Integrand size = 30, antiderivative size = 350 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {c \sqrt {a+b x^4}}{2 b}+\frac {d x \sqrt {a+b x^4}}{3 b}+\frac {e x^2 \sqrt {a+b x^4}}{4 b}+\frac {f x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a f x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {a e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}+\frac {3 a^{5/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {a^{3/4} \left (5 \sqrt {b} d+9 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{30 b^{7/4} \sqrt {a+b x^4}} \] Output:
1/2*c*(b*x^4+a)^(1/2)/b+1/3*d*x*(b*x^4+a)^(1/2)/b+1/4*e*x^2*(b*x^4+a)^(1/2 )/b+1/5*f*x^3*(b*x^4+a)^(1/2)/b-3/5*a*f*x*(b*x^4+a)^(1/2)/b^(3/2)/(a^(1/2) +b^(1/2)*x^2)-1/4*a*e*arctanh(b^(1/2)*x^2/(b*x^4+a)^(1/2))/b^(3/2)+3/5*a^( 5/4)*f*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*Ell ipticE(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)*d+9*a^(1/2)*f)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a) /(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(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 = 0.61 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {30 \sqrt {b} c \left (a+b x^4\right )+20 \sqrt {b} d x \left (a+b x^4\right )+15 \sqrt {b} e x^2 \left (a+b x^4\right )+12 \sqrt {b} f x^3 \left (a+b x^4\right )-15 a e \sqrt {a+b x^4} \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-20 a \sqrt {b} d x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )-12 a \sqrt {b} f x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{60 b^{3/2} \sqrt {a+b x^4}} \] Input:
Integrate[(x^3*(c + d*x + e*x^2 + f*x^3))/Sqrt[a + b*x^4],x]
Output:
(30*Sqrt[b]*c*(a + b*x^4) + 20*Sqrt[b]*d*x*(a + b*x^4) + 15*Sqrt[b]*e*x^2* (a + b*x^4) + 12*Sqrt[b]*f*x^3*(a + b*x^4) - 15*a*e*Sqrt[a + b*x^4]*ArcTan h[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 20*a*Sqrt[b]*d*x*Sqrt[1 + (b*x^4)/a]*Hy pergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] - 12*a*Sqrt[b]*f*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)])/(60*b^(3/2)*Sq rt[a + b*x^4])
Time = 0.93 (sec) , antiderivative size = 350, 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.
| \(\displaystyle \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (\frac {x^3 \left (c+e x^2\right )}{\sqrt {a+b x^4}}+\frac {x^4 \left (d+f x^2\right )}{\sqrt {a+b x^4}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (9 \sqrt {a} f+5 \sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{30 b^{7/4} \sqrt {a+b x^4}}+\frac {3 a^{5/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {a e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}-\frac {3 a f x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {c \sqrt {a+b x^4}}{2 b}+\frac {d x \sqrt {a+b x^4}}{3 b}+\frac {e x^2 \sqrt {a+b x^4}}{4 b}+\frac {f x^3 \sqrt {a+b x^4}}{5 b}\) |
Input:
Int[(x^3*(c + d*x + e*x^2 + f*x^3))/Sqrt[a + b*x^4],x]
Output:
(c*Sqrt[a + b*x^4])/(2*b) + (d*x*Sqrt[a + b*x^4])/(3*b) + (e*x^2*Sqrt[a + b*x^4])/(4*b) + (f*x^3*Sqrt[a + b*x^4])/(5*b) - (3*a*f*x*Sqrt[a + b*x^4])/ (5*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) - (a*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(3/2)) + (3*a^(5/4)*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 ])/(5*b^(7/4)*Sqrt[a + b*x^4]) - (a^(3/4)*(5*Sqrt[b]*d + 9*Sqrt[a]*f)*(Sqr t[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 = 1.84 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.67
| method | result | size |
| risch | \(\frac {\left (12 f \,x^{3}+15 e \,x^{2}+20 d x +30 c \right ) \sqrt {b \,x^{4}+a}}{60 b}-\frac {a \left (\frac {10 d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {18 i f \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {15 e \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}\right )}{30 b}\) | \(235\) |
| default | \(\frac {c \sqrt {b \,x^{4}+a}}{2 b}+d \left (\frac {x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4 b}-\frac {a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}\right )+f \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(269\) |
