Integrand size = 30, antiderivative size = 341 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\frac {2 \sqrt {b} c x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}-\frac {\left (3 c-e x^2\right ) \sqrt {a+b x^4}}{3 x}+\frac {1}{4} \left (2 d+f x^2\right ) \sqrt {a+b x^4}+\frac {a f \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {1}{2} \sqrt {a} d \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {2 \sqrt [4]{a} \sqrt [4]{b} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (3 \sqrt {b} c+\sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a+b x^4}} \] Output:
2*b^(1/2)*c*x*(b*x^4+a)^(1/2)/(a^(1/2)+b^(1/2)*x^2)-1/3*(-e*x^2+3*c)*(b*x^ 4+a)^(1/2)/x+1/4*(f*x^2+2*d)*(b*x^4+a)^(1/2)+1/4*a*f*arctanh(b^(1/2)*x^2/( b*x^4+a)^(1/2))/b^(1/2)-1/2*a^(1/2)*d*arctanh((b*x^4+a)^(1/2)/a^(1/2))-2*a ^(1/4)*b^(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*x^4+a)^( 1/2)+1/3*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)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^(1/4)) ,1/2*2^(1/2))/b^(1/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.26 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.61 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\frac {-4 \sqrt {b} c \sqrt {a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},-\frac {b x^4}{a}\right )+x \left (\sqrt {a} f \sqrt {a+b x^4} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+\sqrt {b} \sqrt {1+\frac {b x^4}{a}} \left (\left (2 d+f x^2\right ) \sqrt {a+b x^4}-2 \sqrt {a} d \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+4 \sqrt {b} e x \sqrt {a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{4 \sqrt {b} x \sqrt {1+\frac {b x^4}{a}}} \] Input:
Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^2,x]
Output:
(-4*Sqrt[b]*c*Sqrt[a + b*x^4]*Hypergeometric2F1[-1/2, -1/4, 3/4, -((b*x^4) /a)] + x*(Sqrt[a]*f*Sqrt[a + b*x^4]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] + Sqrt[ b]*Sqrt[1 + (b*x^4)/a]*((2*d + f*x^2)*Sqrt[a + b*x^4] - 2*Sqrt[a]*d*ArcTan h[Sqrt[a + b*x^4]/Sqrt[a]]) + 4*Sqrt[b]*e*x*Sqrt[a + b*x^4]*Hypergeometric 2F1[-1/2, 1/4, 5/4, -((b*x^4)/a)]))/(4*Sqrt[b]*x*Sqrt[1 + (b*x^4)/a])
Time = 0.90 (sec) , antiderivative size = 341, 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.
| \(\displaystyle \int \frac {\sqrt {a+b x^4} \left (c+d x+e x^2+f x^3\right )}{x^2} \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (\frac {\sqrt {a+b x^4} \left (c+e x^2\right )}{x^2}+\frac {\sqrt {a+b x^4} \left (d+f x^2\right )}{x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {a} e+3 \sqrt {b} c\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a+b x^4}}-\frac {2 \sqrt [4]{a} \sqrt [4]{b} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}-\frac {1}{2} \sqrt {a} d \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+\frac {a f \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {\sqrt {a+b x^4} \left (3 c-e x^2\right )}{3 x}+\frac {2 \sqrt {b} c x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}+\frac {1}{4} \sqrt {a+b x^4} \left (2 d+f x^2\right )\) |
Input:
Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^2,x]
Output:
(2*Sqrt[b]*c*x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2) - ((3*c - e*x^2)*S qrt[a + b*x^4])/(3*x) + ((2*d + f*x^2)*Sqrt[a + b*x^4])/4 + (a*f*ArcTanh[( Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b]) - (Sqrt[a]*d*ArcTanh[Sqrt[a + b *x^4]/Sqrt[a]])/2 - (2*a^(1/4)*b^(1/4)*c*(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])/Sqrt[a + b*x^4] + (a^(1/4)*(3*Sqrt[b]*c + Sqrt[a]*e)*(Sqrt[a] + Sq rt[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])/(3*b^(1/4)*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 = 1.90 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.80
| method | result | size |
| elliptic | \(-\frac {c \sqrt {b \,x^{4}+a}}{x}+\frac {f \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {e x \sqrt {b \,x^{4}+a}}{3}+\frac {d \sqrt {b \,x^{4}+a}}{2}+\frac {2 a e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {a f \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}+\frac {2 i c \sqrt {b}\, \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {\sqrt {a}\, d \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2}\) | \(274\) |
