Integrand size = 15, antiderivative size = 92 \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{a+b x^4}}{13 a x^{13}}+\frac {4 b \sqrt [4]{a+b x^4}}{39 a^2 x^9}-\frac {32 b^2 \sqrt [4]{a+b x^4}}{195 a^3 x^5}+\frac {128 b^3 \sqrt [4]{a+b x^4}}{195 a^4 x} \] Output:
-1/13*(b*x^4+a)^(1/4)/a/x^13+4/39*b*(b*x^4+a)^(1/4)/a^2/x^9-32/195*b^2*(b* x^4+a)^(1/4)/a^3/x^5+128/195*b^3*(b*x^4+a)^(1/4)/a^4/x
Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{a+b x^4} \left (-15 a^3+20 a^2 b x^4-32 a b^2 x^8+128 b^3 x^{12}\right )}{195 a^4 x^{13}} \] Input:
Integrate[1/(x^14*(a + b*x^4)^(3/4)),x]
Output:
((a + b*x^4)^(1/4)*(-15*a^3 + 20*a^2*b*x^4 - 32*a*b^2*x^8 + 128*b^3*x^12)) /(195*a^4*x^13)
Time = 0.36 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.13, 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, 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.
\(\displaystyle \int \frac {1}{x^{14} \left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {12 b \int \frac {1}{x^{10} \left (b x^4+a\right )^{3/4}}dx}{13 a}-\frac {\sqrt [4]{a+b x^4}}{13 a x^{13}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {12 b \left (-\frac {8 b \int \frac {1}{x^6 \left (b x^4+a\right )^{3/4}}dx}{9 a}-\frac {\sqrt [4]{a+b x^4}}{9 a x^9}\right )}{13 a}-\frac {\sqrt [4]{a+b x^4}}{13 a x^{13}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {12 b \left (-\frac {8 b \left (-\frac {4 b \int \frac {1}{x^2 \left (b x^4+a\right )^{3/4}}dx}{5 a}-\frac {\sqrt [4]{a+b x^4}}{5 a x^5}\right )}{9 a}-\frac {\sqrt [4]{a+b x^4}}{9 a x^9}\right )}{13 a}-\frac {\sqrt [4]{a+b x^4}}{13 a x^{13}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {12 b \left (-\frac {8 b \left (\frac {4 b \sqrt [4]{a+b x^4}}{5 a^2 x}-\frac {\sqrt [4]{a+b x^4}}{5 a x^5}\right )}{9 a}-\frac {\sqrt [4]{a+b x^4}}{9 a x^9}\right )}{13 a}-\frac {\sqrt [4]{a+b x^4}}{13 a x^{13}}\) |
Input:
Int[1/(x^14*(a + b*x^4)^(3/4)),x]
Output:
-1/13*(a + b*x^4)^(1/4)/(a*x^13) - (12*b*(-1/9*(a + b*x^4)^(1/4)/(a*x^9) - (8*b*(-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)))/(13*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]
Time = 0.57 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-128 b^{3} x^{12}+32 a \,b^{2} x^{8}-20 a^{2} b \,x^{4}+15 a^{3}\right )}{195 x^{13} a^{4}}\) | \(50\) |
trager | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-128 b^{3} x^{12}+32 a \,b^{2} x^{8}-20 a^{2} b \,x^{4}+15 a^{3}\right )}{195 x^{13} a^{4}}\) | \(50\) |
risch | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-128 b^{3} x^{12}+32 a \,b^{2} x^{8}-20 a^{2} b \,x^{4}+15 a^{3}\right )}{195 x^{13} a^{4}}\) | \(50\) |
pseudoelliptic | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-128 b^{3} x^{12}+32 a \,b^{2} x^{8}-20 a^{2} b \,x^{4}+15 a^{3}\right )}{195 x^{13} a^{4}}\) | \(50\) |
orering | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-128 b^{3} x^{12}+32 a \,b^{2} x^{8}-20 a^{2} b \,x^{4}+15 a^{3}\right )}{195 x^{13} a^{4}}\) | \(50\) |
Input:
int(1/x^14/(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)
Output:
-1/195*(b*x^4+a)^(1/4)*(-128*b^3*x^12+32*a*b^2*x^8-20*a^2*b*x^4+15*a^3)/x^ 13/a^4
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{3/4}} \, dx=\frac {{\left (128 \, b^{3} x^{12} - 32 \, a b^{2} x^{8} + 20 \, a^{2} b x^{4} - 15 \, a^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{195 \, a^{4} x^{13}} \] Input:
integrate(1/x^14/(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
1/195*(128*b^3*x^12 - 32*a*b^2*x^8 + 20*a^2*b*x^4 - 15*a^3)*(b*x^4 + a)^(1 /4)/(a^4*x^13)
Leaf count of result is larger than twice the leaf count of optimal. 692 vs. \(2 (83) = 166\).
