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.18 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{17/4}} \, dx=-\frac {{\left (15 \, b^{3} - \frac {65 \, {\left (b x^{4} + a\right )} b^{2}}{x^{4}} + \frac {117 \, {\left (b x^{4} + a\right )}^{2} b}{x^{8}} - \frac {195 \, {\left (b x^{4} + a\right )}^{3}}{x^{12}}\right )} d x^{13}}{195 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{4}} + \frac {1}{585} \, {\left (\frac {{\left (45 \, b^{4} - \frac {260 \, {\left (b x^{4} + a\right )} b^{3}}{x^{4}} + \frac {702 \, {\left (b x^{4} + a\right )}^{2} b^{2}}{x^{8}} - \frac {2340 \, {\left (b x^{4} + a\right )}^{3} b}{x^{12}}\right )} x^{13}}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{5}} - \frac {195 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{a^{5} x^{3}}\right )} c \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(17/4),x, algorithm="maxima")
Output:
-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)*d*x^13/((b*x^4 + a)^(13/4)*a^4) + 1/585*((45*b^4 - 260*( b*x^4 + a)*b^3/x^4 + 702*(b*x^4 + a)^2*b^2/x^8 - 2340*(b*x^4 + a)^3*b/x^12 )*x^13/((b*x^4 + a)^(13/4)*a^5) - 195*(b*x^4 + a)^(3/4)/(a^5*x^3))*c
\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {17}{4}} x^{4}} \,d x } \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(17/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(17/4)*x^4), x)
Time = 3.60 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{17/4}} \, dx=\frac {x^4\,\left (\frac {12\,a\,d-12\,b\,c}{117\,a^3}-\frac {4\,b\,c}{9\,a^3}\right )-\frac {c}{3\,a^2}}{x^3\,{\left (b\,x^4+a\right )}^{9/4}}+\frac {x\,\left (\frac {d}{13\,a}-\frac {b\,c}{13\,a^2}\right )}{{\left (b\,x^4+a\right )}^{13/4}}+\frac {x\,\left (96\,a\,d-512\,b\,c\right )}{585\,a^4\,{\left (b\,x^4+a\right )}^{5/4}}+\frac {x\,\left (384\,a\,d-2048\,b\,c\right )}{585\,a^5\,{\left (b\,x^4+a\right )}^{1/4}} \] Input:
int((c + d*x^4)/(x^4*(a + b*x^4)^(17/4)),x)
Output:
(x^4*((12*a*d - 12*b*c)/(117*a^3) - (4*b*c)/(9*a^3)) - c/(3*a^2))/(x^3*(a + b*x^4)^(9/4)) + (x*(d/(13*a) - (b*c)/(13*a^2)))/(a + b*x^4)^(13/4) + (x* (96*a*d - 512*b*c))/(585*a^4*(a + b*x^4)^(5/4)) + (x*(384*a*d - 2048*b*c)) /(585*a^5*(a + b*x^4)^(1/4))
\[ \int \frac {c+d x^4}{x^4 \left (a+b x^4\right )^{17/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4} x^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{8}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{12}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{16}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{20}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) d \] Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(17/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*a**4*x**4 + 4*(a + b*x**4)**(1/4)*a**3*b*x**8 + 6*(a + b*x**4)**(1/4)*a**2*b**2*x**12 + 4*(a + b*x**4)**(1/4)*a*b**3*x**1 6 + (a + b*x**4)**(1/4)*b**4*x**20),x)*c + 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 )*d