Integrand size = 22, antiderivative size = 114 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{7/4}} \, dx=-\frac {c}{9 a x^9 \left (a+b x^4\right )^{3/4}}-\frac {4 b c-3 a d}{9 a^2 x^5 \left (a+b x^4\right )^{3/4}}+\frac {8 (4 b c-3 a d) \sqrt [4]{a+b x^4}}{45 a^3 x^5}-\frac {32 b (4 b c-3 a d) \sqrt [4]{a+b x^4}}{45 a^4 x} \] Output:
-1/9*c/a/x^9/(b*x^4+a)^(3/4)-1/9*(-3*a*d+4*b*c)/a^2/x^5/(b*x^4+a)^(3/4)+8/ 45*(-3*a*d+4*b*c)*(b*x^4+a)^(1/4)/a^3/x^5-32/45*b*(-3*a*d+4*b*c)*(b*x^4+a) ^(1/4)/a^4/x
Time = 0.96 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.71 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{7/4}} \, dx=\frac {-128 b^3 c x^{12}+96 a b^2 x^8 \left (-c+d x^4\right )+12 a^2 b x^4 \left (c+6 d x^4\right )-a^3 \left (5 c+9 d x^4\right )}{45 a^4 x^9 \left (a+b x^4\right )^{3/4}} \] Input:
Integrate[(c + d*x^4)/(x^10*(a + b*x^4)^(7/4)),x]
Output:
(-128*b^3*c*x^12 + 96*a*b^2*x^8*(-c + d*x^4) + 12*a^2*b*x^4*(c + 6*d*x^4) - a^3*(5*c + 9*d*x^4))/(45*a^4*x^9*(a + b*x^4)^(3/4))
Time = 0.40 (sec) , antiderivative size = 111, 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.
\(\displaystyle \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(4 b c-3 a d) \int \frac {1}{x^6 \left (b x^4+a\right )^{7/4}}dx}{3 a}-\frac {c}{9 a x^9 \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(4 b c-3 a d) \left (-\frac {8 b \int \frac {1}{x^2 \left (b x^4+a\right )^{7/4}}dx}{5 a}-\frac {1}{5 a x^5 \left (a+b x^4\right )^{3/4}}\right )}{3 a}-\frac {c}{9 a x^9 \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(4 b c-3 a d) \left (-\frac {8 b \left (-\frac {4 b \int \frac {x^2}{\left (b x^4+a\right )^{7/4}}dx}{a}-\frac {1}{a x \left (a+b x^4\right )^{3/4}}\right )}{5 a}-\frac {1}{5 a x^5 \left (a+b x^4\right )^{3/4}}\right )}{3 a}-\frac {c}{9 a x^9 \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {\left (-\frac {8 b \left (-\frac {4 b x^3}{3 a^2 \left (a+b x^4\right )^{3/4}}-\frac {1}{a x \left (a+b x^4\right )^{3/4}}\right )}{5 a}-\frac {1}{5 a x^5 \left (a+b x^4\right )^{3/4}}\right ) (4 b c-3 a d)}{3 a}-\frac {c}{9 a x^9 \left (a+b x^4\right )^{3/4}}\) |
Input:
Int[(c + d*x^4)/(x^10*(a + b*x^4)^(7/4)),x]
Output:
