Integrand size = 22, antiderivative size = 83 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {c}{7 a x^7 \sqrt [4]{a+b x^4}}-\frac {8 b c-7 a d}{7 a^2 x^3 \sqrt [4]{a+b x^4}}+\frac {4 (8 b c-7 a d) \left (a+b x^4\right )^{3/4}}{21 a^3 x^3} \] Output:
-1/7*c/a/x^7/(b*x^4+a)^(1/4)-1/7*(-7*a*d+8*b*c)/a^2/x^3/(b*x^4+a)^(1/4)+4/ 21*(-7*a*d+8*b*c)*(b*x^4+a)^(3/4)/a^3/x^3
Time = 0.58 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{5/4}} \, dx=\frac {-3 a^2 c+8 a b c x^4-7 a^2 d x^4+32 b^2 c x^8-28 a b d x^8}{21 a^3 x^7 \sqrt [4]{a+b x^4}} \] Input:
Integrate[(c + d*x^4)/(x^8*(a + b*x^4)^(5/4)),x]
Output:
(-3*a^2*c + 8*a*b*c*x^4 - 7*a^2*d*x^4 + 32*b^2*c*x^8 - 28*a*b*d*x^8)/(21*a ^3*x^7*(a + b*x^4)^(1/4))
Time = 0.33 (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, 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.
\(\displaystyle \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(8 b c-7 a d) \int \frac {1}{x^4 \left (b x^4+a\right )^{5/4}}dx}{7 a}-\frac {c}{7 a x^7 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(8 b c-7 a d) \left (-\frac {4 b \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx}{3 a}-\frac {1}{3 a x^3 \sqrt [4]{a+b x^4}}\right )}{7 a}-\frac {c}{7 a x^7 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle -\frac {\left (-\frac {4 b x}{3 a^2 \sqrt [4]{a+b x^4}}-\frac {1}{3 a x^3 \sqrt [4]{a+b x^4}}\right ) (8 b c-7 a d)}{7 a}-\frac {c}{7 a x^7 \sqrt [4]{a+b x^4}}\) |
Input:
Int[(c + d*x^4)/(x^8*(a + b*x^4)^(5/4)),x]
Output:
-1/7*c/(a*x^7*(a + b*x^4)^(1/4)) - ((8*b*c - 7*a*d)*(-1/3*1/(a*x^3*(a + b* x^4)^(1/4)) - (4*b*x)/(3*a^2*(a + b*x^4)^(1/4))))/(7*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[(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 = 0.66
method | result | size |
pseudoelliptic | \(-\frac {\left (\frac {7 d \,x^{4}}{3}+c \right ) a^{2}-\frac {8 \left (-\frac {7 d \,x^{4}}{2}+c \right ) b \,x^{4} a}{3}-\frac {32 b^{2} c \,x^{8}}{3}}{7 \left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{7} a^{3}}\) | \(55\) |
gosper | \(-\frac {28 a b d \,x^{8}-32 b^{2} c \,x^{8}+7 a^{2} d \,x^{4}-8 a b c \,x^{4}+3 a^{2} c}{21 x^{7} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}}\) | \(59\) |
trager | \(-\frac {28 a b d \,x^{8}-32 b^{2} c \,x^{8}+7 a^{2} d \,x^{4}-8 a b c \,x^{4}+3 a^{2} c}{21 x^{7} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}}\) | \(59\) |
orering | \(-\frac {28 a b d \,x^{8}-32 b^{2} c \,x^{8}+7 a^{2} d \,x^{4}-8 a b c \,x^{4}+3 a^{2} c}{21 x^{7} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}}\) | \(59\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (7 a d \,x^{4}-11 b c \,x^{4}+3 a c \right )}{21 a^{3} x^{7}}-\frac {x \left (a d -c b \right ) b}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}}\) | \(62\) |
Input:
int((d*x^4+c)/x^8/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)
Output:
-1/7/(b*x^4+a)^(1/4)*((7/3*d*x^4+c)*a^2-8/3*(-7/2*d*x^4+c)*b*x^4*a-32/3*b^ 2*c*x^8)/x^7/a^3
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (4 \, {\left (8 \, b^{2} c - 7 \, a b d\right )} x^{8} + {\left (8 \, a b c - 7 \, a^{2} d\right )} x^{4} - 3 \, a^{2} c\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{21 \, {\left (a^{3} b x^{11} + a^{4} x^{7}\right )}} \] Input:
integrate((d*x^4+c)/x^8/(b*x^4+a)^(5/4),x, algorithm="fricas")
Output:
1/21*(4*(8*b^2*c - 7*a*b*d)*x^8 + (8*a*b*c - 7*a^2*d)*x^4 - 3*a^2*c)*(b*x^ 4 + a)^(3/4)/(a^3*b*x^11 + a^4*x^7)
Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (75) = 150\).
