\(\int \frac {x^6}{\sqrt [4]{a+b x^2} (c+d x^2)} \, dx\) [1521]

Optimal result

Integrand size = 24, antiderivative size = 355 \[ \int \frac {x^6}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\frac {2 \left (15 b^2 c^2+2 a d (3 b c+2 a d)\right ) x}{15 b^2 d^3 \sqrt [4]{a+b x^2}}-\frac {2 (3 b c+2 a d) x \left (a+b x^2\right )^{3/4}}{15 b^2 d^2}+\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b d}-\frac {2 \sqrt {a} \left (15 b^2 c^2+2 a d (3 b c+2 a d)\right ) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{5/2} d^3 \sqrt [4]{a+b x^2}}-\frac {\sqrt [4]{a} c^3 \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{d^{7/2} \sqrt {-b c+a d} x}+\frac {\sqrt [4]{a} c^3 \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{d^{7/2} \sqrt {-b c+a d} x} \] Output:

2/15*(15*b^2*c^2+2*a*d*(2*a*d+3*b*c))*x/b^2/d^3/(b*x^2+a)^(1/4)-2/15*(2*a* 
d+3*b*c)*x*(b*x^2+a)^(3/4)/b^2/d^2+2/9*x^3*(b*x^2+a)^(3/4)/b/d-2/15*a^(1/2 
)*(15*b^2*c^2+2*a*d*(2*a*d+3*b*c))*(1+b*x^2/a)^(1/4)*EllipticE(sin(1/2*arc 
tan(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(5/2)/d^3/(b*x^2+a)^(1/4)-a^(1/4)*c^3*( 
-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^(7/2)/(a*d-b*c)^(1/2)/x+a^(1/4)*c^3*(-b*x^2/a)^(1/2)*Ellipt 
icPi((b*x^2+a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)/d^(7/2)/(a 
*d-b*c)^(1/2)/x
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 7.32 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.79 \[ \int \frac {x^6}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\frac {x \left (-2 \left (a+b x^2\right ) \left (9 b c+6 a d-5 b d x^2\right )+\frac {\left (15 b^2 c^2+6 a b c d+4 a^2 d^2\right ) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c}-\frac {36 a^2 c^2 (3 b c+2 a d) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\left (c+d x^2\right ) \left (-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )}{45 b^2 d^2 \sqrt [4]{a+b x^2}} \] Input:

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

Output:

(x*(-2*(a + b*x^2)*(9*b*c + 6*a*d - 5*b*d*x^2) + ((15*b^2*c^2 + 6*a*b*c*d 
+ 4*a^2*d^2)*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 - (36*a^2*c^2*(3*b*c + 2*a*d)*AppellF1[1/2, 1/4, 1, 
 3/2, -((b*x^2)/a), -((d*x^2)/c)])/((c + d*x^2)*(-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)])))))/(45*b^2*d^2*(a + b*x^2)^(1/4))
 

Rubi [C] (verified)

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.

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

\(\displaystyle \frac {x^7 \sqrt [4]{\frac {b x^2}{a}+1} \operatorname {AppellF1}\left (\frac {7}{2},\frac {1}{4},1,\frac {9}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{7 c \sqrt [4]{a+b x^2}}\)

Input:

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

Output:

(x^7*(1 + (b*x^2)/a)^(1/4)*AppellF1[7/2, 1/4, 1, 9/2, -((b*x^2)/a), -((d*x 
^2)/c)])/(7*c*(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]

\[\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )}d x\]

Input:

int(x^6/(b*x^2+a)^(1/4)/(d*x^2+c),x)
 

Output:

int(x^6/(b*x^2+a)^(1/4)/(d*x^2+c),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^6}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(x^6/(b*x^2+a)^(1/4)/(d*x^2+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate(x**6/(b*x**2+a)**(1/4)/(d*x**2+c),x)
 

Output:

Integral(x**6/((a + b*x**2)**(1/4)*(c + d*x**2)), x)
 

Maxima [F]

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

integrate(x^6/(b*x^2+a)^(1/4)/(d*x^2+c),x, algorithm="maxima")
 

Output:

integrate(x^6/((b*x^2 + a)^(1/4)*(d*x^2 + c)), x)
 

Giac [F]

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

integrate(x^6/(b*x^2+a)^(1/4)/(d*x^2+c),x, algorithm="giac")
 

Output:

integrate(x^6/((b*x^2 + a)^(1/4)*(d*x^2 + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx=\int \frac {x^6}{{\left (b\,x^2+a\right )}^{1/4}\,\left (d\,x^2+c\right )} \,d x \] Input:

int(x^6/((a + b*x^2)^(1/4)*(c + d*x^2)),x)
 

Output:

int(x^6/((a + b*x^2)^(1/4)*(c + d*x^2)), x)
 

Reduce [F]

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

int(x^6/(b*x^2+a)^(1/4)/(d*x^2+c),x)
 

Output:

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