\(\int \frac {x^2 (c+d x^4)}{(a+b x^4)^{13/4}} \, dx\) [174]

Optimal result

Integrand size = 22, antiderivative size = 134 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {(b c-a d) x^3}{9 a b \left (a+b x^4\right )^{9/4}}+\frac {(2 b c+a d) x^3}{15 a^2 b \left (a+b x^4\right )^{5/4}}-\frac {2 (2 b c+a d) \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 a^{5/2} b^{3/2} \sqrt [4]{a+b x^4}} \] Output:

1/9*(-a*d+b*c)*x^3/a/b/(b*x^4+a)^(9/4)+1/15*(a*d+2*b*c)*x^3/a^2/b/(b*x^4+a 
)^(5/4)-2/15*(a*d+2*b*c)*(1+a/b/x^4)^(1/4)*x*EllipticE(sin(1/2*arccot(b^(1 
/2)*x^2/a^(1/2))),2^(1/2))/a^(5/2)/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 = 122, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {x^3 \left (-a^3 d+2 b c \left (a+b x^4\right )^2 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {13}{4},\frac {7}{4},-\frac {b x^4}{a}\right )+a d \left (a+b x^4\right )^2 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {13}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{6 a^3 b \left (a+b x^4\right )^{9/4}} \] Input:

Integrate[(x^2*(c + d*x^4))/(a + b*x^4)^(13/4),x]
 

Output:

(x^3*(-(a^3*d) + 2*b*c*(a + b*x^4)^2*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2 
F1[3/4, 13/4, 7/4, -((b*x^4)/a)] + a*d*(a + b*x^4)^2*(1 + (b*x^4)/a)^(1/4) 
*Hypergeometric2F1[3/4, 13/4, 7/4, -((b*x^4)/a)]))/(6*a^3*b*(a + b*x^4)^(9 
/4))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {957, 819, 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.

\(\displaystyle \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx\)

\(\Big \downarrow \) 957

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

\(\Big \downarrow \) 819

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

\(\Big \downarrow \) 813

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 212

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

Input:

Int[(x^2*(c + d*x^4))/(a + b*x^4)^(13/4),x]
 

Output:

((b*c - a*d)*x^3)/(9*a*b*(a + b*x^4)^(9/4)) + ((2*b*c + a*d)*(x^3/(5*a*(a 
+ b*x^4)^(5/4)) - (2*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqr 
t[b]*x^2)]/2, 2])/(5*a^(3/2)*Sqrt[b]*(a + b*x^4)^(1/4))))/(3*a*b)
 

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 819

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

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 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)]))
 

Maple [F]

\[\int \frac {x^{2} \left (d \,x^{4}+c \right )}{\left (b \,x^{4}+a \right )^{\frac {13}{4}}}d x\]

Input:

int(x^2*(d*x^4+c)/(b*x^4+a)^(13/4),x)
 

Output:

int(x^2*(d*x^4+c)/(b*x^4+a)^(13/4),x)
 

Fricas [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{2}}{{\left (b x^{4} + a\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(x^2*(d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="fricas")
 

Output:

integral((d*x^6 + c*x^2)*(b*x^4 + a)^(3/4)/(b^4*x^16 + 4*a*b^3*x^12 + 6*a^ 
2*b^2*x^8 + 4*a^3*b*x^4 + a^4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 88.59 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.60 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {c x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {13}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {13}{4}} \Gamma \left (\frac {7}{4}\right )} + \frac {d x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {13}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {13}{4}} \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate(x**2*(d*x**4+c)/(b*x**4+a)**(13/4),x)
 

Output:

c*x**3*gamma(3/4)*hyper((3/4, 13/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4* 
a**(13/4)*gamma(7/4)) + d*x**7*gamma(7/4)*hyper((7/4, 13/4), (11/4,), b*x* 
*4*exp_polar(I*pi)/a)/(4*a**(13/4)*gamma(11/4))
 

Maxima [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{2}}{{\left (b x^{4} + a\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(x^2*(d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*x^2/(b*x^4 + a)^(13/4), x)
 

Giac [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{2}}{{\left (b x^{4} + a\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(x^2*(d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^2/(b*x^4 + a)^(13/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int \frac {x^2\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{13/4}} \,d x \] Input:

int((x^2*(c + d*x^4))/(a + b*x^4)^(13/4),x)
 

Output:

int((x^2*(c + d*x^4))/(a + b*x^4)^(13/4), x)
 

Reduce [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) d +\left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) c \] Input:

int(x^2*(d*x^4+c)/(b*x^4+a)^(13/4),x)
 

Output:

int(x**6/((a + b*x**4)**(1/4)*a**3 + 3*(a + b*x**4)**(1/4)*a**2*b*x**4 + 3 
*(a + b*x**4)**(1/4)*a*b**2*x**8 + (a + b*x**4)**(1/4)*b**3*x**12),x)*d + 
int(x**2/((a + b*x**4)**(1/4)*a**3 + 3*(a + b*x**4)**(1/4)*a**2*b*x**4 + 3 
*(a + b*x**4)**(1/4)*a*b**2*x**8 + (a + b*x**4)**(1/4)*b**3*x**12),x)*c