Integrand size = 16, antiderivative size = 133 \[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}+\frac {b \sqrt [4]{a-b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{77 a^2 x^3}+\frac {4 b^{7/2} \left (1-\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{77 a^{5/2} \left (a-b x^4\right )^{3/4}} \] Output:
-1/11*(-b*x^4+a)^(1/4)/x^11+1/77*b*(-b*x^4+a)^(1/4)/a/x^7+2/77*b^2*(-b*x^4 +a)^(1/4)/a^2/x^3+4/77*b^(7/2)*(1-a/b/x^4)^(3/4)*x^3*InverseJacobiAM(1/2*a rccsc(b^(1/2)*x^2/a^(1/2)),2^(1/2))/a^(5/2)/(-b*x^4+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=-\frac {\sqrt [4]{a-b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{4},-\frac {7}{4},\frac {b x^4}{a}\right )}{11 x^{11} \sqrt [4]{1-\frac {b x^4}{a}}} \] Input:
Integrate[(a - b*x^4)^(1/4)/x^12,x]
Output:
-1/11*((a - b*x^4)^(1/4)*Hypergeometric2F1[-11/4, -1/4, -7/4, (b*x^4)/a])/ (x^11*(1 - (b*x^4)/a)^(1/4))
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {809, 847, 847, 768, 858, 807, 230}
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.
\(\displaystyle \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx\) |
\(\Big \downarrow \) 809 |
\(\displaystyle -\frac {1}{11} b \int \frac {1}{x^8 \left (a-b x^4\right )^{3/4}}dx-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle -\frac {1}{11} b \left (\frac {6 b \int \frac {1}{x^4 \left (a-b x^4\right )^{3/4}}dx}{7 a}-\frac {\sqrt [4]{a-b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle -\frac {1}{11} b \left (\frac {6 b \left (\frac {2 b \int \frac {1}{\left (a-b x^4\right )^{3/4}}dx}{3 a}-\frac {\sqrt [4]{a-b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a-b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle -\frac {1}{11} b \left (\frac {6 b \left (\frac {2 b x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} \int \frac {1}{\left (1-\frac {a}{b x^4}\right )^{3/4} x^3}dx}{3 a \left (a-b x^4\right )^{3/4}}-\frac {\sqrt [4]{a-b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a-b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {1}{11} b \left (\frac {6 b \left (-\frac {2 b x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} \int \frac {1}{\left (1-\frac {a}{b x^4}\right )^{3/4} x}d\frac {1}{x}}{3 a \left (a-b x^4\right )^{3/4}}-\frac {\sqrt [4]{a-b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a-b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {1}{11} b \left (\frac {6 b \left (-\frac {b x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} \int \frac {1}{\left (1-\frac {a}{b x^2}\right )^{3/4}}d\frac {1}{x^2}}{3 a \left (a-b x^4\right )^{3/4}}-\frac {\sqrt [4]{a-b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a-b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}\) |
\(\Big \downarrow \) 230 |
\(\displaystyle -\frac {1}{11} b \left (\frac {6 b \left (-\frac {2 b^{3/2} x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{3 a^{3/2} \left (a-b x^4\right )^{3/4}}-\frac {\sqrt [4]{a-b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a-b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}\) |
Input:
Int[(a - b*x^4)^(1/4)/x^12,x]
Output:
-1/11*(a - b*x^4)^(1/4)/x^11 - (b*(-1/7*(a - b*x^4)^(1/4)/(a*x^7) + (6*b*( -1/3*(a - b*x^4)^(1/4)/(a*x^3) - (2*b^(3/2)*(1 - a/(b*x^4))^(3/4)*x^3*Elli pticF[ArcSin[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(3*a^(3/2)*(a - b*x^4)^(3/4)))) /(7*a)))/11
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2] ))*EllipticF[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ [a, 0] && NegQ[b/a]
Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 /4)/(a + b*x^4)^(3/4)) Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ [{a, b}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ [n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntB inomialQ[a, b, c, n, m, p, x]
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] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
\[\int \frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{12}}d x\]
Input:
int((-b*x^4+a)^(1/4)/x^12,x)
Output:
int((-b*x^4+a)^(1/4)/x^12,x)
\[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\int { \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{12}} \,d x } \] Input:
integrate((-b*x^4+a)^(1/4)/x^12,x, algorithm="fricas")
Output:
integral((-b*x^4 + a)^(1/4)/x^12, x)
Result contains complex when optimal does not.
Time = 0.84 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.26 \[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=- \frac {i \sqrt [4]{b} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{10 x^{10}} \] Input:
integrate((-b*x**4+a)**(1/4)/x**12,x)
Output:
-I*b**(1/4)*exp(-I*pi/4)*hyper((-1/4, 5/2), (7/2,), a/(b*x**4))/(10*x**10)
\[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\int { \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{12}} \,d x } \] Input:
integrate((-b*x^4+a)^(1/4)/x^12,x, algorithm="maxima")
Output:
integrate((-b*x^4 + a)^(1/4)/x^12, x)
\[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\int { \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{12}} \,d x } \] Input:
integrate((-b*x^4+a)^(1/4)/x^12,x, algorithm="giac")
Output:
integrate((-b*x^4 + a)^(1/4)/x^12, x)
Timed out. \[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\int \frac {{\left (a-b\,x^4\right )}^{1/4}}{x^{12}} \,d x \] Input:
int((a - b*x^4)^(1/4)/x^12,x)
Output:
int((a - b*x^4)^(1/4)/x^12, x)
\[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\frac {-\left (-b \,x^{4}+a \right )^{\frac {1}{4}}-\left (\int \frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{-b \,x^{16}+a \,x^{12}}d x \right ) a \,x^{11}}{10 x^{11}} \] Input:
int((-b*x^4+a)^(1/4)/x^12,x)
Output:
( - ((a - b*x**4)**(1/4) + int((a - b*x**4)**(1/4)/(a*x**12 - b*x**16),x)* a*x**11))/(10*x**11)