Integrand size = 30, antiderivative size = 369 \[ \int x^2 \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\frac {2 a e x \sqrt {a+b x^4}}{21 b}-\frac {a f x^2 \sqrt {a+b x^4}}{16 b}+\frac {2 a c x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{35} x^3 \left (7 c+5 e x^2\right ) \sqrt {a+b x^4}+\frac {\left (4 d+3 f x^2\right ) \left (a+b x^4\right )^{3/2}}{24 b}-\frac {a^2 f \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 b^{3/2}}-\frac {2 a^{5/4} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a^{5/4} \left (21 \sqrt {b} c-5 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{105 b^{5/4} \sqrt {a+b x^4}} \]
1/24*(3*f*x^2+4*d)*(b*x^4+a)^(3/2)/b-1/16*a^2*f*arctanh(x^2*b^(1/2)/(b*x^4 +a)^(1/2))/b^(3/2)+2/21*a*e*x*(b*x^4+a)^(1/2)/b-1/16*a*f*x^2*(b*x^4+a)^(1/ 2)/b+1/35*x^3*(5*e*x^2+7*c)*(b*x^4+a)^(1/2)+2/5*a*c*x*(b*x^4+a)^(1/2)/b^(1 /2)/(a^(1/2)+x^2*b^(1/2))-2/5*a^(5/4)*c*(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^(3/4)/(b*x^4+a)^(1/2)+1/105*a^(5/4)*(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*arct an(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(-5*e*a^(1/2)+21*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)/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 = 10.69 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.49 \[ \int x^2 \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\frac {1}{336} \sqrt {a+b x^4} \left (\frac {56 d \left (a+b x^4\right )}{b}+\frac {48 e x \left (a+b x^4\right )}{b}+\frac {21 f x^2 \left (a+2 b x^4\right )}{b}-\frac {21 a^{3/2} f \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{b^{3/2} \sqrt {1+\frac {b x^4}{a}}}-\frac {48 a e x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{b \sqrt {1+\frac {b x^4}{a}}}+\frac {112 c x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{\sqrt {1+\frac {b x^4}{a}}}\right ) \]
(Sqrt[a + b*x^4]*((56*d*(a + b*x^4))/b + (48*e*x*(a + b*x^4))/b + (21*f*x^ 2*(a + 2*b*x^4))/b - (21*a^(3/2)*f*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]])/(b^(3/2 )*Sqrt[1 + (b*x^4)/a]) - (48*a*e*x*Hypergeometric2F1[-1/2, 1/4, 5/4, -((b* x^4)/a)])/(b*Sqrt[1 + (b*x^4)/a]) + (112*c*x^3*Hypergeometric2F1[-1/2, 3/4 , 7/4, -((b*x^4)/a)])/Sqrt[1 + (b*x^4)/a]))/336
Time = 0.58 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, 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.
| \(\displaystyle \int x^2 \sqrt {a+b x^4} \left (c+d x+e x^2+f x^3\right ) \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (x^2 \sqrt {a+b x^4} \left (c+e x^2\right )+x^3 \sqrt {a+b x^4} \left (d+f x^2\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (21 \sqrt {b} c-5 \sqrt {a} e\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{105 b^{5/4} \sqrt {a+b x^4}}-\frac {2 a^{5/4} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}-\frac {a^2 f \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 b^{3/2}}+\frac {1}{35} x^3 \sqrt {a+b x^4} \left (7 c+5 e x^2\right )+\frac {2 a c x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {d \left (a+b x^4\right )^{3/2}}{6 b}+\frac {2 a e x \sqrt {a+b x^4}}{21 b}+\frac {f x^2 \left (a+b x^4\right )^{3/2}}{8 b}-\frac {a f x^2 \sqrt {a+b x^4}}{16 b}\) |
(2*a*e*x*Sqrt[a + b*x^4])/(21*b) - (a*f*x^2*Sqrt[a + b*x^4])/(16*b) + (2*a *c*x*Sqrt[a + b*x^4])/(5*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (x^3*(7*c + 5* e*x^2)*Sqrt[a + b*x^4])/35 + (d*(a + b*x^4)^(3/2))/(6*b) + (f*x^2*(a + b*x ^4)^(3/2))/(8*b) - (a^2*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*b^(3 /2)) - (2*a^(5/4)*c*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sq rt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(5*b^(3/4)*Sq rt[a + b*x^4]) + (a^(5/4)*(21*Sqrt[b]*c - 5*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])/(105*b^(5/4)*Sqrt[a + b*x^4])
3.5.97.3.1 Defintions of rubi rules used
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)]
Result contains complex when optimal does not.
