Integrand size = 27, antiderivative size = 275 \[ \int \frac {c+d x+e x^2+f x^3}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {e x \sqrt {a+b x^4}}{2 a \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {a f-b x \left (c+d x+e x^2\right )}{2 a b \sqrt {a+b x^4}}+\frac {e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} c-\sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 a^{5/4} b^{3/4} \sqrt {a+b x^4}} \]
1/2*(-a*f+b*x*(e*x^2+d*x+c))/a/b/(b*x^4+a)^(1/2)-1/2*e*x*(b*x^4+a)^(1/2)/a /b^(1/2)/(a^(1/2)+x^2*b^(1/2))+1/2*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(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)/a^(3/4)/b^(3/4)/(b*x^4+a)^(1/2)+1/4*(cos(2*arctan(b^(1/4)*x/a^( 1/4)))^2)^(1/2)/cos(2*arctan(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)+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)/a^(5/4)/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 = 10.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.42 \[ \int \frac {c+d x+e x^2+f x^3}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-3 a f+3 b c x+3 b d x^2+3 b c x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )+2 b e x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )}{6 a b \sqrt {a+b x^4}} \]
(-3*a*f + 3*b*c*x + 3*b*d*x^2 + 3*b*c*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric 2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] + 2*b*e*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeom etric2F1[3/4, 3/2, 7/4, -((b*x^4)/a)])/(6*a*b*Sqrt[a + b*x^4])
Time = 0.39 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2393, 25, 1512, 27, 761, 1510}
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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3}{\left (a+b x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2393 |
| \(\displaystyle -\frac {\int -\frac {c-e x^2}{\sqrt {b x^4+a}}dx}{2 a}-\frac {a f-b x \left (c+d x+e x^2\right )}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {c-e x^2}{\sqrt {b x^4+a}}dx}{2 a}-\frac {a f-b x \left (c+d x+e x^2\right )}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 1512 |
| \(\displaystyle \frac {\left (c-\frac {\sqrt {a} e}{\sqrt {b}}\right ) \int \frac {1}{\sqrt {b x^4+a}}dx+\frac {\sqrt {a} e \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {a} \sqrt {b x^4+a}}dx}{\sqrt {b}}}{2 a}-\frac {a f-b x \left (c+d x+e x^2\right )}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (c-\frac {\sqrt {a} e}{\sqrt {b}}\right ) \int \frac {1}{\sqrt {b x^4+a}}dx+\frac {e \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}}{2 a}-\frac {a f-b x \left (c+d x+e x^2\right )}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {\frac {e \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}+\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (c-\frac {\sqrt {a} e}{\sqrt {b}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt {a+b x^4}}}{2 a}-\frac {a f-b x \left (c+d x+e x^2\right )}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (c-\frac {\sqrt {a} e}{\sqrt {b}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {e \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^4}}-\frac {x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}\right )}{\sqrt {b}}}{2 a}-\frac {a f-b x \left (c+d x+e x^2\right )}{2 a b \sqrt {a+b x^4}}\) |
-1/2*(a*f - b*x*(c + d*x + e*x^2))/(a*b*Sqrt[a + b*x^4]) + ((e*(-((x*Sqrt[ a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sq rt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a ^(1/4)], 1/2])/(b^(1/4)*Sqrt[a + b*x^4])))/Sqrt[b] + ((c - (Sqrt[a]*e)/Sqr t[b])*(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^(1/4)*b^(1/4)*Sqrt[a + b*x^4]))/(2*a)
3.6.46.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c , d, e}, x] && PosQ[c/a]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[(a*Coeff[Pq, x, q] - b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q , x])*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] + Simp[1/(a*n*(p + 1)) In t[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^( p + 1), x], x] /; q == n - 1] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n , 0] && LtQ[p, -1]
Result contains complex when optimal does not.
Time = 1.73 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.84
| method | result | size |
| elliptic | \(-\frac {2 b \left (-\frac {e \,x^{3}}{4 b a}-\frac {d \,x^{2}}{4 a b}-\frac {c x}{4 b a}+\frac {f}{4 b^{2}}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {i e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\) | \(230\) |
| default | \(c \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {f}{2 b \sqrt {b \,x^{4}+a}}+e \left (\frac {x^{3}}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+\frac {d \,x^{2}}{2 a \sqrt {b \,x^{4}+a}}\) | \(250\) |
-2*b*(-1/4/b/a*e*x^3-1/4/a/b*d*x^2-1/4/b/a*c*x+1/4*f/b^2)/((x^4+a/b)*b)^(1 /2)+1/2*c/a/(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*I/a^(1/2)*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)/b^(1/2)*(Ellipt icF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I ))
Time = 0.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.47 \[ \int \frac {c+d x+e x^2+f x^3}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {{\left (b e x^{4} + a e\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left ({\left (b c + b e\right )} x^{4} + a c + a e\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (b e x^{3} + b d x^{2} + b c x - a f\right )} \sqrt {b x^{4} + a}}{2 \, {\left (a b^{2} x^{4} + a^{2} b\right )}} \]
1/2*((b*e*x^4 + a*e)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a)^(1/4) ), -1) - ((b*c + b*e)*x^4 + a*c + a*e)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arc sin(x*(-b/a)^(1/4)), -1) + (b*e*x^3 + b*d*x^2 + b*c*x - a*f)*sqrt(b*x^4 + a))/(a*b^2*x^4 + a^2*b)
Time = 4.95 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.48 \[ \int \frac {c+d x+e x^2+f x^3}{\left (a+b x^4\right )^{3/2}} \, dx=f \left (\begin {cases} - \frac {1}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {d x^{2}}{2 a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \]
f*Piecewise((-1/(2*b*sqrt(a + b*x**4)), Ne(b, 0)), (x**4/(4*a**(3/2)), Tru e)) + c*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), b*x**4*exp_polar(I*pi)/a)/( 4*a**(3/2)*gamma(5/4)) + d*x**2/(2*a**(3/2)*sqrt(1 + b*x**4/a)) + e*x**3*g amma(3/4)*hyper((3/4, 3/2), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)* gamma(7/4))
\[ \int \frac {c+d x+e x^2+f x^3}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {c+d x+e x^2+f x^3}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {c+d x+e x^2+f x^3}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {f\,x^3+e\,x^2+d\,x+c}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \]