Integrand size = 30, antiderivative size = 345 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\frac {2 a f x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{4} \left (2 c+e x^2\right ) \sqrt {a+b x^4}+\frac {1}{15} x \left (5 d+3 f x^2\right ) \sqrt {a+b x^4}+\frac {a e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {1}{2} \sqrt {a} c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {2 a^{5/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a^{3/4} \left (5 \sqrt {b} d+3 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}} \] Output:
2/5*a*f*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+b^(1/2)*x^2)+1/4*(e*x^2+2*c)*(b *x^4+a)^(1/2)+1/15*x*(3*f*x^2+5*d)*(b*x^4+a)^(1/2)+1/4*a*e*arctanh(b^(1/2) *x^2/(b*x^4+a)^(1/2))/b^(1/2)-1/2*a^(1/2)*c*arctanh((b*x^4+a)^(1/2)/a^(1/2 ))-2/5*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)*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)*d+3*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^(3/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.35 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.60 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\frac {3 a^{3/2} e \sqrt {1+\frac {b x^4}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+3 \sqrt {b} \left (\left (2 c+e x^2\right ) \left (a+b x^4\right )-2 \sqrt {a} c \sqrt {a+b x^4} \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+12 a \sqrt {b} d x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )+4 a \sqrt {b} f x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{12 \sqrt {b} \sqrt {a+b x^4}} \] Input:
Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x,x]
Output:
(3*a^(3/2)*e*Sqrt[1 + (b*x^4)/a]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] + 3*Sqrt[b ]*((2*c + e*x^2)*(a + b*x^4) - 2*Sqrt[a]*c*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]]) + 12*a*Sqrt[b]*d*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F 1[-1/2, 1/4, 5/4, -((b*x^4)/a)] + 4*a*Sqrt[b]*f*x^3*Sqrt[1 + (b*x^4)/a]*Hy pergeometric2F1[-1/2, 3/4, 7/4, -((b*x^4)/a)])/(12*Sqrt[b]*Sqrt[a + b*x^4] )
Time = 0.86 (sec) , antiderivative size = 345, 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.
| \(\displaystyle \int \frac {\sqrt {a+b x^4} \left (c+d x+e x^2+f x^3\right )}{x} \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (\frac {\sqrt {a+b x^4} \left (c+e x^2\right )}{x}+\sqrt {a+b x^4} \left (d+f x^2\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} f+5 \sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}-\frac {2 a^{5/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}-\frac {1}{2} \sqrt {a} c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+\frac {a e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}+\frac {1}{4} \sqrt {a+b x^4} \left (2 c+e x^2\right )+\frac {1}{15} x \sqrt {a+b x^4} \left (5 d+3 f x^2\right )+\frac {2 a f x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}\) |
Input:
Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x,x]
Output:
(2*a*f*x*Sqrt[a + b*x^4])/(5*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + ((2*c + e* x^2)*Sqrt[a + b*x^4])/4 + (x*(5*d + 3*f*x^2)*Sqrt[a + b*x^4])/15 + (a*e*Ar cTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b]) - (Sqrt[a]*c*ArcTanh[Sqr t[a + b*x^4]/Sqrt[a]])/2 - (2*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^(3/4)*Sqrt[a + b*x^4]) + (a^(3/4)*(5*Sqrt[b]*d + 3*Sqrt[a]*f)* (Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ellipt icF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*b^(3/4)*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 = 0.84 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.79
| method | result | size |
| elliptic | \(\frac {f \,x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {e \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {d x \sqrt {b \,x^{4}+a}}{3}+\frac {c \sqrt {b \,x^{4}+a}}{2}+\frac {2 a d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {a e \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}+\frac {2 i a^{\frac {3}{2}} f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}-\frac {\sqrt {a}\, c \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2}\) | \(274\) |
| default | \(d \left (\frac {x \sqrt {b \,x^{4}+a}}{3}+\frac {2 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}\right )+f \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+c \left (\frac {\sqrt {b \,x^{4}+a}}{2}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}\right )\) | \(284\) |
Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x,x,method=_RETURNVERBOSE)
Output:
1/5*f*x^3*(b*x^4+a)^(1/2)+1/4*e*x^2*(b*x^4+a)^(1/2)+1/3*d*x*(b*x^4+a)^(1/2 )+1/2*c*(b*x^4+a)^(1/2)+2/3*a*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)+1/4*a*e*ln(2*b^(1/2)*x^2+2*(b*x^4+a)^(1/2)) /b^(1/2)+2/5*I*a^(3/2)*f/(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)*(Elliptic F(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)) -1/2*a^(1/2)*c*arctanh(a^(1/2)/(b*x^4+a)^(1/2))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x,x, algorithm="fricas")
Output:
integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x, x)
Result contains complex when optimal does not.
Time = 3.96 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=- \frac {\sqrt {a} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2} + \frac {\sqrt {a} d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {a} e x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {\sqrt {a} f x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {a c}{2 \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {a e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 \sqrt {b}} + \frac {\sqrt {b} c x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} \] Input:
integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x,x)
Output:
-sqrt(a)*c*asinh(sqrt(a)/(sqrt(b)*x**2))/2 + sqrt(a)*d*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)*e* x**2*sqrt(1 + b*x**4/a)/4 + sqrt(a)*f*x**3*gamma(3/4)*hyper((-1/2, 3/4), ( 7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4)) + a*c/(2*sqrt(b)*x**2*sqrt (a/(b*x**4) + 1)) + a*e*asinh(sqrt(b)*x**2/sqrt(a))/(4*sqrt(b)) + sqrt(b)* c*x**2/(2*sqrt(a/(b*x**4) + 1))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x,x, algorithm="maxima")
Output:
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x,x, algorithm="giac")
Output:
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x, x)
Timed out. \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\int \frac {\sqrt {b\,x^4+a}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x} \,d x \] Input:
int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x,x)
Output:
int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\frac {60 \sqrt {b \,x^{4}+a}\, b c +40 \sqrt {b \,x^{4}+a}\, b d x +30 \sqrt {b \,x^{4}+a}\, b e \,x^{2}+24 \sqrt {b \,x^{4}+a}\, b f \,x^{3}+30 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {a}\right ) b c -30 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {a}\right ) b c -15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a e +15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a e +80 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a b d +48 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b \,x^{4}+a}d x \right ) a b f}{120 b} \] Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x,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 + 30*sqrt(a)*log(sqrt(a + b*x**4) - sqrt(a))*b*c - 30*sqrt(a)*log(sqrt(a + b*x**4) + sqrt(a))*b*c - 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 + 80*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a*b*d + 48*int((sqrt(a + b*x**4)*x**2)/(a + b*x**4),x)*a*b*f)/(120*b)