Integrand size = 20, antiderivative size = 60 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{11/2}} \, dx=-\frac {2 \left (a-b x^2\right )^{5/4}}{9 a c (c x)^{9/2}}-\frac {8 b \left (a-b x^2\right )^{5/4}}{45 a^2 c^3 (c x)^{5/2}} \] Output:
-2/9*(-b*x^2+a)^(5/4)/a/c/(c*x)^(9/2)-8/45*b*(-b*x^2+a)^(5/4)/a^2/c^3/(c*x )^(5/2)
Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{11/2}} \, dx=-\frac {2 x \sqrt [4]{a-b x^2} \left (5 a^2-a b x^2-4 b^2 x^4\right )}{45 a^2 (c x)^{11/2}} \] Input:
Integrate[(a - b*x^2)^(1/4)/(c*x)^(11/2),x]
Output:
(-2*x*(a - b*x^2)^(1/4)*(5*a^2 - a*b*x^2 - 4*b^2*x^4))/(45*a^2*(c*x)^(11/2 ))
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {246, 242}
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.
\(\displaystyle \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{11/2}} \, dx\) |
\(\Big \downarrow \) 246 |
\(\displaystyle -\frac {4 \int \frac {\left (a-b x^2\right )^{5/4}}{(c x)^{11/2}}dx}{5 a}-\frac {2 \left (a-b x^2\right )^{5/4}}{5 a c (c x)^{9/2}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {8 \left (a-b x^2\right )^{9/4}}{45 a^2 c (c x)^{9/2}}-\frac {2 \left (a-b x^2\right )^{5/4}}{5 a c (c x)^{9/2}}\) |
Input:
Int[(a - b*x^2)^(1/4)/(c*x)^(11/2),x]
Output:
(-2*(a - b*x^2)^(5/4))/(5*a*c*(c*x)^(9/2)) + (8*(a - b*x^2)^(9/4))/(45*a^2 *c*(c*x)^(9/2))
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] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(a*c*2*(p + 1))), x] + Simp[(m + 2*p + 3)/( a*2*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m , p}, x] && ILtQ[Simplify[(m + 1)/2 + p + 1], 0] && NeQ[p, -1]
Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.53
method | result | size |
gosper | \(-\frac {2 x \left (-b \,x^{2}+a \right )^{\frac {5}{4}} \left (4 b \,x^{2}+5 a \right )}{45 a^{2} \left (c x \right )^{\frac {11}{2}}}\) | \(32\) |
orering | \(-\frac {2 x \left (-b \,x^{2}+a \right )^{\frac {5}{4}} \left (4 b \,x^{2}+5 a \right )}{45 a^{2} \left (c x \right )^{\frac {11}{2}}}\) | \(32\) |
risch | \(-\frac {2 \left (-b \,x^{2}+a \right )^{\frac {1}{4}} {\left (\left (-b \,x^{2}+a \right )^{3}\right )}^{\frac {1}{4}} \left (-4 b^{2} x^{4}-a b \,x^{2}+5 a^{2}\right )}{45 \sqrt {c x}\, {\left (-\left (b \,x^{2}-a \right )^{3}\right )}^{\frac {1}{4}} c^{5} x^{4} a^{2}}\) | \(75\) |
Input:
int((-b*x^2+a)^(1/4)/(c*x)^(11/2),x,method=_RETURNVERBOSE)
Output:
-2/45*x*(-b*x^2+a)^(5/4)*(4*b*x^2+5*a)/a^2/(c*x)^(11/2)
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{11/2}} \, dx=\frac {2 \, {\left (4 \, b^{2} x^{4} + a b x^{2} - 5 \, a^{2}\right )} {\left (-b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x}}{45 \, a^{2} c^{6} x^{5}} \] Input:
integrate((-b*x^2+a)^(1/4)/(c*x)^(11/2),x, algorithm="fricas")
Output:
2/45*(4*b^2*x^4 + a*b*x^2 - 5*a^2)*(-b*x^2 + a)^(1/4)*sqrt(c*x)/(a^2*c^6*x ^5)
Result contains complex when optimal does not.
