Integrand size = 22, antiderivative size = 83 \[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{7/4}} \, dx=-\frac {c}{5 a x^5 \left (a+b x^4\right )^{3/4}}-\frac {8 b c-5 a d}{15 a^2 x \left (a+b x^4\right )^{3/4}}+\frac {4 (8 b c-5 a d) \sqrt [4]{a+b x^4}}{15 a^3 x} \] Output:
-1/5*c/a/x^5/(b*x^4+a)^(3/4)-1/15*(-5*a*d+8*b*c)/a^2/x/(b*x^4+a)^(3/4)+4/1 5*(-5*a*d+8*b*c)*(b*x^4+a)^(1/4)/a^3/x
Time = 0.55 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{7/4}} \, dx=\frac {-3 a^2 c+24 a b c x^4-15 a^2 d x^4+32 b^2 c x^8-20 a b d x^8}{15 a^3 x^5 \left (a+b x^4\right )^{3/4}} \] Input:
Integrate[(c + d*x^4)/(x^6*(a + b*x^4)^(7/4)),x]
Output:
(-3*a^2*c + 24*a*b*c*x^4 - 15*a^2*d*x^4 + 32*b^2*c*x^8 - 20*a*b*d*x^8)/(15 *a^3*x^5*(a + b*x^4)^(3/4))
Time = 0.35 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98, 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.
\(\displaystyle \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(8 b c-5 a d) \int \frac {1}{x^2 \left (b x^4+a\right )^{7/4}}dx}{5 a}-\frac {c}{5 a x^5 \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(8 b c-5 a d) \left (-\frac {4 b \int \frac {x^2}{\left (b x^4+a\right )^{7/4}}dx}{a}-\frac {1}{a x \left (a+b x^4\right )^{3/4}}\right )}{5 a}-\frac {c}{5 a x^5 \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {\left (-\frac {4 b x^3}{3 a^2 \left (a+b x^4\right )^{3/4}}-\frac {1}{a x \left (a+b x^4\right )^{3/4}}\right ) (8 b c-5 a d)}{5 a}-\frac {c}{5 a x^5 \left (a+b x^4\right )^{3/4}}\) |
Input:
Int[(c + d*x^4)/(x^6*(a + b*x^4)^(7/4)),x]
Output:
-1/5*c/(a*x^5*(a + b*x^4)^(3/4)) - ((8*b*c - 5*a*d)*(-(1/(a*x*(a + b*x^4)^ (3/4))) - (4*b*x^3)/(3*a^2*(a + b*x^4)^(3/4))))/(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 = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(\frac {\left (-15 d \,x^{4}-3 c \right ) a^{2}+24 b \left (-\frac {5 d \,x^{4}}{6}+c \right ) x^{4} a +32 b^{2} c \,x^{8}}{15 \left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{5} a^{3}}\) | \(57\) |
gosper | \(-\frac {20 a b d \,x^{8}-32 b^{2} c \,x^{8}+15 a^{2} d \,x^{4}-24 a b c \,x^{4}+3 a^{2} c}{15 x^{5} \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{3}}\) | \(59\) |
trager | \(-\frac {20 a b d \,x^{8}-32 b^{2} c \,x^{8}+15 a^{2} d \,x^{4}-24 a b c \,x^{4}+3 a^{2} c}{15 x^{5} \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{3}}\) | \(59\) |
orering | \(-\frac {20 a b d \,x^{8}-32 b^{2} c \,x^{8}+15 a^{2} d \,x^{4}-24 a b c \,x^{4}+3 a^{2} c}{15 x^{5} \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{3}}\) | \(59\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (5 a d \,x^{4}-9 b c \,x^{4}+a c \right )}{5 a^{3} x^{5}}-\frac {b \left (a d -c b \right ) x^{3}}{3 a^{3} \left (b \,x^{4}+a \right )^{\frac {3}{4}}}\) | \(63\) |
Input:
int((d*x^4+c)/x^6/(b*x^4+a)^(7/4),x,method=_RETURNVERBOSE)
Output:
1/15*((-15*d*x^4-3*c)*a^2+24*b*(-5/6*d*x^4+c)*x^4*a+32*b^2*c*x^8)/(b*x^4+a )^(3/4)/x^5/a^3
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{7/4}} \, dx=\frac {{\left (4 \, {\left (8 \, b^{2} c - 5 \, a b d\right )} x^{8} + 3 \, {\left (8 \, a b c - 5 \, a^{2} d\right )} x^{4} - 3 \, a^{2} c\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{15 \, {\left (a^{3} b x^{9} + a^{4} x^{5}\right )}} \] Input:
integrate((d*x^4+c)/x^6/(b*x^4+a)^(7/4),x, algorithm="fricas")
Output:
1/15*(4*(8*b^2*c - 5*a*b*d)*x^8 + 3*(8*a*b*c - 5*a^2*d)*x^4 - 3*a^2*c)*(b* x^4 + a)^(1/4)/(a^3*b*x^9 + a^4*x^5)
Leaf count of result is larger than twice the leaf count of optimal. 398 vs. \(2 (73) = 146\).
