Integrand size = 28, antiderivative size = 303 \[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 a b \sqrt {a+b x^4}}-\frac {e \sqrt {a+b x^4}}{2 a b}-\frac {d x \sqrt {a+b x^4}}{2 a \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a+b x^4}}-\frac {\left (\sqrt {b} d-\sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 a^{3/4} b^{5/4} \sqrt {a+b x^4}} \]
-1/2*x*(-b*e*x^3-b*d*x^2-b*c*x+a*f)/a/b/(b*x^4+a)^(1/2)-1/2*e*(b*x^4+a)^(1 /2)/a/b-1/2*d*x*(b*x^4+a)^(1/2)/a/b^(1/2)/(a^(1/2)+x^2*b^(1/2))+1/2*d*(cos (2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*El lipticE(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))*(-f*a^(1/2 )+d*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^(3/4)/b^(5/4)/(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 = 0.38 \[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-3 a e-3 a f x+3 b c x^2+3 a f x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )+2 b d x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )}{6 a b \sqrt {a+b x^4}} \]
(-3*a*e - 3*a*f*x + 3*b*c*x^2 + 3*a*f*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric 2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] + 2*b*d*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 = 0.48 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2367, 25, 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.
| \(\displaystyle \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2367 |
| \(\displaystyle -\frac {\int -\frac {-2 b e x^3-b d x^2+a f}{\sqrt {b x^4+a}}dx}{2 a b}-\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {-2 b e x^3-b d x^2+a f}{\sqrt {b x^4+a}}dx}{2 a b}-\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \frac {\int \left (\frac {a f-b d x^2}{\sqrt {b x^4+a}}-\frac {2 b e x^3}{\sqrt {b x^4+a}}\right )dx}{2 a b}-\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 a b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {b} d-\sqrt {a} f\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}-\frac {\sqrt {b} d x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}+e \left (-\sqrt {a+b x^4}\right )}{2 a b}-\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 a b \sqrt {a+b x^4}}\) |
-1/2*(x*(a*f - b*c*x - b*d*x^2 - b*e*x^3))/(a*b*Sqrt[a + b*x^4]) + (-(e*Sq rt[a + b*x^4]) - (Sqrt[b]*d*x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2) + ( a^(1/4)*b^(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])/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^(1/4)*Sqrt[a + b*x^4]))/(2*a*b)
3.6.45.3.1 Defintions of rubi rules used
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = PolynomialQuotient[b^(Floor[(q - 1)/n] + 1) *x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*x^ m*Pq, a + b*x^n, x]}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floo r[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) I nt[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0 ] && LtQ[p, -1] && IGtQ[m, 0]
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 = 1.88 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.75
| method | result | size |
| elliptic | \(-\frac {2 b \left (-\frac {d \,x^{3}}{4 a b}-\frac {c \,x^{2}}{4 b a}+\frac {f x}{4 b^{2}}+\frac {e}{4 b^{2}}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {i d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\) | \(227\) |
| default | \(f \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {e}{2 b \sqrt {b \,x^{4}+a}}+d \left (\frac {x^{3}}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+\frac {c \,x^{2}}{2 a \sqrt {b \,x^{4}+a}}\) | \(250\) |
-2*b*(-1/4/a/b*d*x^3-1/4/b/a*c*x^2+1/4*f*x/b^2+1/4*e/b^2)/((x^4+a/b)*b)^(1 /2)+1/2*f/b/(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*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)/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 = 147, normalized size of antiderivative = 0.49 \[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {{\left (b^{2} d x^{4} + a b d\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left ({\left (b^{2} d + a b f\right )} x^{4} + a b d + a^{2} f\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (b^{2} d x^{3} + b^{2} c x^{2} - a b f x - a b e\right )} \sqrt {b x^{4} + a}}{2 \, {\left (a b^{3} x^{4} + a^{2} b^{2}\right )}} \]
1/2*((b^2*d*x^4 + a*b*d)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a)^( 1/4)), -1) - ((b^2*d + a*b*f)*x^4 + a*b*d + a^2*f)*sqrt(a)*(-b/a)^(3/4)*el liptic_f(arcsin(x*(-b/a)^(1/4)), -1) + (b^2*d*x^3 + b^2*c*x^2 - a*b*f*x - a*b*e)*sqrt(b*x^4 + a))/(a*b^3*x^4 + a^2*b^2)
Time = 5.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.44 \[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=e \left (\begin {cases} - \frac {1}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {c x^{2}}{2 a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {d x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {f x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} \]
e*Piecewise((-1/(2*b*sqrt(a + b*x**4)), Ne(b, 0)), (x**4/(4*a**(3/2)), Tru e)) + c*x**2/(2*a**(3/2)*sqrt(1 + b*x**4/a)) + d*x**3*gamma(3/4)*hyper((3/ 4, 3/2), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(7/4)) + f*x** 5*gamma(5/4)*hyper((5/4, 3/2), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/ 2)*gamma(9/4))
\[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \]