Integrand size = 22, antiderivative size = 117 \[ \int \frac {c+d x^4}{x^{16} \sqrt [4]{a+b x^4}} \, dx=-\frac {c \left (a+b x^4\right )^{3/4}}{15 a x^{15}}+\frac {(4 b c-5 a d) \left (a+b x^4\right )^{3/4}}{55 a^2 x^{11}}-\frac {8 b (4 b c-5 a d) \left (a+b x^4\right )^{3/4}}{385 a^3 x^7}+\frac {32 b^2 (4 b c-5 a d) \left (a+b x^4\right )^{3/4}}{1155 a^4 x^3} \] Output:
-1/15*c*(b*x^4+a)^(3/4)/a/x^15+1/55*(-5*a*d+4*b*c)*(b*x^4+a)^(3/4)/a^2/x^1 1-8/385*b*(-5*a*d+4*b*c)*(b*x^4+a)^(3/4)/a^3/x^7+32/1155*b^2*(-5*a*d+4*b*c )*(b*x^4+a)^(3/4)/a^4/x^3
Time = 0.43 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74 \[ \int \frac {c+d x^4}{x^{16} \sqrt [4]{a+b x^4}} \, dx=\frac {\left (a+b x^4\right )^{3/4} \left (-77 a^3 c+84 a^2 b c x^4-105 a^3 d x^4-96 a b^2 c x^8+120 a^2 b d x^8+128 b^3 c x^{12}-160 a b^2 d x^{12}\right )}{1155 a^4 x^{15}} \] Input:
Integrate[(c + d*x^4)/(x^16*(a + b*x^4)^(1/4)),x]
Output:
((a + b*x^4)^(3/4)*(-77*a^3*c + 84*a^2*b*c*x^4 - 105*a^3*d*x^4 - 96*a*b^2* c*x^8 + 120*a^2*b*d*x^8 + 128*b^3*c*x^12 - 160*a*b^2*d*x^12))/(1155*a^4*x^ 15)
Time = 0.39 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {955, 803, 803, 796}
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.
\(\displaystyle \int \frac {c+d x^4}{x^{16} \sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(4 b c-5 a d) \int \frac {1}{x^{12} \sqrt [4]{b x^4+a}}dx}{5 a}-\frac {c \left (a+b x^4\right )^{3/4}}{15 a x^{15}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(4 b c-5 a d) \left (-\frac {8 b \int \frac {1}{x^8 \sqrt [4]{b x^4+a}}dx}{11 a}-\frac {\left (a+b x^4\right )^{3/4}}{11 a x^{11}}\right )}{5 a}-\frac {c \left (a+b x^4\right )^{3/4}}{15 a x^{15}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(4 b c-5 a d) \left (-\frac {8 b \left (-\frac {4 b \int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx}{7 a}-\frac {\left (a+b x^4\right )^{3/4}}{7 a x^7}\right )}{11 a}-\frac {\left (a+b x^4\right )^{3/4}}{11 a x^{11}}\right )}{5 a}-\frac {c \left (a+b x^4\right )^{3/4}}{15 a x^{15}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {\left (-\frac {8 b \left (\frac {4 b \left (a+b x^4\right )^{3/4}}{21 a^2 x^3}-\frac {\left (a+b x^4\right )^{3/4}}{7 a x^7}\right )}{11 a}-\frac {\left (a+b x^4\right )^{3/4}}{11 a x^{11}}\right ) (4 b c-5 a d)}{5 a}-\frac {c \left (a+b x^4\right )^{3/4}}{15 a x^{15}}\) |
Input:
Int[(c + d*x^4)/(x^16*(a + b*x^4)^(1/4)),x]
Output:
-1/15*(c*(a + b*x^4)^(3/4))/(a*x^15) - ((4*b*c - 5*a*d)*(-1/11*(a + b*x^4) ^(3/4)/(a*x^11) - (8*b*(-1/7*(a + b*x^4)^(3/4)/(a*x^7) + (4*b*(a + b*x^4)^ (3/4))/(21*a^2*x^3)))/(11*a)))/(5*a)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 ))) Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
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 = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.63
method | result | size |
pseudoelliptic | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (\left (\frac {15 d \,x^{4}}{11}+c \right ) a^{3}-\frac {12 b \,x^{4} \left (\frac {10 d \,x^{4}}{7}+c \right ) a^{2}}{11}+\frac {96 \left (\frac {5 d \,x^{4}}{3}+c \right ) b^{2} x^{8} a}{77}-\frac {128 b^{3} c \,x^{12}}{77}\right )}{15 x^{15} a^{4}}\) | \(74\) |