-1/9*c/(a*x^9*(a + b*x^4)^(3/4)) - ((4*b*c - 3*a*d)*(-1/5*1/(a*x^5*(a + b* x^4)^(3/4)) - (8*b*(-(1/(a*x*(a + b*x^4)^(3/4))) - (4*b*x^3)/(3*a^2*(a + b *x^4)^(3/4))))/(5*a)))/(3*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 = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {\left (-9 d \,x^{4}-5 c \right ) a^{3}+12 b \,x^{4} \left (6 d \,x^{4}+c \right ) a^{2}-96 b^{2} x^{8} \left (-d \,x^{4}+c \right ) a -128 b^{3} c \,x^{12}}{45 \left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{9} a^{4}}\) | \(76\) |
gosper | \(-\frac {-96 a \,b^{2} d \,x^{12}+128 b^{3} c \,x^{12}-72 a^{2} b d \,x^{8}+96 a \,b^{2} c \,x^{8}+9 a^{3} d \,x^{4}-12 a^{2} b c \,x^{4}+5 c \,a^{3}}{45 x^{9} \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{4}}\) | \(83\) |
trager | \(-\frac {-96 a \,b^{2} d \,x^{12}+128 b^{3} c \,x^{12}-72 a^{2} b d \,x^{8}+96 a \,b^{2} c \,x^{8}+9 a^{3} d \,x^{4}-12 a^{2} b c \,x^{4}+5 c \,a^{3}}{45 x^{9} \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{4}}\) | \(83\) |
orering | \(-\frac {-96 a \,b^{2} d \,x^{12}+128 b^{3} c \,x^{12}-72 a^{2} b d \,x^{8}+96 a \,b^{2} c \,x^{8}+9 a^{3} d \,x^{4}-12 a^{2} b c \,x^{4}+5 c \,a^{3}}{45 x^{9} \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{4}}\) | \(83\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-81 a b d \,x^{8}+113 b^{2} c \,x^{8}+9 a^{2} d \,x^{4}-17 a b c \,x^{4}+5 a^{2} c \right )}{45 a^{4} x^{9}}+\frac {b^{2} \left (a d -c b \right ) x^{3}}{3 a^{4} \left (b \,x^{4}+a \right )^{\frac {3}{4}}}\) | \(88\) |
Input:
int((d*x^4+c)/x^10/(b*x^4+a)^(7/4),x,method=_RETURNVERBOSE)
Output:
1/45*((-9*d*x^4-5*c)*a^3+12*b*x^4*(6*d*x^4+c)*a^2-96*b^2*x^8*(-d*x^4+c)*a- 128*b^3*c*x^12)/(b*x^4+a)^(3/4)/x^9/a^4
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.82 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{7/4}} \, dx=-\frac {{\left (32 \, {\left (4 \, b^{3} c - 3 \, a b^{2} d\right )} x^{12} + 24 \, {\left (4 \, a b^{2} c - 3 \, a^{2} b d\right )} x^{8} - 3 \, {\left (4 \, a^{2} b c - 3 \, a^{3} d\right )} x^{4} + 5 \, a^{3} c\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{45 \, {\left (a^{4} b x^{13} + a^{5} x^{9}\right )}} \] Input:
integrate((d*x^4+c)/x^10/(b*x^4+a)^(7/4),x, algorithm="fricas")
Output:
-1/45*(32*(4*b^3*c - 3*a*b^2*d)*x^12 + 24*(4*a*b^2*c - 3*a^2*b*d)*x^8 - 3* (4*a^2*b*c - 3*a^3*d)*x^4 + 5*a^3*c)*(b*x^4 + a)^(1/4)/(a^4*b*x^13 + a^5*x ^9)
Leaf count of result is larger than twice the leaf count of optimal. 920 vs. \(2 (107) = 214\).
