\(\int \frac {x^5}{(a+b x^2)^{5/4} (c+d x^2)^{5/4}} \, dx\) [1576]

Optimal result

Integrand size = 26, antiderivative size = 227 \[ \int \frac {x^5}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=-\frac {2 a^2}{b^2 (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}-\frac {2 \left (b^2 c^2+a^2 d^2\right ) \left (a+b x^2\right )^{3/4}}{b^2 d (b c-a d)^2 \sqrt [4]{c+d x^2}}+\frac {\left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt [4]{-\frac {b d \left (a+b x^2\right ) \left (c+d x^2\right )}{(b c-a d)^2}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {b c+a d+2 b d x^2}{b c-a d}\right )\right |2\right )}{\sqrt {2} b^2 d^2 (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:

-2*a^2/b^2/(-a*d+b*c)/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)-2*(a^2*d^2+b^2*c^2)* 
(b*x^2+a)^(3/4)/b^2/d/(-a*d+b*c)^2/(d*x^2+c)^(1/4)+1/2*(3*a^2*d^2-2*a*b*c* 
d+3*b^2*c^2)*(-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^2/d^2/(-a*d+ 
b*c)/(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 = 5.49 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.70 \[ \int \frac {x^5}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {-6 b \left (a b c^2+b^2 c^2 x^2+a^2 d \left (c+d x^2\right )\right )+2 \left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \left (a+b x^2\right ) \sqrt [4]{\frac {b \left (c+d x^2\right )}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {7}{4},\frac {d \left (a+b x^2\right )}{-b c+a d}\right )}{3 b^2 d (b c-a d)^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:

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

Output:

(-6*b*(a*b*c^2 + b^2*c^2*x^2 + a^2*d*(c + d*x^2)) + 2*(3*b^2*c^2 - 2*a*b*c 
*d + 3*a^2*d^2)*(a + b*x^2)*((b*(c + d*x^2))/(b*c - a*d))^(1/4)*Hypergeome 
tric2F1[1/4, 3/4, 7/4, (d*(a + b*x^2))/(-(b*c) + a*d)])/(3*b^2*d*(b*c - a* 
d)^2*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.55 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.29, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {354, 100, 27, 87, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 839

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

\(\Big \downarrow \) 813

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 212

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(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] && IntegerQ 
[(m - 1)/2]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(x^5/(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^5/(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^5}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {x^{5}}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a c +\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{4}}d x \] Input:

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

Output:

int(x**5/((c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*a*c + (c + d*x**2)**(1/4 
)*(a + b*x**2)**(1/4)*a*d*x**2 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b 
*c*x**2 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b*d*x**4),x)