Integrand size = 22, antiderivative size = 84 \[ \int \frac {c+d x^4}{x^{12} \sqrt [4]{a+b x^4}} \, dx=-\frac {c \left (a+b x^4\right )^{3/4}}{11 a x^{11}}+\frac {(8 b c-11 a d) \left (a+b x^4\right )^{3/4}}{77 a^2 x^7}-\frac {4 b (8 b c-11 a d) \left (a+b x^4\right )^{3/4}}{231 a^3 x^3} \] Output:
-1/11*c*(b*x^4+a)^(3/4)/a/x^11+1/77*(-11*a*d+8*b*c)*(b*x^4+a)^(3/4)/a^2/x^ 7-4/231*b*(-11*a*d+8*b*c)*(b*x^4+a)^(3/4)/a^3/x^3
Time = 0.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.74 \[ \int \frac {c+d x^4}{x^{12} \sqrt [4]{a+b x^4}} \, dx=\frac {\left (a+b x^4\right )^{3/4} \left (-21 a^2 c+24 a b c x^4-33 a^2 d x^4-32 b^2 c x^8+44 a b d x^8\right )}{231 a^3 x^{11}} \] Input:
Integrate[(c + d*x^4)/(x^12*(a + b*x^4)^(1/4)),x]
Output:
((a + b*x^4)^(3/4)*(-21*a^2*c + 24*a*b*c*x^4 - 33*a^2*d*x^4 - 32*b^2*c*x^8 + 44*a*b*d*x^8))/(231*a^3*x^11)
Time = 0.37 (sec) , antiderivative size = 83, 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, 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.
\(\displaystyle \int \frac {c+d x^4}{x^{12} \sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(8 b c-11 a d) \int \frac {1}{x^8 \sqrt [4]{b x^4+a}}dx}{11 a}-\frac {c \left (a+b x^4\right )^{3/4}}{11 a x^{11}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(8 b c-11 a d) \left (-\frac {4 b \int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx}{7 a}-\frac {\left (a+b x^4\right )^{3/4}}{7 a x^7}\right )}{11 a}-\frac {c \left (a+b x^4\right )^{3/4}}{11 a x^{11}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {\left (\frac {4 b \left (a+b x^4\right )^{3/4}}{21 a^2 x^3}-\frac {\left (a+b x^4\right )^{3/4}}{7 a x^7}\right ) (8 b c-11 a d)}{11 a}-\frac {c \left (a+b x^4\right )^{3/4}}{11 a x^{11}}\) |
Input:
Int[(c + d*x^4)/(x^12*(a + b*x^4)^(1/4)),x]
Output:
-1/11*(c*(a + b*x^4)^(3/4))/(a*x^11) - ((8*b*c - 11*a*d)*(-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)
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 = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (\left (\frac {11 d \,x^{4}}{7}+c \right ) a^{2}-\frac {8 b \left (\frac {11 d \,x^{4}}{6}+c \right ) x^{4} a}{7}+\frac {32 b^{2} c \,x^{8}}{21}\right )}{11 x^{11} a^{3}}\) | \(55\) |
gosper | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (-44 a b d \,x^{8}+32 b^{2} c \,x^{8}+33 a^{2} d \,x^{4}-24 a b c \,x^{4}+21 a^{2} c \right )}{231 x^{11} a^{3}}\) | \(59\) |
trager | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (-44 a b d \,x^{8}+32 b^{2} c \,x^{8}+33 a^{2} d \,x^{4}-24 a b c \,x^{4}+21 a^{2} c \right )}{231 x^{11} a^{3}}\) | \(59\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (-44 a b d \,x^{8}+32 b^{2} c \,x^{8}+33 a^{2} d \,x^{4}-24 a b c \,x^{4}+21 a^{2} c \right )}{231 x^{11} a^{3}}\) | \(59\) |
orering | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (-44 a b d \,x^{8}+32 b^{2} c \,x^{8}+33 a^{2} d \,x^{4}-24 a b c \,x^{4}+21 a^{2} c \right )}{231 x^{11} a^{3}}\) | \(59\) |
Input:
int((d*x^4+c)/x^12/(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
-1/11*(b*x^4+a)^(3/4)*((11/7*d*x^4+c)*a^2-8/7*b*(11/6*d*x^4+c)*x^4*a+32/21 *b^2*c*x^8)/x^11/a^3
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {c+d x^4}{x^{12} \sqrt [4]{a+b x^4}} \, dx=-\frac {{\left (4 \, {\left (8 \, b^{2} c - 11 \, a b d\right )} x^{8} - 3 \, {\left (8 \, a b c - 11 \, a^{2} d\right )} x^{4} + 21 \, a^{2} c\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{231 \, a^{3} x^{11}} \] Input:
integrate((d*x^4+c)/x^12/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
-1/231*(4*(8*b^2*c - 11*a*b*d)*x^8 - 3*(8*a*b*c - 11*a^2*d)*x^4 + 21*a^2*c )*(b*x^4 + a)^(3/4)/(a^3*x^11)
Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (78) = 156\).