Time = 2.08 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.71
| method | result | size |
| risch | \(\frac {\left (210 b f \,x^{6}+240 b e \,x^{5}+280 b d \,x^{4}+336 b c \,x^{3}+105 x^{2} a f +160 a e x +280 a d \right ) \sqrt {b \,x^{4}+a}}{1680 b}-\frac {a \left (\frac {80 a e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {336 i \sqrt {b}\, c \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {105 a f \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}\right )}{840 b}\) | \(262\) |
| default | \(f \left (\frac {x^{2} \left (b \,x^{4}+a \right )^{\frac {3}{2}}}{8 b}-\frac {a \,x^{2} \sqrt {b \,x^{4}+a}}{16 b}-\frac {a^{2} \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{16 b^{\frac {3}{2}}}\right )+e \left (\frac {x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {2 a x \sqrt {b \,x^{4}+a}}{21 b}-\frac {2 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{21 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {d \left (b \,x^{4}+a \right )^{\frac {3}{2}}}{6 b}+c \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )\) | \(303\) |
| elliptic | \(\frac {f \,x^{6} \sqrt {b \,x^{4}+a}}{8}+\frac {e \,x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {d \,x^{4} \sqrt {b \,x^{4}+a}}{6}+\frac {c \,x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {a f \,x^{2} \sqrt {b \,x^{4}+a}}{16 b}+\frac {2 a e x \sqrt {b \,x^{4}+a}}{21 b}+\frac {a d \sqrt {b \,x^{4}+a}}{6 b}-\frac {2 a^{2} e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{21 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a^{2} f \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{16 b^{\frac {3}{2}}}+\frac {2 i a^{\frac {3}{2}} c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\) | \(318\) |
1/1680*(210*b*f*x^6+240*b*e*x^5+280*b*d*x^4+336*b*c*x^3+105*a*f*x^2+160*a* e*x+280*a*d)/b*(b*x^4+a)^(1/2)-1/840*a/b*(80*a*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)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-336*I*b^(1/2)*c*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)-El lipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I))+105/2*a*f*ln(x^2*b^(1/2)+(b*x^4+a) ^(1/2))/b^(1/2))
Time = 0.12 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.52 \[ \int x^2 \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\frac {1344 \, a b^{\frac {3}{2}} c x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 105 \, a^{2} \sqrt {b} f x \log \left (-2 \, b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) - 64 \, {\left (21 \, a b c + 5 \, a b e\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left (210 \, b^{2} f x^{7} + 240 \, b^{2} e x^{6} + 280 \, b^{2} d x^{5} + 336 \, b^{2} c x^{4} + 105 \, a b f x^{3} + 160 \, a b e x^{2} + 280 \, a b d x + 672 \, a b c\right )} \sqrt {b x^{4} + a}}{3360 \, b^{2} x} \]
1/3360*(1344*a*b^(3/2)*c*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) + 105*a^2*sqrt(b)*f*x*log(-2*b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) - 64*(21*a*b*c + 5*a*b*e)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/ b)^(1/4)/x), -1) + 2*(210*b^2*f*x^7 + 240*b^2*e*x^6 + 280*b^2*d*x^5 + 336* b^2*c*x^4 + 105*a*b*f*x^3 + 160*a*b*e*x^2 + 280*a*b*d*x + 672*a*b*c)*sqrt( b*x^4 + a))/(b^2*x)
Time = 3.23 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.57 \[ \int x^2 \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\frac {a^{\frac {3}{2}} f x^{2}}{16 b \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} c x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {a} e x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {3 \sqrt {a} f x^{6}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} - \frac {a^{2} f \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{16 b^{\frac {3}{2}}} + d \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\left (a + b x^{4}\right )^{\frac {3}{2}}}{6 b} & \text {otherwise} \end {cases}\right ) + \frac {b f x^{10}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \]
a**(3/2)*f*x**2/(16*b*sqrt(1 + b*x**4/a)) + sqrt(a)*c*x**3*gamma(3/4)*hype r((-1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4)) + sqrt(a)* e*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4* gamma(9/4)) + 3*sqrt(a)*f*x**6/(16*sqrt(1 + b*x**4/a)) - a**2*f*asinh(sqrt (b)*x**2/sqrt(a))/(16*b**(3/2)) + d*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True)) + b*f*x**10/(8*sqrt(a)*sqrt(1 + b*x**4/ a))
\[ \int x^2 \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\int { \sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )} x^{2} \,d x } \]
\[ \int x^2 \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\int { \sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )} x^{2} \,d x } \]
Timed out. \[ \int x^2 \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\int x^2\,\sqrt {b\,x^4+a}\,\left (f\,x^3+e\,x^2+d\,x+c\right ) \,d x \]