Time = 27.98 (sec) , antiderivative size = 398, normalized size of antiderivative = 4.80 \[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{7/4}} \, dx=c \left (- \frac {3 a^{3} b^{\frac {17}{4}} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {5}{4}\right )}{64 a^{5} b^{4} x^{4} \Gamma \left (\frac {7}{4}\right ) + 128 a^{4} b^{5} x^{8} \Gamma \left (\frac {7}{4}\right ) + 64 a^{3} b^{6} x^{12} \Gamma \left (\frac {7}{4}\right )} + \frac {21 a^{2} b^{\frac {21}{4}} x^{4} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {5}{4}\right )}{64 a^{5} b^{4} x^{4} \Gamma \left (\frac {7}{4}\right ) + 128 a^{4} b^{5} x^{8} \Gamma \left (\frac {7}{4}\right ) + 64 a^{3} b^{6} x^{12} \Gamma \left (\frac {7}{4}\right )} + \frac {56 a b^{\frac {25}{4}} x^{8} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {5}{4}\right )}{64 a^{5} b^{4} x^{4} \Gamma \left (\frac {7}{4}\right ) + 128 a^{4} b^{5} x^{8} \Gamma \left (\frac {7}{4}\right ) + 64 a^{3} b^{6} x^{12} \Gamma \left (\frac {7}{4}\right )} + \frac {32 b^{\frac {29}{4}} x^{12} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {5}{4}\right )}{64 a^{5} b^{4} x^{4} \Gamma \left (\frac {7}{4}\right ) + 128 a^{4} b^{5} x^{8} \Gamma \left (\frac {7}{4}\right ) + 64 a^{3} b^{6} x^{12} \Gamma \left (\frac {7}{4}\right )}\right ) + d \left (\frac {3 \Gamma \left (- \frac {1}{4}\right )}{16 a b^{\frac {3}{4}} x^{4} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt [4]{b} \Gamma \left (- \frac {1}{4}\right )}{4 a^{2} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )}\right ) \] Input:
integrate((d*x**4+c)/x**6/(b*x**4+a)**(7/4),x)
Output:
c*(-3*a**3*b**(17/4)*(a/(b*x**4) + 1)**(1/4)*gamma(-5/4)/(64*a**5*b**4*x** 4*gamma(7/4) + 128*a**4*b**5*x**8*gamma(7/4) + 64*a**3*b**6*x**12*gamma(7/ 4)) + 21*a**2*b**(21/4)*x**4*(a/(b*x**4) + 1)**(1/4)*gamma(-5/4)/(64*a**5* b**4*x**4*gamma(7/4) + 128*a**4*b**5*x**8*gamma(7/4) + 64*a**3*b**6*x**12* gamma(7/4)) + 56*a*b**(25/4)*x**8*(a/(b*x**4) + 1)**(1/4)*gamma(-5/4)/(64* a**5*b**4*x**4*gamma(7/4) + 128*a**4*b**5*x**8*gamma(7/4) + 64*a**3*b**6*x **12*gamma(7/4)) + 32*b**(29/4)*x**12*(a/(b*x**4) + 1)**(1/4)*gamma(-5/4)/ (64*a**5*b**4*x**4*gamma(7/4) + 128*a**4*b**5*x**8*gamma(7/4) + 64*a**3*b* *6*x**12*gamma(7/4))) + d*(3*gamma(-1/4)/(16*a*b**(3/4)*x**4*(a/(b*x**4) + 1)**(3/4)*gamma(7/4)) + b**(1/4)*gamma(-1/4)/(4*a**2*(a/(b*x**4) + 1)**(3 /4)*gamma(7/4)))
Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.18 \[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{7/4}} \, dx=\frac {1}{15} \, {\left (\frac {5 \, b^{2} x^{3}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{3}} + \frac {3 \, {\left (\frac {10 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{x} - \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{5}}\right )}}{a^{3}}\right )} c - \frac {1}{3} \, {\left (\frac {b x^{3}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2}} + \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{2} x}\right )} d \] Input:
integrate((d*x^4+c)/x^6/(b*x^4+a)^(7/4),x, algorithm="maxima")
Output:
1/15*(5*b^2*x^3/((b*x^4 + a)^(3/4)*a^3) + 3*(10*(b*x^4 + a)^(1/4)*b/x - (b *x^4 + a)^(5/4)/x^5)/a^3)*c - 1/3*(b*x^3/((b*x^4 + a)^(3/4)*a^2) + 3*(b*x^ 4 + a)^(1/4)/(a^2*x))*d
\[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {7}{4}} x^{6}} \,d x } \] Input:
integrate((d*x^4+c)/x^6/(b*x^4+a)^(7/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(7/4)*x^6), x)
Time = 4.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{7/4}} \, dx=-\frac {15\,d\,a^2\,x^4+3\,c\,a^2+20\,d\,a\,b\,x^8-24\,c\,a\,b\,x^4-32\,c\,b^2\,x^8}{15\,a^3\,x^5\,{\left (b\,x^4+a\right )}^{3/4}} \] Input:
int((c + d*x^4)/(x^6*(a + b*x^4)^(7/4)),x)
Output:
-(3*a^2*c + 15*a^2*d*x^4 - 32*b^2*c*x^8 - 24*a*b*c*x^4 + 20*a*b*d*x^8)/(15 *a^3*x^5*(a + b*x^4)^(3/4))
\[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{7/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a \,x^{6}+\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{10}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a \,x^{2}+\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{6}}d x \right ) d \] Input:
int((d*x^4+c)/x^6/(b*x^4+a)^(7/4),x)
Output:
int(1/((a + b*x**4)**(3/4)*a*x**6 + (a + b*x**4)**(3/4)*b*x**10),x)*c + in t(1/((a + b*x**4)**(3/4)*a*x**2 + (a + b*x**4)**(3/4)*b*x**6),x)*d