\(\int \frac {(c+d x^4)^2}{(a+b x^4)^{7/4}} \, dx\) [100]

Optimal result

Integrand size = 21, antiderivative size = 141 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {(b c-a d)^2 x}{3 a b^2 \left (a+b x^4\right )^{3/4}}+\frac {d^2 x \sqrt [4]{a+b x^4}}{2 b^2}-\frac {\left (4 b^2 c^2+4 a b c d-5 a^2 d^2\right ) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{6 a^{3/2} b^{3/2} \left (a+b x^4\right )^{3/4}} \] Output:

1/3*(-a*d+b*c)^2*x/a/b^2/(b*x^4+a)^(3/4)+1/2*d^2*x*(b*x^4+a)^(1/4)/b^2-1/6 
*(-5*a^2*d^2+4*a*b*c*d+4*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^(3/2)/b^(3/2)/(b*x^4+a)^(3/4)
 

Mathematica [C] (warning: unable to verify)

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

Time = 13.19 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.21 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {x \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Gamma}\left (\frac {3}{4}\right ) \left (13 a \left (45 c^2+18 c d x^4+5 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {7}{4},\frac {13}{4},-\frac {b x^4}{a}\right )-14 b x^4 \left (7 c^2+10 c d x^4+3 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {11}{4},\frac {17}{4},-\frac {b x^4}{a}\right )-28 b x^4 \left (c+d x^4\right )^2 \, _3F_2\left (\frac {5}{4},2,\frac {11}{4};1,\frac {17}{4};-\frac {b x^4}{a}\right )\right )}{780 a^2 \left (a+b x^4\right )^{3/4} \operatorname {Gamma}\left (\frac {7}{4}\right )} \] Input:

Integrate[(c + d*x^4)^2/(a + b*x^4)^(7/4),x]
 

Output:

(x*(1 + (b*x^4)/a)^(3/4)*Gamma[3/4]*(13*a*(45*c^2 + 18*c*d*x^4 + 5*d^2*x^8 
)*Hypergeometric2F1[1/4, 7/4, 13/4, -((b*x^4)/a)] - 14*b*x^4*(7*c^2 + 10*c 
*d*x^4 + 3*d^2*x^8)*Hypergeometric2F1[5/4, 11/4, 17/4, -((b*x^4)/a)] - 28* 
b*x^4*(c + d*x^4)^2*HypergeometricPFQ[{5/4, 2, 11/4}, {1, 17/4}, -((b*x^4) 
/a)]))/(780*a^2*(a + b*x^4)^(3/4)*Gamma[7/4])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {930, 913, 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.

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

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 913

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 229

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

Input:

Int[(c + d*x^4)^2/(a + b*x^4)^(7/4),x]
 

Output:

((b*c - a*d)*x*(c + d*x^4))/(3*a*b*(a + b*x^4)^(3/4)) + (-1/2*(d*(2*b*c - 
5*a*d)*x*(a + b*x^4)^(1/4))/b - ((4*b^2*c^2 + 4*a*b*c*d - 5*a^2*d^2)*(1 + 
a/(b*x^4))^(3/4)*x^3*EllipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*Sqr 
t[a]*Sqrt[b]*(a + b*x^4)^(3/4)))/(3*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 913

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

\[\int \frac {\left (d \,x^{4}+c \right )^{2}}{\left (b \,x^{4}+a \right )^{\frac {7}{4}}}d x\]

Input:

int((d*x^4+c)^2/(b*x^4+a)^(7/4),x)
 

Output:

int((d*x^4+c)^2/(b*x^4+a)^(7/4),x)
 

Fricas [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate((d*x^4+c)^2/(b*x^4+a)^(7/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{7/4}} \, dx=\int \frac {\left (c + d x^{4}\right )^{2}}{\left (a + b x^{4}\right )^{\frac {7}{4}}}\, dx \] Input:

integrate((d*x**4+c)**2/(b*x**4+a)**(7/4),x)
 

Output:

Integral((c + d*x**4)**2/(a + b*x**4)**(7/4), x)
 

Maxima [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate((d*x^4+c)^2/(b*x^4+a)^(7/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)^2/(b*x^4 + a)^(7/4), x)
 

Giac [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate((d*x^4+c)^2/(b*x^4+a)^(7/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)^2/(b*x^4 + a)^(7/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{7/4}} \, dx=\int \frac {{\left (d\,x^4+c\right )}^2}{{\left (b\,x^4+a\right )}^{7/4}} \,d x \] Input:

int((c + d*x^4)^2/(a + b*x^4)^(7/4),x)
 

Output:

int((c + d*x^4)^2/(a + b*x^4)^(7/4), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{7/4}} \, dx=\left (\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) d^{2}+2 \left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) c^{2} \] Input:

int((d*x^4+c)^2/(b*x^4+a)^(7/4),x)
                                                                                    
                                                                                    
 

Output:

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