Integrand size = 20, antiderivative size = 159 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}+\frac {2 b \sqrt [4]{a-b x^2}}{77 a c^3 (c x)^{7/2}}+\frac {4 b^2 \sqrt [4]{a-b x^2}}{77 a^2 c^5 (c x)^{3/2}}+\frac {8 b^{7/2} \left (1-\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{77 a^{5/2} c^8 \left (a-b x^2\right )^{3/4}} \]
-2/11*(-b*x^2+a)^(1/4)/c/(c*x)^(11/2)+2/77*b*(-b*x^2+a)^(1/4)/a/c^3/(c*x)^ (7/2)+4/77*b^2*(-b*x^2+a)^(1/4)/a^2/c^5/(c*x)^(3/2)+8/77*b^(7/2)*(1-a/b/x^ 2)^(3/4)*(c*x)^(3/2)*(cos(1/2*arccsc(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2* arccsc(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccsc(x*b^(1/2)/a^(1/2))),2^ (1/2))/a^(5/2)/c^8/(-b*x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=-\frac {2 x \sqrt [4]{a-b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{4},-\frac {7}{4},\frac {b x^2}{a}\right )}{11 (c x)^{13/2} \sqrt [4]{1-\frac {b x^2}{a}}} \]
(-2*x*(a - b*x^2)^(1/4)*Hypergeometric2F1[-11/4, -1/4, -7/4, (b*x^2)/a])/( 11*(c*x)^(13/2)*(1 - (b*x^2)/a)^(1/4))
Time = 0.33 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {247, 264, 264, 266, 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.
\(\displaystyle \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -\frac {b \int \frac {1}{(c x)^{9/2} \left (a-b x^2\right )^{3/4}}dx}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {b \left (\frac {6 b \int \frac {1}{(c x)^{5/2} \left (a-b x^2\right )^{3/4}}dx}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {b \left (\frac {6 b \left (\frac {2 b \int \frac {1}{\sqrt {c x} \left (a-b x^2\right )^{3/4}}dx}{3 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {b \left (\frac {6 b \left (\frac {4 b \int \frac {1}{\left (a-b x^2\right )^{3/4}}d\sqrt {c x}}{3 a c^3}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle -\frac {b \left (\frac {6 b \left (\frac {4 b (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \int \frac {1}{\left (1-\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}}d\sqrt {c x}}{3 a c^3 \left (a-b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {b \left (\frac {6 b \left (-\frac {4 b (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \int \frac {1}{\sqrt {c x} \left (1-\frac {a c^4 x^2}{b}\right )^{3/4}}d\frac {1}{\sqrt {c x}}}{3 a c^3 \left (a-b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {b \left (\frac {6 b \left (-\frac {2 b (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \int \frac {1}{\left (1-\frac {a c^3 x}{b}\right )^{3/4}}d(c x)}{3 a c^3 \left (a-b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\) |
\(\Big \downarrow \) 230 |
\(\displaystyle -\frac {b \left (\frac {6 b \left (-\frac {4 b^{3/2} (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a} c^2 x}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} c^4 \left (a-b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\) |
(-2*(a - b*x^2)^(1/4))/(11*c*(c*x)^(11/2)) - (b*((-2*(a - b*x^2)^(1/4))/(7 *a*c*(c*x)^(7/2)) + (6*b*((-2*(a - b*x^2)^(1/4))/(3*a*c*(c*x)^(3/2)) - (4* b^(3/2)*(1 - a/(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcSin[(Sqrt[a]*c^2*x) /Sqrt[b]]/2, 2])/(3*a^(3/2)*c^4*(a - b*x^2)^(3/4))))/(7*a*c^2)))/(11*c^2)
3.10.36.3.1 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)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ [a, 0] && NegQ[b/a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
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[(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]
\[\int \frac {\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (c x \right )^{\frac {13}{2}}}d x\]
\[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {13}{2}}} \,d x } \]
Result contains complex when optimal does not.
Time = 148.44 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.25 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=- \frac {i \sqrt [4]{b} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a}{b x^{2}}} \right )}}{5 c^{\frac {13}{2}} x^{5}} \]
\[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {13}{2}}} \,d x } \]
\[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {13}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=\int \frac {{\left (a-b\,x^2\right )}^{1/4}}{{\left (c\,x\right )}^{13/2}} \,d x \]