\(\int x^2 (a+c x^4)^{3/2} \, dx\) [348]

Optimal result

Integrand size = 15, antiderivative size = 255 \[ \int x^2 \left (a+c x^4\right )^{3/2} \, dx=\frac {2}{15} a x^3 \sqrt {a+c x^4}+\frac {4 a^2 x \sqrt {a+c x^4}}{15 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{9} x^3 \left (a+c x^4\right )^{3/2}-\frac {4 a^{9/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}}+\frac {2 a^{9/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 c^{3/4} \sqrt {a+c x^4}} \] Output:

2/15*a*x^3*(c*x^4+a)^(1/2)+4/15*a^2*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+c^( 
1/2)*x^2)+1/9*x^3*(c*x^4+a)^(3/2)-4/15*a^(9/4)*(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)+2/15*a^(9/4)*(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 = 6.80 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.20 \[ \int x^2 \left (a+c x^4\right )^{3/2} \, dx=\frac {a x^3 \sqrt {a+c x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^4}{a}\right )}{3 \sqrt {1+\frac {c x^4}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {811, 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.

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

\(\Big \downarrow \) 811

\(\displaystyle \frac {2}{3} a \int x^2 \sqrt {c x^4+a}dx+\frac {1}{9} x^3 \left (a+c x^4\right )^{3/2}\)

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

(x^3*(a + c*x^4)^(3/2))/9 + (2*a*((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] + Sqrt[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*A 
rcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*c^(3/4)*Sqrt[a + c*x^4])))/5))/3
 

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 = 0.65 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.48

method	result	size
risch	\(\frac {x^{3} \left (5 c \,x^{4}+11 a \right ) \sqrt {c \,x^{4}+a}}{45}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{15 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\)	\(122\)
default	\(\frac {x^{7} \sqrt {c \,x^{4}+a}\, c}{9}+\frac {11 a \,x^{3} \sqrt {c \,x^{4}+a}}{45}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{15 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\)	\(128\)
elliptic	\(\frac {x^{7} \sqrt {c \,x^{4}+a}\, c}{9}+\frac {11 a \,x^{3} \sqrt {c \,x^{4}+a}}{45}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{15 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\)	\(128\)

Input:

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

Output:

1/45*x^3*(5*c*x^4+11*a)*(c*x^4+a)^(1/2)+4/15*I*a^(5/2)/(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 = 104, normalized size of antiderivative = 0.41 \[ \int x^2 \left (a+c x^4\right )^{3/2} \, dx=\frac {12 \, a^{2} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 12 \, a^{2} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (5 \, c^{2} x^{8} + 11 \, a c x^{4} + 12 \, a^{2}\right )} \sqrt {c x^{4} + a}}{45 \, c x} \] Input:

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

Output:

1/45*(12*a^2*sqrt(c)*x*(-a/c)^(3/4)*elliptic_e(arcsin((-a/c)^(1/4)/x), -1) 
 - 12*a^2*sqrt(c)*x*(-a/c)^(3/4)*elliptic_f(arcsin((-a/c)^(1/4)/x), -1) + 
(5*c^2*x^8 + 11*a*c*x^4 + 12*a^2)*sqrt(c*x^4 + a))/(c*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.15 \[ \int x^2 \left (a+c x^4\right )^{3/2} \, dx=\frac {a^{\frac {3}{2}} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^2 \left (a+c x^4\right )^{3/2} \, dx=\frac {11 \sqrt {c \,x^{4}+a}\, a \,x^{3}}{45}+\frac {\sqrt {c \,x^{4}+a}\, c \,x^{7}}{9}+\frac {4 \left (\int \frac {\sqrt {c \,x^{4}+a}\, x^{2}}{c \,x^{4}+a}d x \right ) a^{2}}{15} \] Input:

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

Output:

(11*sqrt(a + c*x**4)*a*x**3 + 5*sqrt(a + c*x**4)*c*x**7 + 12*int((sqrt(a + 
 c*x**4)*x**2)/(a + c*x**4),x)*a**2)/45