\(\int \frac {x^6}{(a+b x^4)^{3/2}} \, dx\) [392]

Optimal result

Integrand size = 15, antiderivative size = 236 \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {x^3}{2 b \sqrt {a+b x^4}}+\frac {3 x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {3 \sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 b^{7/4} \sqrt {a+b x^4}} \] Output:

-1/2*x^3/b/(b*x^4+a)^(1/2)+3/2*x*(b*x^4+a)^(1/2)/b^(3/2)/(a^(1/2)+b^(1/2)* 
x^2)-3/2*a^(1/4)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2) 
^(1/2)*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))/b^(7/4)/(b* 
x^4+a)^(1/2)+3/4*a^(1/4)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2) 
*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/b^ 
(7/4)/(b*x^4+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 9.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.24 \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x^3-x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )}{b \sqrt {a+b x^4}} \] Input:

Integrate[x^6/(a + b*x^4)^(3/2),x]
 

Output:

(x^3 - x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[3/4, 3/2, 7/4, -((b*x^4)/ 
a)])/(b*Sqrt[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {817, 834, 27, 761, 1510}

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.

\(\displaystyle \int \frac {x^6}{\left (a+b x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {3 \int \frac {x^2}{\sqrt {b x^4+a}}dx}{2 b}-\frac {x^3}{2 b \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

Int[x^6/(a + b*x^4)^(3/2),x]
 

Output:

-1/2*x^3/(b*Sqrt[a + b*x^4]) + (3*(-((-((x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqr 
t[b]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + 
Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(b^(1/4)*Sq 
rt[a + b*x^4]))/Sqrt[b]) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^ 
4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2 
])/(2*b^(3/4)*Sqrt[a + b*x^4])))/(2*b)
 

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 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 817

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( 
n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.50

method	result	size
default	\(-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(119\)
elliptic	\(-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(119\)

Input:

int(x^6/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/b*x^3/((x^4+a/b)*b)^(1/2)+3/2*I/b^(3/2)*a^(1/2)/(I/a^(1/2)*b^(1/2))^( 
1/2)*(1-I*b^(1/2)*x^2/a^(1/2))^(1/2)*(1+I*b^(1/2)*x^2/a^(1/2))^(1/2)/(b*x^ 
4+a)^(1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2 
)*b^(1/2))^(1/2),I))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.48 \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {3 \, {\left (b x^{5} + a x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 3 \, {\left (b x^{5} + a x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (2 \, b x^{4} + 3 \, a\right )} \sqrt {b x^{4} + a}}{2 \, {\left (b^{3} x^{5} + a b^{2} x\right )}} \] Input:

integrate(x^6/(b*x^4+a)^(3/2),x, algorithm="fricas")
 

Output:

1/2*(3*(b*x^5 + a*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x 
), -1) - 3*(b*x^5 + a*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/ 
4)/x), -1) + (2*b*x^4 + 3*a)*sqrt(b*x^4 + a))/(b^3*x^5 + a*b^2*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.16 \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate(x**6/(b*x**4+a)**(3/2),x)
 

Output:

x**7*gamma(7/4)*hyper((3/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*a* 
*(3/2)*gamma(11/4))
 

Maxima [F]

\[ \int \frac {x^6}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{6}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^6/(b*x^4+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^6/(b*x^4 + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {x^6}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{6}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^6/(b*x^4+a)^(3/2),x, algorithm="giac")
 

Output:

integrate(x^6/(b*x^4 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^6}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:

int(x^6/(a + b*x^4)^(3/2),x)
 

Output:

int(x^6/(a + b*x^4)^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^6}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {\sqrt {b \,x^{4}+a}\, x^{3}-3 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2}-3 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a b \,x^{4}}{b \left (b \,x^{4}+a \right )} \] Input:

int(x^6/(b*x^4+a)^(3/2),x)
 

Output:

(sqrt(a + b*x**4)*x**3 - 3*int((sqrt(a + b*x**4)*x**2)/(a**2 + 2*a*b*x**4 
+ b**2*x**8),x)*a**2 - 3*int((sqrt(a + b*x**4)*x**2)/(a**2 + 2*a*b*x**4 + 
b**2*x**8),x)*a*b*x**4)/(b*(a + b*x**4))