\(\int \frac {x^3}{(a+b x^2)^{3/4} (c+d x^2)^{3/4}} \, dx\) [1564]

Optimal result

Integrand size = 26, antiderivative size = 141 \[ \int \frac {x^3}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}{b d}+\frac {(b c+a d) \left (a+b x^2\right )^{3/4} \left (\frac {b \left (c+d x^2\right )}{d \left (a+b x^2\right )}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b c-a d}}{\sqrt {d} \sqrt {a+b x^2}}\right ),2\right )}{b^2 \sqrt {d} \sqrt {b c-a d} \left (c+d x^2\right )^{3/4}} \] Output:

(b*x^2+a)^(1/4)*(d*x^2+c)^(1/4)/b/d+(a*d+b*c)*(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^2/d^(1/2)/(-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 = 5.50 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.71 \[ \int \frac {x^3}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{a+b x^2} \left (b \left (c+d x^2\right )-(b c+a d) \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},\frac {d \left (a+b x^2\right )}{-b c+a d}\right )\right )}{b^2 d \left (c+d x^2\right )^{3/4}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {354, 90, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 229

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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