\(\int \frac {x^2 (c+d x^4)}{(a+b x^4)^{17/4}} \, dx\) [188]

Optimal result

Integrand size = 22, antiderivative size = 169 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {(b c-a d) x^3}{13 a b \left (a+b x^4\right )^{13/4}}+\frac {(10 b c+3 a d) x^3}{117 a^2 b \left (a+b x^4\right )^{9/4}}+\frac {2 (10 b c+3 a d) x^3}{195 a^3 b \left (a+b x^4\right )^{5/4}}-\frac {4 (10 b c+3 a d) \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{195 a^{7/2} b^{3/2} \sqrt [4]{a+b x^4}} \] Output:

1/13*(-a*d+b*c)*x^3/a/b/(b*x^4+a)^(13/4)+1/117*(3*a*d+10*b*c)*x^3/a^2/b/(b 
*x^4+a)^(9/4)+2/195*(3*a*d+10*b*c)*x^3/a^3/b/(b*x^4+a)^(5/4)-4/195*(3*a*d+ 
10*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^(7/2)/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 = 10.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.73 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {x^3 \left (-3 a^4 d+10 b c \left (a+b x^4\right )^3 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {17}{4},\frac {7}{4},-\frac {b x^4}{a}\right )+3 a d \left (a+b x^4\right )^3 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {17}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{30 a^4 b \left (a+b x^4\right )^{13/4}} \] Input:

Integrate[(x^2*(c + d*x^4))/(a + b*x^4)^(17/4),x]
 

Output:

(x^3*(-3*a^4*d + 10*b*c*(a + b*x^4)^3*(1 + (b*x^4)/a)^(1/4)*Hypergeometric 
2F1[3/4, 17/4, 7/4, -((b*x^4)/a)] + 3*a*d*(a + b*x^4)^3*(1 + (b*x^4)/a)^(1 
/4)*Hypergeometric2F1[3/4, 17/4, 7/4, -((b*x^4)/a)]))/(30*a^4*b*(a + b*x^4 
)^(13/4))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {957, 819, 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.

\(\displaystyle \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {(3 a d+10 b c) \int \frac {x^2}{\left (b x^4+a\right )^{13/4}}dx}{13 a b}+\frac {x^3 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 819

\(\displaystyle \frac {(3 a d+10 b c) \left (\frac {2 \int \frac {x^2}{\left (b x^4+a\right )^{9/4}}dx}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^3 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 819

\(\displaystyle \frac {(3 a d+10 b c) \left (\frac {2 \left (\frac {2 \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^3 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {(3 a d+10 b c) \left (\frac {2 \left (\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{5 a b \sqrt [4]{a+b x^4}}+\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^3 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {(3 a d+10 b c) \left (\frac {2 \left (\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}-\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{5 a b \sqrt [4]{a+b x^4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^3 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(3 a d+10 b c) \left (\frac {2 \left (\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}-\frac {x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{5 a b \sqrt [4]{a+b x^4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^3 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {(3 a d+10 b c) \left (\frac {2 \left (\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}-\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{5 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^3 (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

Input:

Int[(x^2*(c + d*x^4))/(a + b*x^4)^(17/4),x]
 

Output:

((b*c - a*d)*x^3)/(13*a*b*(a + b*x^4)^(13/4)) + ((10*b*c + 3*a*d)*(x^3/(9* 
a*(a + b*x^4)^(9/4)) + (2*(x^3/(5*a*(a + b*x^4)^(5/4)) - (2*(1 + a/(b*x^4) 
)^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(5*a^(3/2)*Sqrt[b 
]*(a + b*x^4)^(1/4))))/(3*a)))/(13*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]

\[\int \frac {x^{2} \left (d \,x^{4}+c \right )}{\left (b \,x^{4}+a \right )^{\frac {17}{4}}}d x\]

Input:

int(x^2*(d*x^4+c)/(b*x^4+a)^(17/4),x)
 

Output:

int(x^2*(d*x^4+c)/(b*x^4+a)^(17/4),x)
 

Fricas [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{2}}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:

integrate(x^2*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="fricas")
 

Output:

integral((d*x^6 + c*x^2)*(b*x^4 + a)^(3/4)/(b^5*x^20 + 5*a*b^4*x^16 + 10*a 
^2*b^3*x^12 + 10*a^3*b^2*x^8 + 5*a^4*b*x^4 + a^5), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\text {Timed out} \] Input:

integrate(x**2*(d*x**4+c)/(b*x**4+a)**(17/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{2}}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:

integrate(x^2*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*x^2/(b*x^4 + a)^(17/4), x)
 

Giac [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{2}}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:

integrate(x^2*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^2/(b*x^4 + a)^(17/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int \frac {x^2\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{17/4}} \,d x \] Input:

int((x^2*(c + d*x^4))/(a + b*x^4)^(17/4),x)
 

Output:

int((x^2*(c + d*x^4))/(a + b*x^4)^(17/4), x)
 

Reduce [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) d +\left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) c \] Input:

int(x^2*(d*x^4+c)/(b*x^4+a)^(17/4),x)
                                                                                    
                                                                                    
 

Output:

int(x**6/((a + b*x**4)**(1/4)*a**4 + 4*(a + b*x**4)**(1/4)*a**3*b*x**4 + 6 
*(a + b*x**4)**(1/4)*a**2*b**2*x**8 + 4*(a + b*x**4)**(1/4)*a*b**3*x**12 + 
 (a + b*x**4)**(1/4)*b**4*x**16),x)*d + int(x**2/((a + b*x**4)**(1/4)*a**4 
 + 4*(a + b*x**4)**(1/4)*a**3*b*x**4 + 6*(a + b*x**4)**(1/4)*a**2*b**2*x** 
8 + 4*(a + b*x**4)**(1/4)*a*b**3*x**12 + (a + b*x**4)**(1/4)*b**4*x**16),x 
)*c