Integrand size = 30, antiderivative size = 322 \[ \int \frac {x^2 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {d \sqrt {a+b x^4}}{2 b}+\frac {e x \sqrt {a+b x^4}}{3 b}+\frac {f x^2 \sqrt {a+b x^4}}{4 b}+\frac {c x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {a f \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}-\frac {\sqrt [4]{a} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (3 \sqrt {b} c-\sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 b^{5/4} \sqrt {a+b x^4}} \] Output:
1/2*d*(b*x^4+a)^(1/2)/b+1/3*e*x*(b*x^4+a)^(1/2)/b+1/4*f*x^2*(b*x^4+a)^(1/2 )/b+c*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+b^(1/2)*x^2)-1/4*a*f*arctanh(b^(1 /2)*x^2/(b*x^4+a)^(1/2))/b^(3/2)-a^(1/4)*c*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a )/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4)) ),1/2*2^(1/2))/b^(3/4)/(b*x^4+a)^(1/2)+1/6*a^(1/4)*(3*b^(1/2)*c-a^(1/2)*e) *(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJa cobiAM(2*arctan(b^(1/4)*x/a^(1/4)),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 = 10.14 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.60 \[ \int \frac {x^2 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {6 \sqrt {b} d \left (a+b x^4\right )+4 \sqrt {b} e x \left (a+b x^4\right )+3 \sqrt {b} f x^2 \left (a+b x^4\right )-3 a f \sqrt {a+b x^4} \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-4 a \sqrt {b} e x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )+4 b^{3/2} c x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{12 b^{3/2} \sqrt {a+b x^4}} \] Input:
Integrate[(x^2*(c + d*x + e*x^2 + f*x^3))/Sqrt[a + b*x^4],x]
Output:
(6*Sqrt[b]*d*(a + b*x^4) + 4*Sqrt[b]*e*x*(a + b*x^4) + 3*Sqrt[b]*f*x^2*(a + b*x^4) - 3*a*f*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 4*a*Sqrt[b]*e*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b* x^4)/a)] + 4*b^(3/2)*c*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)])/(12*b^(3/2)*Sqrt[a + b*x^4])
Time = 0.86 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, 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.
| \(\displaystyle \int \frac {x^2 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (\frac {x^2 \left (c+e x^2\right )}{\sqrt {a+b x^4}}+\frac {x^3 \left (d+f x^2\right )}{\sqrt {a+b x^4}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {b} c-\sqrt {a} e\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 b^{5/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{a} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}-\frac {a f \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}+\frac {c x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {d \sqrt {a+b x^4}}{2 b}+\frac {e x \sqrt {a+b x^4}}{3 b}+\frac {f x^2 \sqrt {a+b x^4}}{4 b}\) |
Input:
Int[(x^2*(c + d*x + e*x^2 + f*x^3))/Sqrt[a + b*x^4],x]
Output:
(d*Sqrt[a + b*x^4])/(2*b) + (e*x*Sqrt[a + b*x^4])/(3*b) + (f*x^2*Sqrt[a + b*x^4])/(4*b) + (c*x*Sqrt[a + b*x^4])/(Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) - (a*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(3/2)) - (a^(1/4)*c*(Sqr t[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)* (3*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])/(6*b^(5 /4)*Sqrt[a + b*x^4])
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 = 1.64 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.71
| method | result | size |
| risch | \(\frac {\left (3 f \,x^{2}+4 e x +6 d \right ) \sqrt {b \,x^{4}+a}}{12 b}-\frac {\frac {2 a e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 a f \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}-\frac {6 i c \sqrt {b}\, \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}}{6 b}\) | \(230\) |
| default | \(\frac {i c \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {d \sqrt {b \,x^{4}+a}}{2 b}+e \left (\frac {x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+f \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4 b}-\frac {a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}\right )\) | \(250\) |
| elliptic | \(\frac {f \,x^{2} \sqrt {b \,x^{4}+a}}{4 b}+\frac {e x \sqrt {b \,x^{4}+a}}{3 b}+\frac {d \sqrt {b \,x^{4}+a}}{2 b}-\frac {a e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a f \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}+\frac {i c \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\) | \(251\) |
Input:
