\(\int \frac {x^4}{\sqrt [4]{a+b x^2} (c+d x^2)^2} \, dx\) [1536]

Optimal result

Integrand size = 24, antiderivative size = 349 \[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^2} \, dx=\frac {(5 b c-4 a d) x}{2 d^2 (b c-a d) \sqrt [4]{a+b x^2}}-\frac {c x \left (a+b x^2\right )^{3/4}}{2 d (b c-a d) \left (c+d x^2\right )}-\frac {\sqrt {a} (5 b c-4 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {b} d^2 (b c-a d) \sqrt [4]{a+b x^2}}+\frac {\sqrt [4]{a} c (5 b c-6 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 d^{5/2} (-b c+a d)^{3/2} x}-\frac {\sqrt [4]{a} c (5 b c-6 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 d^{5/2} (-b c+a d)^{3/2} x} \] Output:

1/2*(-4*a*d+5*b*c)*x/d^2/(-a*d+b*c)/(b*x^2+a)^(1/4)-1/2*c*x*(b*x^2+a)^(3/4 
)/d/(-a*d+b*c)/(d*x^2+c)-1/2*a^(1/2)*(-4*a*d+5*b*c)*(1+b*x^2/a)^(1/4)*Elli 
pticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(1/2)/d^2/(-a*d+b*c)/( 
b*x^2+a)^(1/4)+1/4*a^(1/4)*c*(-6*a*d+5*b*c)*(-b*x^2/a)^(1/2)*EllipticPi((b 
*x^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)/d^(5/2)/(a*d-b*c 
)^(3/2)/x-1/4*a^(1/4)*c*(-6*a*d+5*b*c)*(-b*x^2/a)^(1/2)*EllipticPi((b*x^2+ 
a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)/d^(5/2)/(a*d-b*c)^(3/2 
)/x
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 9.21 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.72 \[ \int \frac {x^4}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^2} \, dx=\frac {x \left (\frac {(-5 b c+4 a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c d}+\frac {6 c \left (a+b x^2+\frac {6 a^2 c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}\right )}{d \left (c+d x^2\right )}\right )}{12 (-b c+a d) \sqrt [4]{a+b x^2}} \] Input:

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

Output:

(x*(((-5*b*c + 4*a*d)*x^2*(1 + (b*x^2)/a)^(1/4)*AppellF1[3/2, 1/4, 1, 5/2, 
 -((b*x^2)/a), -((d*x^2)/c)])/(c*d) + (6*c*(a + b*x^2 + (6*a^2*c*AppellF1[ 
1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/(-6*a*c*AppellF1[1/2, 1/4, 
1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(4*a*d*AppellF1[3/2, 1/4, 2, 5/2 
, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*x^2)/a 
), -((d*x^2)/c)]))))/(d*(c + d*x^2))))/(12*(-(b*c) + a*d)*(a + b*x^2)^(1/4 
))
 

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.18, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {395, 394}

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.

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

\(\displaystyle \frac {x^5 \sqrt [4]{\frac {b x^2}{a}+1} \operatorname {AppellF1}\left (\frac {5}{2},\frac {1}{4},2,\frac {7}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{5 c^2 \sqrt [4]{a+b x^2}}\)

Input:

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

Output:

(x^5*(1 + (b*x^2)/a)^(1/4)*AppellF1[5/2, 1/4, 2, 7/2, -((b*x^2)/a), -((d*x 
^2)/c)])/(5*c^2*(a + b*x^2)^(1/4))
 

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ 
FracPart[p])   Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ 
[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 
1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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