Integrand size = 15, antiderivative size = 149 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{16}} \, dx=-\frac {b \sqrt [4]{a+b x^4}}{33 x^{11}}-\frac {b^2 \sqrt [4]{a+b x^4}}{231 a x^7}+\frac {2 b^3 \sqrt [4]{a+b x^4}}{231 a^2 x^3}-\frac {\left (a+b x^4\right )^{5/4}}{15 x^{15}}-\frac {4 b^{9/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{231 a^{5/2} \left (a+b x^4\right )^{3/4}} \]
-1/33*b*(b*x^4+a)^(1/4)/x^11-1/231*b^2*(b*x^4+a)^(1/4)/a/x^7+2/231*b^3*(b* x^4+a)^(1/4)/a^2/x^3-1/15*(b*x^4+a)^(5/4)/x^15-4/231*b^(9/2)*(1+a/b/x^4)^( 3/4)*x^3*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2 *b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccot(x^2*b^(1/2)/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 = 0.35 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{16}} \, dx=-\frac {a \sqrt [4]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {15}{4},-\frac {5}{4},-\frac {11}{4},-\frac {b x^4}{a}\right )}{15 x^{15} \sqrt [4]{1+\frac {b x^4}{a}}} \]
-1/15*(a*(a + b*x^4)^(1/4)*Hypergeometric2F1[-15/4, -5/4, -11/4, -((b*x^4) /a)])/(x^15*(1 + (b*x^4)/a)^(1/4))
Time = 0.32 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {809, 809, 847, 847, 768, 858, 807, 229}
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.
\(\displaystyle \int \frac {\left (a+b x^4\right )^{5/4}}{x^{16}} \, dx\) |
\(\Big \downarrow \) 809 |
\(\displaystyle \frac {1}{3} b \int \frac {\sqrt [4]{b x^4+a}}{x^{12}}dx-\frac {\left (a+b x^4\right )^{5/4}}{15 x^{15}}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle \frac {1}{3} b \left (\frac {1}{11} b \int \frac {1}{x^8 \left (b x^4+a\right )^{3/4}}dx-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}\right )-\frac {\left (a+b x^4\right )^{5/4}}{15 x^{15}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {1}{3} b \left (\frac {1}{11} b \left (-\frac {6 b \int \frac {1}{x^4 \left (b x^4+a\right )^{3/4}}dx}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}\right )-\frac {\left (a+b x^4\right )^{5/4}}{15 x^{15}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {1}{3} b \left (\frac {1}{11} b \left (-\frac {6 b \left (-\frac {2 b \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx}{3 a}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}\right )-\frac {\left (a+b x^4\right )^{5/4}}{15 x^{15}}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {1}{3} b \left (\frac {1}{11} b \left (-\frac {6 b \left (-\frac {2 b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{3 a \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}\right )-\frac {\left (a+b x^4\right )^{5/4}}{15 x^{15}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {1}{3} b \left (\frac {1}{11} b \left (-\frac {6 b \left (\frac {2 b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{3 a \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}\right )-\frac {\left (a+b x^4\right )^{5/4}}{15 x^{15}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{3} b \left (\frac {1}{11} b \left (-\frac {6 b \left (\frac {b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{3 a \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}\right )-\frac {\left (a+b x^4\right )^{5/4}}{15 x^{15}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {1}{3} b \left (\frac {1}{11} b \left (-\frac {6 b \left (\frac {2 b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{3 a^{3/2} \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}\right )-\frac {\left (a+b x^4\right )^{5/4}}{15 x^{15}}\) |
-1/15*(a + b*x^4)^(5/4)/x^15 + (b*(-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*EllipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(3*a^(3/2)*(a + b*x^4)^(3/4))))/(7*a)))/11))/3
3.11.78.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[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]
\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}}}{x^{16}}d x\]
\[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{16}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{16}} \,d x } \]
Result contains complex when optimal does not.
Time = 1.38 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.31 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{16}} \, dx=\frac {a^{\frac {5}{4}} \Gamma \left (- \frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {15}{4}, - \frac {5}{4} \\ - \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{15} \Gamma \left (- \frac {11}{4}\right )} \]
a**(5/4)*gamma(-15/4)*hyper((-15/4, -5/4), (-11/4,), b*x**4*exp_polar(I*pi )/a)/(4*x**15*gamma(-11/4))
\[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{16}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{16}} \,d x } \]
\[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{16}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{16}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{16}} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{5/4}}{x^{16}} \,d x \]