Time = 30.63 (sec) , antiderivative size = 396, normalized size of antiderivative = 4.77 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{5/4}} \, dx=c \left (- \frac {3 a^{3} b^{\frac {19}{4}} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{64 a^{5} b^{4} x^{4} \Gamma \left (\frac {5}{4}\right ) + 128 a^{4} b^{5} x^{8} \Gamma \left (\frac {5}{4}\right ) + 64 a^{3} b^{6} x^{12} \Gamma \left (\frac {5}{4}\right )} + \frac {5 a^{2} b^{\frac {23}{4}} x^{4} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{64 a^{5} b^{4} x^{4} \Gamma \left (\frac {5}{4}\right ) + 128 a^{4} b^{5} x^{8} \Gamma \left (\frac {5}{4}\right ) + 64 a^{3} b^{6} x^{12} \Gamma \left (\frac {5}{4}\right )} + \frac {40 a b^{\frac {27}{4}} x^{8} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{64 a^{5} b^{4} x^{4} \Gamma \left (\frac {5}{4}\right ) + 128 a^{4} b^{5} x^{8} \Gamma \left (\frac {5}{4}\right ) + 64 a^{3} b^{6} x^{12} \Gamma \left (\frac {5}{4}\right )} + \frac {32 b^{\frac {31}{4}} x^{12} \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{64 a^{5} b^{4} x^{4} \Gamma \left (\frac {5}{4}\right ) + 128 a^{4} b^{5} x^{8} \Gamma \left (\frac {5}{4}\right ) + 64 a^{3} b^{6} x^{12} \Gamma \left (\frac {5}{4}\right )}\right ) + d \left (\frac {\Gamma \left (- \frac {3}{4}\right )}{16 a \sqrt [4]{b} x^{4} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (\frac {5}{4}\right )} + \frac {b^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{4 a^{2} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (\frac {5}{4}\right )}\right ) \] Input:
integrate((d*x**4+c)/x**8/(b*x**4+a)**(5/4),x)
Output:
c*(-3*a**3*b**(19/4)*(a/(b*x**4) + 1)**(3/4)*gamma(-7/4)/(64*a**5*b**4*x** 4*gamma(5/4) + 128*a**4*b**5*x**8*gamma(5/4) + 64*a**3*b**6*x**12*gamma(5/ 4)) + 5*a**2*b**(23/4)*x**4*(a/(b*x**4) + 1)**(3/4)*gamma(-7/4)/(64*a**5*b **4*x**4*gamma(5/4) + 128*a**4*b**5*x**8*gamma(5/4) + 64*a**3*b**6*x**12*g amma(5/4)) + 40*a*b**(27/4)*x**8*(a/(b*x**4) + 1)**(3/4)*gamma(-7/4)/(64*a **5*b**4*x**4*gamma(5/4) + 128*a**4*b**5*x**8*gamma(5/4) + 64*a**3*b**6*x* *12*gamma(5/4)) + 32*b**(31/4)*x**12*(a/(b*x**4) + 1)**(3/4)*gamma(-7/4)/( 64*a**5*b**4*x**4*gamma(5/4) + 128*a**4*b**5*x**8*gamma(5/4) + 64*a**3*b** 6*x**12*gamma(5/4))) + d*(gamma(-3/4)/(16*a*b**(1/4)*x**4*(a/(b*x**4) + 1) **(1/4)*gamma(5/4)) + b**(3/4)*gamma(-3/4)/(4*a**2*(a/(b*x**4) + 1)**(1/4) *gamma(5/4)))
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{3} \, d {\left (\frac {3 \, b x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}} + \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{a^{2} x^{3}}\right )} + \frac {1}{21} \, c {\left (\frac {21 \, b^{2} x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}} + \frac {\frac {14 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b}{x^{3}} - \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}}}{x^{7}}}{a^{3}}\right )} \] Input:
integrate((d*x^4+c)/x^8/(b*x^4+a)^(5/4),x, algorithm="maxima")
Output:
-1/3*d*(3*b*x/((b*x^4 + a)^(1/4)*a^2) + (b*x^4 + a)^(3/4)/(a^2*x^3)) + 1/2 1*c*(21*b^2*x/((b*x^4 + a)^(1/4)*a^3) + (14*(b*x^4 + a)^(3/4)*b/x^3 - 3*(b *x^4 + a)^(7/4)/x^7)/a^3)
\[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{8}} \,d x } \] Input:
integrate((d*x^4+c)/x^8/(b*x^4+a)^(5/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(5/4)*x^8), x)
Time = 3.80 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.16 \[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {32\,c\,{\left (b\,x^4+a\right )}^2+21\,a^2\,c+21\,a^2\,d\,x^4-56\,a\,c\,\left (b\,x^4+a\right )-28\,a\,d\,x^4\,\left (b\,x^4+a\right )}{\left (\frac {21\,a^4\,x^3}{b}-\frac {21\,a^3\,x^3\,\left (b\,x^4+a\right )}{b}\right )\,{\left (b\,x^4+a\right )}^{1/4}} \] Input:
int((c + d*x^4)/(x^8*(a + b*x^4)^(5/4)),x)
Output:
-(32*c*(a + b*x^4)^2 + 21*a^2*c + 21*a^2*d*x^4 - 56*a*c*(a + b*x^4) - 28*a *d*x^4*(a + b*x^4))/(((21*a^4*x^3)/b - (21*a^3*x^3*(a + b*x^4))/b)*(a + b* x^4)^(1/4))
\[ \int \frac {c+d x^4}{x^8 \left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{12}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{8}}d x \right ) d \] Input:
int((d*x^4+c)/x^8/(b*x^4+a)^(5/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*a*x**8 + (a + b*x**4)**(1/4)*b*x**12),x)*c + in t(1/((a + b*x**4)**(1/4)*a*x**4 + (a + b*x**4)**(1/4)*b*x**8),x)*d