3.6.1 \(\int \frac {(c+d x+e x^2+f x^3) \sqrt {a+b x^4}}{x^2} \, dx\) [501]

3.6.1.1 Optimal result
3.6.1.2 Mathematica [C] (verified)
3.6.1.3 Rubi [A] (verified)
3.6.1.4 Maple [C] (verified)
3.6.1.5 Fricas [F]
3.6.1.6 Sympy [C] (verification not implemented)
3.6.1.7 Maxima [F]
3.6.1.8 Giac [F]
3.6.1.9 Mupad [F(-1)]

3.6.1.1 Optimal result

Integrand size = 30, antiderivative size = 341 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\frac {2 \sqrt {b} c x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}-\frac {\left (3 c-e x^2\right ) \sqrt {a+b x^4}}{3 x}+\frac {1}{4} \left (2 d+f x^2\right ) \sqrt {a+b x^4}+\frac {a f \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {1}{2} \sqrt {a} d \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {2 \sqrt [4]{a} \sqrt [4]{b} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (3 \sqrt {b} c+\sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a+b x^4}} \]

output
-1/2*d*arctanh((b*x^4+a)^(1/2)/a^(1/2))*a^(1/2)+1/4*a*f*arctanh(x^2*b^(1/2 
)/(b*x^4+a)^(1/2))/b^(1/2)-1/3*(-e*x^2+3*c)*(b*x^4+a)^(1/2)/x+1/4*(f*x^2+2 
*d)*(b*x^4+a)^(1/2)+2*c*x*b^(1/2)*(b*x^4+a)^(1/2)/(a^(1/2)+x^2*b^(1/2))-2* 
a^(1/4)*b^(1/4)*c*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan( 
b^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2) 
)*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/(b*x^4+a 
)^(1/2)+1/3*a^(1/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arcta 
n(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/ 
2))*(e*a^(1/2)+3*c*b^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2* 
b^(1/2))^2)^(1/2)/b^(1/4)/(b*x^4+a)^(1/2)
 
3.6.1.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.36 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.61 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\frac {-4 \sqrt {b} c \sqrt {a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},-\frac {b x^4}{a}\right )+x \left (\sqrt {a} f \sqrt {a+b x^4} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+\sqrt {b} \sqrt {1+\frac {b x^4}{a}} \left (\left (2 d+f x^2\right ) \sqrt {a+b x^4}-2 \sqrt {a} d \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+4 \sqrt {b} e x \sqrt {a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{4 \sqrt {b} x \sqrt {1+\frac {b x^4}{a}}} \]

input
Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^2,x]
 
output
(-4*Sqrt[b]*c*Sqrt[a + b*x^4]*Hypergeometric2F1[-1/2, -1/4, 3/4, -((b*x^4) 
/a)] + x*(Sqrt[a]*f*Sqrt[a + b*x^4]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] + Sqrt[ 
b]*Sqrt[1 + (b*x^4)/a]*((2*d + f*x^2)*Sqrt[a + b*x^4] - 2*Sqrt[a]*d*ArcTan 
h[Sqrt[a + b*x^4]/Sqrt[a]]) + 4*Sqrt[b]*e*x*Sqrt[a + b*x^4]*Hypergeometric 
2F1[-1/2, 1/4, 5/4, -((b*x^4)/a)]))/(4*Sqrt[b]*x*Sqrt[1 + (b*x^4)/a])
 
3.6.1.3 Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 2372

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

\(\Big \downarrow \) 2009

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

input
Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^2,x]
 
output
(2*Sqrt[b]*c*x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2) - ((3*c - e*x^2)*S 
qrt[a + b*x^4])/(3*x) + ((2*d + f*x^2)*Sqrt[a + b*x^4])/4 + (a*f*ArcTanh[( 
Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b]) - (Sqrt[a]*d*ArcTanh[Sqrt[a + b 
*x^4]/Sqrt[a]])/2 - (2*a^(1/4)*b^(1/4)*c*(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])/Sqrt[a + b*x^4] + (a^(1/4)*(3*Sqrt[b]*c + Sqrt[a]*e)*(Sqrt[a] + Sq 
rt[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])/(3*b^(1/4)*Sqrt[a + b*x^4])
 

3.6.1.3.1 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)]
 
3.6.1.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.40 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.80

method result size
elliptic \(-\frac {c \sqrt {b \,x^{4}+a}}{x}+\frac {f \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {e x \sqrt {b \,x^{4}+a}}{3}+\frac {d \sqrt {b \,x^{4}+a}}{2}+\frac {2 a e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {a f \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}+\frac {2 i \sqrt {b}\, c \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {\sqrt {a}\, d \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2}\) \(274\)
risch \(-\frac {c \sqrt {b \,x^{4}+a}}{x}+\frac {2 a e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {f \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {a f \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}+\frac {e x \sqrt {b \,x^{4}+a}}{3}+\frac {d \sqrt {b \,x^{4}+a}}{2}-\frac {\sqrt {a}\, d \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}+\frac {2 i \sqrt {b}\, c \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) \(280\)
default \(e \left (\frac {x \sqrt {b \,x^{4}+a}}{3}+\frac {2 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+f \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {a \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}\right )+d \left (\frac {\sqrt {b \,x^{4}+a}}{2}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}\right )+c \left (-\frac {\sqrt {b \,x^{4}+a}}{x}+\frac {2 i \sqrt {b}\, \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) \(284\)

input
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^2,x,method=_RETURNVERBOSE)
 
output
-c*(b*x^4+a)^(1/2)/x+1/4*f*x^2*(b*x^4+a)^(1/2)+1/3*e*x*(b*x^4+a)^(1/2)+1/2 
*d*(b*x^4+a)^(1/2)+2/3*a*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)+1/4*a*f*ln(2*x^2*b^(1/2)+2*(b*x^4+a)^(1/2))/b^(1 
/2)+2*I*b^(1/2)*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)*(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*a^ 
(1/2)*d*arctanh(a^(1/2)/(b*x^4+a)^(1/2))
 
3.6.1.5 Fricas [F]

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

input
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^2,x, algorithm="fricas")
 
output
integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^2, x)
 
3.6.1.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.92 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.60 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\frac {\sqrt {a} c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {\sqrt {a} d \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2} + \frac {\sqrt {a} e x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {a} f x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {a d}{2 \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {a f \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 \sqrt {b}} + \frac {\sqrt {b} d x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} \]

input
integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x**2,x)
 
output
sqrt(a)*c*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**4*exp_polar(I*pi)/a 
)/(4*x*gamma(3/4)) - sqrt(a)*d*asinh(sqrt(a)/(sqrt(b)*x**2))/2 + sqrt(a)*e 
*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamm 
a(5/4)) + sqrt(a)*f*x**2*sqrt(1 + b*x**4/a)/4 + a*d/(2*sqrt(b)*x**2*sqrt(a 
/(b*x**4) + 1)) + a*f*asinh(sqrt(b)*x**2/sqrt(a))/(4*sqrt(b)) + sqrt(b)*d* 
x**2/(2*sqrt(a/(b*x**4) + 1))
 
3.6.1.7 Maxima [F]

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

input
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^2,x, algorithm="maxima")
 
output
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^2, x)
 
3.6.1.8 Giac [F]

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

input
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^2,x, algorithm="giac")
 
output
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^2, x)
 
3.6.1.9 Mupad [F(-1)]

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

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