Integrand size = 19, antiderivative size = 120 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx=-\frac {2 \left (a+b x^2\right )^{5/4}}{17 a c (c x)^{17/2}}+\frac {24 b \left (a+b x^2\right )^{5/4}}{221 a^2 c^3 (c x)^{13/2}}-\frac {64 b^2 \left (a+b x^2\right )^{5/4}}{663 a^3 c^5 (c x)^{9/2}}+\frac {256 b^3 \left (a+b x^2\right )^{5/4}}{3315 a^4 c^7 (c x)^{5/2}} \] Output:
-2/17*(b*x^2+a)^(5/4)/a/c/(c*x)^(17/2)+24/221*b*(b*x^2+a)^(5/4)/a^2/c^3/(c *x)^(13/2)-64/663*b^2*(b*x^2+a)^(5/4)/a^3/c^5/(c*x)^(9/2)+256/3315*b^3*(b* x^2+a)^(5/4)/a^4/c^7/(c*x)^(5/2)
Time = 1.40 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.48 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx=-\frac {2 x \left (a+b x^2\right )^{5/4} \left (195 a^3-180 a^2 b x^2+160 a b^2 x^4-128 b^3 x^6\right )}{3315 a^4 (c x)^{19/2}} \] Input:
Integrate[(a + b*x^2)^(1/4)/(c*x)^(19/2),x]
Output:
(-2*x*(a + b*x^2)^(5/4)*(195*a^3 - 180*a^2*b*x^2 + 160*a*b^2*x^4 - 128*b^3 *x^6))/(3315*a^4*(c*x)^(19/2))
Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {246, 246, 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.
\(\displaystyle \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx\) |
\(\Big \downarrow \) 246 |
\(\displaystyle -\frac {12 \int \frac {\left (b x^2+a\right )^{5/4}}{(c x)^{19/2}}dx}{5 a}-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}}\) |
\(\Big \downarrow \) 246 |
\(\displaystyle -\frac {12 \left (-\frac {8 \int \frac {\left (b x^2+a\right )^{9/4}}{(c x)^{19/2}}dx}{9 a}-\frac {2 \left (a+b x^2\right )^{9/4}}{9 a c (c x)^{17/2}}\right )}{5 a}-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}}\) |
\(\Big \downarrow \) 246 |
\(\displaystyle -\frac {12 \left (-\frac {8 \left (-\frac {4 \int \frac {\left (b x^2+a\right )^{13/4}}{(c x)^{19/2}}dx}{13 a}-\frac {2 \left (a+b x^2\right )^{13/4}}{13 a c (c x)^{17/2}}\right )}{9 a}-\frac {2 \left (a+b x^2\right )^{9/4}}{9 a c (c x)^{17/2}}\right )}{5 a}-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle -\frac {12 \left (-\frac {8 \left (\frac {8 \left (a+b x^2\right )^{17/4}}{221 a^2 c (c x)^{17/2}}-\frac {2 \left (a+b x^2\right )^{13/4}}{13 a c (c x)^{17/2}}\right )}{9 a}-\frac {2 \left (a+b x^2\right )^{9/4}}{9 a c (c x)^{17/2}}\right )}{5 a}-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a c (c x)^{17/2}}\) |
Input:
Int[(a + b*x^2)^(1/4)/(c*x)^(19/2),x]
Output:
(-2*(a + b*x^2)^(5/4))/(5*a*c*(c*x)^(17/2)) - (12*((-2*(a + b*x^2)^(9/4))/ (9*a*c*(c*x)^(17/2)) - (8*((-2*(a + b*x^2)^(13/4))/(13*a*c*(c*x)^(17/2)) + (8*(a + b*x^2)^(17/4))/(221*a^2*c*(c*x)^(17/2))))/(9*a)))/(5*a)
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 = 53, normalized size of antiderivative = 0.44
method | result | size |
