Integrand size = 30, antiderivative size = 342 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^3} \, dx=\frac {2 \sqrt {b} d x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}-\frac {\left (c-e x^2\right ) \sqrt {a+b x^4}}{2 x^2}-\frac {\left (3 d-f x^2\right ) \sqrt {a+b x^4}}{3 x}+\frac {1}{2} \sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {1}{2} \sqrt {a} e \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {2 \sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (3 \sqrt {b} d+\sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a+b x^4}} \] Output:
2*b^(1/2)*d*x*(b*x^4+a)^(1/2)/(a^(1/2)+b^(1/2)*x^2)-1/2*(-e*x^2+c)*(b*x^4+ a)^(1/2)/x^2-1/3*(-f*x^2+3*d)*(b*x^4+a)^(1/2)/x+1/2*b^(1/2)*c*arctanh(b^(1 /2)*x^2/(b*x^4+a)^(1/2))-1/2*a^(1/2)*e*arctanh((b*x^4+a)^(1/2)/a^(1/2))-2* a^(1/4)*b^(1/4)*d*(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)*d+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^(1/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.16 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.60 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^3} \, dx=\frac {-a c+a e x^2-b c x^4+b e x^6+\sqrt {a} \sqrt {b} c x^2 \sqrt {1+\frac {b x^4}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )-\sqrt {a} e x^2 \sqrt {a+b x^4} \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-2 a d x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},-\frac {b x^4}{a}\right )+2 a f x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{2 x^2 \sqrt {a+b x^4}} \] Input:
Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^3,x]
Output:
(-(a*c) + a*e*x^2 - b*c*x^4 + b*e*x^6 + Sqrt[a]*Sqrt[b]*c*x^2*Sqrt[1 + (b* x^4)/a]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] - Sqrt[a]*e*x^2*Sqrt[a + b*x^4]*Arc Tanh[Sqrt[a + b*x^4]/Sqrt[a]] - 2*a*d*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric 2F1[-1/2, -1/4, 3/4, -((b*x^4)/a)] + 2*a*f*x^3*Sqrt[1 + (b*x^4)/a]*Hyperge ometric2F1[-1/2, 1/4, 5/4, -((b*x^4)/a)])/(2*x^2*Sqrt[a + b*x^4])
Time = 0.87 (sec) , antiderivative size = 342, 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.
| \(\displaystyle \int \frac {\sqrt {a+b x^4} \left (c+d x+e x^2+f x^3\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (\frac {\sqrt {a+b x^4} \left (c+e x^2\right )}{x^3}+\frac {\sqrt {a+b x^4} \left (d+f x^2\right )}{x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {a} f+3 \sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a+b x^4}}-\frac {2 \sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}+\frac {1}{2} \sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {1}{2} \sqrt {a} e \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {\sqrt {a+b x^4} \left (c-e x^2\right )}{2 x^2}-\frac {\sqrt {a+b x^4} \left (3 d-f x^2\right )}{3 x}+\frac {2 \sqrt {b} d x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}\) |
Input:
Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^3,x]
Output:
(2*Sqrt[b]*d*x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2) - ((c - e*x^2)*Sqr t[a + b*x^4])/(2*x^2) - ((3*d - f*x^2)*Sqrt[a + b*x^4])/(3*x) + (Sqrt[b]*c *ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (Sqrt[a]*e*ArcTanh[Sqrt[a + b *x^4]/Sqrt[a]])/2 - (2*a^(1/4)*b^(1/4)*d*(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]*d + Sqrt[a]*f)*(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 = 1.64 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.79
| method | result | size |
| risch | \(-\frac {\sqrt {b \,x^{4}+a}\, \left (2 d x +c \right )}{2 x^{2}}+\frac {2 a f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {\sqrt {a}\, e \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}+\frac {e \sqrt {b \,x^{4}+a}}{2}+\frac {c \sqrt {b}\, \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2}+\frac {f x \sqrt {b \,x^{4}+a}}{3}+\frac {2 i \sqrt {b}\, d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(269\) |
| elliptic | \(-\frac {c \sqrt {b \,x^{4}+a}}{2 x^{2}}-\frac {d \sqrt {b \,x^{4}+a}}{x}+\frac {f x \sqrt {b \,x^{4}+a}}{3}+\frac {e \sqrt {b \,x^{4}+a}}{2}+\frac {2 a f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {c \sqrt {b}\, \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{2}+\frac {2 i \sqrt {b}\, d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {\sqrt {a}\, e \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2}\) | \(273\) |
