\(\int \frac {x}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx\) [1552]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 106 \[ \int \frac {x}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\frac {(b c-a d) \sqrt [4]{-\frac {b d \left (a+b x^2\right ) \left (c+d x^2\right )}{(b c-a d)^2}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {b c+a d+2 b d x^2}{b c-a d}\right )\right |2\right )}{\sqrt {2} b d \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:

1/2*(-a*d+b*c)*(-b*d*(b*x^2+a)*(d*x^2+c)/(-a*d+b*c)^2)^(1/4)*EllipticE(sin 
(1/2*arcsin((2*b*d*x^2+a*d+b*c)/(-a*d+b*c))),2^(1/2))*2^(1/2)/b/d/(b*x^2+a 
)^(1/4)/(d*x^2+c)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 1.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.76 \[ \int \frac {x}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\frac {2 \left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {b \left (c+d x^2\right )}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {7}{4},\frac {d \left (a+b x^2\right )}{-b c+a d}\right )}{3 b \sqrt [4]{c+d x^2}} \] Input:

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

Output:

(2*(a + b*x^2)^(3/4)*((b*(c + d*x^2))/(b*c - a*d))^(1/4)*Hypergeometric2F1 
[1/4, 3/4, 7/4, (d*(a + b*x^2))/(-(b*c) + a*d)])/(3*b*(c + d*x^2)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.38, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {353, 73, 839, 813, 858, 807, 212}

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.

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 \int \frac {x^4}{\sqrt [4]{\frac {d x^8}{b}+c-\frac {a d}{b}}}d\sqrt [4]{b x^2+a}}{b}\)

\(\Big \downarrow \) 839

\(\displaystyle \frac {2 \left (\frac {x^6}{2 \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}-\frac {1}{2} \left (c-\frac {a d}{b}\right ) \int \frac {x^4}{\left (\frac {d x^8}{b}+c-\frac {a d}{b}\right )^{5/4}}d\sqrt [4]{b x^2+a}\right )}{b}\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {2 \left (\frac {x^6}{2 \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}-\frac {b \sqrt [4]{a+b x^2} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d x^8}+1} \int \frac {1}{\left (\frac {b c-a d}{d x^8}+1\right )^{5/4} x^6}d\sqrt [4]{b x^2+a}}{2 d \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}\right )}{b}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {2 \left (\frac {b \sqrt [4]{a+b x^2} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d x^8}+1} \int \frac {1}{x^2 \left (\frac {(b c-a d) x^8}{d}+1\right )^{5/4}}d\frac {1}{x^2}}{2 d \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}+\frac {x^6}{2 \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}\right )}{b}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {2 \left (\frac {b \sqrt [4]{a+b x^2} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d x^8}+1} \int \frac {1}{\left (\frac {(b c-a d) x^4}{d}+1\right )^{5/4}}dx^4}{4 d \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}+\frac {x^6}{2 \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}\right )}{b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {2 \left (\frac {b \sqrt [4]{a+b x^2} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d x^8}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} x^4}{\sqrt {d}}\right )\right |2\right )}{2 \sqrt {d} \sqrt {b c-a d} \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}+\frac {x^6}{2 \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}\right )}{b}\)

Input:

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

Output:

(2*(x^6/(2*(c - (a*d)/b + (d*x^8)/b)^(1/4)) + (b*(c - (a*d)/b)*(1 + (b*c - 
 a*d)/(d*x^8))^(1/4)*(a + b*x^2)^(1/4)*EllipticE[ArcTan[(Sqrt[b*c - a*d]*x 
^4)/Sqrt[d]]/2, 2])/(2*Sqrt[d]*Sqrt[b*c - a*d]*(c - (a*d)/b + (d*x^8)/b)^( 
1/4))))/b
 

Defintions of rubi rules used

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 212
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 353
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

rule 807
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 813
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) 
)^(1/4)/(b*(a + b*x^4)^(1/4)))   Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] 
/; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 839
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 
)^(1/4)), x] - Simp[a/2   Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} 
, x] && PosQ[b/a]
 

rule 858
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)*x/(b*d*x^4 + (b*c + a*d)*x^2 
+ a*c), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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