\(\int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx\) [369]

Optimal result

Integrand size = 15, antiderivative size = 261 \[ \int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx=-\frac {7 a x^3 \sqrt {a+b x^4}}{45 b^2}+\frac {x^7 \sqrt {a+b x^4}}{9 b}+\frac {7 a^2 x \sqrt {a+b x^4}}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {7 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^4}}+\frac {7 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{30 b^{11/4} \sqrt {a+b x^4}} \] Output:

-7/45*a*x^3*(b*x^4+a)^(1/2)/b^2+1/9*x^7*(b*x^4+a)^(1/2)/b+7/15*a^2*x*(b*x^ 
4+a)^(1/2)/b^(5/2)/(a^(1/2)+b^(1/2)*x^2)-7/15*a^(9/4)*(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^(11/4)/(b*x^4+a)^(1/2)+7/30*a^(9/4)*(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*ar 
ctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/b^(11/4)/(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 = 10.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.31 \[ \int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx=\frac {x^3 \left (-7 a^2-2 a b x^4+5 b^2 x^8+7 a^2 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{45 b^2 \sqrt {a+b x^4}} \] Input:

Integrate[x^10/Sqrt[a + b*x^4],x]
 

Output:

(x^3*(-7*a^2 - 2*a*b*x^4 + 5*b^2*x^8 + 7*a^2*Sqrt[1 + (b*x^4)/a]*Hypergeom 
etric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)]))/(45*b^2*Sqrt[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {843, 843, 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.

\(\displaystyle \int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \int \frac {x^6}{\sqrt {b x^4+a}}dx}{9 b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a}}dx}{5 b}\right )}{9 b}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {a} \sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^4}}-\frac {x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\)

Input:

Int[x^10/Sqrt[a + b*x^4],x]
 

Output:

(x^7*Sqrt[a + b*x^4])/(9*b) - (7*a*((x^3*Sqrt[a + b*x^4])/(5*b) - (3*a*(-( 
(-((x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[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)*Sqrt[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]*Ellipti 
cF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*b^(3/4)*Sqrt[a + b*x^4])))/(5*b 
)))/(9*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 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 843

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*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

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 = 1.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.48

method	result	size
risch	\(-\frac {x^{3} \left (-5 b \,x^{4}+7 a \right ) \sqrt {b \,x^{4}+a}}{45 b^{2}}+\frac {7 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(125\)
default	\(\frac {x^{7} \sqrt {b \,x^{4}+a}}{9 b}-\frac {7 a \,x^{3} \sqrt {b \,x^{4}+a}}{45 b^{2}}+\frac {7 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(133\)
elliptic	\(\frac {x^{7} \sqrt {b \,x^{4}+a}}{9 b}-\frac {7 a \,x^{3} \sqrt {b \,x^{4}+a}}{45 b^{2}}+\frac {7 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(133\)

Input:

int(x^10/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/45*x^3*(-5*b*x^4+7*a)/b^2*(b*x^4+a)^(1/2)+7/15*I*a^(5/2)/b^(5/2)/(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)-Ellipt 
icE(x*(I/a^(1/2)*b^(1/2))^(1/2),I))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.40 \[ \int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx=\frac {21 \, a^{2} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 21 \, a^{2} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (5 \, b^{2} x^{8} - 7 \, a b x^{4} + 21 \, a^{2}\right )} \sqrt {b x^{4} + a}}{45 \, b^{3} x} \] Input:

integrate(x^10/(b*x^4+a)^(1/2),x, algorithm="fricas")
 

Output:

1/45*(21*a^2*sqrt(b)*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) 
 - 21*a^2*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + 
(5*b^2*x^8 - 7*a*b*x^4 + 21*a^2)*sqrt(b*x^4 + a))/(b^3*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.14 \[ \int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx=\frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {15}{4}\right )} \] Input:

integrate(x**10/(b*x**4+a)**(1/2),x)
 

Output:

x**11*gamma(11/4)*hyper((1/2, 11/4), (15/4,), b*x**4*exp_polar(I*pi)/a)/(4 
*sqrt(a)*gamma(15/4))
 

Maxima [F]

\[ \int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx=\int { \frac {x^{10}}{\sqrt {b x^{4} + a}} \,d x } \] Input:

integrate(x^10/(b*x^4+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate(x^10/sqrt(b*x^4 + a), x)
 

Giac [F]

\[ \int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx=\int { \frac {x^{10}}{\sqrt {b x^{4} + a}} \,d x } \] Input:

integrate(x^10/(b*x^4+a)^(1/2),x, algorithm="giac")
 

Output:

integrate(x^10/sqrt(b*x^4 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx=\int \frac {x^{10}}{\sqrt {b\,x^4+a}} \,d x \] Input:

int(x^10/(a + b*x^4)^(1/2),x)
 

Output:

int(x^10/(a + b*x^4)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^{10}}{\sqrt {a+b x^4}} \, dx=\frac {-7 \sqrt {b \,x^{4}+a}\, a \,x^{3}+5 \sqrt {b \,x^{4}+a}\, b \,x^{7}+21 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) a^{2}}{45 b^{2}} \] Input:

int(x^10/(b*x^4+a)^(1/2),x)
 

Output:

( - 7*sqrt(a + b*x**4)*a*x**3 + 5*sqrt(a + b*x**4)*b*x**7 + 21*int((sqrt(a 
 + b*x**4)*x**2)/(a + b*x**4),x)*a**2)/(45*b**2)