Integrand size = 30, antiderivative size = 326 \[ \int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x^4}} \, dx=-\frac {c \sqrt {a+b x^4}}{4 a x^4}-\frac {d \sqrt {a+b x^4}}{3 a x^3}-\frac {e \sqrt {a+b x^4}}{2 a x^2}-\frac {f \sqrt {a+b x^4}}{\sqrt {a} x \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {b c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {\sqrt [4]{b} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} \left (\sqrt {b} d-3 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+b x^4}} \] Output:
-1/4*c*(b*x^4+a)^(1/2)/a/x^4-1/3*d*(b*x^4+a)^(1/2)/a/x^3-1/2*e*(b*x^4+a)^( 1/2)/a/x^2-f*(b*x^4+a)^(1/2)/a^(1/2)/x/(a^(1/2)+b^(1/2)*x^2)+1/4*b*c*arcta nh((b*x^4+a)^(1/2)/a^(1/2))/a^(3/2)-b^(1/4)*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))/a^(3/4)/(b*x^4+a)^(1/2)-1/6*b^(1/4)*(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)*Invers eJacobiAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(5/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.45 \[ \int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x^4}} \, dx=-\frac {\sqrt {a+b x^4} \left (3 a c \sqrt {1+\frac {b x^4}{a}}+6 a e x^2 \sqrt {1+\frac {b x^4}{a}}-3 b c x^4 \text {arctanh}\left (\sqrt {1+\frac {b x^4}{a}}\right )+4 a d x \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},-\frac {b x^4}{a}\right )+12 a f x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {b x^4}{a}\right )\right )}{12 a^2 x^4 \sqrt {1+\frac {b x^4}{a}}} \] Input:
Integrate[(c + d*x + e*x^2 + f*x^3)/(x^5*Sqrt[a + b*x^4]),x]
Output:
-1/12*(Sqrt[a + b*x^4]*(3*a*c*Sqrt[1 + (b*x^4)/a] + 6*a*e*x^2*Sqrt[1 + (b* x^4)/a] - 3*b*c*x^4*ArcTanh[Sqrt[1 + (b*x^4)/a]] + 4*a*d*x*Hypergeometric2 F1[-3/4, 1/2, 1/4, -((b*x^4)/a)] + 12*a*f*x^3*Hypergeometric2F1[-1/4, 1/2, 3/4, -((b*x^4)/a)]))/(a^2*x^4*Sqrt[1 + (b*x^4)/a])
Time = 0.93 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.06, 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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x^4}} \, dx\) |
| \(\Big \downarrow \) 2372 |
| \(\displaystyle \int \left (\frac {c+e x^2}{x^5 \sqrt {a+b x^4}}+\frac {d+f x^2}{x^4 \sqrt {a+b x^4}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {b} d-3 \sqrt {a} f\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}+\frac {b c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {c \sqrt {a+b x^4}}{4 a x^4}-\frac {d \sqrt {a+b x^4}}{3 a x^3}-\frac {e \sqrt {a+b x^4}}{2 a x^2}-\frac {f \sqrt {a+b x^4}}{a x}+\frac {\sqrt {b} f x \sqrt {a+b x^4}}{a \left (\sqrt {a}+\sqrt {b} x^2\right )}\) |
Input:
Int[(c + d*x + e*x^2 + f*x^3)/(x^5*Sqrt[a + b*x^4]),x]
Output:
-1/4*(c*Sqrt[a + b*x^4])/(a*x^4) - (d*Sqrt[a + b*x^4])/(3*a*x^3) - (e*Sqrt [a + b*x^4])/(2*a*x^2) - (f*Sqrt[a + b*x^4])/(a*x) + (Sqrt[b]*f*x*Sqrt[a + b*x^4])/(a*(Sqrt[a] + Sqrt[b]*x^2)) + (b*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a ]])/(4*a^(3/2)) - (b^(1/4)*f*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqr t[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(a^( 3/4)*Sqrt[a + b*x^4]) - (b^(1/4)*(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]*EllipticF[2*ArcTan[(b ^(1/4)*x)/a^(1/4)], 1/2])/(6*a^(5/4)*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 = 1.77 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.75
