Integrand size = 27, antiderivative size = 331 \[ \int \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\frac {1}{4} d x^2 \sqrt {a+b x^4}+\frac {2 a e x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{15} x \left (5 c+3 e x^2\right ) \sqrt {a+b x^4}+\frac {f \left (a+b x^4\right )^{3/2}}{6 b}+\frac {a d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {2 a^{5/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a^{3/4} \left (5 \sqrt {b} c+3 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}} \] Output:
1/4*d*x^2*(b*x^4+a)^(1/2)+2/5*a*e*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+b^(1/ 2)*x^2)+1/15*x*(3*e*x^2+5*c)*(b*x^4+a)^(1/2)+1/6*f*(b*x^4+a)^(3/2)/b+1/4*a *d*arctanh(b^(1/2)*x^2/(b*x^4+a)^(1/2))/b^(1/2)-2/5*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^(3/4)/(b*x^4+a)^(1/2)+1/15*a^(3/4)*(5* b^(1/2)*c+3*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)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/b^(3 /4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.10 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.52 \[ \int \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\frac {\sqrt {a+b x^4} \left (2 a f \sqrt {1+\frac {b x^4}{a}}+3 b d x^2 \sqrt {1+\frac {b x^4}{a}}+2 b f x^4 \sqrt {1+\frac {b x^4}{a}}+3 \sqrt {a} \sqrt {b} d \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+12 b c x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )+4 b e x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{12 b \sqrt {1+\frac {b x^4}{a}}} \] Input:
Integrate[(c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4],x]
Output:
(Sqrt[a + b*x^4]*(2*a*f*Sqrt[1 + (b*x^4)/a] + 3*b*d*x^2*Sqrt[1 + (b*x^4)/a ] + 2*b*f*x^4*Sqrt[1 + (b*x^4)/a] + 3*Sqrt[a]*Sqrt[b]*d*ArcSinh[(Sqrt[b]*x ^2)/Sqrt[a]] + 12*b*c*x*Hypergeometric2F1[-1/2, 1/4, 5/4, -((b*x^4)/a)] + 4*b*e*x^3*Hypergeometric2F1[-1/2, 3/4, 7/4, -((b*x^4)/a)]))/(12*b*Sqrt[1 + (b*x^4)/a])
Time = 0.75 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {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.
| \(\displaystyle \int \sqrt {a+b x^4} \left (c+d x+e x^2+f x^3\right ) \, dx\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \int \left (\sqrt {a+b x^4} \left (c+e x^2\right )+x \sqrt {a+b x^4} \left (d+f x^2\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} e+5 \sqrt {b} c\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}-\frac {2 a^{5/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}+\frac {1}{15} x \sqrt {a+b x^4} \left (5 c+3 e x^2\right )+\frac {1}{4} d x^2 \sqrt {a+b x^4}+\frac {2 a e x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {f \left (a+b x^4\right )^{3/2}}{6 b}\) |
Input:
Int[(c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4],x]
Output:
(d*x^2*Sqrt[a + b*x^4])/4 + (2*a*e*x*Sqrt[a + b*x^4])/(5*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (x*(5*c + 3*e*x^2)*Sqrt[a + b*x^4])/15 + (f*(a + b*x^4)^ (3/2))/(6*b) + (a*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b]) - (2*a^(5/4)*e*(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^(3/4)*Sqrt[a + b*x^4]) + (a^(3/4)*(5*Sqrt[b]*c + 3*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqr t[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^ (1/4)], 1/2])/(15*b^(3/4)*Sqrt[a + b*x^4])
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 = 1.21 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.73
| method | result | size |
| risch | \(\frac {\left (10 b f \,x^{4}+12 b e \,x^{3}+15 b d \,x^{2}+20 c b x +10 a f \right ) \sqrt {b \,x^{4}+a}}{60 b}+\frac {a \left (\frac {20 c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {15 d \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {12 i e \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )}{30}\) | \(243\) |
| default | \(c \left (\frac {x \sqrt {b \,x^{4}+a}}{3}+\frac {2 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}\right )+e \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+\frac {f \left (b \,x^{4}+a \right )^{\frac {3}{2}}}{6 b}\) | \(257\) |
| elliptic | \(\frac {f \,x^{4} \sqrt {b \,x^{4}+a}}{6}+\frac {x^{3} e \sqrt {b \,x^{4}+a}}{5}+\frac {d \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {c x \sqrt {b \,x^{4}+a}}{3}+\frac {a f \sqrt {b \,x^{4}+a}}{6 b}+\frac {2 a c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {a d \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}+\frac {2 i a^{\frac {3}{2}} e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\) | \(273\) |
Input:
int((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+10*a*f)/b*(b*x^4+a)^(1/2)+ 1/30*a*(20*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)+12*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 = 0.49 \[ \int \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\frac {48 \, a \sqrt {b} e x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 15 \, a \sqrt {b} d x \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 16 \, {\left (5 \, b c - 3 \, a e\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left (10 \, b f x^{5} + 12 \, b e x^{4} + 15 \, b d x^{3} + 20 \, b c x^{2} + 10 \, a f x + 24 \, a e\right )} \sqrt {b x^{4} + a}}{120 \, b x} \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2),x, algorithm="fricas")
Output:
1/120*(48*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) + 1 6*(5*b*c - 3*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 + 10*a*f*x + 24*a*e)*sqrt(b*x^4 + a))/(b*x)
Time = 2.03 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.47 \[ \int \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\frac {\sqrt {a} c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {a} d x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {\sqrt {a} e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {a d \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 \sqrt {b}} + f \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\left (a + b x^{4}\right )^{\frac {3}{2}}}{6 b} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2),x)
Output:
sqrt(a)*c*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a )/(4*gamma(5/4)) + sqrt(a)*d*x**2*sqrt(1 + b*x**4/a)/4 + sqrt(a)*e*x**3*ga mma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4 )) + a*d*asinh(sqrt(b)*x**2/sqrt(a))/(4*sqrt(b)) + f*Piecewise((sqrt(a)*x* *4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True))
\[ \int \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\int { \sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c), x)
\[ \int \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\int { \sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c), x)
Timed out. \[ \int \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\int \sqrt {b\,x^4+a}\,\left (f\,x^3+e\,x^2+d\,x+c\right ) \,d x \] Input:
int((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3),x)
Output:
int((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3), x)
\[ \int \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx=\frac {20 \sqrt {b \,x^{4}+a}\, a f +40 \sqrt {b \,x^{4}+a}\, b c x +30 \sqrt {b \,x^{4}+a}\, b d \,x^{2}+24 \sqrt {b \,x^{4}+a}\, b e \,x^{3}+20 \sqrt {b \,x^{4}+a}\, b f \,x^{4}-15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a d +15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a d +80 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a b c +48 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) a b e}{120 b} \] Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2),x)
Output:
(20*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(sqrt( a + b*x**4) + sqrt(b)*x**2)*a*d + 80*int(sqrt(a + b*x**4)/(a + b*x**4),x)* a*b*c + 48*int((sqrt(a + b*x**4)*x**2)/(a + b*x**4),x)*a*b*e)/(120*b)