Integrand size = 30, antiderivative size = 302 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-c-d x-e x^2-f x^3}{2 b \sqrt {a+b x^4}}+\frac {3 f x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {3 \sqrt [4]{a} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} d+3 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}} \]
1/2*e*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(3/2)+1/2*(-f*x^3-e*x^2-d*x-c )/b/(b*x^4+a)^(1/2)+3/2*f*x*(b*x^4+a)^(1/2)/b^(3/2)/(a^(1/2)+x^2*b^(1/2))- 3/2*a^(1/4)*f*(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^(7/4)/(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))* (3*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^(1/4)/b^(7/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.13 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.60 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-\sqrt {b} c-\sqrt {b} d x-\sqrt {b} e x^2+2 \sqrt {b} f x^3+\sqrt {a} e \sqrt {1+\frac {b x^4}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+\sqrt {b} d x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )-2 \sqrt {b} f x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )}{2 b^{3/2} \sqrt {a+b x^4}} \]
(-(Sqrt[b]*c) - Sqrt[b]*d*x - Sqrt[b]*e*x^2 + 2*Sqrt[b]*f*x^3 + Sqrt[a]*e* Sqrt[1 + (b*x^4)/a]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] + Sqrt[b]*d*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] - 2*Sqrt[b]*f*x^ 3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[3/4, 3/2, 7/4, -((b*x^4)/a)])/(2*b ^(3/2)*Sqrt[a + b*x^4])
Time = 0.49 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2363, 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.
| \(\displaystyle \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2363 |
| \(\displaystyle \frac {\int \frac {3 f x^2+2 e x+d}{\sqrt {b x^4+a}}dx}{2 b}-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \frac {\int \left (\frac {2 e x}{\sqrt {b x^4+a}}+\frac {3 f x^2+d}{\sqrt {b x^4+a}}\right )dx}{2 b}-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} f+\sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} b^{3/4} \sqrt {a+b x^4}}-\frac {3 \sqrt [4]{a} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{\sqrt {b}}+\frac {3 f x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}}{2 b}-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}\) |
-1/2*(c + d*x + e*x^2 + f*x^3)/(b*Sqrt[a + b*x^4]) + ((3*f*x*Sqrt[a + b*x^ 4])/(Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/Sqrt[b] - (3*a^(1/4)*f*(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]) + ((Sqrt[b]*d + 3*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*a^(1/4)*b^(3/4)*Sqrt[a + b*x^4]))/(2*b)
3.6.43.3.1 Defintions of rubi rules used
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Pq*(( a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[1/(b*n*(p + 1)) Int[D[Pq, x] *(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, m, n}, x] && PolyQ[Pq, x] && E qQ[m - n + 1, 0] && LtQ[p, -1]
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 = 1.80 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.82
| method | result | size |
| elliptic | \(-\frac {2 b \left (\frac {f \,x^{3}}{4 b^{2}}+\frac {e \,x^{2}}{4 b^{2}}+\frac {d x}{4 b^{2}}+\frac {c}{4 b^{2}}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {e \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {3 i f \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(248\) |
| default | \(f \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (-\frac {x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {\ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}\right )+d \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {c}{2 b \sqrt {b \,x^{4}+a}}\) | \(275\) |
-2*b*(1/4*f*x^3/b^2+1/4*e/b^2*x^2+1/4*d*x/b^2+1/4*c/b^2)/((x^4+a/b)*b)^(1/ 2)+1/2*d/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/b^(3/2)*e*ln(2*x^2*b^(1/2)+2*(b*x^4+a)^(1/2))+3/2*I*f/b^(3/2 )*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))
Time = 0.13 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.71 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {6 \, {\left (a b f x^{5} + a^{2} f x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left ({\left (b^{2} d - 3 \, a b f\right )} x^{5} + {\left (a b d - 3 \, a^{2} f\right )} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (a b e x^{5} + a^{2} e x\right )} \sqrt {b} \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, {\left (2 \, a b f x^{4} - a b e x^{3} - a b d x^{2} - a b c x + 3 \, a^{2} f\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a b^{3} x^{5} + a^{2} b^{2} x\right )}} \]
1/4*(6*(a*b*f*x^5 + a^2*f*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b) ^(1/4)/x), -1) + 2*((b^2*d - 3*a*b*f)*x^5 + (a*b*d - 3*a^2*f)*x)*sqrt(b)*( -a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + (a*b*e*x^5 + a^2*e*x) *sqrt(b)*log(-2*b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 2*(2*a*b*f*x^ 4 - a*b*e*x^3 - a*b*d*x^2 - a*b*c*x + 3*a^2*f)*sqrt(b*x^4 + a))/(a*b^3*x^5 + a^2*b^2*x)
Time = 6.39 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.52 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=c \left (\begin {cases} - \frac {1}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + e \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} - \frac {x^{2}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}}\right ) + \frac {d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {f x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \]
c*Piecewise((-1/(2*b*sqrt(a + b*x**4)), Ne(b, 0)), (x**4/(4*a**(3/2)), Tru e)) + e*(asinh(sqrt(b)*x**2/sqrt(a))/(2*b**(3/2)) - x**2/(2*sqrt(a)*b*sqrt (1 + b*x**4/a))) + d*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)) + f*x**7*gamma(7/4)*hyper((3/2, 7/4 ), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(11/4))
\[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{3}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{3}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^3\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \]