\(\int \frac {x}{(a+b x^2)^{3/4} (c+d x^2)^{3/4}} \, dx\) [1565]

Optimal result

Integrand size = 24, antiderivative size = 105 \[ \int \frac {x}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=-\frac {2 \sqrt {d} \left (a+b x^2\right )^{3/4} \left (\frac {b \left (c+d x^2\right )}{d \left (a+b x^2\right )}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d}}{\sqrt {d} \sqrt {a+b x^2}}\right ),2\right )}{b \sqrt {b c-a d} \left (c+d x^2\right )^{3/4}} \] Output:

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

Mathematica [C] (verified)

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

Time = 4.87 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.75 \[ \int \frac {x}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\frac {2 \sqrt [4]{a+b x^2} \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},\frac {d \left (a+b x^2\right )}{-b c+a d}\right )}{b \left (c+d x^2\right )^{3/4}} \] Input:

Integrate[x/((a + b*x^2)^(3/4)*(c + d*x^2)^(3/4)),x]
 

Output:

(2*(a + b*x^2)^(1/4)*((b*(c + d*x^2))/(b*c - a*d))^(3/4)*Hypergeometric2F1 
[1/4, 3/4, 5/4, (d*(a + b*x^2))/(-(b*c) + a*d)])/(b*(c + d*x^2)^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {353, 73, 768, 858, 807, 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.

\(\displaystyle \int \frac {x}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {1}{2} \int \frac {1}{\left (b x^2+a\right )^{3/4} \left (d x^2+c\right )^{3/4}}dx^2\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 \int \frac {1}{\left (\frac {d x^8}{b}+c-\frac {a d}{b}\right )^{3/4}}d\sqrt [4]{b x^2+a}}{b}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {2 x^6 \left (\frac {b c-a d}{d x^8}+1\right )^{3/4} \int \frac {1}{\left (\frac {b c-a d}{d x^8}+1\right )^{3/4} x^6}d\sqrt [4]{b x^2+a}}{b \left (-\frac {a d}{b}+\frac {d x^8}{b}+c\right )^{3/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {2 x^6 \left (\frac {b c-a d}{d x^8}+1\right )^{3/4} \int \frac {1}{x^2 \left (\frac {(b c-a d) x^8}{d}+1\right )^{3/4}}d\frac {1}{x^2}}{b \left (-\frac {a d}{b}+\frac {d x^8}{b}+c\right )^{3/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {x^6 \left (\frac {b c-a d}{d x^8}+1\right )^{3/4} \int \frac {1}{\left (\frac {(b c-a d) x^4}{d}+1\right )^{3/4}}dx^4}{b \left (-\frac {a d}{b}+\frac {d x^8}{b}+c\right )^{3/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle -\frac {2 \sqrt {d} x^6 \left (\frac {b c-a d}{d x^8}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d} x^4}{\sqrt {d}}\right ),2\right )}{b \sqrt {b c-a d} \left (-\frac {a d}{b}+\frac {d x^8}{b}+c\right )^{3/4}}\)

Input:

Int[x/((a + b*x^2)^(3/4)*(c + d*x^2)^(3/4)),x]
 

Output:

(-2*Sqrt[d]*(1 + (b*c - a*d)/(d*x^8))^(3/4)*x^6*EllipticF[ArcTan[(Sqrt[b*c 
 - a*d]*x^4)/Sqrt[d]]/2, 2])/(b*Sqrt[b*c - a*d]*(c - (a*d)/b + (d*x^8)/b)^ 
(3/4))
 

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 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 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 768

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

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 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]

\[\int \frac {x}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (x^{2} d +c \right )^{\frac {3}{4}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int { \frac {x}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(x/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int \frac {x}{\left (a + b x^{2}\right )^{\frac {3}{4}} \left (c + d x^{2}\right )^{\frac {3}{4}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int { \frac {x}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(x/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int { \frac {x}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(x/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int \frac {x}{{\left (b\,x^2+a\right )}^{3/4}\,{\left (d\,x^2+c\right )}^{3/4}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int \frac {x}{\left (d \,x^{2}+c \right )^{\frac {3}{4}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \] Input:

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

Output:

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