\(\int \frac {\sqrt [4]{a+b x^2} (c+d x^2)}{x^4} \, dx\) [341]

Optimal result

Integrand size = 22, antiderivative size = 118 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^4} \, dx=\frac {(b c-6 a d) \sqrt [4]{a+b x^2}}{6 a x}-\frac {c \left (a+b x^2\right )^{5/4}}{3 a x^3}-\frac {\sqrt {b} (b c-6 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 \sqrt {a} \left (a+b x^2\right )^{3/4}} \] Output:

1/6*(-6*a*d+b*c)*(b*x^2+a)^(1/4)/a/x-1/3*c*(b*x^2+a)^(5/4)/a/x^3-1/6*b^(1/ 
2)*(-6*a*d+b*c)*(1+b*x^2/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x/a^( 
1/2)),2^(1/2))/a^(1/2)/(b*x^2+a)^(3/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^4} \, dx=\frac {\sqrt [4]{a+b x^2} \left (-c \left (a+b x^2\right )+\frac {(b c-6 a d) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {1}{2},-\frac {b x^2}{a}\right )}{2 \sqrt [4]{1+\frac {b x^2}{a}}}\right )}{3 a x^3} \] Input:

Integrate[((a + b*x^2)^(1/4)*(c + d*x^2))/x^4,x]
 

Output:

((a + b*x^2)^(1/4)*(-(c*(a + b*x^2)) + ((b*c - 6*a*d)*x^2*Hypergeometric2F 
1[-1/2, -1/4, 1/2, -((b*x^2)/a)])/(2*(1 + (b*x^2)/a)^(1/4))))/(3*a*x^3)
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {359, 247, 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.

\(\displaystyle \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^4} \, dx\)

\(\Big \downarrow \) 359

\(\displaystyle -\frac {(b c-6 a d) \int \frac {\sqrt [4]{b x^2+a}}{x^2}dx}{6 a}-\frac {c \left (a+b x^2\right )^{5/4}}{3 a x^3}\)

\(\Big \downarrow \) 247

\(\displaystyle -\frac {(b c-6 a d) \left (\frac {1}{2} b \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx-\frac {\sqrt [4]{a+b x^2}}{x}\right )}{6 a}-\frac {c \left (a+b x^2\right )^{5/4}}{3 a x^3}\)

\(\Big \downarrow \) 231

\(\displaystyle -\frac {(b c-6 a d) \left (\frac {b \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{2 \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a+b x^2}}{x}\right )}{6 a}-\frac {c \left (a+b x^2\right )^{5/4}}{3 a x^3}\)

\(\Big \downarrow \) 229

\(\displaystyle -\frac {(b c-6 a d) \left (\frac {\sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a+b x^2}}{x}\right )}{6 a}-\frac {c \left (a+b x^2\right )^{5/4}}{3 a x^3}\)

Input:

Int[((a + b*x^2)^(1/4)*(c + d*x^2))/x^4,x]
 

Output:

-1/3*(c*(a + b*x^2)^(5/4))/(a*x^3) - ((b*c - 6*a*d)*(-((a + b*x^2)^(1/4)/x 
) + (Sqrt[a]*Sqrt[b]*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sq 
rt[a]]/2, 2])/(a + b*x^2)^(3/4)))/(6*a)
 

Defintions of rubi rules used

rule 229

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]
 

rule 231

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]
 

rule 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 359

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]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )}{x^{4}}d x\]

Input:

int((b*x^2+a)^(1/4)*(d*x^2+c)/x^4,x)
 

Output:

int((b*x^2+a)^(1/4)*(d*x^2+c)/x^4,x)
 

Fricas [F]

\[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}}{x^{4}} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)*(d*x^2+c)/x^4,x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(1/4)*(d*x^2 + c)/x^4, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^4} \, dx=- \frac {\sqrt [4]{a} c {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 x^{3}} - \frac {\sqrt [4]{a} d {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} \] Input:

integrate((b*x**2+a)**(1/4)*(d*x**2+c)/x**4,x)
 

Output:

-a**(1/4)*c*hyper((-3/2, -1/4), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*x**3 
) - a**(1/4)*d*hyper((-1/2, -1/4), (1/2,), b*x**2*exp_polar(I*pi)/a)/x
 

Maxima [F]

\[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}}{x^{4}} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)*(d*x^2+c)/x^4,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(1/4)*(d*x^2 + c)/x^4, x)
 

Giac [F]

\[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}}{x^{4}} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)*(d*x^2+c)/x^4,x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(1/4)*(d*x^2 + c)/x^4, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^4} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (d\,x^2+c\right )}{x^4} \,d x \] Input:

int(((a + b*x^2)^(1/4)*(c + d*x^2))/x^4,x)
 

Output:

int(((a + b*x^2)^(1/4)*(c + d*x^2))/x^4, x)
 

Reduce [F]

\[ \int \frac {\sqrt [4]{a+b x^2} \left (c+d x^2\right )}{x^4} \, dx=\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d -2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c -10 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{2}+6 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{6}+a \,x^{4}}d x \right ) a^{2} d \,x^{3}-\left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{6}+a \,x^{4}}d x \right ) a b c \,x^{3}}{5 b \,x^{3}} \] Input:

int((b*x^2+a)^(1/4)*(d*x^2+c)/x^4,x)
 

Output:

(2*(a + b*x**2)**(1/4)*a*d - 2*(a + b*x**2)**(1/4)*b*c - 10*(a + b*x**2)** 
(1/4)*b*d*x**2 + 6*int((a + b*x**2)**(1/4)/(a*x**4 + b*x**6),x)*a**2*d*x** 
3 - int((a + b*x**2)**(1/4)/(a*x**4 + b*x**6),x)*a*b*c*x**3)/(5*b*x**3)