Integrand size = 22, antiderivative size = 84 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx=-\frac {c \sqrt [4]{a+b x^4}}{9 a x^9}+\frac {(8 b c-9 a d) \sqrt [4]{a+b x^4}}{45 a^2 x^5}-\frac {4 b (8 b c-9 a d) \sqrt [4]{a+b x^4}}{45 a^3 x} \] Output:
-1/9*c*(b*x^4+a)^(1/4)/a/x^9+1/45*(-9*a*d+8*b*c)*(b*x^4+a)^(1/4)/a^2/x^5-4 /45*b*(-9*a*d+8*b*c)*(b*x^4+a)^(1/4)/a^3/x
Time = 0.50 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.74 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{a+b x^4} \left (-5 a^2 c+8 a b c x^4-9 a^2 d x^4-32 b^2 c x^8+36 a b d x^8\right )}{45 a^3 x^9} \] Input:
Integrate[(c + d*x^4)/(x^10*(a + b*x^4)^(3/4)),x]
Output:
((a + b*x^4)^(1/4)*(-5*a^2*c + 8*a*b*c*x^4 - 9*a^2*d*x^4 - 32*b^2*c*x^8 + 36*a*b*d*x^8))/(45*a^3*x^9)
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.
\(\displaystyle \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(8 b c-9 a d) \int \frac {1}{x^6 \left (b x^4+a\right )^{3/4}}dx}{9 a}-\frac {c \sqrt [4]{a+b x^4}}{9 a x^9}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(8 b c-9 a d) \left (-\frac {4 b \int \frac {1}{x^2 \left (b x^4+a\right )^{3/4}}dx}{5 a}-\frac {\sqrt [4]{a+b x^4}}{5 a x^5}\right )}{9 a}-\frac {c \sqrt [4]{a+b x^4}}{9 a x^9}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {\left (\frac {4 b \sqrt [4]{a+b x^4}}{5 a^2 x}-\frac {\sqrt [4]{a+b x^4}}{5 a x^5}\right ) (8 b c-9 a d)}{9 a}-\frac {c \sqrt [4]{a+b x^4}}{9 a x^9}\) |
Input:
Int[(c + d*x^4)/(x^10*(a + b*x^4)^(3/4)),x]
Output:
-1/9*(c*(a + b*x^4)^(1/4))/(a*x^9) - ((8*b*c - 9*a*d)*(-1/5*(a + b*x^4)^(1 /4)/(a*x^5) + (4*b*(a + b*x^4)^(1/4))/(5*a^2*x)))/(9*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 = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (\left (\frac {9 d \,x^{4}}{5}+c \right ) a^{2}-\frac {8 \left (\frac {9 d \,x^{4}}{2}+c \right ) b \,x^{4} a}{5}+\frac {32 b^{2} c \,x^{8}}{5}\right )}{9 x^{9} a^{3}}\) | \(55\) |
gosper | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-36 a b d \,x^{8}+32 b^{2} c \,x^{8}+9 a^{2} d \,x^{4}-8 a b c \,x^{4}+5 a^{2} c \right )}{45 x^{9} a^{3}}\) | \(59\) |
trager | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-36 a b d \,x^{8}+32 b^{2} c \,x^{8}+9 a^{2} d \,x^{4}-8 a b c \,x^{4}+5 a^{2} c \right )}{45 x^{9} a^{3}}\) | \(59\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-36 a b d \,x^{8}+32 b^{2} c \,x^{8}+9 a^{2} d \,x^{4}-8 a b c \,x^{4}+5 a^{2} c \right )}{45 x^{9} a^{3}}\) | \(59\) |
orering | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-36 a b d \,x^{8}+32 b^{2} c \,x^{8}+9 a^{2} d \,x^{4}-8 a b c \,x^{4}+5 a^{2} c \right )}{45 x^{9} a^{3}}\) | \(59\) |
Input:
int((d*x^4+c)/x^10/(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)
Output:
-1/9*(b*x^4+a)^(1/4)*((9/5*d*x^4+c)*a^2-8/5*(9/2*d*x^4+c)*b*x^4*a+32/5*b^2 *c*x^8)/x^9/a^3
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx=-\frac {{\left (4 \, {\left (8 \, b^{2} c - 9 \, a b d\right )} x^{8} - {\left (8 \, a b c - 9 \, a^{2} d\right )} x^{4} + 5 \, a^{2} c\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{45 \, a^{3} x^{9}} \] Input:
integrate((d*x^4+c)/x^10/(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
-1/45*(4*(8*b^2*c - 9*a*b*d)*x^8 - (8*a*b*c - 9*a^2*d)*x^4 + 5*a^2*c)*(b*x ^4 + a)^(1/4)/(a^3*x^9)
Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (76) = 152\).
