Integrand size = 22, antiderivative size = 116 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^2} \, dx=\frac {(3 b c+2 a d) x \sqrt [4]{a+b x^2}}{3 a}-\frac {c \left (a+b x^2\right )^{5/4}}{a x}+\frac {\sqrt {a} (3 b c+2 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {b} \left (a+b x^2\right )^{3/4}} \] Output:
1/3*(2*a*d+3*b*c)*x*(b*x^2+a)^(1/4)/a-c*(b*x^2+a)^(5/4)/a/x+1/3*a^(1/2)*(2 *a*d+3*b*c)*(1+b*x^2/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2) ),2^(1/2))/b^(1/2)/(b*x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 6.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^2} \, dx=\frac {\sqrt [4]{a+b x^2} \left (-c \left (a+b x^2\right )+\frac {(3 b c+2 a d) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{2 \sqrt [4]{1+\frac {b x^2}{a}}}\right )}{a x} \] Input:
Integrate[((a + b*x^2)^(1/4)*(c + d*x^2))/x^2,x]
Output:
((a + b*x^2)^(1/4)*(-(c*(a + b*x^2)) + ((3*b*c + 2*a*d)*x^2*Hypergeometric 2F1[-1/4, 1/2, 3/2, -((b*x^2)/a)])/(2*(1 + (b*x^2)/a)^(1/4))))/(a*x)
Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {359, 211, 231, 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.
\(\displaystyle \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {(2 a d+3 b c) \int \sqrt [4]{b x^2+a}dx}{2 a}-\frac {c \left (a+b x^2\right )^{5/4}}{a x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {(2 a d+3 b c) \left (\frac {1}{3} a \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{2 a}-\frac {c \left (a+b x^2\right )^{5/4}}{a x}\) |
\(\Big \downarrow \) 231 |
\(\displaystyle \frac {(2 a d+3 b c) \left (\frac {a \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{2 a}-\frac {c \left (a+b x^2\right )^{5/4}}{a x}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {(2 a d+3 b c) \left (\frac {2 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{2 a}-\frac {c \left (a+b x^2\right )^{5/4}}{a x}\) |
Input:
Int[((a + b*x^2)^(1/4)*(c + d*x^2))/x^2,x]
Output:
-((c*(a + b*x^2)^(5/4))/(a*x)) + ((3*b*c + 2*a*d)*((2*x*(a + b*x^2)^(1/4)) /3 + (2*a^(3/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a] ]/2, 2])/(3*Sqrt[b]*(a + b*x^2)^(3/4))))/(2*a)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( a + b*x^2)^(3/4) Int[1/(1 + b*(x^2/a))^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )}{x^{2}}d x\]
Input:
int((b*x^2+a)^(1/4)*(d*x^2+c)/x^2,x)
Output:
int((b*x^2+a)^(1/4)*(d*x^2+c)/x^2,x)
\[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}}{x^{2}} \,d x } \] Input:
integrate((b*x^2+a)^(1/4)*(d*x^2+c)/x^2,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/4)*(d*x^2 + c)/x^2, x)
Result contains complex when optimal does not.
Time = 1.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^2} \, dx=- \frac {\sqrt [4]{a} c {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} + \sqrt [4]{a} d x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} \] Input:
integrate((b*x**2+a)**(1/4)*(d*x**2+c)/x**2,x)
Output:
-a**(1/4)*c*hyper((-1/2, -1/4), (1/2,), b*x**2*exp_polar(I*pi)/a)/x + a**( 1/4)*d*x*hyper((-1/4, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a)
\[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}}{x^{2}} \,d x } \] Input:
integrate((b*x^2+a)^(1/4)*(d*x^2+c)/x^2,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(1/4)*(d*x^2 + c)/x^2, x)
\[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}}{x^{2}} \,d x } \] Input:
integrate((b*x^2+a)^(1/4)*(d*x^2+c)/x^2,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(1/4)*(d*x^2 + c)/x^2, x)
Time = 0.91 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^2} \, dx=\frac {d\,x\,{\left (b\,x^2+a\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{1/4}}-\frac {2\,c\,{\left (b\,x^2+a\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {a}{b\,x^2}\right )}{x\,{\left (\frac {a}{b\,x^2}+1\right )}^{1/4}} \] Input:
int(((a + b*x^2)^(1/4)*(c + d*x^2))/x^2,x)
Output:
(d*x*(a + b*x^2)^(1/4)*hypergeom([-1/4, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(1/4) - (2*c*(a + b*x^2)^(1/4)*hypergeom([-1/4, 1/4], 5/4, -a/(b*x^2 )))/(x*(a/(b*x^2) + 1)^(1/4))
\[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^2} \, dx=\frac {-2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d -6 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c +2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{2}-2 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{4}+a \,x^{2}}d x \right ) a^{2} d x -3 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{4}+a \,x^{2}}d x \right ) a b c x}{3 b x} \] Input:
int((b*x^2+a)^(1/4)*(d*x^2+c)/x^2,x)
Output:
( - 2*(a + b*x**2)**(1/4)*a*d - 6*(a + b*x**2)**(1/4)*b*c + 2*(a + b*x**2) **(1/4)*b*d*x**2 - 2*int((a + b*x**2)**(1/4)/(a*x**2 + b*x**4),x)*a**2*d*x - 3*int((a + b*x**2)**(1/4)/(a*x**2 + b*x**4),x)*a*b*c*x)/(3*b*x)