\(\int \frac {x^3}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx\) [1551]

Optimal result

Integrand size = 26, antiderivative size = 149 \[ \int \frac {x^3}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\frac {\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}}{3 b d}-\frac {(b c-a d) (b c+a d) \sqrt [4]{-\frac {b d \left (a+b x^2\right ) \left (c+d x^2\right )}{(b c-a d)^2}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {b c+a d+2 b d x^2}{b c-a d}\right )\right |2\right )}{2 \sqrt {2} b^2 d^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:

1/3*(b*x^2+a)^(3/4)*(d*x^2+c)^(3/4)/b/d-1/4*(-a*d+b*c)*(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^2/d^2/(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 = 2.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\frac {\left (a+b x^2\right )^{3/4} \left (b \left (c+d x^2\right )-(b c+a d) \sqrt [4]{\frac {b \left (c+d x^2\right )}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {7}{4},\frac {d \left (a+b x^2\right )}{-b c+a d}\right )\right )}{3 b^2 d \sqrt [4]{c+d x^2}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.39 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.30, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {354, 90, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 839

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

\(\Big \downarrow \) 813

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 212

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

Input:

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

Output:

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

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 90

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

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]

\[\int \frac {x^{3}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )^{\frac {1}{4}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

\[ \int \frac {x^3}{\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=\int \frac {x^{3}}{\sqrt [4]{a + b x^{2}} \sqrt [4]{c + d x^{2}}}\, dx \] Input:

integrate(x**3/(b*x**2+a)**(1/4)/(d*x**2+c)**(1/4),x)
 

Output:

Integral(x**3/((a + b*x**2)**(1/4)*(c + d*x**2)**(1/4)), x)
 

Maxima [F]

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

integrate(x^3/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^3/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x**3/((c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)),x)