| default | \(f \left (\frac {x \sqrt {b \,x^{4}+a}}{3}+\frac {2 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+c \left (-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \,x^{2} \sqrt {b \,x^{4}+a}}{2 a}+\frac {\sqrt {b}\, \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2}\right )+d \left (-\frac {\sqrt {b \,x^{4}+a}}{x}+\frac {2 i \sqrt {b}\, \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (\frac {\sqrt {b \,x^{4}+a}}{2}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}\right )\) | \(304\) |
Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^3,x,method=_RETURNVERBOSE)
Output:
-1/2*(b*x^4+a)^(1/2)*(2*d*x+c)/x^2+2/3*a*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) *EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-1/2*a^(1/2)*e*ln((2*a+2*a^(1/2)* (b*x^4+a)^(1/2))/x^2)+1/2*e*(b*x^4+a)^(1/2)+1/2*c*b^(1/2)*ln(b^(1/2)*x^2+( b*x^4+a)^(1/2))+1/3*f*x*(b*x^4+a)^(1/2)+2*I*b^(1/2)*d*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))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^3} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{3}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^3,x, algorithm="fricas")
Output:
integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^3, x)
Result contains complex when optimal does not.
Time = 2.70 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.67 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^3} \, dx=- \frac {\sqrt {a} c}{2 x^{2} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} d \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {\sqrt {a} e \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2} + \frac {\sqrt {a} f x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a e}{2 \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {\sqrt {b} c \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2} + \frac {\sqrt {b} e x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {b c x^{2}}{2 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \] Input:
integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x**3,x)
Output:
-sqrt(a)*c/(2*x**2*sqrt(1 + b*x**4/a)) + sqrt(a)*d*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)*e*as inh(sqrt(a)/(sqrt(b)*x**2))/2 + sqrt(a)*f*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + a*e/(2*sqrt(b)*x**2*sqr t(a/(b*x**4) + 1)) + sqrt(b)*c*asinh(sqrt(b)*x**2/sqrt(a))/2 + sqrt(b)*e*x **2/(2*sqrt(a/(b*x**4) + 1)) - b*c*x**2/(2*sqrt(a)*sqrt(1 + b*x**4/a))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^3} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{3}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^3,x, algorithm="maxima")
Output:
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^3, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^3} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{3}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^3,x, algorithm="giac")
Output:
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^3, x)
Timed out. \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^3} \, dx=\int \frac {\sqrt {b\,x^4+a}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x^3} \,d x \] Input:
int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^3,x)
Output:
int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^3, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^3} \, dx=\frac {-6 \sqrt {b \,x^{4}+a}\, c +12 \sqrt {b \,x^{4}+a}\, d x +6 \sqrt {b \,x^{4}+a}\, e \,x^{2}+4 \sqrt {b \,x^{4}+a}\, f \,x^{3}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {a}\right ) e \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {a}\right ) e \,x^{2}-3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) c \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) c \,x^{2}+24 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{6}+a \,x^{2}}d x \right ) a d \,x^{2}+8 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a f \,x^{2}}{12 x^{2}} \] Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^3,x)
Output:
( - 6*sqrt(a + b*x**4)*c + 12*sqrt(a + b*x**4)*d*x + 6*sqrt(a + b*x**4)*e* x**2 + 4*sqrt(a + b*x**4)*f*x**3 + 3*sqrt(a)*log(sqrt(a + b*x**4) - sqrt(a ))*e*x**2 - 3*sqrt(a)*log(sqrt(a + b*x**4) + sqrt(a))*e*x**2 - 3*sqrt(b)*l og(sqrt(a + b*x**4) - sqrt(b)*x**2)*c*x**2 + 3*sqrt(b)*log(sqrt(a + b*x**4 ) + sqrt(b)*x**2)*c*x**2 + 24*int(sqrt(a + b*x**4)/(a*x**2 + b*x**6),x)*a* d*x**2 + 8*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a*f*x**2)/(12*x**2)