\(\int \frac {c+d x^2+e x^4+f x^6}{x^6 (a+b x^2)} \, dx\) [120]

Optimal result

Integrand size = 30, antiderivative size = 104 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \left (a+b x^2\right )} \, dx=-\frac {c}{5 a x^5}+\frac {b c-a d}{3 a^2 x^3}-\frac {b^2 c-a b d+a^2 e}{a^3 x}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2} \sqrt {b}} \]

[Out]

-1/5*c/a/x^5+1/3*(-a*d+b*c)/a^2/x^3+(-a^2*e+a*b*d-b^2*c)/a^3/x-(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*arctan(x*b^(1/2)
/a^(1/2))/a^(7/2)/b^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1816, 211} \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \left (a+b x^2\right )} \, dx=\frac {b c-a d}{3 a^2 x^3}-\frac {a^2 e-a b d+b^2 c}{a^3 x}-\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{a^{7/2} \sqrt {b}}-\frac {c}{5 a x^5} \]

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/(x^6*(a + b*x^2)),x]

[Out]

-1/5*c/(a*x^5) + (b*c - a*d)/(3*a^2*x^3) - (b^2*c - a*b*d + a^2*e)/(a^3*x) - ((b^3*c - a*b^2*d + a^2*b*e - a^3
*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(7/2)*Sqrt[b])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c}{a x^6}+\frac {-b c+a d}{a^2 x^4}+\frac {b^2 c-a b d+a^2 e}{a^3 x^2}+\frac {-b^3 c+a b^2 d-a^2 b e+a^3 f}{a^3 \left (a+b x^2\right )}\right ) \, dx \\ & = -\frac {c}{5 a x^5}+\frac {b c-a d}{3 a^2 x^3}-\frac {b^2 c-a b d+a^2 e}{a^3 x}+\frac {\left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \int \frac {1}{a+b x^2} \, dx}{a^3} \\ & = -\frac {c}{5 a x^5}+\frac {b c-a d}{3 a^2 x^3}-\frac {b^2 c-a b d+a^2 e}{a^3 x}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2} \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \left (a+b x^2\right )} \, dx=-\frac {c}{5 a x^5}+\frac {b c-a d}{3 a^2 x^3}+\frac {-b^2 c+a b d-a^2 e}{a^3 x}+\frac {\left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2} \sqrt {b}} \]

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(x^6*(a + b*x^2)),x]

[Out]

-1/5*c/(a*x^5) + (b*c - a*d)/(3*a^2*x^3) + (-(b^2*c) + a*b*d - a^2*e)/(a^3*x) + ((-(b^3*c) + a*b^2*d - a^2*b*e
+ a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(7/2)*Sqrt[b])

Maple [A] (verified)

Time = 3.42 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {c}{5 a \,x^{5}}-\frac {a d -b c}{3 a^{2} x^{3}}-\frac {a^{2} e -a b d +b^{2} c}{a^{3} x}+\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}\)	\(94\)
risch	\(\frac {-\frac {\left (a^{2} e -a b d +b^{2} c \right ) x^{4}}{a^{3}}-\frac {\left (a d -b c \right ) x^{2}}{3 a^{2}}-\frac {c}{5 a}}{x^{5}}-\frac {\ln \left (-\sqrt {-a b}\, x +a \right ) f}{2 \sqrt {-a b}}+\frac {\ln \left (-\sqrt {-a b}\, x +a \right ) b e}{2 \sqrt {-a b}\, a}-\frac {\ln \left (-\sqrt {-a b}\, x +a \right ) b^{2} d}{2 \sqrt {-a b}\, a^{2}}+\frac {\ln \left (-\sqrt {-a b}\, x +a \right ) b^{3} c}{2 \sqrt {-a b}\, a^{3}}+\frac {\ln \left (-\sqrt {-a b}\, x -a \right ) f}{2 \sqrt {-a b}}-\frac {\ln \left (-\sqrt {-a b}\, x -a \right ) b e}{2 \sqrt {-a b}\, a}+\frac {\ln \left (-\sqrt {-a b}\, x -a \right ) b^{2} d}{2 \sqrt {-a b}\, a^{2}}-\frac {\ln \left (-\sqrt {-a b}\, x -a \right ) b^{3} c}{2 \sqrt {-a b}\, a^{3}}\)	\(261\)

