Integrand size = 30, antiderivative size = 370 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=-\frac {a f \sqrt {a+b x^4}}{2 b^2}+\frac {c x \sqrt {a+b x^4}}{3 b}+\frac {d x^2 \sqrt {a+b x^4}}{4 b}+\frac {e x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a e x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {f \left (a+b x^4\right )^{3/2}}{6 b^2}-\frac {a d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}+\frac {3 a^{5/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {a^{3/4} \left (5 \sqrt {b} c+9 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{30 b^{7/4} \sqrt {a+b x^4}} \] Output:
-1/2*a*f*(b*x^4+a)^(1/2)/b^2+1/3*c*x*(b*x^4+a)^(1/2)/b+1/4*d*x^2*(b*x^4+a) ^(1/2)/b+1/5*e*x^3*(b*x^4+a)^(1/2)/b-3/5*a*e*x*(b*x^4+a)^(1/2)/b^(3/2)/(a^ (1/2)+b^(1/2)*x^2)+1/6*f*(b*x^4+a)^(3/2)/b^2-1/4*a*d*arctanh(b^(1/2)*x^2/( b*x^4+a)^(1/2))/b^(3/2)+3/5*a^(5/4)*e*(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^(7/4)/(b*x^4+a)^(1/2)-1/30*a^(3/4)*(5*b^(1/2)*c+9*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)*InverseJaco biAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/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 = 10.12 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.57 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {-20 a^2 f+20 a b c x+15 a b d x^2+12 a b e x^3-10 a b f x^4+20 b^2 c x^5+15 b^2 d x^6+12 b^2 e x^7+10 b^2 f x^8-15 a \sqrt {b} d \sqrt {a+b x^4} \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-20 a b c x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )-12 a b e x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{60 b^2 \sqrt {a+b x^4}} \] Input:
Integrate[(x^4*(c + d*x + e*x^2 + f*x^3))/Sqrt[a + b*x^4],x]
Output:
(-20*a^2*f + 20*a*b*c*x + 15*a*b*d*x^2 + 12*a*b*e*x^3 - 10*a*b*f*x^4 + 20* b^2*c*x^5 + 15*b^2*d*x^6 + 12*b^2*e*x^7 + 10*b^2*f*x^8 - 15*a*Sqrt[b]*d*Sq rt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 20*a*b*c*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] - 12*a*b*e*x^3* Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)])/(60*b^ 2*Sqrt[a + b*x^4])
Time = 0.99 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (\frac {x^4 \left (c+e x^2\right )}{\sqrt {a+b x^4}}+\frac {x^5 \left (d+f x^2\right )}{\sqrt {a+b x^4}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (9 \sqrt {a} e+5 \sqrt {b} c\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{30 b^{7/4} \sqrt {a+b x^4}}+\frac {3 a^{5/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {a d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}-\frac {3 a e x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {a f \sqrt {a+b x^4}}{3 b^2}+\frac {c x \sqrt {a+b x^4}}{3 b}+\frac {d x^2 \sqrt {a+b x^4}}{4 b}+\frac {e x^3 \sqrt {a+b x^4}}{5 b}+\frac {f x^4 \sqrt {a+b x^4}}{6 b}\) |
Input:
Int[(x^4*(c + d*x + e*x^2 + f*x^3))/Sqrt[a + b*x^4],x]
Output:
-1/3*(a*f*Sqrt[a + b*x^4])/b^2 + (c*x*Sqrt[a + b*x^4])/(3*b) + (d*x^2*Sqrt [a + b*x^4])/(4*b) + (e*x^3*Sqrt[a + b*x^4])/(5*b) + (f*x^4*Sqrt[a + b*x^4 ])/(6*b) - (3*a*e*x*Sqrt[a + b*x^4])/(5*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) - (a*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(3/2)) + (3*a^(5/4)*e*( Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ellipti cE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(5*b^(7/4)*Sqrt[a + b*x^4]) - (a^( 3/4)*(5*Sqrt[b]*c + 9*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])/ (30*b^(7/4)*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 = 2.03 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.66
| method | result | size |
| risch | \(-\frac {\left (-10 b f \,x^{4}-12 b e \,x^{3}-15 b d \,x^{2}-20 c b x +20 a f \right ) \sqrt {b \,x^{4}+a}}{60 b^{2}}-\frac {a \left (\frac {10 c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {15 d \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {18 i e \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )}{30 b}\) | \(246\) |
| default | \(c \left (\frac {x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4 b}-\frac {a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}\right )+e \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {f \sqrt {b \,x^{4}+a}\, \left (-b \,x^{4}+2 a \right )}{6 b^{2}}\) | \(279\) |
