\(\int \frac {c+d x+e x^2+f x^3}{x \sqrt {a+b x^4}} \, dx\) [77]

Optimal result

Integrand size = 30, antiderivative size = 286 \[ \int \frac {c+d x+e x^2+f x^3}{x \sqrt {a+b x^4}} \, dx=\frac {f x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}-\frac {c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt [4]{a} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} d+\sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} b^{3/4} \sqrt {a+b x^4}} \] Output:

f*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+b^(1/2)*x^2)+1/2*e*arctanh(b^(1/2)*x^ 
2/(b*x^4+a)^(1/2))/b^(1/2)-1/2*c*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1/2)- 
a^(1/4)*f*(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))/b^(3/4)/(b*x^4+a)^ 
(1/2)+1/2*(b^(1/2)*d+a^(1/2)*f)*(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^(1/4)/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 = 10.16 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.56 \[ \int \frac {c+d x+e x^2+f x^3}{x \sqrt {a+b x^4}} \, dx=\frac {e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}-\frac {c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}}+\frac {d x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt {a+b x^4}}+\frac {f x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 \sqrt {a+b x^4}} \] Input:

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

Output:

(e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*Sqrt[b]) - (c*ArcTanh[Sqrt[a 
 + b*x^4]/Sqrt[a]])/(2*Sqrt[a]) + (d*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2 
F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/Sqrt[a + b*x^4] + (f*x^3*Sqrt[1 + (b*x^4) 
/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)])/(3*Sqrt[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2371, 798, 73, 221, 2424, 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.

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

\(\Big \downarrow \) 2371

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

\(\Big \downarrow \) 798

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {c \int \frac {1}{\frac {x^8}{b}-\frac {a}{b}}d\sqrt {b x^4+a}}{2 b}+\int \frac {f x^2+e x+d}{\sqrt {b x^4+a}}dx\)

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 2424

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2009

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

rule 2371

Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Simp[Coeff[Pq, 
x, 0]   Int[1/(x*Sqrt[a + b*x^n]), x], x] + Int[ExpandToSum[(Pq - Coeff[Pq, 
 x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IG 
tQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]
 

rule 2424

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, 
 x], j, k}, Int[Sum[x^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, 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 = 0.74 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.76

method	result	size
elliptic	\(\frac {d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {e \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {i f \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}-\frac {c \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2 \sqrt {a}}\)	\(216\)
default	\(\frac {d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {e \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {i f \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}-\frac {c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 \sqrt {a}}\)	\(222\)

Input:

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

Output:

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

Fricas [F]

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

integrate((f*x^3+e*x^2+d*x+c)/x/(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^5 + a*x), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.44 \[ \int \frac {c+d x+e x^2+f x^3}{x \sqrt {a+b x^4}} \, dx=\frac {e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} - \frac {c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2 \sqrt {a}} + \frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {f x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Output:

e*asinh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b)) - c*asinh(sqrt(a)/(sqrt(b)*x**2) 
)/(2*sqrt(a)) + d*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)) + f*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]

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

integrate((f*x^3+e*x^2+d*x+c)/x/(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), x)
 

Giac [F]

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

integrate((f*x^3+e*x^2+d*x+c)/x/(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), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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