| method | result | size |
| risch | \(-\frac {\sqrt {b \,x^{4}+a}\, \left (12 f \,x^{3}+6 e \,x^{2}+4 d x +3 c \right )}{12 a \,x^{4}}-\frac {b \left (\frac {2 d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {3 c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 \sqrt {a}}-\frac {6 i f \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )}{6 a}\) | \(243\) |
| elliptic | \(-\frac {c \sqrt {b \,x^{4}+a}}{4 a \,x^{4}}-\frac {d \sqrt {b \,x^{4}+a}}{3 a \,x^{3}}-\frac {e \sqrt {b \,x^{4}+a}}{2 a \,x^{2}}-\frac {f \sqrt {b \,x^{4}+a}}{a x}-\frac {d b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {i \sqrt {b}\, f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {b c \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{4 a^{\frac {3}{2}}}\) | \(267\) |
| default | \(c \left (-\frac {\sqrt {b \,x^{4}+a}}{4 a \,x^{4}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}\right )+d \left (-\frac {\sqrt {b \,x^{4}+a}}{3 a \,x^{3}}-\frac {b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {e \sqrt {b \,x^{4}+a}}{2 a \,x^{2}}+f \left (-\frac {\sqrt {b \,x^{4}+a}}{a x}+\frac {i \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(279\) |
Input:
int((f*x^3+e*x^2+d*x+c)/x^5/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/12*(b*x^4+a)^(1/2)*(12*f*x^3+6*e*x^2+4*d*x+3*c)/a/x^4-1/6*b/a*(2*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)-3/2*c /a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^2)-6*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)-Ell ipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)))
Time = 0.14 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.46 \[ \int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x^4}} \, dx=-\frac {24 \, a^{\frac {3}{2}} f x^{4} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 3 \, \sqrt {a} b c x^{4} \log \left (-\frac {b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 8 \, {\left (a d + 3 \, a f\right )} \sqrt {a} x^{4} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + 2 \, {\left (12 \, a f x^{3} + 6 \, a e x^{2} + 4 \, a d x + 3 \, a c\right )} \sqrt {b x^{4} + a}}{24 \, a^{2} x^{4}} \] Input:
integrate((f*x^3+e*x^2+d*x+c)/x^5/(b*x^4+a)^(1/2),x, algorithm="fricas")
Output:
-1/24*(24*a^(3/2)*f*x^4*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a)^(1/4)), -1 ) - 3*sqrt(a)*b*c*x^4*log(-(b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(a) + 2*a)/x^4) - 8*(a*d + 3*a*f)*sqrt(a)*x^4*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4 )), -1) + 2*(12*a*f*x^3 + 6*a*e*x^2 + 4*a*d*x + 3*a*c)*sqrt(b*x^4 + a))/(a ^2*x^4)
Result contains complex when optimal does not.
Time = 2.34 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.48 \[ \int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x^4}} \, dx=- \frac {\sqrt {b} c \sqrt {\frac {a}{b x^{4}} + 1}}{4 a x^{2}} - \frac {\sqrt {b} e \sqrt {\frac {a}{b x^{4}} + 1}}{2 a} + \frac {d \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {f \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x \Gamma \left (\frac {3}{4}\right )} + \frac {b c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 a^{\frac {3}{2}}} \] Input:
