\(\int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx\) [78]

Optimal result

Integrand size = 30, antiderivative size = 289 \[ \int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx=-\frac {c \sqrt {a+b x^4}}{\sqrt {a} x \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {f \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}-\frac {d \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt [4]{b} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} c+\sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 a^{3/4} \sqrt [4]{b} \sqrt {a+b x^4}} \] Output:

-c*(b*x^4+a)^(1/2)/a^(1/2)/x/(a^(1/2)+b^(1/2)*x^2)+1/2*f*arctanh(b^(1/2)*x 
^2/(b*x^4+a)^(1/2))/b^(1/2)-1/2*d*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1/2) 
-b^(1/4)*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*x^4+a) 
^(1/2)+1/2*(b^(1/2)*c+a^(1/2)*e)*(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^(1/4)/(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 = 10.17 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.54 \[ \int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx=\frac {f \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}-\frac {d \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {c \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {b x^4}{a}\right )}{x \sqrt {a+b x^4}}+\frac {e x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt {a+b x^4}} \] Input:

Integrate[(c + d*x + e*x^2 + f*x^3)/(x^2*Sqrt[a + b*x^4]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2372, 2009}

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.

\(\displaystyle \int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx\)

\(\Big \downarrow \) 2372

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

\(\Big \downarrow \) 2009

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

Input:

Int[(c + d*x + e*x^2 + f*x^3)/(x^2*Sqrt[a + b*x^4]),x]
 

Output:

-((c*Sqrt[a + b*x^4])/(a*x)) + (Sqrt[b]*c*x*Sqrt[a + b*x^4])/(a*(Sqrt[a] + 
 Sqrt[b]*x^2)) + (f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*Sqrt[b]) - 
(d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(2*Sqrt[a]) - (b^(1/4)*c*(Sqrt[a] + S 
qrt[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])/(a^(3/4)*Sqrt[a + b*x^4]) + ((Sqrt[b]*c + Sqr 
t[a]*e)*(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*a^(3/4)*b^(1/4)*Sqrt[a 
 + b*x^4])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2372

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Mo 
dule[{q = Expon[Pq, x], j, k}, Int[Sum[((c*x)^(m + j)/c^j)*Sum[Coeff[Pq, x, 
 j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2*((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, 
 n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0 
] &&  !PolyQ[Pq, x^(n/2)]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.20 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.81

method	result	size
elliptic	\(-\frac {c \sqrt {b \,x^{4}+a}}{a x}+\frac {e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {f \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {i \sqrt {b}\, c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2 \sqrt {a}}\)	\(234\)
default	\(\frac {e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+c \left (-\frac {\sqrt {b \,x^{4}+a}}{a x}+\frac {i \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {d \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 \sqrt {a}}+\frac {f \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}\)	\(241\)
risch	\(-\frac {c \sqrt {b \,x^{4}+a}}{a x}+\frac {\frac {a e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {i c \sqrt {b}\, \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {\sqrt {a}\, d \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}+\frac {a f \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}}{a}\)	\(247\)

Input:

int((f*x^3+e*x^2+d*x+c)/x^2/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-c*(b*x^4+a)^(1/2)/a/x+e/(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)+1/2*f*ln(2*b^(1/2)*x^2+2*(b*x^4+a)^(1/2))/b^(1/2)+ 
I*b^(1/2)/a^(1/2)*c/(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))-1/2*d/a^(1/2 
)*arctanh(a^(1/2)/(b*x^4+a)^(1/2))
 

Fricas [F]

\[ \int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a} x^{2}} \,d x } \] Input:

integrate((f*x^3+e*x^2+d*x+c)/x^2/(b*x^4+a)^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/(b*x^6 + a*x^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.75 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.44 \[ \int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx=\frac {f \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} + \frac {c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x \Gamma \left (\frac {3}{4}\right )} - \frac {d \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2 \sqrt {a}} + \frac {e x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((f*x**3+e*x**2+d*x+c)/x**2/(b*x**4+a)**(1/2),x)
 

Output:

f*asinh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b)) + 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*asinh(sq 
rt(a)/(sqrt(b)*x**2))/(2*sqrt(a)) + e*x*gamma(1/4)*hyper((1/4, 1/2), (5/4, 
), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4))
 

Maxima [F]

\[ \int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a} x^{2}} \,d x } \] Input:

integrate((f*x^3+e*x^2+d*x+c)/x^2/(b*x^4+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate((f*x^3 + e*x^2 + d*x + c)/(sqrt(b*x^4 + a)*x^2), x)
 

Giac [F]

\[ \int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a} x^{2}} \,d x } \] Input:

integrate((f*x^3+e*x^2+d*x+c)/x^2/(b*x^4+a)^(1/2),x, algorithm="giac")
 

Output:

integrate((f*x^3 + e*x^2 + d*x + c)/(sqrt(b*x^4 + a)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx=\int \frac {f\,x^3+e\,x^2+d\,x+c}{x^2\,\sqrt {b\,x^4+a}} \,d x \] Input:

int((c + d*x + e*x^2 + f*x^3)/(x^2*(a + b*x^4)^(1/2)),x)
 

Output:

int((c + d*x + e*x^2 + f*x^3)/(x^2*(a + b*x^4)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {c+d x+e x^2+f x^3}{x^2 \sqrt {a+b x^4}} \, dx=\frac {\sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {a}\right ) b d -\sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {a}\right ) b d -\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a f +\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a f +4 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{6}+a \,x^{2}}d x \right ) a b c +4 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a b e}{4 a b} \] Input:

int((f*x^3+e*x^2+d*x+c)/x^2/(b*x^4+a)^(1/2),x)
 

Output:

(sqrt(a)*log(sqrt(a + b*x**4) - sqrt(a))*b*d - sqrt(a)*log(sqrt(a + b*x**4 
) + sqrt(a))*b*d - sqrt(b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a*f + sqrt 
(b)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*a*f + 4*int(sqrt(a + b*x**4)/(a*x 
**2 + b*x**6),x)*a*b*c + 4*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a*b*e)/(4* 
a*b)