Time = 41.65 (sec) , antiderivative size = 462, normalized size of antiderivative = 7.70 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{11/2}} \, dx=\begin {cases} - \frac {5 \sqrt [4]{b} \sqrt [4]{\frac {a}{b x^{2}} - 1} \Gamma \left (- \frac {9}{4}\right )}{8 c^{\frac {11}{2}} x^{4} \Gamma \left (- \frac {1}{4}\right )} + \frac {b^{\frac {5}{4}} \sqrt [4]{\frac {a}{b x^{2}} - 1} \Gamma \left (- \frac {9}{4}\right )}{8 a c^{\frac {11}{2}} x^{2} \Gamma \left (- \frac {1}{4}\right )} + \frac {b^{\frac {9}{4}} \sqrt [4]{\frac {a}{b x^{2}} - 1} \Gamma \left (- \frac {9}{4}\right )}{2 a^{2} c^{\frac {11}{2}} \Gamma \left (- \frac {1}{4}\right )} & \text {for}\: \left |{\frac {a}{b x^{2}}}\right | > 1 \\\frac {5 a^{3} b^{\frac {5}{4}} \sqrt [4]{- \frac {a}{b x^{2}} + 1} e^{\frac {i \pi }{4}} \Gamma \left (- \frac {9}{4}\right )}{x^{2} \left (- 8 a^{3} b c^{\frac {11}{2}} x^{2} \Gamma \left (- \frac {1}{4}\right ) + 8 a^{2} b^{2} c^{\frac {11}{2}} x^{4} \Gamma \left (- \frac {1}{4}\right )\right )} - \frac {6 a^{2} b^{\frac {9}{4}} \sqrt [4]{- \frac {a}{b x^{2}} + 1} e^{\frac {i \pi }{4}} \Gamma \left (- \frac {9}{4}\right )}{- 8 a^{3} b c^{\frac {11}{2}} x^{2} \Gamma \left (- \frac {1}{4}\right ) + 8 a^{2} b^{2} c^{\frac {11}{2}} x^{4} \Gamma \left (- \frac {1}{4}\right )} - \frac {3 a b^{\frac {13}{4}} x^{2} \sqrt [4]{- \frac {a}{b x^{2}} + 1} e^{\frac {i \pi }{4}} \Gamma \left (- \frac {9}{4}\right )}{- 8 a^{3} b c^{\frac {11}{2}} x^{2} \Gamma \left (- \frac {1}{4}\right ) + 8 a^{2} b^{2} c^{\frac {11}{2}} x^{4} \Gamma \left (- \frac {1}{4}\right )} + \frac {4 b^{\frac {17}{4}} x^{4} \sqrt [4]{- \frac {a}{b x^{2}} + 1} e^{\frac {i \pi }{4}} \Gamma \left (- \frac {9}{4}\right )}{- 8 a^{3} b c^{\frac {11}{2}} x^{2} \Gamma \left (- \frac {1}{4}\right ) + 8 a^{2} b^{2} c^{\frac {11}{2}} x^{4} \Gamma \left (- \frac {1}{4}\right )} & \text {otherwise} \end {cases} \] Input:
integrate((-b*x**2+a)**(1/4)/(c*x)**(11/2),x)
Output:
Piecewise((-5*b**(1/4)*(a/(b*x**2) - 1)**(1/4)*gamma(-9/4)/(8*c**(11/2)*x* *4*gamma(-1/4)) + b**(5/4)*(a/(b*x**2) - 1)**(1/4)*gamma(-9/4)/(8*a*c**(11 /2)*x**2*gamma(-1/4)) + b**(9/4)*(a/(b*x**2) - 1)**(1/4)*gamma(-9/4)/(2*a* *2*c**(11/2)*gamma(-1/4)), Abs(a/(b*x**2)) > 1), (5*a**3*b**(5/4)*(-a/(b*x **2) + 1)**(1/4)*exp(I*pi/4)*gamma(-9/4)/(x**2*(-8*a**3*b*c**(11/2)*x**2*g amma(-1/4) + 8*a**2*b**2*c**(11/2)*x**4*gamma(-1/4))) - 6*a**2*b**(9/4)*(- a/(b*x**2) + 1)**(1/4)*exp(I*pi/4)*gamma(-9/4)/(-8*a**3*b*c**(11/2)*x**2*g amma(-1/4) + 8*a**2*b**2*c**(11/2)*x**4*gamma(-1/4)) - 3*a*b**(13/4)*x**2* (-a/(b*x**2) + 1)**(1/4)*exp(I*pi/4)*gamma(-9/4)/(-8*a**3*b*c**(11/2)*x**2 *gamma(-1/4) + 8*a**2*b**2*c**(11/2)*x**4*gamma(-1/4)) + 4*b**(17/4)*x**4* (-a/(b*x**2) + 1)**(1/4)*exp(I*pi/4)*gamma(-9/4)/(-8*a**3*b*c**(11/2)*x**2 *gamma(-1/4) + 8*a**2*b**2*c**(11/2)*x**4*gamma(-1/4)), True))
\[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{11/2}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {11}{2}}} \,d x } \] Input:
integrate((-b*x^2+a)^(1/4)/(c*x)^(11/2),x, algorithm="maxima")
Output:
integrate((-b*x^2 + a)^(1/4)/(c*x)^(11/2), x)
\[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{11/2}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {11}{2}}} \,d x } \] Input:
integrate((-b*x^2+a)^(1/4)/(c*x)^(11/2),x, algorithm="giac")
Output:
integrate((-b*x^2 + a)^(1/4)/(c*x)^(11/2), x)
Time = 0.39 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{11/2}} \, dx=\frac {{\left (a-b\,x^2\right )}^{1/4}\,\left (\frac {2\,b\,x^2}{45\,a\,c^5}-\frac {2}{9\,c^5}+\frac {8\,b^2\,x^4}{45\,a^2\,c^5}\right )}{x^4\,\sqrt {c\,x}} \] Input:
int((a - b*x^2)^(1/4)/(c*x)^(11/2),x)
Output:
((a - b*x^2)^(1/4)*((2*b*x^2)/(45*a*c^5) - 2/(9*c^5) + (8*b^2*x^4)/(45*a^2 *c^5)))/(x^4*(c*x)^(1/2))
Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{11/2}} \, dx=\frac {2 \sqrt {c}\, \left (-b \,x^{2}+a \right )^{\frac {1}{4}} \left (4 b^{2} x^{4}+a b \,x^{2}-5 a^{2}\right )}{45 \sqrt {x}\, a^{2} c^{6} x^{4}} \] Input:
int((-b*x^2+a)^(1/4)/(c*x)^(11/2),x)
Output:
(2*sqrt(c)*(a - b*x**2)**(1/4)*( - 5*a**2 + a*b*x**2 + 4*b**2*x**4))/(45*s qrt(x)*a**2*c**6*x**4)