\(\int \frac {x}{(a+b x^2)^{5/4} (c+d x^2)^{5/4}} \, dx\) [1578]

Optimal result

Integrand size = 24, antiderivative size = 137 \[ \int \frac {x}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=-\frac {2}{(b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}+\frac {4 \sqrt {d} \sqrt [4]{a+b x^2} \sqrt [4]{\frac {b \left (c+d x^2\right )}{d \left (a+b x^2\right )}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d}}{\sqrt {d} \sqrt {a+b x^2}}\right )\right |2\right )}{(b c-a d)^{3/2} \sqrt [4]{c+d x^2}} \] Output:

-2/(-a*d+b*c)/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)+4*d^(1/2)*(b*x^2+a)^(1/4)*(b 
*(d*x^2+c)/d/(b*x^2+a))^(1/4)*EllipticE(sin(1/2*arctan((-a*d+b*c)^(1/2)/d^ 
(1/2)/(b*x^2+a)^(1/2))),2^(1/2))/(-a*d+b*c)^(3/2)/(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 = 4.87 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.58 \[ \int \frac {x}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=-\frac {2 \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4},\frac {3}{4},\frac {d \left (a+b x^2\right )}{-b c+a d}\right )}{b \sqrt [4]{a+b x^2} \left (c+d x^2\right )^{5/4}} \] Input:

Integrate[x/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.41 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.75, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {353, 61, 61, 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.

\(\displaystyle \int \frac {x}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {1}{2} \int \frac {1}{\left (b x^2+a\right )^{5/4} \left (d x^2+c\right )^{5/4}}dx^2\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 839

\(\displaystyle \frac {1}{2} \left (-\frac {2 d \left (\frac {4 \left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2} (b c-a d)}-\frac {8 \left (\frac {x^6}{2 \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}-\frac {1}{2} \left (c-\frac {a d}{b}\right ) \int \frac {x^4}{\left (\frac {d x^8}{b}+c-\frac {a d}{b}\right )^{5/4}}d\sqrt [4]{b x^2+a}\right )}{b c-a d}\right )}{b c-a d}-\frac {4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} (b c-a d)}\right )\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {1}{2} \left (-\frac {2 d \left (\frac {4 \left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2} (b c-a d)}-\frac {8 \left (\frac {x^6}{2 \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}-\frac {b \sqrt [4]{a+b x^2} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d x^8}+1} \int \frac {1}{\left (\frac {b c-a d}{d x^8}+1\right )^{5/4} x^6}d\sqrt [4]{b x^2+a}}{2 d \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}\right )}{b c-a d}\right )}{b c-a d}-\frac {4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} (b c-a d)}\right )\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {1}{2} \left (-\frac {2 d \left (\frac {4 \left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2} (b c-a d)}-\frac {8 \left (\frac {b \sqrt [4]{a+b x^2} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d x^8}+1} \int \frac {1}{x^2 \left (\frac {(b c-a d) x^8}{d}+1\right )^{5/4}}d\frac {1}{x^2}}{2 d \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}+\frac {x^6}{2 \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}\right )}{b c-a d}\right )}{b c-a d}-\frac {4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} (b c-a d)}\right )\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {1}{2} \left (-\frac {2 d \left (\frac {4 \left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2} (b c-a d)}-\frac {8 \left (\frac {b \sqrt [4]{a+b x^2} \left (c-\frac {a d}{b}\right ) \sqrt [4]{\frac {b c-a d}{d x^8}+1} \int \frac {1}{\left (\frac {(b c-a d) x^4}{d}+1\right )^{5/4}}dx^4}{4 d \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}+\frac {x^6}{2 \sqrt [4]{-\frac {a d}{b}+\frac {d x^8}{b}+c}}\right )}{b c-a d}\right )}{b c-a d}-\frac {4}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} (b c-a d)}\right )\)

\(\Big \downarrow \) 212

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

Input:

Int[x/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
 

Output:

(-4/((b*c - a*d)*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4)) - (2*d*((4*(a + b*x^ 
2)^(3/4))/((b*c - a*d)*(c + d*x^2)^(1/4)) - (8*(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*c - a*d)))/(b*c - a 
*d))/2
 

Defintions of rubi rules used

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

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]

\[\int \frac {x}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )^{\frac {5}{4}}}d x\]

Input:

int(x/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Output:

int(x/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Fricas [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {x}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)*x/(b^2*d^2*x^8 + 2*(b^2*c*d + 
 a*b*d^2)*x^6 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4 + a^2*c^2 + 2*(a*b*c^2 
 + a^2*c*d)*x^2), x)
 

Sympy [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {x}{\left (a + b x^{2}\right )^{\frac {5}{4}} \left (c + d x^{2}\right )^{\frac {5}{4}}}\, dx \] Input:

integrate(x/(b*x**2+a)**(5/4)/(d*x**2+c)**(5/4),x)
 

Output:

Integral(x/((a + b*x**2)**(5/4)*(c + d*x**2)**(5/4)), x)
 

Maxima [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {x}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x, algorithm="maxima")
 

Output:

integrate(x/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(5/4)), x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {x}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {x}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x, algorithm="giac")
 

Output:

integrate(x/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(5/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {x}{{\left (b\,x^2+a\right )}^{5/4}\,{\left (d\,x^2+c\right )}^{5/4}} \,d x \] Input:

int(x/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x)
 

Output:

int(x/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int \frac {x}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{4}} \sqrt {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +a +b \,x^{2}}\, \sqrt {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}\, \left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +a +b \,x^{2}\right )}{a \left (a b \,d^{2} x^{4}-b^{2} c d \,x^{4}+a^{2} d^{2} x^{2}-b^{2} c^{2} x^{2}+a^{2} c d -a b \,c^{2}\right )} \] Input:

int(x/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Output:

(2*(c + d*x**2)**(3/4)*sqrt(sqrt(b)*sqrt(a + b*x**2)*x + a + b*x**2)*sqrt( 
sqrt(a + b*x**2) + sqrt(b)*x)*( - sqrt(b)*sqrt(a + b*x**2)*x + a + b*x**2) 
)/(a*(a**2*c*d + a**2*d**2*x**2 - a*b*c**2 + a*b*d**2*x**4 - b**2*c**2*x** 
2 - b**2*c*d*x**4))