int(x^2*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/12*(3*f*x^2+4*e*x+6*d)/b*(b*x^4+a)^(1/2)-1/6/b*(2*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)+3/2*a*f*ln(b^(1/2)* x^2+(b*x^4+a)^(1/2))/b^(1/2)-6*I*c*b^(1/2)*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 = 0.15 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.46 \[ \int \frac {x^2 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {24 \, b^{\frac {3}{2}} c x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 3 \, a \sqrt {b} f x \log \left (-2 \, b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) - 8 \, {\left (3 \, b c + b e\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left (3 \, b f x^{3} + 4 \, b e x^{2} + 6 \, b d x + 12 \, b c\right )} \sqrt {b x^{4} + a}}{24 \, b^{2} x} \] Input:
integrate(x^2*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x, algorithm="fricas")
Output:
1/24*(24*b^(3/2)*c*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) + 3*a*sqrt(b)*f*x*log(-2*b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) - 8*(3* b*c + b*e)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + 2*(3*b*f*x^3 + 4*b*e*x^2 + 6*b*d*x + 12*b*c)*sqrt(b*x^4 + a))/(b^2*x)
Time = 2.47 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.48 \[ \int \frac {x^2 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {\sqrt {a} f x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4 b} - \frac {a f \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {3}{2}}} + d \left (\begin {cases} \frac {x^{4}}{4 \sqrt {a}} & \text {for}\: b = 0 \\\frac {\sqrt {a + b x^{4}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {c x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {e x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate(x**2*(f*x**3+e*x**2+d*x+c)/(b*x**4+a)**(1/2),x)
Output:
sqrt(a)*f*x**2*sqrt(1 + b*x**4/a)/(4*b) - a*f*asinh(sqrt(b)*x**2/sqrt(a))/ (4*b**(3/2)) + d*Piecewise((x**4/(4*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**4) /(2*b), True)) + c*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp_po lar(I*pi)/a)/(4*sqrt(a)*gamma(7/4)) + e*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(9/4))
\[ \int \frac {x^2 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{2}}{\sqrt {b x^{4} + a}} \,d x } \] Input:
integrate(x^2*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x, algorithm="maxima")
Output:
integrate((f*x^3 + e*x^2 + d*x + c)*x^2/sqrt(b*x^4 + a), x)
\[ \int \frac {x^2 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{2}}{\sqrt {b x^{4} + a}} \,d x } \] Input:
integrate(x^2*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x, algorithm="giac")
Output:
integrate((f*x^3 + e*x^2 + d*x + c)*x^2/sqrt(b*x^4 + a), x)
Timed out. \[ \int \frac {x^2 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\int \frac {x^2\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{\sqrt {b\,x^4+a}} \,d x \] Input:
int((x^2*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(1/2),x)
Output:
int((x^2*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(1/2), x)
\[ \int \frac {x^2 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx =\text {Too large to display} \] Input:
int(x^2*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x)
Output:
(3*sqrt(b)*sqrt(a + b*x**4)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a**2*f + 12*sqrt(b)*sqrt(a + b*x**4)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a*b*f*x** 4 - 3*sqrt(b)*sqrt(a + b*x**4)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*a**2*f - 12*sqrt(b)*sqrt(a + b*x**4)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*a*b*f* x**4 + 36*sqrt(b)*sqrt(a + b*x**4)*a*b*d*x**2 + 24*sqrt(b)*sqrt(a + b*x**4 )*a*b*e*x**3 + 18*sqrt(b)*sqrt(a + b*x**4)*a*b*f*x**4 + 48*sqrt(b)*sqrt(a + b*x**4)*b**2*d*x**6 + 32*sqrt(b)*sqrt(a + b*x**4)*b**2*e*x**7 + 24*sqrt( b)*sqrt(a + b*x**4)*b**2*f*x**8 - 8*sqrt(a + b*x**4)*int(sqrt(a + b*x**4)/ (a + b*x**4),x)*a**2*b*e - 32*sqrt(a + b*x**4)*int(sqrt(a + b*x**4)/(a + b *x**4),x)*a*b**2*e*x**4 + 24*sqrt(a + b*x**4)*int((sqrt(a + b*x**4)*x**2)/ (a + b*x**4),x)*a*b**2*c + 96*sqrt(a + b*x**4)*int((sqrt(a + b*x**4)*x**2) /(a + b*x**4),x)*b**3*c*x**4 - 24*sqrt(b)*int(sqrt(a + b*x**4)/(a + b*x**4 ),x)*a**2*b*e*x**2 - 32*sqrt(b)*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a*b** 2*e*x**6 + 72*sqrt(b)*int((sqrt(a + b*x**4)*x**2)/(a + b*x**4),x)*a*b**2*c *x**2 + 96*sqrt(b)*int((sqrt(a + b*x**4)*x**2)/(a + b*x**4),x)*b**3*c*x**6 + 9*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a**2*b*f*x**2 + 12*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a*b**2*f*x**6 - 9*log(sqrt(a + b*x**4) + sqrt(b)*x **2)*a**2*b*f*x**2 - 12*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*a*b**2*f*x**6 + 12*a**2*b*d + 8*a**2*b*e*x + 6*a**2*b*f*x**2 + 60*a*b**2*d*x**4 + 40*a* b**2*e*x**5 + 30*a*b**2*f*x**6 + 48*b**3*d*x**8 + 32*b**3*e*x**9 + 24*b...