\(\int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx\) [84]

Optimal result

Integrand size = 22, antiderivative size = 123 \[ \int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx=\frac {2 b c-5 a d}{5 a x \sqrt [4]{a+b x^4}}-\frac {c \left (a+b x^4\right )^{3/4}}{5 a x^5}-\frac {\sqrt {b} (2 b c-5 a d) \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt [4]{a+b x^4}} \] Output:

1/5*(-5*a*d+2*b*c)/a/x/(b*x^4+a)^(1/4)-1/5*c*(b*x^4+a)^(3/4)/a/x^5-1/5*b^( 
1/2)*(-5*a*d+2*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^(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 = 10.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63 \[ \int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx=\frac {-c \left (a+b x^4\right )+(2 b c-5 a d) x^4 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},-\frac {b x^4}{a}\right )}{5 a x^5 \sqrt [4]{a+b x^4}} \] Input:

Integrate[(c + d*x^4)/(x^6*(a + b*x^4)^(1/4)),x]
 

Output:

(-(c*(a + b*x^4)) + (2*b*c - 5*a*d)*x^4*(1 + (b*x^4)/a)^(1/4)*Hypergeometr 
ic2F1[-1/4, 1/4, 3/4, -((b*x^4)/a)])/(5*a*x^5*(a + b*x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {955, 841, 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.

\(\displaystyle \int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx\)

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 841

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

\(\Big \downarrow \) 813

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 212

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

Input:

Int[(c + d*x^4)/(x^6*(a + b*x^4)^(1/4)),x]
 

Output:

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

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 841

Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> -Simp[(x*(a + b*x^ 
4)^(1/4))^(-1), x] - Simp[b   Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a 
, b}, x] && PosQ[b/a]
 

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 955

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

Maple [F]

\[\int \frac {d \,x^{4}+c}{x^{6} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x\]

Input:

int((d*x^4+c)/x^6/(b*x^4+a)^(1/4),x)
 

Output:

int((d*x^4+c)/x^6/(b*x^4+a)^(1/4),x)
 

Fricas [F]

\[ \int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{6}} \,d x } \] Input:

integrate((d*x^4+c)/x^6/(b*x^4+a)^(1/4),x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(3/4)*(d*x^4 + c)/(b*x^10 + a*x^6), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.62 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.58 \[ \int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{6 \sqrt [4]{b} x^{6}} + \frac {d \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} x \Gamma \left (\frac {3}{4}\right )} \] Input:

integrate((d*x**4+c)/x**6/(b*x**4+a)**(1/4),x)
 

Output:

-c*hyper((1/4, 3/2), (5/2,), a*exp_polar(I*pi)/(b*x**4))/(6*b**(1/4)*x**6) 
 + d*gamma(-1/4)*hyper((-1/4, 1/4), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*a 
**(1/4)*x*gamma(3/4))
 

Maxima [F]

\[ \int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{6}} \,d x } \] Input:

integrate((d*x^4+c)/x^6/(b*x^4+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(1/4)*x^6), x)
 

Giac [F]

\[ \int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{6}} \,d x } \] Input:

integrate((d*x^4+c)/x^6/(b*x^4+a)^(1/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(1/4)*x^6), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx=\int \frac {d\,x^4+c}{x^6\,{\left (b\,x^4+a\right )}^{1/4}} \,d x \] Input:

int((c + d*x^4)/(x^6*(a + b*x^4)^(1/4)),x)
 

Output:

int((c + d*x^4)/(x^6*(a + b*x^4)^(1/4)), x)
 

Reduce [F]

\[ \int \frac {c+d x^4}{x^6 \sqrt [4]{a+b x^4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{6}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{2}}d x \right ) d \] Input:

int((d*x^4+c)/x^6/(b*x^4+a)^(1/4),x)
 

Output:

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