\(\int \frac {x^2 (c+d x^4)}{\sqrt {a+b x^4}} \, dx\) [32]

Optimal result

Integrand size = 22, antiderivative size = 264 \[ \int \frac {x^2 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {d x^3 \sqrt {a+b x^4}}{5 b}+\frac {(5 b c-3 a d) x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\sqrt [4]{a} (5 b c-3 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} (5 b c-3 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{10 b^{7/4} \sqrt {a+b x^4}} \] Output:

1/5*d*x^3*(b*x^4+a)^(1/2)/b+1/5*(-3*a*d+5*b*c)*x*(b*x^4+a)^(1/2)/b^(3/2)/( 
a^(1/2)+b^(1/2)*x^2)-1/5*a^(1/4)*(-3*a*d+5*b*c)*(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)+1/10*a^(1/4)*(-3*a*d+5*b*c)*(a 
^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacob 
iAM(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 = 10.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.28 \[ \int \frac {x^2 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {x^3 \left (3 d \left (a+b x^4\right )+(5 b c-3 a d) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{15 b \sqrt {a+b x^4}} \] Input:

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

Output:

(x^3*(3*d*(a + b*x^4) + (5*b*c - 3*a*d)*Sqrt[1 + (b*x^4)/a]*Hypergeometric 
2F1[1/2, 3/4, 7/4, -((b*x^4)/a)]))/(15*b*Sqrt[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {959, 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.

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

\(\Big \downarrow \) 959

\(\displaystyle \frac {(5 b c-3 a d) \int \frac {x^2}{\sqrt {b x^4+a}}dx}{5 b}+\frac {d x^3 \sqrt {a+b x^4}}{5 b}\)

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

(d*x^3*Sqrt[a + b*x^4])/(5*b) + ((5*b*c - 3*a*d)*(-((-((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]*EllipticF[2*ArcTan[(b^(1/4)*x 
)/a^(1/4)], 1/2])/(2*b^(3/4)*Sqrt[a + b*x^4])))/(5*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 959

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

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 = 0.88 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.47

method	result	size
risch	\(\frac {d \,x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {i \left (3 a d -5 c b \right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(125\)
elliptic	\(\frac {d \,x^{3} \sqrt {b \,x^{4}+a}}{5 b}+\frac {i \left (c -\frac {3 a d}{5 b}\right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(125\)
default	\(\frac {i c \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+d \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(215\)

Input:

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

Output:

1/5*d*x^3*(b*x^4+a)^(1/2)/b-1/5*I*(3*a*d-5*b*c)/b^(3/2)*a^(1/2)/(I/a^(1/2) 
*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^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 = 109, normalized size of antiderivative = 0.41 \[ \int \frac {x^2 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {{\left (5 \, b c - 3 \, a d\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (5 \, b c - 3 \, a d\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (b d x^{4} + 5 \, b c - 3 \, a d\right )} \sqrt {b x^{4} + a}}{5 \, b^{2} x} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.45 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.30 \[ \int \frac {x^2 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {c x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {d x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \] Input:

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

Output:

c*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*s 
qrt(a)*gamma(7/4)) + d*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**4*e 
xp_polar(I*pi)/a)/(4*sqrt(a)*gamma(11/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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