gosper | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (160 a \,b^{2} d \,x^{12}-128 b^{3} c \,x^{12}-120 a^{2} b d \,x^{8}+96 a \,b^{2} c \,x^{8}+105 a^{3} d \,x^{4}-84 a^{2} b c \,x^{4}+77 c \,a^{3}\right )}{1155 x^{15} a^{4}}\) | \(83\) |
trager | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (160 a \,b^{2} d \,x^{12}-128 b^{3} c \,x^{12}-120 a^{2} b d \,x^{8}+96 a \,b^{2} c \,x^{8}+105 a^{3} d \,x^{4}-84 a^{2} b c \,x^{4}+77 c \,a^{3}\right )}{1155 x^{15} a^{4}}\) | \(83\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (160 a \,b^{2} d \,x^{12}-128 b^{3} c \,x^{12}-120 a^{2} b d \,x^{8}+96 a \,b^{2} c \,x^{8}+105 a^{3} d \,x^{4}-84 a^{2} b c \,x^{4}+77 c \,a^{3}\right )}{1155 x^{15} a^{4}}\) | \(83\) |
orering | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (160 a \,b^{2} d \,x^{12}-128 b^{3} c \,x^{12}-120 a^{2} b d \,x^{8}+96 a \,b^{2} c \,x^{8}+105 a^{3} d \,x^{4}-84 a^{2} b c \,x^{4}+77 c \,a^{3}\right )}{1155 x^{15} a^{4}}\) | \(83\) |
Input:
int((d*x^4+c)/x^16/(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
-1/15*(b*x^4+a)^(3/4)*((15/11*d*x^4+c)*a^3-12/11*b*x^4*(10/7*d*x^4+c)*a^2+ 96/77*(5/3*d*x^4+c)*b^2*x^8*a-128/77*b^3*c*x^12)/x^15/a^4
Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.70 \[ \int \frac {c+d x^4}{x^{16} \sqrt [4]{a+b x^4}} \, dx=\frac {{\left (32 \, {\left (4 \, b^{3} c - 5 \, a b^{2} d\right )} x^{12} - 24 \, {\left (4 \, a b^{2} c - 5 \, a^{2} b d\right )} x^{8} + 21 \, {\left (4 \, a^{2} b c - 5 \, a^{3} d\right )} x^{4} - 77 \, a^{3} c\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{1155 \, a^{4} x^{15}} \] Input:
integrate((d*x^4+c)/x^16/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
1/1155*(32*(4*b^3*c - 5*a*b^2*d)*x^12 - 24*(4*a*b^2*c - 5*a^2*b*d)*x^8 + 2 1*(4*a^2*b*c - 5*a^3*d)*x^4 - 77*a^3*c)*(b*x^4 + a)^(3/4)/(a^4*x^15)
Leaf count of result is larger than twice the leaf count of optimal. 1120 vs. \(2 (112) = 224\).
Time = 6.68 (sec) , antiderivative size = 1120, normalized size of antiderivative = 9.57 \[ \int \frac {c+d x^4}{x^{16} \sqrt [4]{a+b x^4}} \, dx=\text {Too large to display} \] Input:
integrate((d*x**4+c)/x**16/(b*x**4+a)**(1/4),x)
Output:
-231*a**6*b**(39/4)*c*(a/(b*x**4) + 1)**(3/4)*gamma(-15/4)/(256*a**7*b**9* x**12*gamma(1/4) + 768*a**6*b**10*x**16*gamma(1/4) + 768*a**5*b**11*x**20* gamma(1/4) + 256*a**4*b**12*x**24*gamma(1/4)) - 441*a**5*b**(43/4)*c*x**4* (a/(b*x**4) + 1)**(3/4)*gamma(-15/4)/(256*a**7*b**9*x**12*gamma(1/4) + 768 *a**6*b**10*x**16*gamma(1/4) + 768*a**5*b**11*x**20*gamma(1/4) + 256*a**4* b**12*x**24*gamma(1/4)) - 225*a**4*b**(47/4)*c*x**8*(a/(b*x**4) + 1)**(3/4 )*gamma(-15/4)/(256*a**7*b**9*x**12*gamma(1/4) + 768*a**6*b**10*x**16*gamm a(1/4) + 768*a**5*b**11*x**20*gamma(1/4) + 256*a**4*b**12*x**24*gamma(1/4) ) + 21*a**4*b**(19/4)*d*(a/(b*x**4) + 1)**(3/4)*gamma(-11/4)/(64*a**5*b**4 *x**8*gamma(1/4) + 128*a**4*b**5*x**12*gamma(1/4) + 64*a**3*b**6*x**16*gam ma(1/4)) + 