| elliptic | \(\frac {f \,x^{4} \sqrt {b \,x^{4}+a}}{6 b}+\frac {e \,x^{3} \sqrt {b \,x^{4}+a}}{5 b}+\frac {d \,x^{2} \sqrt {b \,x^{4}+a}}{4 b}+\frac {c x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a f \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {a c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a d \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}-\frac {3 i a^{\frac {3}{2}} e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(288\) |
Input:
int(x^4*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/60*(-10*b*f*x^4-12*b*e*x^3-15*b*d*x^2-20*b*c*x+20*a*f)/b^2*(b*x^4+a)^(1 /2)-1/30*a/b*(10*c/(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)+15/2*d*ln(b^(1/2)*x^2+(b*x^4+a)^(1/2))/b^(1/2)+18*I*e*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)*(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 = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.44 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=-\frac {72 \, a \sqrt {b} e x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 15 \, a \sqrt {b} d x \log \left (-2 \, b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 8 \, {\left (5 \, b c - 9 \, a e\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 2 \, {\left (10 \, b f x^{5} + 12 \, b e x^{4} + 15 \, b d x^{3} + 20 \, b c x^{2} - 20 \, a f x - 36 \, a e\right )} \sqrt {b x^{4} + a}}{120 \, b^{2} x} \] Input:
integrate(x^4*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x, algorithm="fricas")
Output:
-1/120*(72*a*sqrt(b)*e*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), - 1) - 15*a*sqrt(b)*d*x*log(-2*b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 8*(5*b*c - 9*a*e)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x) , -1) - 2*(10*b*f*x^5 + 12*b*e*x^4 + 15*b*d*x^3 + 20*b*c*x^2 - 20*a*f*x - 36*a*e)*sqrt(b*x^4 + a))/(b^2*x)
Time = 2.67 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.48 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {\sqrt {a} d x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4 b} - \frac {a d \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {3}{2}}} + f \left (\begin {cases} - \frac {a \sqrt {a + b x^{4}}}{3 b^{2}} + \frac {x^{4} \sqrt {a + b x^{4}}}{6 b} & \text {for}\: b \neq 0 \\\frac {x^{8}}{8 \sqrt {a}} & \text {otherwise} \end {cases}\right ) + \frac {c x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {e x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \] Input:
integrate(x**4*(f*x**3+e*x**2+d*x+c)/(b*x**4+a)**(1/2),x)
Output:
sqrt(a)*d*x**2*sqrt(1 + b*x**4/a)/(4*b) - a*d*asinh(sqrt(b)*x**2/sqrt(a))/ (4*b**(3/2)) + f*Piecewise((-a*sqrt(a + b*x**4)/(3*b**2) + x**4*sqrt(a + b *x**4)/(6*b), Ne(b, 0)), (x**8/(8*sqrt(a)), True)) + c*x**5*gamma(5/4)*hyp er((1/2, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + e*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4* sqrt(a)*gamma(11/4))
\[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{4}}{\sqrt {b x^{4} + a}} \,d x } \] Input:
integrate(x^4*(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^4/sqrt(b*x^4 + a), x)
\[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{4}}{\sqrt {b x^{4} + a}} \,d x } \] Input:
integrate(x^4*(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^4/sqrt(b*x^4 + a), x)
Timed out. \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\int \frac {x^4\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{\sqrt {b\,x^4+a}} \,d x \] Input:
int((x^4*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(1/2),x)
Output:
int((x^4*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(1/2), x)
\[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {-40 \sqrt {b \,x^{4}+a}\, a f +40 \sqrt {b \,x^{4}+a}\, b c x +30 \sqrt {b \,x^{4}+a}\, b d \,x^{2}+24 \sqrt {b \,x^{4}+a}\, b e \,x^{3}+20 \sqrt {b \,x^{4}+a}\, b f \,x^{4}+15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a d -15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a d -40 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a b c -72 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) a b e}{120 b^{2}} \] Input:
int(x^4*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x)
Output:
( - 40*sqrt(a + b*x**4)*a*f + 40*sqrt(a + b*x**4)*b*c*x + 30*sqrt(a + b*x* *4)*b*d*x**2 + 24*sqrt(a + b*x**4)*b*e*x**3 + 20*sqrt(a + b*x**4)*b*f*x**4 + 15*sqrt(b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a*d - 15*sqrt(b)*log(sq rt(a + b*x**4) + sqrt(b)*x**2)*a*d - 40*int(sqrt(a + b*x**4)/(a + b*x**4), x)*a*b*c - 72*int((sqrt(a + b*x**4)*x**2)/(a + b*x**4),x)*a*b*e)/(120*b**2 )