\(\int x^2 \sqrt {a+c x^4} \, dx\) [328]

Optimal result

Integrand size = 15, antiderivative size = 234 \[ \int x^2 \sqrt {a+c x^4} \, dx=\frac {1}{5} x^3 \sqrt {a+c x^4}+\frac {2 a x \sqrt {a+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {2 a^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{3/4} \sqrt {a+c x^4}}+\frac {a^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 c^{3/4} \sqrt {a+c x^4}} \] Output:

1/5*x^3*(c*x^4+a)^(1/2)+2/5*a*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+c^(1/2)*x 
^2)-2/5*a^(5/4)*(a^(1/2)+c^(1/2)*x^2)*((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^ 
(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))/c^(3/4)/(c*x 
^4+a)^(1/2)+1/5*a^(5/4)*(a^(1/2)+c^(1/2)*x^2)*((c*x^4+a)/(a^(1/2)+c^(1/2)* 
x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4)*x/a^(1/4)),1/2*2^(1/2))/c^( 
3/4)/(c*x^4+a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 4.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.22 \[ \int x^2 \sqrt {a+c x^4} \, dx=\frac {x^3 \sqrt {a+c x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^4}{a}\right )}{3 \sqrt {1+\frac {c x^4}{a}}} \] Input:

Integrate[x^2*Sqrt[a + c*x^4],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {811, 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.

\(\displaystyle \int x^2 \sqrt {a+c x^4} \, dx\)

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

Int[x^2*Sqrt[a + c*x^4],x]
 

Output:

(x^3*Sqrt[a + c*x^4])/5 + (2*a*(-((-((x*Sqrt[a + c*x^4])/(Sqrt[a] + Sqrt[c 
]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqr 
t[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c^(1/4)*Sqrt[ 
a + c*x^4]))/Sqrt[c]) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/ 
(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/ 
(2*c^(3/4)*Sqrt[a + c*x^4])))/5
 

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 811

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 
))   Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 = 112, normalized size of antiderivative = 0.48

method	result	size
default	\(\frac {x^{3} \sqrt {c \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\)	\(112\)
risch	\(\frac {x^{3} \sqrt {c \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\)	\(112\)
elliptic	\(\frac {x^{3} \sqrt {c \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\)	\(112\)

Input:

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

Output:

1/5*x^3*(c*x^4+a)^(1/2)+2/5*I*a^(3/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/ 
2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)/c^(1 
/2)*(EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*c^(1/ 
2))^(1/2),I))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.38 \[ \int x^2 \sqrt {a+c x^4} \, dx=\frac {2 \, a \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 2 \, a \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (c x^{4} + 2 \, a\right )} \sqrt {c x^{4} + a}}{5 \, c x} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.17 \[ \int x^2 \sqrt {a+c x^4} \, dx=\frac {\sqrt {a} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Output:

sqrt(a)*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/ 
a)/(4*gamma(7/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^2 \sqrt {a+c x^4} \, dx=\int x^2\,\sqrt {c\,x^4+a} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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