\(\int \frac {(c+d x^4)^2}{(a+b x^4)^{11/4}} \, dx\) [101]

Optimal result

Integrand size = 21, antiderivative size = 158 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{11/4}} \, dx=\frac {(b c-a d)^2 x}{7 a b^2 \left (a+b x^4\right )^{7/4}}+\frac {2 (b c-a d) (3 b c+4 a d) x}{21 a^2 b^2 \left (a+b x^4\right )^{3/4}}-\frac {\left (12 b^2 c^2+4 a b c d+5 a^2 d^2\right ) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{21 a^{5/2} b^{3/2} \left (a+b x^4\right )^{3/4}} \] Output:

1/7*(-a*d+b*c)^2*x/a/b^2/(b*x^4+a)^(7/4)+2/21*(-a*d+b*c)*(4*a*d+3*b*c)*x/a 
^2/b^2/(b*x^4+a)^(3/4)-1/21*(5*a^2*d^2+4*a*b*c*d+12*b^2*c^2)*(1+a/b/x^4)^( 
3/4)*x^3*InverseJacobiAM(1/2*arccot(b^(1/2)*x^2/a^(1/2)),2^(1/2))/a^(5/2)/ 
b^(3/2)/(b*x^4+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 15.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{11/4}} \, dx=\frac {x \left (3 a (b c-a d)^2+2 \left (3 b^2 c^2+a b c d-4 a^2 d^2\right ) \left (a+b x^4\right )+\left (12 b^2 c^2+4 a b c d+5 a^2 d^2\right ) \left (a+b x^4\right ) \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{21 a^2 b^2 \left (a+b x^4\right )^{7/4}} \] Input:

Integrate[(c + d*x^4)^2/(a + b*x^4)^(11/4),x]
 

Output:

(x*(3*a*(b*c - a*d)^2 + 2*(3*b^2*c^2 + a*b*c*d - 4*a^2*d^2)*(a + b*x^4) + 
(12*b^2*c^2 + 4*a*b*c*d + 5*a^2*d^2)*(a + b*x^4)*(1 + (b*x^4)/a)^(3/4)*Hyp 
ergeometric2F1[1/4, 3/4, 5/4, -((b*x^4)/a)]))/(21*a^2*b^2*(a + b*x^4)^(7/4 
))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {930, 910, 768, 858, 807, 229}

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.

\(\displaystyle \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{11/4}} \, dx\)

\(\Big \downarrow \) 930

\(\displaystyle \frac {\int \frac {d (2 b c+5 a d) x^4+c (6 b c+a d)}{\left (b x^4+a\right )^{7/4}}dx}{7 a b}+\frac {x \left (c+d x^4\right ) (b c-a d)}{7 a b \left (a+b x^4\right )^{7/4}}\)

\(\Big \downarrow \) 910

\(\displaystyle \frac {\frac {1}{3} \left (\frac {12 b c^2}{a}+\frac {5 a d^2}{b}+4 c d\right ) \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx+\frac {x \left (\frac {6 b c^2}{a}-\frac {5 a d^2}{b}-c d\right )}{3 \left (a+b x^4\right )^{3/4}}}{7 a b}+\frac {x \left (c+d x^4\right ) (b c-a d)}{7 a b \left (a+b x^4\right )^{7/4}}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \left (\frac {12 b c^2}{a}+\frac {5 a d^2}{b}+4 c d\right ) \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{3 \left (a+b x^4\right )^{3/4}}+\frac {x \left (\frac {6 b c^2}{a}-\frac {5 a d^2}{b}-c d\right )}{3 \left (a+b x^4\right )^{3/4}}}{7 a b}+\frac {x \left (c+d x^4\right ) (b c-a d)}{7 a b \left (a+b x^4\right )^{7/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {\frac {x \left (\frac {6 b c^2}{a}-\frac {5 a d^2}{b}-c d\right )}{3 \left (a+b x^4\right )^{3/4}}-\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \left (\frac {12 b c^2}{a}+\frac {5 a d^2}{b}+4 c d\right ) \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{3 \left (a+b x^4\right )^{3/4}}}{7 a b}+\frac {x \left (c+d x^4\right ) (b c-a d)}{7 a b \left (a+b x^4\right )^{7/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\frac {x \left (\frac {6 b c^2}{a}-\frac {5 a d^2}{b}-c d\right )}{3 \left (a+b x^4\right )^{3/4}}-\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \left (\frac {12 b c^2}{a}+\frac {5 a d^2}{b}+4 c d\right ) \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{6 \left (a+b x^4\right )^{3/4}}}{7 a b}+\frac {x \left (c+d x^4\right ) (b c-a d)}{7 a b \left (a+b x^4\right )^{7/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\frac {x \left (\frac {6 b c^2}{a}-\frac {5 a d^2}{b}-c d\right )}{3 \left (a+b x^4\right )^{3/4}}-\frac {\sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \left (\frac {12 b c^2}{a}+\frac {5 a d^2}{b}+4 c d\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{3 \sqrt {a} \left (a+b x^4\right )^{3/4}}}{7 a b}+\frac {x \left (c+d x^4\right ) (b c-a d)}{7 a b \left (a+b x^4\right )^{7/4}}\)

