\(\int \frac {c+d x^4}{(a+b x^4)^{15/4}} \, dx\) [90]

Optimal result

Integrand size = 19, antiderivative size = 162 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{15/4}} \, dx=\frac {(b c-a d) x}{11 a b \left (a+b x^4\right )^{11/4}}+\frac {(10 b c+a d) x}{77 a^2 b \left (a+b x^4\right )^{7/4}}+\frac {2 (10 b c+a d) x}{77 a^3 b \left (a+b x^4\right )^{3/4}}-\frac {4 (10 b c+a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{77 a^{7/2} \sqrt {b} \left (a+b x^4\right )^{3/4}} \] Output:

1/11*(-a*d+b*c)*x/a/b/(b*x^4+a)^(11/4)+1/77*(a*d+10*b*c)*x/a^2/b/(b*x^4+a) 
^(7/4)+2/77*(a*d+10*b*c)*x/a^3/b/(b*x^4+a)^(3/4)-4/77*(a*d+10*b*c)*(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^ 
(7/2)/b^(1/2)/(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 = 10.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.68 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{15/4}} \, dx=\frac {-77 a^3 d x+(10 b c+a d) x \left (7 a^2+10 a \left (a+b x^4\right )+20 \left (a+b x^4\right )^2+40 \left (a+b x^4\right )^2 \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{770 a^3 b \left (a+b x^4\right )^{11/4}} \] Input:

Integrate[(c + d*x^4)/(a + b*x^4)^(15/4),x]
 

Output:

(-77*a^3*d*x + (10*b*c + a*d)*x*(7*a^2 + 10*a*(a + b*x^4) + 20*(a + b*x^4) 
^2 + 40*(a + b*x^4)^2*(1 + (b*x^4)/a)^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/ 
4, -((b*x^4)/a)]))/(770*a^3*b*(a + b*x^4)^(11/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.54 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {910, 749, 749, 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.

\(\displaystyle \int \frac {c+d x^4}{\left (a+b x^4\right )^{15/4}} \, dx\)

\(\Big \downarrow \) 910

\(\displaystyle \frac {(a d+10 b c) \int \frac {1}{\left (b x^4+a\right )^{11/4}}dx}{11 a b}+\frac {x (b c-a d)}{11 a b \left (a+b x^4\right )^{11/4}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {(a d+10 b c) \left (\frac {6 \int \frac {1}{\left (b x^4+a\right )^{7/4}}dx}{7 a}+\frac {x}{7 a \left (a+b x^4\right )^{7/4}}\right )}{11 a b}+\frac {x (b c-a d)}{11 a b \left (a+b x^4\right )^{11/4}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {(a d+10 b c) \left (\frac {6 \left (\frac {2 \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx}{3 a}+\frac {x}{3 a \left (a+b x^4\right )^{3/4}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^4\right )^{7/4}}\right )}{11 a b}+\frac {x (b c-a d)}{11 a b \left (a+b x^4\right )^{11/4}}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {(a d+10 b c) \left (\frac {6 \left (\frac {2 x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{3 a \left (a+b x^4\right )^{3/4}}+\frac {x}{3 a \left (a+b x^4\right )^{3/4}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^4\right )^{7/4}}\right )}{11 a b}+\frac {x (b c-a d)}{11 a b \left (a+b x^4\right )^{11/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {(a d+10 b c) \left (\frac {6 \left (\frac {x}{3 a \left (a+b x^4\right )^{3/4}}-\frac {2 x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{3 a \left (a+b x^4\right )^{3/4}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^4\right )^{7/4}}\right )}{11 a b}+\frac {x (b c-a d)}{11 a b \left (a+b x^4\right )^{11/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(a d+10 b c) \left (\frac {6 \left (\frac {x}{3 a \left (a+b x^4\right )^{3/4}}-\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{3 a \left (a+b x^4\right )^{3/4}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^4\right )^{7/4}}\right )}{11 a b}+\frac {x (b c-a d)}{11 a b \left (a+b x^4\right )^{11/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {(a d+10 b c) \left (\frac {6 \left (\frac {x}{3 a \left (a+b x^4\right )^{3/4}}-\frac {2 \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{3 a^{3/2} \left (a+b x^4\right )^{3/4}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^4\right )^{7/4}}\right )}{11 a b}+\frac {x (b c-a d)}{11 a b \left (a+b x^4\right )^{11/4}}\)

Input:

Int[(c + d*x^4)/(a + b*x^4)^(15/4),x]
 

Output:

((b*c - a*d)*x)/(11*a*b*(a + b*x^4)^(11/4)) + ((10*b*c + a*d)*(x/(7*a*(a + 
 b*x^4)^(7/4)) + (6*(x/(3*a*(a + b*x^4)^(3/4)) - (2*Sqrt[b]*(1 + a/(b*x^4) 
)^(3/4)*x^3*EllipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(3*a^(3/2)*(a + 
 b*x^4)^(3/4))))/(7*a)))/(11*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 749

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 
 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1))   Int[(a + b*x^ 
n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte 
gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
 

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])
 

Maple [F]

\[\int \frac {d \,x^{4}+c}{\left (b \,x^{4}+a \right )^{\frac {15}{4}}}d x\]

Input:

int((d*x^4+c)/(b*x^4+a)^(15/4),x)
 

Output:

int((d*x^4+c)/(b*x^4+a)^(15/4),x)
 

Fricas [F]

\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{15/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((d*x^4+c)/(b*x^4+a)^(15/4),x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(1/4)*(d*x^4 + c)/(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 = 134.88 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.48 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{15/4}} \, dx=\frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {15}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {15}{4}} \Gamma \left (\frac {5}{4}\right )} + \frac {d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {15}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {15}{4}} \Gamma \left (\frac {9}{4}\right )} \] Input:

integrate((d*x**4+c)/(b*x**4+a)**(15/4),x)
 

Output:

c*x*gamma(1/4)*hyper((1/4, 15/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a** 
(15/4)*gamma(5/4)) + d*x**5*gamma(5/4)*hyper((5/4, 15/4), (9/4,), b*x**4*e 
xp_polar(I*pi)/a)/(4*a**(15/4)*gamma(9/4))
 

Maxima [F]

\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{15/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((d*x^4+c)/(b*x^4+a)^(15/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)/(b*x^4 + a)^(15/4), x)
 

Giac [F]

\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{15/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((d*x^4+c)/(b*x^4+a)^(15/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)/(b*x^4 + a)^(15/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{15/4}} \, dx=\int \frac {d\,x^4+c}{{\left (b\,x^4+a\right )}^{15/4}} \,d x \] Input:

int((c + d*x^4)/(a + b*x^4)^(15/4),x)
 

Output:

int((c + d*x^4)/(a + b*x^4)^(15/4), x)
 

Reduce [F]

\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{15/4}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{3} x^{12}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{3} x^{12}}d x \right ) c \] Input:

int((d*x^4+c)/(b*x^4+a)^(15/4),x)
 

Output:

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