Time = 4.93 (sec) , antiderivative size = 490, normalized size of antiderivative = 5.83 \[ \int \frac {c+d x^4}{x^{12} \sqrt [4]{a+b x^4}} \, dx=\frac {21 a^{4} b^{\frac {19}{4}} c \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {1}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {1}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {1}{4}\right )} + \frac {18 a^{3} b^{\frac {23}{4}} c x^{4} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {1}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {1}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {1}{4}\right )} + \frac {5 a^{2} b^{\frac {27}{4}} c x^{8} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {1}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {1}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {1}{4}\right )} + \frac {40 a b^{\frac {31}{4}} c x^{12} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {1}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {1}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {1}{4}\right )} + \frac {32 b^{\frac {35}{4}} c x^{16} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {11}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {1}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {1}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {1}{4}\right )} - \frac {3 b^{\frac {3}{4}} d \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{16 a x^{4} \Gamma \left (\frac {1}{4}\right )} + \frac {b^{\frac {7}{4}} d \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{4 a^{2} \Gamma \left (\frac {1}{4}\right )} \] Input:
integrate((d*x**4+c)/x**12/(b*x**4+a)**(1/4),x)
Output:
21*a**4*b**(19/4)*c*(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*gamma(1 /4)) + 18*a**3*b**(23/4)*c*x**4*(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*gamma(1/4)) + 5*a**2*b**(27/4)*c*x**8*(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*gamma(1/4)) + 40*a*b**(31/4)*c*x**12*(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*gamma(1/4)) + 32*b**(35/4)*c*x**16*(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**1 2*gamma(1/4) + 64*a**3*b**6*x**16*gamma(1/4)) - 3*b**(3/4)*d*(a/(b*x**4) + 1)**(3/4)*gamma(-7/4)/(16*a*x**4*gamma(1/4)) + b**(7/4)*d*(a/(b*x**4) + 1 )**(3/4)*gamma(-7/4)/(4*a**2*gamma(1/4))
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int \frac {c+d x^4}{x^{12} \sqrt [4]{a+b x^4}} \, dx=\frac {d {\left (\frac {7 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b}{x^{3}} - \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}}}{x^{7}}\right )}}{21 \, a^{2}} - \frac {c {\left (\frac {77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{2}}{x^{3}} - \frac {66 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b}{x^{7}} + \frac {21 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}}}{x^{11}}\right )}}{231 \, a^{3}} \] Input:
integrate((d*x^4+c)/x^12/(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
1/21*d*(7*(b*x^4 + a)^(3/4)*b/x^3 - 3*(b*x^4 + a)^(7/4)/x^7)/a^2 - 1/231*c *(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
\[ \int \frac {c+d x^4}{x^{12} \sqrt [4]{a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{12}} \,d x } \] Input:
integrate((d*x^4+c)/x^12/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(1/4)*x^12), x)
Time = 4.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {c+d x^4}{x^{12} \sqrt [4]{a+b x^4}} \, dx=-\frac {{\left (b\,x^4+a\right )}^{3/4}\,\left (33\,d\,a^2\,x^4+21\,c\,a^2-44\,d\,a\,b\,x^8-24\,c\,a\,b\,x^4+32\,c\,b^2\,x^8\right )}{231\,a^3\,x^{11}} \] Input:
int((c + d*x^4)/(x^12*(a + b*x^4)^(1/4)),x)
Output:
-((a + b*x^4)^(3/4)*(21*a^2*c + 33*a^2*d*x^4 + 32*b^2*c*x^8 - 24*a*b*c*x^4 - 44*a*b*d*x^8))/(231*a^3*x^11)
\[ \int \frac {c+d x^4}{x^{12} \sqrt [4]{a+b x^4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{12}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{8}}d x \right ) d \] Input:
int((d*x^4+c)/x^12/(b*x^4+a)^(1/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*x**12),x)*c + int(1/((a + b*x**4)**(1/4)*x**8), x)*d