gosper | \(-\frac {2 x \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (-128 b^{3} x^{6}+160 a \,b^{2} x^{4}-180 a^{2} b \,x^{2}+195 a^{3}\right )}{3315 a^{4} \left (c x \right )^{\frac {19}{2}}}\) | \(53\) |
orering | \(-\frac {2 x \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (-128 b^{3} x^{6}+160 a \,b^{2} x^{4}-180 a^{2} b \,x^{2}+195 a^{3}\right )}{3315 a^{4} \left (c x \right )^{\frac {19}{2}}}\) | \(53\) |
risch | \(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (-128 b^{4} x^{8}+32 a \,b^{3} x^{6}-20 a^{2} b^{2} x^{4}+15 a^{3} b \,x^{2}+195 a^{4}\right )}{3315 c^{9} \sqrt {c x}\, x^{8} a^{4}}\) | \(69\) |
Input:
int((b*x^2+a)^(1/4)/(c*x)^(19/2),x,method=_RETURNVERBOSE)
Output:
-2/3315*x*(b*x^2+a)^(5/4)*(-128*b^3*x^6+160*a*b^2*x^4-180*a^2*b*x^2+195*a^ 3)/a^4/(c*x)^(19/2)
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx=\frac {2 \, {\left (128 \, b^{4} x^{8} - 32 \, a b^{3} x^{6} + 20 \, a^{2} b^{2} x^{4} - 15 \, a^{3} b x^{2} - 195 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x}}{3315 \, a^{4} c^{10} x^{9}} \] Input:
integrate((b*x^2+a)^(1/4)/(c*x)^(19/2),x, algorithm="fricas")
Output:
2/3315*(128*b^4*x^8 - 32*a*b^3*x^6 + 20*a^2*b^2*x^4 - 15*a^3*b*x^2 - 195*a ^4)*(b*x^2 + a)^(1/4)*sqrt(c*x)/(a^4*c^10*x^9)
Timed out. \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx=\text {Timed out} \] Input:
integrate((b*x**2+a)**(1/4)/(c*x)**(19/2),x)
Output:
Timed out
\[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {19}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/4)/(c*x)^(19/2),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(1/4)/(c*x)^(19/2), x)
\[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {19}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/4)/(c*x)^(19/2),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(1/4)/(c*x)^(19/2), x)
Time = 0.46 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx=-\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2}{17\,c^9}+\frac {2\,b\,x^2}{221\,a\,c^9}-\frac {8\,b^2\,x^4}{663\,a^2\,c^9}+\frac {64\,b^3\,x^6}{3315\,a^3\,c^9}-\frac {256\,b^4\,x^8}{3315\,a^4\,c^9}\right )}{x^8\,\sqrt {c\,x}} \] Input:
int((a + b*x^2)^(1/4)/(c*x)^(19/2),x)
Output:
-((a + b*x^2)^(1/4)*(2/(17*c^9) + (2*b*x^2)/(221*a*c^9) - (8*b^2*x^4)/(663 *a^2*c^9) + (64*b^3*x^6)/(3315*a^3*c^9) - (256*b^4*x^8)/(3315*a^4*c^9)))/( x^8*(c*x)^(1/2))
Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{19/2}} \, dx=\frac {2 \sqrt {c}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (128 b^{4} x^{8}-32 a \,b^{3} x^{6}+20 a^{2} b^{2} x^{4}-15 a^{3} b \,x^{2}-195 a^{4}\right )}{3315 \sqrt {x}\, a^{4} c^{10} x^{8}} \] Input:
int((b*x^2+a)^(1/4)/(c*x)^(19/2),x)
Output:
(2*sqrt(c)*(a + b*x**2)**(1/4)*( - 195*a**4 - 15*a**3*b*x**2 + 20*a**2*b** 2*x**4 - 32*a*b**3*x**6 + 128*b**4*x**8))/(3315*sqrt(x)*a**4*c**10*x**8)