Input:

Int[(c + d*x^4)^2/(a + b*x^4)^(11/4),x]
 

Output:

((b*c - a*d)*x*(c + d*x^4))/(7*a*b*(a + b*x^4)^(7/4)) + ((((6*b*c^2)/a - c 
*d - (5*a*d^2)/b)*x)/(3*(a + b*x^4)^(3/4)) - (Sqrt[b]*((12*b*c^2)/a + 4*c* 
d + (5*a*d^2)/b)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcTan[Sqrt[a]/(Sqrt[ 
b]*x^2)]/2, 2])/(3*Sqrt[a]*(a + b*x^4)^(3/4)))/(7*a*b)
 

Defintions of rubi rules used

rule 229

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

rule 768

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

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 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 910

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - 
 b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1))   Int[(a + b*x^n)^(p + 1), x], x] /; 
FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ 
n + p, 0])
 

rule 930

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[(a*d - c*b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 
1))), x] - Simp[1/(a*b*n*(p + 1))   Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q 
- 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*( 
p + q) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q, x]
 

Maple [F]

\[\int \frac {\left (d \,x^{4}+c \right )^{2}}{\left (b \,x^{4}+a \right )^{\frac {11}{4}}}d x\]

Input:

int((d*x^4+c)^2/(b*x^4+a)^(11/4),x)
 

Output:

int((d*x^4+c)^2/(b*x^4+a)^(11/4),x)
 

Fricas [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{11/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{{\left (b x^{4} + a\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate((d*x^4+c)^2/(b*x^4+a)^(11/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{11/4}} \, dx=\int \frac {\left (c + d x^{4}\right )^{2}}{\left (a + b x^{4}\right )^{\frac {11}{4}}}\, dx \] Input:

integrate((d*x**4+c)**2/(b*x**4+a)**(11/4),x)
 

Output:

Integral((c + d*x**4)**2/(a + b*x**4)**(11/4), x)
 

Maxima [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{11/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{{\left (b x^{4} + a\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate((d*x^4+c)^2/(b*x^4+a)^(11/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)^2/(b*x^4 + a)^(11/4), x)
 

Giac [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{11/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{{\left (b x^{4} + a\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate((d*x^4+c)^2/(b*x^4+a)^(11/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)^2/(b*x^4 + a)^(11/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{11/4}} \, dx=\int \frac {{\left (d\,x^4+c\right )}^2}{{\left (b\,x^4+a\right )}^{11/4}} \,d x \] Input:

int((c + d*x^4)^2/(a + b*x^4)^(11/4),x)
 

Output:

int((c + d*x^4)^2/(a + b*x^4)^(11/4), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{11/4}} \, dx=\left (\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{2} x^{8}}d x \right ) d^{2}+2 \left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{2} x^{8}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{2} x^{8}}d x \right ) c^{2} \] Input:

int((d*x^4+c)^2/(b*x^4+a)^(11/4),x)
                                                                                    
                                                                                    
 

Output:

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