45*a**3*b**(51/4)*c*x**12*(a/(b*x**4) + 1)**(3/4)*gamma(-15/4)/ (256*a**7*b**9*x**12*gamma(1/4) + 768*a**6*b**10*x**16*gamma(1/4) + 768*a* *5*b**11*x**20*gamma(1/4) + 256*a**4*b**12*x**24*gamma(1/4)) + 18*a**3*b** (23/4)*d*x**4*(a/(b*x**4) + 1)**(3/4)*gamma(-11/4)/(64*a**5*b**4*x**8*gamm a(1/4) + 128*a**4*b**5*x**12*gamma(1/4) + 64*a**3*b**6*x**16*gamma(1/4)) + 540*a**2*b**(55/4)*c*x**16*(a/(b*x**4) + 1)**(3/4)*gamma(-15/4)/(256*a**7 *b**9*x**12*gamma(1/4) + 768*a**6*b**10*x**16*gamma(1/4) + 768*a**5*b**11* x**20*gamma(1/4) + 256*a**4*b**12*x**24*gamma(1/4)) + 5*a**2*b**(27/4)*d*x **8*(a/(b*x**4) + 1)**(3/4)*gamma(-11/4)/(64*a**5*b**4*x**8*gamma(1/4) + 1 28*a**4*b**5*x**12*gamma(1/4) + 64*a**3*b**6*x**16*gamma(1/4)) + 864*a*...
Time = 0.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x^4}{x^{16} \sqrt [4]{a+b x^4}} \, dx=-\frac {d {\left (\frac {77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{2}}{x^{3}} - \frac {66 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b}{x^{7}} + \frac {21 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}}}{x^{11}}\right )}}{231 \, a^{3}} + \frac {c {\left (\frac {385 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{3}}{x^{3}} - \frac {495 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{2}}{x^{7}} + \frac {315 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} b}{x^{11}} - \frac {77 \, {\left (b x^{4} + a\right )}^{\frac {15}{4}}}{x^{15}}\right )}}{1155 \, a^{4}} \] Input:
integrate((d*x^4+c)/x^16/(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
-1/231*d*(77*(b*x^4 + a)^(3/4)*b^2/x^3 - 66*(b*x^4 + a)^(7/4)*b/x^7 + 21*( b*x^4 + a)^(11/4)/x^11)/a^3 + 1/1155*c*(385*(b*x^4 + a)^(3/4)*b^3/x^3 - 49 5*(b*x^4 + a)^(7/4)*b^2/x^7 + 315*(b*x^4 + a)^(11/4)*b/x^11 - 77*(b*x^4 + a)^(15/4)/x^15)/a^4
\[ \int \frac {c+d x^4}{x^{16} \sqrt [4]{a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{16}} \,d x } \] Input:
integrate((d*x^4+c)/x^16/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(1/4)*x^16), x)
Time = 4.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.90 \[ \int \frac {c+d x^4}{x^{16} \sqrt [4]{a+b x^4}} \, dx=\frac {{\left (b\,x^4+a\right )}^{3/4}\,\left (128\,b^3\,c-160\,a\,b^2\,d\right )}{1155\,a^4\,x^3}-\frac {{\left (b\,x^4+a\right )}^{3/4}\,\left (32\,b^2\,c-40\,a\,b\,d\right )}{385\,a^3\,x^7}-\frac {{\left (b\,x^4+a\right )}^{3/4}\,\left (5\,a\,d-4\,b\,c\right )}{55\,a^2\,x^{11}}-\frac {c\,{\left (b\,x^4+a\right )}^{3/4}}{15\,a\,x^{15}} \] Input:
int((c + d*x^4)/(x^16*(a + b*x^4)^(1/4)),x)
Output:
((a + b*x^4)^(3/4)*(128*b^3*c - 160*a*b^2*d))/(1155*a^4*x^3) - ((a + b*x^4 )^(3/4)*(32*b^2*c - 40*a*b*d))/(385*a^3*x^7) - ((a + b*x^4)^(3/4)*(5*a*d - 4*b*c))/(55*a^2*x^11) - (c*(a + b*x^4)^(3/4))/(15*a*x^15)
\[ \int \frac {c+d x^4}{x^{16} \sqrt [4]{a+b x^4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{16}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{12}}d x \right ) d \] Input:
int((d*x^4+c)/x^16/(b*x^4+a)^(1/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*x**16),x)*c + int(1/((a + b*x**4)**(1/4)*x**12) ,x)*d