Integrand size = 27, antiderivative size = 276 \[ \int \frac {c+d x+e x^2+f x^3}{\sqrt {a+b x^4}} \, dx=\frac {f \sqrt {a+b x^4}}{2 b}+\frac {e x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}-\frac {\sqrt [4]{a} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (\frac {\sqrt {b} c}{\sqrt {a}}+e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}} \]
1/2*d*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(1/2)+1/2*f*(b*x^4+a)^(1/2)/b +e*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+x^2*b^(1/2))-a^(1/4)*e*(cos(2*arctan (b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticE(s in(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^(3/4)/(b*x^4+a)^(1/2)+1/2*a^(1/4)*(co s(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*E llipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2 ))*(e+c*b^(1/2)/a^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/b^(3/4) /(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.10 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.54 \[ \int \frac {c+d x+e x^2+f x^3}{\sqrt {a+b x^4}} \, dx=\frac {f \sqrt {a+b x^4}}{2 b}+\frac {d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}+\frac {c x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt {a+b x^4}}+\frac {e x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 \sqrt {a+b x^4}} \]
(f*Sqrt[a + b*x^4])/(2*b) + (d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2* Sqrt[b]) + (c*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b* x^4)/a)])/Sqrt[a + b*x^4] + (e*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])
Time = 0.39 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3}{\sqrt {a+b x^4}} \, dx\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \int \left (\frac {c+e x^2}{\sqrt {a+b x^4}}+\frac {x \left (d+f x^2\right )}{\sqrt {a+b x^4}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\frac {\sqrt {b} c}{\sqrt {a}}+e\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{a} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}+\frac {e x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {f \sqrt {a+b x^4}}{2 b}\) |
(f*Sqrt[a + b*x^4])/(2*b) + (e*x*Sqrt[a + b*x^4])/(Sqrt[b]*(Sqrt[a] + Sqrt [b]*x^2)) + (d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*Sqrt[b]) - (a^(1 /4)*e*(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)*((Sqrt[b]*c)/Sqrt[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*b^(3/4)*Sqrt[a + b*x^4])
3.6.33.3.1 Defintions of rubi rules used
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)]
Result contains complex when optimal does not.
Time = 1.90 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(\frac {c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {f \sqrt {b \,x^{4}+a}}{2 b}+\frac {i e \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {d \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}\) | \(208\) |
| risch | \(\frac {c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {f \sqrt {b \,x^{4}+a}}{2 b}+\frac {i e \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {d \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}\) | \(208\) |
| elliptic | \(\frac {f \sqrt {b \,x^{4}+a}}{2 b}+\frac {c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {d \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {i e \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\) | \(211\) |
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) +1/2*f*(b*x^4+a)^(1/2)/b+I*e*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))+1/2*d*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2))/b^(1/2)
Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.49 \[ \int \frac {c+d x+e x^2+f x^3}{\sqrt {a+b x^4}} \, dx=\frac {4 \, a \sqrt {b} e x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + a \sqrt {b} d x \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 4 \, {\left (b c - a e\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, \sqrt {b x^{4} + a} {\left (a f x + 2 \, a e\right )}}{4 \, a b x} \]
1/4*(4*a*sqrt(b)*e*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) + a*sqrt(b)*d*x*log(-2*b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 4*(b*c - a*e)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + 2*s qrt(b*x^4 + a)*(a*f*x + 2*a*e))/(a*b*x)
Time = 1.51 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.46 \[ \int \frac {c+d x+e x^2+f x^3}{\sqrt {a+b x^4}} \, dx=f \left (\begin {cases} \frac {x^{4}}{4 \sqrt {a}} & \text {for}\: b = 0 \\\frac {\sqrt {a + b x^{4}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {d \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} + \frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \]
f*Piecewise((x**4/(4*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**4)/(2*b), True)) + d*asinh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b)) + c*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)) + e*x**3*gam ma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gam ma(7/4))
\[ \int \frac {c+d x+e x^2+f x^3}{\sqrt {a+b x^4}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a}} \,d x } \]
\[ \int \frac {c+d x+e x^2+f x^3}{\sqrt {a+b x^4}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a}} \,d x } \]
Timed out. \[ \int \frac {c+d x+e x^2+f x^3}{\sqrt {a+b x^4}} \, dx=\int \frac {f\,x^3+e\,x^2+d\,x+c}{\sqrt {b\,x^4+a}} \,d x \]