Integrand size = 19, antiderivative size = 94 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{5/2}} \, dx=-\frac {2 \sqrt [4]{a+b x^2}}{3 c (c x)^{3/2}}-\frac {2 b^{3/2} \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {a} c^4 \left (a+b x^2\right )^{3/4}} \] Output:
-2/3*(b*x^2+a)^(1/4)/c/(c*x)^(3/2)-2/3*b^(3/2)*(1+a/b/x^2)^(3/4)*(c*x)^(3/ 2)*InverseJacobiAM(1/2*arccot(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(1/2)/c^4/(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 = 56, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{5/2}} \, dx=-\frac {2 x \sqrt [4]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4},\frac {1}{4},-\frac {b x^2}{a}\right )}{3 (c x)^{5/2} \sqrt [4]{1+\frac {b x^2}{a}}} \] Input:
Integrate[(a + b*x^2)^(1/4)/(c*x)^(5/2),x]
Output:
(-2*x*(a + b*x^2)^(1/4)*Hypergeometric2F1[-3/4, -1/4, 1/4, -((b*x^2)/a)])/ (3*(c*x)^(5/2)*(1 + (b*x^2)/a)^(1/4))
Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {247, 266, 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.
\(\displaystyle \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {b \int \frac {1}{\sqrt {c x} \left (b x^2+a\right )^{3/4}}dx}{3 c^2}-\frac {2 \sqrt [4]{a+b x^2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 b \int \frac {1}{\left (b x^2+a\right )^{3/4}}d\sqrt {c x}}{3 c^3}-\frac {2 \sqrt [4]{a+b x^2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {2 b (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4} (c x)^{3/2}}d\sqrt {c x}}{3 c^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {2 b (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\sqrt {c x} \left (\frac {a x^2 c^4}{b}+1\right )^{3/4}}d\frac {1}{\sqrt {c x}}}{3 c^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {b (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a x c^3}{b}+1\right )^{3/4}}d(c x)}{3 c^3 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle -\frac {2 b^{3/2} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} c^2 x}{\sqrt {b}}\right ),2\right )}{3 \sqrt {a} c^4 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 c (c x)^{3/2}}\) |
Input:
Int[(a + b*x^2)^(1/4)/(c*x)^(5/2),x]
Output:
(-2*(a + b*x^2)^(1/4))/(3*c*(c*x)^(3/2)) - (2*b^(3/2)*(1 + a/(b*x^2))^(3/4 )*(c*x)^(3/2)*EllipticF[ArcTan[(Sqrt[a]*c^2*x)/Sqrt[b]]/2, 2])/(3*Sqrt[a]* c^4*(a + b*x^2)^(3/4))
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[((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] :> 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]
\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (c x \right )^{\frac {5}{2}}}d x\]
Input:
int((b*x^2+a)^(1/4)/(c*x)^(5/2),x)
Output:
int((b*x^2+a)^(1/4)/(c*x)^(5/2),x)
\[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/4)/(c*x)^(5/2),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/4)*sqrt(c*x)/(c^3*x^3), x)
Result contains complex when optimal does not.
Time = 1.67 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.34 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{5/2}} \, dx=- \frac {\sqrt [4]{b} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{c^{\frac {5}{2}} x} \] Input:
integrate((b*x**2+a)**(1/4)/(c*x)**(5/2),x)
Output:
-b**(1/4)*hyper((-1/4, 1/2), (3/2,), a*exp_polar(I*pi)/(b*x**2))/(c**(5/2) *x)
\[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/4)/(c*x)^(5/2),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(1/4)/(c*x)^(5/2), x)
\[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/4)/(c*x)^(5/2),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(1/4)/(c*x)^(5/2), x)
Timed out. \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{5/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{1/4}}{{\left (c\,x\right )}^{5/2}} \,d x \] Input:
int((a + b*x^2)^(1/4)/(c*x)^(5/2),x)
Output:
int((a + b*x^2)^(1/4)/(c*x)^(5/2), x)
\[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-2 \left (b \,x^{2}+a \right )^{\frac {1}{4}}-\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{5}+a \,x^{3}}d x \right ) a x \right )}{2 \sqrt {x}\, c^{3} x} \] Input:
int((b*x^2+a)^(1/4)/(c*x)^(5/2),x)
Output:
(sqrt(c)*( - 2*(a + b*x**2)**(1/4) - sqrt(x)*int((sqrt(x)*(a + b*x**2)**(1 /4))/(a*x**3 + b*x**5),x)*a*x))/(2*sqrt(x)*c**3*x)