Integrand size = 19, antiderivative size = 84 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {2}{a c (c x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {16 \left (a+b x^2\right )^{3/4}}{7 a^2 c (c x)^{7/2}}+\frac {64 b \left (a+b x^2\right )^{3/4}}{21 a^3 c^3 (c x)^{3/2}} \] Output:
2/a/c/(c*x)^(7/2)/(b*x^2+a)^(1/4)-16/7*(b*x^2+a)^(3/4)/a^2/c/(c*x)^(7/2)+6 4/21*b*(b*x^2+a)^(3/4)/a^3/c^3/(c*x)^(3/2)
Time = 0.37 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {2 x \left (3 a^2-8 a b x^2-32 b^2 x^4\right )}{21 a^3 (c x)^{9/2} \sqrt [4]{a+b x^2}} \] Input:
Integrate[1/((c*x)^(9/2)*(a + b*x^2)^(5/4)),x]
Output:
(-2*x*(3*a^2 - 8*a*b*x^2 - 32*b^2*x^4))/(21*a^3*(c*x)^(9/2)*(a + b*x^2)^(1 /4))
Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.
\(\displaystyle \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 246 |
\(\displaystyle \frac {8 \int \frac {1}{(c x)^{9/2} \sqrt [4]{b x^2+a}}dx}{a}+\frac {2}{a c (c x)^{7/2} \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 246 |
\(\displaystyle \frac {8 \left (-\frac {4 \int \frac {\left (b x^2+a\right )^{3/4}}{(c x)^{9/2}}dx}{3 a}-\frac {2 \left (a+b x^2\right )^{3/4}}{3 a c (c x)^{7/2}}\right )}{a}+\frac {2}{a c (c x)^{7/2} \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {8 \left (\frac {8 \left (a+b x^2\right )^{7/4}}{21 a^2 c (c x)^{7/2}}-\frac {2 \left (a+b x^2\right )^{3/4}}{3 a c (c x)^{7/2}}\right )}{a}+\frac {2}{a c (c x)^{7/2} \sqrt [4]{a+b x^2}}\) |
Input:
Int[1/((c*x)^(9/2)*(a + b*x^2)^(5/4)),x]
Output:
2/(a*c*(c*x)^(7/2)*(a + b*x^2)^(1/4)) + (8*((-2*(a + b*x^2)^(3/4))/(3*a*c* (c*x)^(7/2)) + (8*(a + b*x^2)^(7/4))/(21*a^2*c*(c*x)^(7/2))))/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 = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.50
method | result | size |
gosper | \(-\frac {2 x \left (-32 b^{2} x^{4}-8 a b \,x^{2}+3 a^{2}\right )}{21 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{3} \left (c x \right )^{\frac {9}{2}}}\) | \(42\) |
orering | \(-\frac {2 x \left (-32 b^{2} x^{4}-8 a b \,x^{2}+3 a^{2}\right )}{21 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{3} \left (c x \right )^{\frac {9}{2}}}\) | \(42\) |
risch | \(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (-11 b \,x^{2}+3 a \right )}{21 a^{3} x^{3} c^{4} \sqrt {c x}}+\frac {2 b^{2} x}{a^{3} c^{4} \sqrt {c x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}}}\) | \(63\) |
Input:
int(1/(c*x)^(9/2)/(b*x^2+a)^(5/4),x,method=_RETURNVERBOSE)
Output:
-2/21*x*(-32*b^2*x^4-8*a*b*x^2+3*a^2)/(b*x^2+a)^(1/4)/a^3/(c*x)^(9/2)
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {2 \, {\left (32 \, b^{2} x^{4} + 8 \, a b x^{2} - 3 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x}}{21 \, {\left (a^{3} b c^{5} x^{6} + a^{4} c^{5} x^{4}\right )}} \] Input:
integrate(1/(c*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")
Output:
2/21*(32*b^2*x^4 + 8*a*b*x^2 - 3*a^2)*(b*x^2 + a)^(3/4)*sqrt(c*x)/(a^3*b*c ^5*x^6 + a^4*c^5*x^4)
Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (75) = 150\).