integrate((f*x**3+e*x**2+d*x+c)/x**5/(b*x**4+a)**(1/2),x)
Output:
-sqrt(b)*c*sqrt(a/(b*x**4) + 1)/(4*a*x**2) - sqrt(b)*e*sqrt(a/(b*x**4) + 1 )/(2*a) + d*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), b*x**4*exp_polar(I*pi)/ a)/(4*sqrt(a)*x**3*gamma(1/4)) + f*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*x*gamma(3/4)) + b*c*asinh(sqrt(a)/(sq rt(b)*x**2))/(4*a**(3/2))
\[ \int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x^4}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a} x^{5}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)/x^5/(b*x^4+a)^(1/2),x, algorithm="maxima")
Output:
-1/8*c*(2*sqrt(b*x^4 + a)*b/((b*x^4 + a)*a - a^2) + b*log((sqrt(b*x^4 + a) - sqrt(a))/(sqrt(b*x^4 + a) + sqrt(a)))/a^(3/2)) + integrate((f*x^2 + e*x + d)/(sqrt(b*x^4 + a)*x^4), x)
\[ \int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x^4}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a} x^{5}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)/x^5/(b*x^4+a)^(1/2),x, algorithm="giac")
Output:
integrate((f*x^3 + e*x^2 + d*x + c)/(sqrt(b*x^4 + a)*x^5), x)
Timed out. \[ \int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x^4}} \, dx=\int \frac {f\,x^3+e\,x^2+d\,x+c}{x^5\,\sqrt {b\,x^4+a}} \,d x \] Input:
int((c + d*x + e*x^2 + f*x^3)/(x^5*(a + b*x^4)^(1/2)),x)
Output:
int((c + d*x + e*x^2 + f*x^3)/(x^5*(a + b*x^4)^(1/2)), x)
\[ \int \frac {c+d x+e x^2+f x^3}{x^5 \sqrt {a+b x^4}} \, dx =\text {Too large to display} \] Input:
int((f*x^3+e*x^2+d*x+c)/x^5/(b*x^4+a)^(1/2),x)
Output:
( - 6*sqrt(b)*sqrt(a)*sqrt(a + b*x**4)*a*c*x**2 - 16*sqrt(b)*sqrt(a)*sqrt( a + b*x**4)*a*e*x**4 - 8*sqrt(b)*sqrt(a)*sqrt(a + b*x**4)*b*c*x**6 - 32*sq rt(b)*sqrt(a)*sqrt(a + b*x**4)*b*e*x**8 + 8*sqrt(a)*sqrt(a + b*x**4)*int(s qrt(a + b*x**4)/(a*x**4 + b*x**8),x)*a**2*d*x**4 + 32*sqrt(a)*sqrt(a + b*x **4)*int(sqrt(a + b*x**4)/(a*x**4 + b*x**8),x)*a*b*d*x**8 + 8*sqrt(a)*sqrt (a + b*x**4)*int(sqrt(a + b*x**4)/(a*x**2 + b*x**6),x)*a**2*f*x**4 + 32*sq rt(a)*sqrt(a + b*x**4)*int(sqrt(a + b*x**4)/(a*x**2 + b*x**6),x)*a*b*f*x** 8 - sqrt(a + b*x**4)*log(sqrt(a + b*x**4) - sqrt(a))*a*b*c*x**4 - 4*sqrt(a + b*x**4)*log(sqrt(a + b*x**4) - sqrt(a))*b**2*c*x**8 + sqrt(a + b*x**4)* log(sqrt(a + b*x**4) + sqrt(a))*a*b*c*x**4 + 4*sqrt(a + b*x**4)*log(sqrt(a + b*x**4) + sqrt(a))*b**2*c*x**8 + 24*sqrt(b)*sqrt(a)*int(sqrt(a + b*x**4 )/(a*x**4 + b*x**8),x)*a**2*d*x**6 + 32*sqrt(b)*sqrt(a)*int(sqrt(a + b*x** 4)/(a*x**4 + b*x**8),x)*a*b*d*x**10 + 24*sqrt(b)*sqrt(a)*int(sqrt(a + b*x* *4)/(a*x**2 + b*x**6),x)*a**2*f*x**6 + 32*sqrt(b)*sqrt(a)*int(sqrt(a + b*x **4)/(a*x**2 + b*x**6),x)*a*b*f*x**10 - 2*sqrt(a)*a**2*c - 4*sqrt(a)*a**2* e*x**2 - 10*sqrt(a)*a*b*c*x**4 - 32*sqrt(a)*a*b*e*x**6 - 8*sqrt(a)*b**2*c* x**8 - 32*sqrt(a)*b**2*e*x**10 - 3*sqrt(b)*log(sqrt(a + b*x**4) - sqrt(a)) *a*b*c*x**6 - 4*sqrt(b)*log(sqrt(a + b*x**4) - sqrt(a))*b**2*c*x**10 + 3*s qrt(b)*log(sqrt(a + b*x**4) + sqrt(a))*a*b*c*x**6 + 4*sqrt(b)*log(sqrt(a + b*x**4) + sqrt(a))*b**2*c*x**10)/(8*sqrt(a)*a*x**4*(sqrt(a + b*x**4)*a...