\(\int \frac {c+d x^4}{x^6 (a+b x^4)^{5/4}} \, dx\) [120]

Optimal result

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

-1/5*c/a/x^5/(b*x^4+a)^(1/4)+1/5*(-5*a*d+6*b*c)/a^2/x/(b*x^4+a)^(1/4)-2/5* 
b^(1/2)*(-5*a*d+6*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^(5/2)/(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 = 72, normalized size of antiderivative = 0.59 \[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\frac {-a c+(6 b c-5 a d) x^4 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4},\frac {3}{4},-\frac {b x^4}{a}\right )}{5 a^2 x^5 \sqrt [4]{a+b x^4}} \] Input:

Integrate[(c + d*x^4)/(x^6*(a + b*x^4)^(5/4)),x]
 

Output:

(-(a*c) + (6*b*c - 5*a*d)*x^4*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-1/4 
, 5/4, 3/4, -((b*x^4)/a)])/(5*a^2*x^5*(a + b*x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {955, 816, 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.

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

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 816

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

\(\Big \downarrow \) 813

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 212

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

Input:

Int[(c + d*x^4)/(x^6*(a + b*x^4)^(5/4)),x]
 

Output:

-1/5*c/(a*x^5*(a + b*x^4)^(1/4)) - ((6*b*c - 5*a*d)*(-(1/(a*x*(a + b*x^4)^ 
(1/4))) + (2*Sqrt[b]*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqr 
t[b]*x^2)]/2, 2])/(a^(3/2)*(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 816

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x^(m + 1)/(a*( 
m + 1)*(a + b*x^4)^(1/4)), x] - Simp[b*(m/(a*(m + 1)))   Int[x^(m + 4)/(a + 
 b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a] && ILtQ[(m - 2)/4, 0 
]
 

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]

\[\int \frac {d \,x^{4}+c}{x^{6} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]

Input:

int((d*x^4+c)/x^6/(b*x^4+a)^(5/4),x)
 

Output:

int((d*x^4+c)/x^6/(b*x^4+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{6}} \,d x } \] Input:

integrate((d*x^4+c)/x^6/(b*x^4+a)^(5/4),x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(3/4)*(d*x^4 + c)/(b^2*x^14 + 2*a*b*x^10 + a^2*x^6), 
x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 22.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.72 \[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\frac {c \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {5}{4} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} x^{5} \Gamma \left (- \frac {1}{4}\right )} + \frac {d \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} x \Gamma \left (\frac {3}{4}\right )} \] Input:

integrate((d*x**4+c)/x**6/(b*x**4+a)**(5/4),x)
 

Output:

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

Maxima [F]

\[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{6}} \,d x } \] Input:

integrate((d*x^4+c)/x^6/(b*x^4+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(5/4)*x^6), x)
 

Giac [F]

\[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{6}} \,d x } \] Input:

integrate((d*x^4+c)/x^6/(b*x^4+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(5/4)*x^6), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\int \frac {d\,x^4+c}{x^6\,{\left (b\,x^4+a\right )}^{5/4}} \,d x \] Input:

int((c + d*x^4)/(x^6*(a + b*x^4)^(5/4)),x)
 

Output:

int((c + d*x^4)/(x^6*(a + b*x^4)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {c+d x^4}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{6}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{10}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{2}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{6}}d x \right ) d \] Input:

int((d*x^4+c)/x^6/(b*x^4+a)^(5/4),x)
 

Output:

int(1/((a + b*x**4)**(1/4)*a*x**6 + (a + b*x**4)**(1/4)*b*x**10),x)*c + in 
t(1/((a + b*x**4)**(1/4)*a*x**2 + (a + b*x**4)**(1/4)*b*x**6),x)*d