Time = 68.28 (sec) , antiderivative size = 920, normalized size of antiderivative = 8.07 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{7/4}} \, dx=\text {Too large to display} \] Input:
integrate((d*x**4+c)/x**10/(b*x**4+a)**(7/4),x)
Output:
c*(15*a**5*b**(37/4)*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(256*a**7*b**9*x* *8*gamma(7/4) + 768*a**6*b**10*x**12*gamma(7/4) + 768*a**5*b**11*x**16*gam ma(7/4) + 256*a**4*b**12*x**20*gamma(7/4)) - 6*a**4*b**(41/4)*x**4*(a/(b*x **4) + 1)**(1/4)*gamma(-9/4)/(256*a**7*b**9*x**8*gamma(7/4) + 768*a**6*b** 10*x**12*gamma(7/4) + 768*a**5*b**11*x**16*gamma(7/4) + 256*a**4*b**12*x** 20*gamma(7/4)) + 231*a**3*b**(45/4)*x**8*(a/(b*x**4) + 1)**(1/4)*gamma(-9/ 4)/(256*a**7*b**9*x**8*gamma(7/4) + 768*a**6*b**10*x**12*gamma(7/4) + 768* a**5*b**11*x**16*gamma(7/4) + 256*a**4*b**12*x**20*gamma(7/4)) + 924*a**2* b**(49/4)*x**12*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(256*a**7*b**9*x**8*ga mma(7/4) + 768*a**6*b**10*x**12*gamma(7/4) + 768*a**5*b**11*x**16*gamma(7/ 4) + 256*a**4*b**12*x**20*gamma(7/4)) + 1056*a*b**(53/4)*x**16*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(256*a**7*b**9*x**8*gamma(7/4) + 768*a**6*b**10*x **12*gamma(7/4) + 768*a**5*b**11*x**16*gamma(7/4) + 256*a**4*b**12*x**20*g amma(7/4)) + 384*b**(57/4)*x**20*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(256* a**7*b**9*x**8*gamma(7/4) + 768*a**6*b**10*x**12*gamma(7/4) + 768*a**5*b** 11*x**16*gamma(7/4) + 256*a**4*b**12*x**20*gamma(7/4))) + d*(-3*a**3*b**(1 7/4)*(a/(b*x**4) + 1)**(1/4)*gamma(-5/4)/(64*a**5*b**4*x**4*gamma(7/4) + 1 28*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)) +...
Time = 0.04 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.18 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{7/4}} \, dx=-\frac {1}{45} \, {\left (\frac {15 \, b^{3} x^{3}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{4}} + \frac {\frac {135 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2}}{x} - \frac {27 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b}{x^{5}} + \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}}}{x^{9}}}{a^{4}}\right )} c + \frac {1}{15} \, {\left (\frac {5 \, b^{2} x^{3}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{3}} + \frac {3 \, {\left (\frac {10 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{x} - \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{5}}\right )}}{a^{3}}\right )} d \] Input:
integrate((d*x^4+c)/x^10/(b*x^4+a)^(7/4),x, algorithm="maxima")
Output:
-1/45*(15*b^3*x^3/((b*x^4 + a)^(3/4)*a^4) + (135*(b*x^4 + a)^(1/4)*b^2/x - 27*(b*x^4 + a)^(5/4)*b/x^5 + 5*(b*x^4 + a)^(9/4)/x^9)/a^4)*c + 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)*d
\[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {7}{4}} x^{10}} \,d x } \] Input:
integrate((d*x^4+c)/x^10/(b*x^4+a)^(7/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(7/4)*x^10), x)
Time = 3.93 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.72 \[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{7/4}} \, dx=-\frac {9\,d\,a^3\,x^4+5\,c\,a^3-72\,d\,a^2\,b\,x^8-12\,c\,a^2\,b\,x^4-96\,d\,a\,b^2\,x^{12}+96\,c\,a\,b^2\,x^8+128\,c\,b^3\,x^{12}}{45\,a^4\,x^9\,{\left (b\,x^4+a\right )}^{3/4}} \] Input:
int((c + d*x^4)/(x^10*(a + b*x^4)^(7/4)),x)
Output:
-(5*a^3*c + 9*a^3*d*x^4 + 128*b^3*c*x^12 - 12*a^2*b*c*x^4 + 96*a*b^2*c*x^8 - 72*a^2*b*d*x^8 - 96*a*b^2*d*x^12)/(45*a^4*x^9*(a + b*x^4)^(3/4))
\[ \int \frac {c+d x^4}{x^{10} \left (a+b x^4\right )^{7/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a \,x^{10}+\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{14}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a \,x^{6}+\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{10}}d x \right ) d \] Input:
int((d*x^4+c)/x^10/(b*x^4+a)^(7/4),x)
Output:
int(1/((a + b*x**4)**(3/4)*a*x**10 + (a + b*x**4)**(3/4)*b*x**14),x)*c + i nt(1/((a + b*x**4)**(3/4)*a*x**6 + (a + b*x**4)**(3/4)*b*x**10),x)*d