| elliptic | \(\frac {f \,x^{3} \sqrt {b \,x^{4}+a}}{5 b}+\frac {e \,x^{2} \sqrt {b \,x^{4}+a}}{4 b}+\frac {d x \sqrt {b \,x^{4}+a}}{3 b}+\frac {c \sqrt {b \,x^{4}+a}}{2 b}-\frac {a d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a e \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}-\frac {3 i a^{\frac {3}{2}} f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(269\) |
Input:
int(x^3*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/60*(12*f*x^3+15*e*x^2+20*d*x+30*c)/b*(b*x^4+a)^(1/2)-1/30*a/b*(10*d/(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)+18*I* f*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))+15/2*e*ln(b^(1 /2)*x^2+(b*x^4+a)^(1/2))/b^(1/2))
Time = 0.16 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.45 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=-\frac {72 \, a \sqrt {b} f x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 15 \, a \sqrt {b} e x \log \left (-2 \, b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 8 \, {\left (5 \, b d - 9 \, a f\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 2 \, {\left (12 \, b f x^{4} + 15 \, b e x^{3} + 20 \, b d x^{2} + 30 \, b c x - 36 \, a f\right )} \sqrt {b x^{4} + a}}{120 \, b^{2} x} \] Input:
integrate(x^3*(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)*f*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), - 1) - 15*a*sqrt(b)*e*x*log(-2*b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 8*(5*b*d - 9*a*f)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x) , -1) - 2*(12*b*f*x^4 + 15*b*e*x^3 + 20*b*d*x^2 + 30*b*c*x - 36*a*f)*sqrt( b*x^4 + a))/(b^2*x)
Time = 2.59 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.45 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {\sqrt {a} e x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4 b} - \frac {a e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {3}{2}}} + c \left (\begin {cases} \frac {x^{4}}{4 \sqrt {a}} & \text {for}\: b = 0 \\\frac {\sqrt {a + b x^{4}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {f x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \] Input:
integrate(x**3*(f*x**3+e*x**2+d*x+c)/(b*x**4+a)**(1/2),x)
Output:
sqrt(a)*e*x**2*sqrt(1 + b*x**4/a)/(4*b) - a*e*asinh(sqrt(b)*x**2/sqrt(a))/ (4*b**(3/2)) + c*Piecewise((x**4/(4*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**4) /(2*b), True)) + d*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), b*x**4*exp_po lar(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + f*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))
\[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{3}}{\sqrt {b x^{4} + a}} \,d x } \] Input:
integrate(x^3*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x, algorithm="maxima")
Output:
1/2*sqrt(b*x^4 + a)*c/b + integrate((f*x^6 + e*x^5 + d*x^4)/sqrt(b*x^4 + a ), x)
\[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{3}}{\sqrt {b x^{4} + a}} \,d x } \] Input:
integrate(x^3*(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^3/sqrt(b*x^4 + a), x)
Timed out. \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\int \frac {x^3\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{\sqrt {b\,x^4+a}} \,d x \] Input:
int((x^3*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(1/2),x)
Output:
int((x^3*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(1/2), x)
\[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx=\frac {60 \sqrt {b \,x^{4}+a}\, b c +40 \sqrt {b \,x^{4}+a}\, b d x +30 \sqrt {b \,x^{4}+a}\, b e \,x^{2}+24 \sqrt {b \,x^{4}+a}\, b f \,x^{3}+15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a e -15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a e -40 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a b d -72 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) a b f}{120 b^{2}} \] Input:
int(x^3*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x)
Output:
(60*sqrt(a + b*x**4)*b*c + 40*sqrt(a + b*x**4)*b*d*x + 30*sqrt(a + b*x**4) *b*e*x**2 + 24*sqrt(a + b*x**4)*b*f*x**3 + 15*sqrt(b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a*e - 15*sqrt(b)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*a*e - 40*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a*b*d - 72*int((sqrt(a + b*x**4 )*x**2)/(a + b*x**4),x)*a*b*f)/(120*b**2)