\(\int \frac {c+d x^4}{x^6 \sqrt {a+b x^4}} \, dx\) [34]

Optimal result

Integrand size = 22, antiderivative size = 297 \[ \int \frac {c+d x^4}{x^6 \sqrt {a+b x^4}} \, dx=-\frac {c \sqrt {a+b x^4}}{5 a x^5}+\frac {(3 b c-5 a d) \sqrt {a+b x^4}}{5 a^2 x}-\frac {\sqrt {b} (3 b c-5 a d) x \sqrt {a+b x^4}}{5 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {\sqrt [4]{b} (3 b c-5 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} (3 b c-5 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{10 a^{7/4} \sqrt {a+b x^4}} \] Output:

-1/5*c*(b*x^4+a)^(1/2)/a/x^5+1/5*(-5*a*d+3*b*c)*(b*x^4+a)^(1/2)/a^2/x-1/5* 
b^(1/2)*(-5*a*d+3*b*c)*x*(b*x^4+a)^(1/2)/a^2/(a^(1/2)+b^(1/2)*x^2)+1/5*b^( 
1/4)*(-5*a*d+3*b*c)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2) 
^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))/a^(7/4)/ 
(b*x^4+a)^(1/2)-1/10*b^(1/4)*(-5*a*d+3*b*c)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+ 
a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^(1/ 
4)),1/2*2^(1/2))/a^(7/4)/(b*x^4+a)^(1/2)
 

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 = 0.26 \[ \int \frac {c+d x^4}{x^6 \sqrt {a+b x^4}} \, dx=\frac {-c \left (a+b x^4\right )+(3 b c-5 a d) x^4 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {b x^4}{a}\right )}{5 a x^5 \sqrt {a+b x^4}} \] Input:

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

Output:

(-(c*(a + b*x^4)) + (3*b*c - 5*a*d)*x^4*Sqrt[1 + (b*x^4)/a]*Hypergeometric 
2F1[-1/4, 1/2, 3/4, -((b*x^4)/a)])/(5*a*x^5*Sqrt[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {955, 847, 834, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

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]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.05 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.48

method	result	size
risch	\(-\frac {\sqrt {b \,x^{4}+a}\, \left (5 a d \,x^{4}-3 b c \,x^{4}+a c \right )}{5 a^{2} x^{5}}+\frac {i \sqrt {b}\, \left (5 a d -3 c b \right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(142\)
elliptic	\(-\frac {c \sqrt {b \,x^{4}+a}}{5 a \,x^{5}}-\frac {\left (5 a d -3 c b \right ) \sqrt {b \,x^{4}+a}}{5 a^{2} x}+\frac {i \sqrt {b}\, \left (5 a d -3 c b \right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(151\)
default	\(c \left (-\frac {\sqrt {b \,x^{4}+a}}{5 a \,x^{5}}+\frac {3 b \sqrt {b \,x^{4}+a}}{5 a^{2} x}-\frac {3 i b^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (-\frac {\sqrt {b \,x^{4}+a}}{a x}+\frac {i \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(252\)

Input:

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

Output:

-1/5*(b*x^4+a)^(1/2)*(5*a*d*x^4-3*b*c*x^4+a*c)/a^2/x^5+1/5*I*b^(1/2)*(5*a* 
d-3*b*c)/a^(3/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2) 
*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*(EllipticF(x*(I/a^(1/2)*b 
^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.38 \[ \int \frac {c+d x^4}{x^6 \sqrt {a+b x^4}} \, dx=\frac {{\left (3 \, b c - 5 \, a d\right )} \sqrt {a} x^{5} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (3 \, b c - 5 \, a d\right )} \sqrt {a} x^{5} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left ({\left (3 \, b c - 5 \, a d\right )} x^{4} - a c\right )} \sqrt {b x^{4} + a}}{5 \, a^{2} x^{5}} \] Input:

integrate((d*x^4+c)/x^6/(b*x^4+a)^(1/2),x, algorithm="fricas")
 

Output:

1/5*((3*b*c - 5*a*d)*sqrt(a)*x^5*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a)^( 
1/4)), -1) - (3*b*c - 5*a*d)*sqrt(a)*x^5*(-b/a)^(3/4)*elliptic_f(arcsin(x* 
(-b/a)^(1/4)), -1) + ((3*b*c - 5*a*d)*x^4 - a*c)*sqrt(b*x^4 + a))/(a^2*x^5 
)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.30 \[ \int \frac {c+d x^4}{x^6 \sqrt {a+b x^4}} \, dx=\frac {c \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x^{5} \Gamma \left (- \frac {1}{4}\right )} + \frac {d \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x \Gamma \left (\frac {3}{4}\right )} \] Input:

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

Output:

c*gamma(-5/4)*hyper((-5/4, 1/2), (-1/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqr 
t(a)*x**5*gamma(-1/4)) + d*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), b*x**4*e 
xp_polar(I*pi)/a)/(4*sqrt(a)*x*gamma(3/4))
 

Maxima [F]

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

integrate((d*x^4+c)/x^6/(b*x^4+a)^(1/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

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

Giac [F]

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

integrate((d*x^4+c)/x^6/(b*x^4+a)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {c+d x^4}{x^6 \sqrt {a+b x^4}} \, dx=\frac {-\sqrt {b \,x^{4}+a}\, d -5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{10}+a \,x^{6}}d x \right ) a d \,x^{5}+3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{10}+a \,x^{6}}d x \right ) b c \,x^{5}}{3 b \,x^{5}} \] Input:

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

Output:

( - sqrt(a + b*x**4)*d - 5*int(sqrt(a + b*x**4)/(a*x**6 + b*x**10),x)*a*d* 
x**5 + 3*int(sqrt(a + b*x**4)/(a*x**6 + b*x**10),x)*b*c*x**5)/(3*b*x**5)