\(\int \frac {x^3}{(a+b x^2)^{5/4} (c+d x^2)^{5/4}} \, dx\) [1577]

Optimal result

Integrand size = 26, antiderivative size = 151 \[ \int \frac {x^3}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {2 a}{b (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}-\frac {2 (b c+a d) \sqrt [4]{a+b x^2} \sqrt [4]{\frac {b \left (c+d x^2\right )}{d \left (a+b x^2\right )}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d}}{\sqrt {d} \sqrt {a+b x^2}}\right )\right |2\right )}{b \sqrt {d} (b c-a d)^{3/2} \sqrt [4]{c+d x^2}} \] Output:

2*a/b/(-a*d+b*c)/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)-2*(a*d+b*c)*(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))/b/d^(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 = 5.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.70 \[ \int \frac {x^3}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {-2 b c+2 (b c+a d) \sqrt [4]{\frac {b \left (c+d x^2\right )}{b c-a d}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},\frac {d \left (a+b x^2\right )}{-b c+a d}\right )}{b d (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:

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

Output:

(-2*b*c + 2*(b*c + a*d)*((b*(c + d*x^2))/(b*c - a*d))^(1/4)*Hypergeometric 
2F1[-1/4, 1/4, 3/4, (d*(a + b*x^2))/(-(b*c) + a*d)])/(b*d*(b*c - a*d)*(a + 
 b*x^2)^(1/4)*(c + d*x^2)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.41 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.70, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {354, 87, 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.

\(\displaystyle \int \frac {x^3}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 839

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

\(\Big \downarrow \) 813

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 212

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

Input:

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

Output:

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

\[\int \frac {x^{3}}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )^{\frac {5}{4}}}d x\]

Input:

int(x^3/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Output:

int(x^3/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Fricas [F]

\[ \int \frac {x^3}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {x^{3}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x^3/(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^3/(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]

\[ \int \frac {x^3}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {x^{3}}{\left (a + b x^{2}\right )^{\frac {5}{4}} \left (c + d x^{2}\right )^{\frac {5}{4}}}\, dx \] Input:

integrate(x**3/(b*x**2+a)**(5/4)/(d*x**2+c)**(5/4),x)
 

Output:

Integral(x**3/((a + b*x**2)**(5/4)*(c + d*x**2)**(5/4)), x)
 

Maxima [F]

\[ \int \frac {x^3}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {x^{3}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x^3/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate(x^3/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(5/4)), x)
 

Giac [F]

\[ \int \frac {x^3}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {x^{3}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x^3/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x, algorithm="giac")
 

Output:

integrate(x^3/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(5/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {x^3}{{\left (b\,x^2+a\right )}^{5/4}\,{\left (d\,x^2+c\right )}^{5/4}} \,d x \] Input:

int(x^3/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x)
 

Output:

int(x^3/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {x^3}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {x^{3}}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a c +\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{4}}d x \] Input:

int(x^3/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Output:

int(x**3/((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)