Integrand size = 15, antiderivative size = 86 \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{5/4}} \, dx=\frac {1}{a x^{11} \sqrt [4]{a+b x^4}}-\frac {12 \left (a+b x^4\right )^{3/4}}{11 a^2 x^{11}}+\frac {96 b \left (a+b x^4\right )^{3/4}}{77 a^3 x^7}-\frac {128 b^2 \left (a+b x^4\right )^{3/4}}{77 a^4 x^3} \] Output:
1/a/x^11/(b*x^4+a)^(1/4)-12/11*(b*x^4+a)^(3/4)/a^2/x^11+96/77*b*(b*x^4+a)^ (3/4)/a^3/x^7-128/77*b^2*(b*x^4+a)^(3/4)/a^4/x^3
Time = 0.49 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{5/4}} \, dx=\frac {-7 a^3+12 a^2 b x^4-32 a b^2 x^8-128 b^3 x^{12}}{77 a^4 x^{11} \sqrt [4]{a+b x^4}} \] Input:
Integrate[1/(x^12*(a + b*x^4)^(5/4)),x]
Output:
(-7*a^3 + 12*a^2*b*x^4 - 32*a*b^2*x^8 - 128*b^3*x^12)/(77*a^4*x^11*(a + b* x^4)^(1/4))
Time = 0.34 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {803, 803, 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.
\(\displaystyle \int \frac {1}{x^{12} \left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {12 b \int \frac {1}{x^8 \left (b x^4+a\right )^{5/4}}dx}{11 a}-\frac {1}{11 a x^{11} \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {12 b \left (-\frac {8 b \int \frac {1}{x^4 \left (b x^4+a\right )^{5/4}}dx}{7 a}-\frac {1}{7 a x^7 \sqrt [4]{a+b x^4}}\right )}{11 a}-\frac {1}{11 a x^{11} \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {12 b \left (-\frac {8 b \left (-\frac {4 b \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx}{3 a}-\frac {1}{3 a x^3 \sqrt [4]{a+b x^4}}\right )}{7 a}-\frac {1}{7 a x^7 \sqrt [4]{a+b x^4}}\right )}{11 a}-\frac {1}{11 a x^{11} \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle -\frac {12 b \left (-\frac {8 b \left (-\frac {4 b x}{3 a^2 \sqrt [4]{a+b x^4}}-\frac {1}{3 a x^3 \sqrt [4]{a+b x^4}}\right )}{7 a}-\frac {1}{7 a x^7 \sqrt [4]{a+b x^4}}\right )}{11 a}-\frac {1}{11 a x^{11} \sqrt [4]{a+b x^4}}\) |
Input:
Int[1/(x^12*(a + b*x^4)^(5/4)),x]
Output:
-1/11*1/(a*x^11*(a + b*x^4)^(1/4)) - (12*b*(-1/7*1/(a*x^7*(a + b*x^4)^(1/4 )) - (8*b*(-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)))/(11*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]
Time = 0.58 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(-\frac {128 b^{3} x^{12}+32 a \,b^{2} x^{8}-12 a^{2} b \,x^{4}+7 a^{3}}{77 x^{11} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}}\) | \(50\) |
trager | \(-\frac {128 b^{3} x^{12}+32 a \,b^{2} x^{8}-12 a^{2} b \,x^{4}+7 a^{3}}{77 x^{11} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}}\) | \(50\) |
pseudoelliptic | \(\frac {-128 b^{3} x^{12}-32 a \,b^{2} x^{8}+12 a^{2} b \,x^{4}-7 a^{3}}{77 x^{11} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}}\) | \(50\) |
orering | \(-\frac {128 b^{3} x^{12}+32 a \,b^{2} x^{8}-12 a^{2} b \,x^{4}+7 a^{3}}{77 x^{11} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}}\) | \(50\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (51 b^{2} x^{8}-19 a b \,x^{4}+7 a^{2}\right )}{77 a^{4} x^{11}}-\frac {x \,b^{3}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}}\) | \(58\) |
Input:
int(1/x^12/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)
Output:
-1/77*(128*b^3*x^12+32*a*b^2*x^8-12*a^2*b*x^4+7*a^3)/x^11/(b*x^4+a)^(1/4)/ a^4
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{5/4}} \, dx=-\frac {{\left (128 \, b^{3} x^{12} + 32 \, a b^{2} x^{8} - 12 \, a^{2} b x^{4} + 7 \, a^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{77 \, {\left (a^{4} b x^{15} + a^{5} x^{11}\right )}} \] Input:
integrate(1/x^12/(b*x^4+a)^(5/4),x, algorithm="fricas")
Output:
-1/77*(128*b^3*x^12 + 32*a*b^2*x^8 - 12*a^2*b*x^4 + 7*a^3)*(b*x^4 + a)^(3/ 4)/(a^4*b*x^15 + a^5*x^11)
Leaf count of result is larger than twice the leaf count of optimal. 592 vs. \(2 (82) = 164\).