Time = 1.24 (sec) , antiderivative size = 692, normalized size of antiderivative = 7.52 \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{3/4}} \, dx =\text {Too large to display} \] Input:
integrate(1/x**14/(b*x**4+a)**(3/4),x)
Output:
-45*a**6*b**(37/4)*(a/(b*x**4) + 1)**(1/4)*gamma(-13/4)/(256*a**7*b**9*x** 12*gamma(3/4) + 768*a**6*b**10*x**16*gamma(3/4) + 768*a**5*b**11*x**20*gam ma(3/4) + 256*a**4*b**12*x**24*gamma(3/4)) - 75*a**5*b**(41/4)*x**4*(a/(b* x**4) + 1)**(1/4)*gamma(-13/4)/(256*a**7*b**9*x**12*gamma(3/4) + 768*a**6* b**10*x**16*gamma(3/4) + 768*a**5*b**11*x**20*gamma(3/4) + 256*a**4*b**12* x**24*gamma(3/4)) - 51*a**4*b**(45/4)*x**8*(a/(b*x**4) + 1)**(1/4)*gamma(- 13/4)/(256*a**7*b**9*x**12*gamma(3/4) + 768*a**6*b**10*x**16*gamma(3/4) + 768*a**5*b**11*x**20*gamma(3/4) + 256*a**4*b**12*x**24*gamma(3/4)) + 231*a **3*b**(49/4)*x**12*(a/(b*x**4) + 1)**(1/4)*gamma(-13/4)/(256*a**7*b**9*x* *12*gamma(3/4) + 768*a**6*b**10*x**16*gamma(3/4) + 768*a**5*b**11*x**20*ga mma(3/4) + 256*a**4*b**12*x**24*gamma(3/4)) + 924*a**2*b**(53/4)*x**16*(a/ (b*x**4) + 1)**(1/4)*gamma(-13/4)/(256*a**7*b**9*x**12*gamma(3/4) + 768*a* *6*b**10*x**16*gamma(3/4) + 768*a**5*b**11*x**20*gamma(3/4) + 256*a**4*b** 12*x**24*gamma(3/4)) + 1056*a*b**(57/4)*x**20*(a/(b*x**4) + 1)**(1/4)*gamm a(-13/4)/(256*a**7*b**9*x**12*gamma(3/4) + 768*a**6*b**10*x**16*gamma(3/4) + 768*a**5*b**11*x**20*gamma(3/4) + 256*a**4*b**12*x**24*gamma(3/4)) + 38 4*b**(61/4)*x**24*(a/(b*x**4) + 1)**(1/4)*gamma(-13/4)/(256*a**7*b**9*x**1 2*gamma(3/4) + 768*a**6*b**10*x**16*gamma(3/4) + 768*a**5*b**11*x**20*gamm a(3/4) + 256*a**4*b**12*x**24*gamma(3/4))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{3/4}} \, dx=\frac {\frac {195 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}}{x} - \frac {117 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{2}}{x^{5}} + \frac {65 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} b}{x^{9}} - \frac {15 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}}}{x^{13}}}{195 \, a^{4}} \] Input:
integrate(1/x^14/(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
1/195*(195*(b*x^4 + a)^(1/4)*b^3/x - 117*(b*x^4 + a)^(5/4)*b^2/x^5 + 65*(b *x^4 + a)^(9/4)*b/x^9 - 15*(b*x^4 + a)^(13/4)/x^13)/a^4
\[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{14}} \,d x } \] Input:
integrate(1/x^14/(b*x^4+a)^(3/4),x, algorithm="giac")
Output:
integrate(1/((b*x^4 + a)^(3/4)*x^14), x)
Time = 0.55 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{3/4}} \, dx=\frac {4\,b\,{\left (b\,x^4+a\right )}^{1/4}}{39\,a^2\,x^9}-\frac {{\left (b\,x^4+a\right )}^{1/4}}{13\,a\,x^{13}}+\frac {128\,b^3\,{\left (b\,x^4+a\right )}^{1/4}}{195\,a^4\,x}-\frac {32\,b^2\,{\left (b\,x^4+a\right )}^{1/4}}{195\,a^3\,x^5} \] Input:
int(1/(x^14*(a + b*x^4)^(3/4)),x)
Output:
(4*b*(a + b*x^4)^(1/4))/(39*a^2*x^9) - (a + b*x^4)^(1/4)/(13*a*x^13) + (12 8*b^3*(a + b*x^4)^(1/4))/(195*a^4*x) - (32*b^2*(a + b*x^4)^(1/4))/(195*a^3 *x^5)
\[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{3/4}} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{14}}d x \] Input:
int(1/x^14/(b*x^4+a)^(3/4),x)
Output:
int(1/((a + b*x**4)**(3/4)*x**14),x)