Time = 65.83 (sec) , antiderivative size = 384, normalized size of antiderivative = 4.57 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx=- \frac {3 a^{3} b^{\frac {19}{4}} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{32 a^{5} b^{4} c^{\frac {9}{2}} x^{2} \Gamma \left (\frac {5}{4}\right ) + 64 a^{4} b^{5} c^{\frac {9}{2}} x^{4} \Gamma \left (\frac {5}{4}\right ) + 32 a^{3} b^{6} c^{\frac {9}{2}} x^{6} \Gamma \left (\frac {5}{4}\right )} + \frac {5 a^{2} b^{\frac {23}{4}} x^{2} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{32 a^{5} b^{4} c^{\frac {9}{2}} x^{2} \Gamma \left (\frac {5}{4}\right ) + 64 a^{4} b^{5} c^{\frac {9}{2}} x^{4} \Gamma \left (\frac {5}{4}\right ) + 32 a^{3} b^{6} c^{\frac {9}{2}} x^{6} \Gamma \left (\frac {5}{4}\right )} + \frac {40 a b^{\frac {27}{4}} x^{4} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{32 a^{5} b^{4} c^{\frac {9}{2}} x^{2} \Gamma \left (\frac {5}{4}\right ) + 64 a^{4} b^{5} c^{\frac {9}{2}} x^{4} \Gamma \left (\frac {5}{4}\right ) + 32 a^{3} b^{6} c^{\frac {9}{2}} x^{6} \Gamma \left (\frac {5}{4}\right )} + \frac {32 b^{\frac {31}{4}} x^{6} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{32 a^{5} b^{4} c^{\frac {9}{2}} x^{2} \Gamma \left (\frac {5}{4}\right ) + 64 a^{4} b^{5} c^{\frac {9}{2}} x^{4} \Gamma \left (\frac {5}{4}\right ) + 32 a^{3} b^{6} c^{\frac {9}{2}} x^{6} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate(1/(c*x)**(9/2)/(b*x**2+a)**(5/4),x)
Output:
-3*a**3*b**(19/4)*(a/(b*x**2) + 1)**(3/4)*gamma(-7/4)/(32*a**5*b**4*c**(9/ 2)*x**2*gamma(5/4) + 64*a**4*b**5*c**(9/2)*x**4*gamma(5/4) + 32*a**3*b**6* c**(9/2)*x**6*gamma(5/4)) + 5*a**2*b**(23/4)*x**2*(a/(b*x**2) + 1)**(3/4)* gamma(-7/4)/(32*a**5*b**4*c**(9/2)*x**2*gamma(5/4) + 64*a**4*b**5*c**(9/2) *x**4*gamma(5/4) + 32*a**3*b**6*c**(9/2)*x**6*gamma(5/4)) + 40*a*b**(27/4) *x**4*(a/(b*x**2) + 1)**(3/4)*gamma(-7/4)/(32*a**5*b**4*c**(9/2)*x**2*gamm a(5/4) + 64*a**4*b**5*c**(9/2)*x**4*gamma(5/4) + 32*a**3*b**6*c**(9/2)*x** 6*gamma(5/4)) + 32*b**(31/4)*x**6*(a/(b*x**2) + 1)**(3/4)*gamma(-7/4)/(32* a**5*b**4*c**(9/2)*x**2*gamma(5/4) + 64*a**4*b**5*c**(9/2)*x**4*gamma(5/4) + 32*a**3*b**6*c**(9/2)*x**6*gamma(5/4))
\[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(9/2)), x)
\[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(9/2)), x)
Time = 0.56 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {16\,x^2}{21\,a^2\,c^4}-\frac {2}{7\,a\,b\,c^4}+\frac {64\,b\,x^4}{21\,a^3\,c^4}\right )}{x^5\,\sqrt {c\,x}+\frac {a\,x^3\,\sqrt {c\,x}}{b}} \] Input:
int(1/((c*x)^(9/2)*(a + b*x^2)^(5/4)),x)
Output:
((a + b*x^2)^(3/4)*((16*x^2)/(21*a^2*c^4) - 2/(7*a*b*c^4) + (64*b*x^4)/(21 *a^3*c^4)))/(x^5*(c*x)^(1/2) + (a*x^3*(c*x)^(1/2))/b)
Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(c x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {2 \sqrt {c}\, \left (32 b^{2} x^{4}+8 a b \,x^{2}-3 a^{2}\right )}{21 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \sqrt {x}\, a^{3} c^{5} x^{3}} \] Input:
int(1/(c*x)^(9/2)/(b*x^2+a)^(5/4),x)
Output:
(2*sqrt(c)*(a + b*x**2)**(3/4)*( - 3*a**2 + 8*a*b*x**2 + 32*b**2*x**4))/(2 1*sqrt(x)*a**3*c**5*x**3*(a + b*x**2))