\(\int \frac {x^2 (c+d x^4)}{(a+b x^4)^{3/2}} \, dx\) [42]

Optimal result

Integrand size = 22, antiderivative size = 274 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {(b c-a d) x^3}{2 a b \sqrt {a+b x^4}}-\frac {(b c-3 a d) x \sqrt {a+b x^4}}{2 a b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {(b c-3 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{7/4} \sqrt {a+b x^4}}-\frac {(b c-3 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 a^{3/4} b^{7/4} \sqrt {a+b x^4}} \] Output:

1/2*(-a*d+b*c)*x^3/a/b/(b*x^4+a)^(1/2)-1/2*(-3*a*d+b*c)*x*(b*x^4+a)^(1/2)/ 
a/b^(3/2)/(a^(1/2)+b^(1/2)*x^2)+1/2*(-3*a*d+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))/a^(3/4)/b^(7/4)/(b*x^4+a)^(1/2)-1/4*(-3*a*d+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)*InverseJacobiA 
M(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(3/4)/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 = 10.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.26 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x^3 \left (3 a d+(b c-3 a d) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{3 a b \sqrt {a+b x^4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {957, 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.

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

\(\Big \downarrow \) 957

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

((b*c - a*d)*x^3)/(2*a*b*Sqrt[a + b*x^4]) - ((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)*Sqr 
t[(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] + Sq 
rt[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*a*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 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

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 = 0.90 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.55

method	result	size
elliptic	\(-\frac {x^{3} \left (a d -c b \right )}{2 b a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {i \left (\frac {d}{b}+\frac {a d -c b}{2 a b}\right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(152\)
default	\(c \left (\frac {x^{3}}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+d \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(242\)

Input:

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

Output:

-1/2/b/a*x^3*(a*d-b*c)/((x^4+a/b)*b)^(1/2)+I*(d/b+1/2*(a*d-b*c)/a/b)*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)/b^(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 = 0.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.61 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {{\left ({\left (b^{2} c - 3 \, a b d\right )} x^{5} + {\left (a b c - 3 \, a^{2} d\right )} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left ({\left (b^{2} c - 3 \, a b d\right )} x^{5} + {\left (a b c - 3 \, a^{2} d\right )} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (2 \, a b d x^{4} - a b c + 3 \, a^{2} d\right )} \sqrt {b x^{4} + a}}{2 \, {\left (a b^{3} x^{5} + a^{2} b^{2} x\right )}} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.29 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {c x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {d x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \] Input:

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

Output:

c*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a 
**(3/2)*gamma(7/4)) + d*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]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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