Time = 4.14 (sec) , antiderivative size = 488, normalized size of antiderivative = 5.81 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx=\frac {5 a^{4} b^{\frac {17}{4}} c \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} + \frac {2 a^{3} b^{\frac {21}{4}} c x^{4} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} + \frac {21 a^{2} b^{\frac {25}{4}} c x^{8} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} + \frac {56 a b^{\frac {29}{4}} c x^{12} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} + \frac {32 b^{\frac {33}{4}} c x^{16} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} - \frac {\sqrt [4]{b} d \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {5}{4}\right )}{16 a x^{4} \Gamma \left (\frac {3}{4}\right )} + \frac {b^{\frac {5}{4}} d \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {5}{4}\right )}{4 a^{2} \Gamma \left (\frac {3}{4}\right )} \] Input:
integrate((d*x**4+c)/x**10/(b*x**4+a)**(3/4),x)
Output:
5*a**4*b**(17/4)*c*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(64*a**5*b**4*x**8* gamma(3/4) + 128*a**4*b**5*x**12*gamma(3/4) + 64*a**3*b**6*x**16*gamma(3/4 )) + 2*a**3*b**(21/4)*c*x**4*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(64*a**5* b**4*x**8*gamma(3/4) + 128*a**4*b**5*x**12*gamma(3/4) + 64*a**3*b**6*x**16 *gamma(3/4)) + 21*a**2*b**(25/4)*c*x**8*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4 )/(64*a**5*b**4*x**8*gamma(3/4) + 128*a**4*b**5*x**12*gamma(3/4) + 64*a**3 *b**6*x**16*gamma(3/4)) + 56*a*b**(29/4)*c*x**12*(a/(b*x**4) + 1)**(1/4)*g amma(-9/4)/(64*a**5*b**4*x**8*gamma(3/4) + 128*a**4*b**5*x**12*gamma(3/4) + 64*a**3*b**6*x**16*gamma(3/4)) + 32*b**(33/4)*c*x**16*(a/(b*x**4) + 1)** (1/4)*gamma(-9/4)/(64*a**5*b**4*x**8*gamma(3/4) + 128*a**4*b**5*x**12*gamm a(3/4) + 64*a**3*b**6*x**16*gamma(3/4)) - b**(1/4)*d*(a/(b*x**4) + 1)**(1/ 4)*gamma(-5/4)/(16*a*x**4*gamma(3/4)) + b**(5/4)*d*(a/(b*x**4) + 1)**(1/4) *gamma(-5/4)/(4*a**2*gamma(3/4))
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx=\frac {d {\left (\frac {5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{x} - \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{5}}\right )}}{5 \, a^{2}} - \frac {{\left (\frac {45 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2}}{x} - \frac {18 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b}{x^{5}} + \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}}}{x^{9}}\right )} c}{45 \, a^{3}} \] Input:
integrate((d*x^4+c)/x^10/(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
1/5*d*(5*(b*x^4 + a)^(1/4)*b/x - (b*x^4 + a)^(5/4)/x^5)/a^2 - 1/45*(45*(b* x^4 + a)^(1/4)*b^2/x - 18*(b*x^4 + a)^(5/4)*b/x^5 + 5*(b*x^4 + a)^(9/4)/x^ 9)*c/a^3
\[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{10}} \,d x } \] Input:
integrate((d*x^4+c)/x^10/(b*x^4+a)^(3/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(3/4)*x^10), x)
Time = 3.40 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx=-\frac {{\left (b\,x^4+a\right )}^{1/4}\,\left (9\,d\,a^2\,x^4+5\,c\,a^2-36\,d\,a\,b\,x^8-8\,c\,a\,b\,x^4+32\,c\,b^2\,x^8\right )}{45\,a^3\,x^9} \] Input:
int((c + d*x^4)/(x^10*(a + b*x^4)^(3/4)),x)
Output:
-((a + b*x^4)^(1/4)*(5*a^2*c + 9*a^2*d*x^4 + 32*b^2*c*x^8 - 8*a*b*c*x^4 - 36*a*b*d*x^8))/(45*a^3*x^9)
\[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{10}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{6}}d x \right ) d \] Input:
int((d*x^4+c)/x^10/(b*x^4+a)^(3/4),x)
Output:
int(1/((a + b*x**4)**(3/4)*x**10),x)*c + int(1/((a + b*x**4)**(3/4)*x**6), x)*d