| risch | \(-\frac {c \sqrt {b \,x^{4}+a}}{x}+\frac {2 a e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {\sqrt {a}\, d \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}+\frac {a f \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}+\frac {d \sqrt {b \,x^{4}+a}}{2}+\frac {e x \sqrt {b \,x^{4}+a}}{3}+\frac {f \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {2 i c \sqrt {b}\, \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(280\) |
| default | \(e \left (\frac {x \sqrt {b \,x^{4}+a}}{3}+\frac {2 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+c \left (-\frac {\sqrt {b \,x^{4}+a}}{x}+\frac {2 i \sqrt {b}\, \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {\sqrt {b \,x^{4}+a}}{2}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}\right )+f \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}\right )\) | \(284\) |
Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-1/x*c*(b*x^4+a)^(1/2)+1/4*f*x^2*(b*x^4+a)^(1/2)+1/3*e*x*(b*x^4+a)^(1/2)+1 /2*d*(b*x^4+a)^(1/2)+2/3*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)+1/4*a*f*ln(2*b^(1/2)*x^2+2*(b*x^4+a)^(1/2))/b^ (1/2)+2*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))-1/2* a^(1/2)*d*arctanh(a^(1/2)/(b*x^4+a)^(1/2))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{2}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^2,x, algorithm="fricas")
Output:
integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^2, x)
Result contains complex when optimal does not.
Time = 2.85 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.60 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\frac {\sqrt {a} c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {\sqrt {a} d \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2} + \frac {\sqrt {a} e x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {a} f x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {a d}{2 \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {a f \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 \sqrt {b}} + \frac {\sqrt {b} d x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} \] Input:
integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x**2,x)
Output:
sqrt(a)*c*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**4*exp_polar(I*pi)/a )/(4*x*gamma(3/4)) - sqrt(a)*d*asinh(sqrt(a)/(sqrt(b)*x**2))/2 + sqrt(a)*e *x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamm a(5/4)) + sqrt(a)*f*x**2*sqrt(1 + b*x**4/a)/4 + a*d/(2*sqrt(b)*x**2*sqrt(a /(b*x**4) + 1)) + a*f*asinh(sqrt(b)*x**2/sqrt(a))/(4*sqrt(b)) + sqrt(b)*d* x**2/(2*sqrt(a/(b*x**4) + 1))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{2}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^2,x, algorithm="maxima")
Output:
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^2, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{2}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^2,x, algorithm="giac")
Output:
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^2, x)
Timed out. \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\int \frac {\sqrt {b\,x^4+a}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x^2} \,d x \] Input:
int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^2,x)
Output:
int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^2, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^2} \, dx=\frac {24 \sqrt {b \,x^{4}+a}\, b c +12 \sqrt {b \,x^{4}+a}\, b d x +8 \sqrt {b \,x^{4}+a}\, b e \,x^{2}+6 \sqrt {b \,x^{4}+a}\, b f \,x^{3}+6 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {a}\right ) b d x -6 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {a}\right ) b d x -3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a f x +3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a f x +48 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{6}+a \,x^{2}}d x \right ) a b c x +16 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a b e x}{24 b x} \] Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^2,x)
Output:
(24*sqrt(a + b*x**4)*b*c + 12*sqrt(a + b*x**4)*b*d*x + 8*sqrt(a + b*x**4)* b*e*x**2 + 6*sqrt(a + b*x**4)*b*f*x**3 + 6*sqrt(a)*log(sqrt(a + b*x**4) - sqrt(a))*b*d*x - 6*sqrt(a)*log(sqrt(a + b*x**4) + sqrt(a))*b*d*x - 3*sqrt( b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a*f*x + 3*sqrt(b)*log(sqrt(a + b*x **4) + sqrt(b)*x**2)*a*f*x + 48*int(sqrt(a + b*x**4)/(a*x**2 + b*x**6),x)* a*b*c*x + 16*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a*b*e*x)/(24*b*x)