Time = 1.26 (sec) , antiderivative size = 592, normalized size of antiderivative = 6.88 \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{5/4}} \, dx =\text {Too large to display} \] Input:
integrate(1/x**12/(b*x**4+a)**(5/4),x)
Output:
21*a**5*b**(39/4)*(a/(b*x**4) + 1)**(3/4)*gamma(-11/4)/(256*a**7*b**9*x**8 *gamma(5/4) + 768*a**6*b**10*x**12*gamma(5/4) + 768*a**5*b**11*x**16*gamma (5/4) + 256*a**4*b**12*x**20*gamma(5/4)) + 6*a**4*b**(43/4)*x**4*(a/(b*x** 4) + 1)**(3/4)*gamma(-11/4)/(256*a**7*b**9*x**8*gamma(5/4) + 768*a**6*b**1 0*x**12*gamma(5/4) + 768*a**5*b**11*x**16*gamma(5/4) + 256*a**4*b**12*x**2 0*gamma(5/4)) + 45*a**3*b**(47/4)*x**8*(a/(b*x**4) + 1)**(3/4)*gamma(-11/4 )/(256*a**7*b**9*x**8*gamma(5/4) + 768*a**6*b**10*x**12*gamma(5/4) + 768*a **5*b**11*x**16*gamma(5/4) + 256*a**4*b**12*x**20*gamma(5/4)) + 540*a**2*b **(51/4)*x**12*(a/(b*x**4) + 1)**(3/4)*gamma(-11/4)/(256*a**7*b**9*x**8*ga mma(5/4) + 768*a**6*b**10*x**12*gamma(5/4) + 768*a**5*b**11*x**16*gamma(5/ 4) + 256*a**4*b**12*x**20*gamma(5/4)) + 864*a*b**(55/4)*x**16*(a/(b*x**4) + 1)**(3/4)*gamma(-11/4)/(256*a**7*b**9*x**8*gamma(5/4) + 768*a**6*b**10*x **12*gamma(5/4) + 768*a**5*b**11*x**16*gamma(5/4) + 256*a**4*b**12*x**20*g amma(5/4)) + 384*b**(59/4)*x**20*(a/(b*x**4) + 1)**(3/4)*gamma(-11/4)/(256 *a**7*b**9*x**8*gamma(5/4) + 768*a**6*b**10*x**12*gamma(5/4) + 768*a**5*b* *11*x**16*gamma(5/4) + 256*a**4*b**12*x**20*gamma(5/4))
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{5/4}} \, dx=-\frac {b^{3} x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4}} - \frac {\frac {77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} b^{2}}{x^{3}} - \frac {33 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b}{x^{7}} + \frac {7 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}}}{x^{11}}}{77 \, a^{4}} \] Input:
integrate(1/x^12/(b*x^4+a)^(5/4),x, algorithm="maxima")
Output:
-b^3*x/((b*x^4 + a)^(1/4)*a^4) - 1/77*(77*(b*x^4 + a)^(3/4)*b^2/x^3 - 33*( b*x^4 + a)^(7/4)*b/x^7 + 7*(b*x^4 + a)^(11/4)/x^11)/a^4
\[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{12}} \,d x } \] Input:
integrate(1/x^12/(b*x^4+a)^(5/4),x, algorithm="giac")
Output:
integrate(1/((b*x^4 + a)^(5/4)*x^12), x)
Time = 0.68 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{5/4}} \, dx=\frac {19\,b\,{\left (b\,x^4+a\right )}^{3/4}}{77\,a^3\,x^7}-\frac {b^3\,x}{a^4\,{\left (b\,x^4+a\right )}^{1/4}}-\frac {{\left (b\,x^4+a\right )}^{3/4}}{11\,a^2\,x^{11}}-\frac {51\,b^2\,{\left (b\,x^4+a\right )}^{3/4}}{77\,a^4\,x^3} \] Input:
int(1/(x^12*(a + b*x^4)^(5/4)),x)
Output:
(19*b*(a + b*x^4)^(3/4))/(77*a^3*x^7) - (b^3*x)/(a^4*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(11*a^2*x^11) - (51*b^2*(a + b*x^4)^(3/4))/(77*a^4*x^3)
\[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{5/4}} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{16}}d x \] Input:
int(1/x^12/(b*x^4+a)^(5/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*a*x**12 + (a + b*x**4)**(1/4)*b*x**16),x)