Integrand size = 21, antiderivative size = 134 \[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^4} \, dx=-\frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (a+b x^2\right )}{3 a x^3}+\frac {5 b \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (a+b x^2\right )}{6 a^2 x}+\frac {5 b^{3/2} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 a^{3/2}} \] Output:
-1/3*(c/(b*x^2+a)^(1/2))^(3/2)*(b*x^2+a)/a/x^3+5/6*b*(c/(b*x^2+a)^(1/2))^( 3/2)*(b*x^2+a)/a^2/x+5/6*b^(3/2)*(c/(b*x^2+a)^(1/2))^(3/2)*(1+b*x^2/a)^(3/ 4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.43 \[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^4} \, dx=-\frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3} \] Input:
Integrate[(c/Sqrt[a + b*x^2])^(3/2)/x^4,x]
Output:
-1/3*((c/Sqrt[a + b*x^2])^(3/2)*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[-3 /2, 3/4, -1/2, -((b*x^2)/a)])/x^3
Time = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2045, 264, 264, 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.
\(\displaystyle \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^4} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \int \frac {1}{x^4 \left (\frac {b x^2}{a}+1\right )^{3/4}}dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (-\frac {5 b \int \frac {1}{x^2 \left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{6 a}-\frac {\sqrt [4]{\frac {b x^2}{a}+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (-\frac {5 b \left (-\frac {b \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{2 a}-\frac {\sqrt [4]{\frac {b x^2}{a}+1}}{x}\right )}{6 a}-\frac {\sqrt [4]{\frac {b x^2}{a}+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (-\frac {5 b \left (-\frac {\sqrt {b} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a}}-\frac {\sqrt [4]{\frac {b x^2}{a}+1}}{x}\right )}{6 a}-\frac {\sqrt [4]{\frac {b x^2}{a}+1}}{3 x^3}\right )\) |
Input:
Int[(c/Sqrt[a + b*x^2])^(3/2)/x^4,x]
Output:
(c/Sqrt[a + b*x^2])^(3/2)*(1 + (b*x^2)/a)^(3/4)*(-1/3*(1 + (b*x^2)/a)^(1/4 )/x^3 - (5*b*(-((1 + (b*x^2)/a)^(1/4)/x) - (Sqrt[b]*EllipticF[ArcTan[(Sqrt [b]*x)/Sqrt[a]]/2, 2])/Sqrt[a]))/(6*a))
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 + 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[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)] Int[u*(1 + b*(x^n/a))^(p*q) , x], x] /; FreeQ[{a, b, c, n, p, q}, x] && !GeQ[a, 0]
\[\int \frac {{\left (\frac {c}{\sqrt {b \,x^{2}+a}}\right )}^{\frac {3}{2}}}{x^{4}}d x\]
Input:
int((c/(b*x^2+a)^(1/2))^(3/2)/x^4,x)
Output:
int((c/(b*x^2+a)^(1/2))^(3/2)/x^4,x)
\[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^4} \, dx=\int { \frac {\left (\frac {c}{\sqrt {b x^{2} + a}}\right )^{\frac {3}{2}}}{x^{4}} \,d x } \] Input:
integrate((c/(b*x^2+a)^(1/2))^(3/2)/x^4,x, algorithm="fricas")
Output:
integral(sqrt(b*x^2 + a)*c*sqrt(c/sqrt(b*x^2 + a))/(b*x^6 + a*x^4), x)
\[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^4} \, dx=\int \frac {\left (\frac {c}{\sqrt {a + b x^{2}}}\right )^{\frac {3}{2}}}{x^{4}}\, dx \] Input:
integrate((c/(b*x**2+a)**(1/2))**(3/2)/x**4,x)
Output:
Integral((c/sqrt(a + b*x**2))**(3/2)/x**4, x)
\[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^4} \, dx=\int { \frac {\left (\frac {c}{\sqrt {b x^{2} + a}}\right )^{\frac {3}{2}}}{x^{4}} \,d x } \] Input:
integrate((c/(b*x^2+a)^(1/2))^(3/2)/x^4,x, algorithm="maxima")
Output:
integrate((c/sqrt(b*x^2 + a))^(3/2)/x^4, x)
Exception generated. \[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^4} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c/(b*x^2+a)^(1/2))^(3/2)/x^4,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{1,[0,1,1,2,0,0]%%%}+%%%{-1,[0,1,0,0,1,0]%%%} / %%%{1,[0,0, 2,0,0,2]%
Timed out. \[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (\frac {c}{\sqrt {b\,x^2+a}}\right )}^{3/2}}{x^4} \,d x \] Input:
int((c/(a + b*x^2)^(1/2))^(3/2)/x^4,x)
Output:
int((c/(a + b*x^2)^(1/2))^(3/2)/x^4, x)
\[ \int \frac {\left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}}{x^4} \, dx=\sqrt {c}\, \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{6}+a \,x^{4}}d x \right ) c \] Input:
int((c/(b*x^2+a)^(1/2))^(3/2)/x^4,x)
Output:
sqrt(c)*int((a + b*x**2)**(1/4)/(a*x**4 + b*x**6),x)*c