\(\int \frac {x^2 (c+d x^4)}{(a+b x^4)^{5/4}} \, dx\) [118]

Optimal result

Integrand size = 22, antiderivative size = 93 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {d x^3}{2 b \sqrt [4]{a+b x^4}}-\frac {(2 b c-3 a d) \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {a} b^{3/2} \sqrt [4]{a+b x^4}} \] Output:

1/2*d*x^3/b/(b*x^4+a)^(1/4)-1/2*(-3*a*d+2*b*c)*(1+a/b/x^4)^(1/4)*x*Ellipti 
cE(sin(1/2*arccot(b^(1/2)*x^2/a^(1/2))),2^(1/2))/a^(1/2)/b^(3/2)/(b*x^4+a) 
^(1/4)
 

Mathematica [C] (verified)

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

Time = 10.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^3 \left (3 a d+(2 b c-3 a d) \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{6 a b \sqrt [4]{a+b x^4}} \] Input:

Integrate[(x^2*(c + d*x^4))/(a + b*x^4)^(5/4),x]
 

Output:

(x^3*(3*a*d + (2*b*c - 3*a*d)*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[3/4, 
 5/4, 7/4, -((b*x^4)/a)]))/(6*a*b*(a + b*x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {959, 813, 858, 807, 212}

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.

\(\displaystyle \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 813

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 212

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

Input:

Int[(x^2*(c + d*x^4))/(a + b*x^4)^(5/4),x]
 

Output:

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

Defintions of rubi rules used

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 813

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) 
)^(1/4)/(b*(a + b*x^4)^(1/4)))   Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] 
/; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

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]
 

Maple [F]

\[\int \frac {x^{2} \left (d \,x^{4}+c \right )}{\left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]

Input:

int(x^2*(d*x^4+c)/(b*x^4+a)^(5/4),x)
 

Output:

int(x^2*(d*x^4+c)/(b*x^4+a)^(5/4),x)
 

Fricas [F]

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

integrate(x^2*(d*x^4+c)/(b*x^4+a)^(5/4),x, algorithm="fricas")
 

Output:

integral((d*x^6 + c*x^2)*(b*x^4 + a)^(3/4)/(b^2*x^8 + 2*a*b*x^4 + a^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {c x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {7}{4}\right )} + \frac {d x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate(x**2*(d*x**4+c)/(b*x**4+a)**(5/4),x)
 

Output:

c*x**3*gamma(3/4)*hyper((3/4, 5/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a 
**(5/4)*gamma(7/4)) + d*x**7*gamma(7/4)*hyper((5/4, 7/4), (11/4,), b*x**4* 
exp_polar(I*pi)/a)/(4*a**(5/4)*gamma(11/4))
 

Maxima [F]

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

integrate(x^2*(d*x^4+c)/(b*x^4+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*x^2/(b*x^4 + a)^(5/4), x)
 

Giac [F]

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

integrate(x^2*(d*x^4+c)/(b*x^4+a)^(5/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x^4 + c)*x^2/(b*x^4 + a)^(5/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\int \frac {x^2\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \] Input:

int((x^2*(c + d*x^4))/(a + b*x^4)^(5/4),x)
 

Output:

int((x^2*(c + d*x^4))/(a + b*x^4)^(5/4), x)
 

Reduce [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) c \] Input:

int(x^2*(d*x^4+c)/(b*x^4+a)^(5/4),x)
 

Output:

int(x**6/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*d + int(x 
**2/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*c