\(\int \frac {(c+d x^4)^2}{(a+b x^4)^{3/4}} \, dx\) [99]

Optimal result

Integrand size = 21, antiderivative size = 140 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {d (12 b c-5 a d) x \sqrt [4]{a+b x^4}}{12 b^2}+\frac {d^2 x^5 \sqrt [4]{a+b x^4}}{6 b}-\frac {\left (12 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 \sqrt {a} b^{3/2} \left (a+b x^4\right )^{3/4}} \] Output:

1/12*d*(-5*a*d+12*b*c)*x*(b*x^4+a)^(1/4)/b^2+1/6*d^2*x^5*(b*x^4+a)^(1/4)/b 
-1/12*(5*a^2*d^2-12*a*b*c*d+12*b^2*c^2)*(1+a/b/x^4)^(3/4)*x^3*InverseJacob 
iAM(1/2*arccot(b^(1/2)*x^2/a^(1/2)),2^(1/2))/a^(1/2)/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 = 13.10 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.15 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {x \left (1+\frac {b x^4}{a}\right )^{3/4} \left (13 a \left (45 c^2+18 c d x^4+5 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {13}{4},-\frac {b x^4}{a}\right )-6 b x^4 \left (7 c^2+10 c d x^4+3 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {7}{4},\frac {17}{4},-\frac {b x^4}{a}\right )-12 b x^4 \left (c+d x^4\right )^2 \, _3F_2\left (\frac {5}{4},\frac {7}{4},2;1,\frac {17}{4};-\frac {b x^4}{a}\right )\right )}{585 a \left (a+b x^4\right )^{3/4}} \] Input:

Integrate[(c + d*x^4)^2/(a + b*x^4)^(3/4),x]
 

Output:

(x*(1 + (b*x^4)/a)^(3/4)*(13*a*(45*c^2 + 18*c*d*x^4 + 5*d^2*x^8)*Hypergeom 
etric2F1[1/4, 3/4, 13/4, -((b*x^4)/a)] - 6*b*x^4*(7*c^2 + 10*c*d*x^4 + 3*d 
^2*x^8)*Hypergeometric2F1[5/4, 7/4, 17/4, -((b*x^4)/a)] - 12*b*x^4*(c + d* 
x^4)^2*HypergeometricPFQ[{5/4, 7/4, 2}, {1, 17/4}, -((b*x^4)/a)]))/(585*a* 
(a + b*x^4)^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.50 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {933, 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.

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

\(\Big \downarrow \) 933

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

\(\Big \downarrow \) 913

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 229

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

Input:

Int[(c + d*x^4)^2/(a + b*x^4)^(3/4),x]
 

Output:

(d*x*(a + b*x^4)^(1/4)*(c + d*x^4))/(6*b) + ((5*d*(2*b*c - a*d)*x*(a + b*x 
^4)^(1/4))/(2*b) - ((12*b^2*c^2 - 12*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*Sqrt[a]*Sqrt[b 
]*(a + b*x^4)^(3/4)))/(6*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 933

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

Maple [F]

\[\int \frac {\left (d \,x^{4}+c \right )^{2}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((d*x^4+c)^2/(b*x^4+a)^(3/4),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.77 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {c^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} + \frac {c d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {9}{4}\right )} + \frac {d^{2} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate((d*x**4+c)**2/(b*x**4+a)**(3/4),x)
 

Output:

c**2*x*gamma(1/4)*hyper((1/4, 3/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a 
**(3/4)*gamma(5/4)) + c*d*x**5*gamma(5/4)*hyper((3/4, 5/4), (9/4,), b*x**4 
*exp_polar(I*pi)/a)/(2*a**(3/4)*gamma(9/4)) + d**2*x**9*gamma(9/4)*hyper(( 
3/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/4)*gamma(13/4))
 

Maxima [F]

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

integrate((d*x^4+c)^2/(b*x^4+a)^(3/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)^2/(b*x^4 + a)^(3/4), x)
 

Giac [F]

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

integrate((d*x^4+c)^2/(b*x^4+a)^(3/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)^2/(b*x^4 + a)^(3/4), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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