\(\int \frac {c+d x^4}{x^2 \sqrt {a+b x^4}} \, dx\) [33]

Optimal result

Integrand size = 22, antiderivative size = 254 \[ \int \frac {c+d x^4}{x^2 \sqrt {a+b x^4}} \, dx=-\frac {c \sqrt {a+b x^4}}{a x}+\frac {(b c+a d) x \sqrt {a+b x^4}}{a \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {(b c+a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} b^{3/4} \sqrt {a+b x^4}}+\frac {(b c+a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a+b x^4}} \] Output:

-c*(b*x^4+a)^(1/2)/a/x+(a*d+b*c)*x*(b*x^4+a)^(1/2)/a/b^(1/2)/(a^(1/2)+b^(1 
/2)*x^2)-(a*d+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^(3/4)/b 
^(3/4)/(b*x^4+a)^(1/2)+1/2*(a*d+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^(3/4)/b^(3/4)/(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 = 10.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.30 \[ \int \frac {c+d x^4}{x^2 \sqrt {a+b x^4}} \, dx=\frac {-3 c \left (a+b x^4\right )+(b c+a d) x^4 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 a x \sqrt {a+b x^4}} \] Input:

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

Output:

(-3*c*(a + b*x^4) + (b*c + a*d)*x^4*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[ 
1/2, 3/4, 7/4, -((b*x^4)/a)])/(3*a*x*Sqrt[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {955, 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.

\(\displaystyle \int \frac {c+d x^4}{x^2 \sqrt {a+b x^4}} \, dx\)

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

-((c*Sqrt[a + b*x^4])/(a*x)) + ((b*c + a*d)*(-((-((x*Sqrt[a + b*x^4])/(Sqr 
t[a] + Sqrt[b]*x^2)) + (a^(1/4)*(Sqrt[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
 

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 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 = 0.63 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.48

method	result	size
risch	\(-\frac {c \sqrt {b \,x^{4}+a}}{a x}+\frac {i \left (a d +c b \right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(123\)
elliptic	\(-\frac {c \sqrt {b \,x^{4}+a}}{a x}+\frac {i \left (d +\frac {b c}{a}\right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(124\)
default	\(c \left (-\frac {\sqrt {b \,x^{4}+a}}{a x}+\frac {i \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(215\)

Input:

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

Output:

-c*(b*x^4+a)^(1/2)/a/x+I*(a*d+b*c)/a^(1/2)/(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) 
/b^(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 [F]

\[ \int \frac {c+d x^4}{x^2 \sqrt {a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{\sqrt {b x^{4} + a} x^{2}} \,d x } \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.32 \[ \int \frac {c+d x^4}{x^2 \sqrt {a+b x^4}} \, dx=\frac {c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x \Gamma \left (\frac {3}{4}\right )} + \frac {d x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Output:

c*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt 
(a)*x*gamma(3/4)) + d*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp 
_polar(I*pi)/a)/(4*sqrt(a)*gamma(7/4))
 

Maxima [F]

\[ \int \frac {c+d x^4}{x^2 \sqrt {a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{\sqrt {b x^{4} + a} x^{2}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {c+d x^4}{x^2 \sqrt {a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{\sqrt {b x^{4} + a} x^{2}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{x^2 \sqrt {a+b x^4}} \, dx=\int \frac {d\,x^4+c}{x^2\,\sqrt {b\,x^4+a}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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