\(\int \frac {x (c+d x+e x^2+f x^3)}{\sqrt {a+b x^4}} \, dx\) [75]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 299 \[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {e \sqrt {a+b x^4}}{2 b}+\frac {f x \sqrt {a+b x^4}}{3 b}+\frac {d x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {c \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}-\frac {\sqrt [4]{a} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (3 \sqrt {b} d-\sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 b^{5/4} \sqrt {a+b x^4}} \] Output:

1/2*e*(b*x^4+a)^(1/2)/b+1/3*f*x*(b*x^4+a)^(1/2)/b+d*x*(b*x^4+a)^(1/2)/b^(1 
/2)/(a^(1/2)+b^(1/2)*x^2)+1/2*c*arctanh(b^(1/2)*x^2/(b*x^4+a)^(1/2))/b^(1/ 
2)-a^(1/4)*d*(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)+1/6*a^(1/4)*(3*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))/b^(5/4)/(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 = 10.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.54 \[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {3 a e+2 a f x+3 b e x^4+2 b f x^5+3 \sqrt {b} c \sqrt {a+b x^4} \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-2 a f x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )+2 b d x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{6 b \sqrt {a+b x^4}} \] Input:

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

Output:

(3*a*e + 2*a*f*x + 3*b*e*x^4 + 2*b*f*x^5 + 3*Sqrt[b]*c*Sqrt[a + b*x^4]*Arc 
Tanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 2*a*f*x*Sqrt[1 + (b*x^4)/a]*Hypergeo 
metric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] + 2*b*d*x^3*Sqrt[1 + (b*x^4)/a]*Hyp 
ergeometric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)])/(6*b*Sqrt[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, 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.

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

\(\Big \downarrow \) 2372

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {b} d-\sqrt {a} f\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 b^{5/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{a} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {c \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}+\frac {d x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {e \sqrt {a+b x^4}}{2 b}+\frac {f x \sqrt {a+b x^4}}{3 b}\)

Input:

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

Output:

(e*Sqrt[a + b*x^4])/(2*b) + (f*x*Sqrt[a + b*x^4])/(3*b) + (d*x*Sqrt[a + b* 
x^4])/(Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (c*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a 
+ b*x^4]])/(2*Sqrt[b]) - (a^(1/4)*d*(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^(3/4)*Sqrt[a + b*x^4]) + (a^(1/4)*(3*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*Ar 
cTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(6*b^(5/4)*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 = 1.58 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.74

method result size
risch \(\frac {\left (2 f x +3 e \right ) \sqrt {b \,x^{4}+a}}{6 b}-\frac {\frac {a f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {3 c \sqrt {b}\, \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2}-\frac {3 i \sqrt {b}\, d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}}{3 b}\) \(222\)
default \(\frac {c \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+f \left (\frac {x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {e \sqrt {b \,x^{4}+a}}{2 b}\) \(230\)
elliptic \(\frac {f x \sqrt {b \,x^{4}+a}}{3 b}+\frac {e \sqrt {b \,x^{4}+a}}{2 b}-\frac {a f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {c \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\) \(232\)

Input:

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

Output:

1/6*(2*f*x+3*e)/b*(b*x^4+a)^(1/2)-1/3/b*(a*f/(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)-3/2*c*b^(1/2)*ln(b^(1/2)*x^2+( 
b*x^4+a)^(1/2))-3*I*b^(1/2)*d*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)*(Ell 
ipticF(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 = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.44 \[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {12 \, \sqrt {b} d x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 4 \, \sqrt {b} {\left (3 \, d + f\right )} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 3 \, \sqrt {b} c x \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, \sqrt {b x^{4} + a} {\left (2 \, f x^{2} + 3 \, e x + 6 \, d\right )}}{12 \, b x} \] Input:

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

Output:

1/12*(12*sqrt(b)*d*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) - 
 4*sqrt(b)*(3*d + f)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) 
 + 3*sqrt(b)*c*x*log(-2*b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 2*sqr 
t(b*x^4 + a)*(2*f*x^2 + 3*e*x + 6*d))/(b*x)
 

Sympy [A] (verification not implemented)

Time = 1.82 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.43 \[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=e \left (\begin {cases} \frac {x^{4}}{4 \sqrt {a}} & \text {for}\: b = 0 \\\frac {\sqrt {a + b x^{4}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {c \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} + \frac {d x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {f x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} \] Input:

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

Output:

e*Piecewise((x**4/(4*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**4)/(2*b), True)) 
+ c*asinh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b)) + 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)) + f*x**5* 
gamma(5/4)*hyper((1/2, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)* 
gamma(9/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {-3 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) c +3 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) c +6 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, e \,x^{2}+4 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, f \,x^{3}-4 \sqrt {b \,x^{4}+a}\, \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a f +12 \sqrt {b \,x^{4}+a}\, \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) b d -4 \sqrt {b}\, \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a f \,x^{2}+12 \sqrt {b}\, \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) b d \,x^{2}-3 \,\mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) b c \,x^{2}+3 \,\mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) b c \,x^{2}+6 a e +4 a f x +6 b e \,x^{4}+4 b f \,x^{5}}{12 b \left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right )} \] Input:

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

Output:

( - 3*sqrt(b)*sqrt(a + b*x**4)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*c + 3* 
sqrt(b)*sqrt(a + b*x**4)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*c + 6*sqrt(b 
)*sqrt(a + b*x**4)*e*x**2 + 4*sqrt(b)*sqrt(a + b*x**4)*f*x**3 - 4*sqrt(a + 
 b*x**4)*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a*f + 12*sqrt(a + b*x**4)*in 
t((sqrt(a + b*x**4)*x**2)/(a + b*x**4),x)*b*d - 4*sqrt(b)*int(sqrt(a + b*x 
**4)/(a + b*x**4),x)*a*f*x**2 + 12*sqrt(b)*int((sqrt(a + b*x**4)*x**2)/(a 
+ b*x**4),x)*b*d*x**2 - 3*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*b*c*x**2 + 
3*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*b*c*x**2 + 6*a*e + 4*a*f*x + 6*b*e* 
x**4 + 4*b*f*x**5)/(12*b*(sqrt(a + b*x**4) + sqrt(b)*x**2))