[In]

int((f*x^6+e*x^4+d*x^2+c)/x^6/(b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

-1/5*c/a/x^5-1/3*(a*d-b*c)/a^2/x^3-(a^2*e-a*b*d+b^2*c)/a^3/x+(a^3*f-a^2*b*e+a*b^2*d-b^3*c)/a^3/(a*b)^(1/2)*arc
tan(b*x/(a*b)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.37 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \left (a+b x^2\right )} \, dx=\left [\frac {15 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \sqrt {-a b} x^{5} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 6 \, a^{3} b c - 30 \, {\left (a b^{3} c - a^{2} b^{2} d + a^{3} b e\right )} x^{4} + 10 \, {\left (a^{2} b^{2} c - a^{3} b d\right )} x^{2}}{30 \, a^{4} b x^{5}}, -\frac {15 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \sqrt {a b} x^{5} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 3 \, a^{3} b c + 15 \, {\left (a b^{3} c - a^{2} b^{2} d + a^{3} b e\right )} x^{4} - 5 \, {\left (a^{2} b^{2} c - a^{3} b d\right )} x^{2}}{15 \, a^{4} b x^{5}}\right ] \]

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^6/(b*x^2+a),x, algorithm="fricas")

[Out]

[1/30*(15*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*sqrt(-a*b)*x^5*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 6
*a^3*b*c - 30*(a*b^3*c - a^2*b^2*d + a^3*b*e)*x^4 + 10*(a^2*b^2*c - a^3*b*d)*x^2)/(a^4*b*x^5), -1/15*(15*(b^3*
c - a*b^2*d + a^2*b*e - a^3*f)*sqrt(a*b)*x^5*arctan(sqrt(a*b)*x/a) + 3*a^3*b*c + 15*(a*b^3*c - a^2*b^2*d + a^3
*b*e)*x^4 - 5*(a^2*b^2*c - a^3*b*d)*x^2)/(a^4*b*x^5)]

Sympy [A] (verification not implemented)

Time = 2.25 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.61 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \left (a+b x^2\right )} \, dx=- \frac {\sqrt {- \frac {1}{a^{7} b}} \left (a^{3} f - a^{2} b e + a b^{2} d - b^{3} c\right ) \log {\left (- a^{4} \sqrt {- \frac {1}{a^{7} b}} + x \right )}}{2} + \frac {\sqrt {- \frac {1}{a^{7} b}} \left (a^{3} f - a^{2} b e + a b^{2} d - b^{3} c\right ) \log {\left (a^{4} \sqrt {- \frac {1}{a^{7} b}} + x \right )}}{2} + \frac {- 3 a^{2} c + x^{4} \left (- 15 a^{2} e + 15 a b d - 15 b^{2} c\right ) + x^{2} \left (- 5 a^{2} d + 5 a b c\right )}{15 a^{3} x^{5}} \]

[In]

integrate((f*x**6+e*x**4+d*x**2+c)/x**6/(b*x**2+a),x)

[Out]

-sqrt(-1/(a**7*b))*(a**3*f - a**2*b*e + a*b**2*d - b**3*c)*log(-a**4*sqrt(-1/(a**7*b)) + x)/2 + sqrt(-1/(a**7*
b))*(a**3*f - a**2*b*e + a*b**2*d - b**3*c)*log(a**4*sqrt(-1/(a**7*b)) + x)/2 + (-3*a**2*c + x**4*(-15*a**2*e
+ 15*a*b*d - 15*b**2*c) + x**2*(-5*a**2*d + 5*a*b*c))/(15*a**3*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \left (a+b x^2\right )} \, dx=-\frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {15 \, {\left (b^{2} c - a b d + a^{2} e\right )} x^{4} + 3 \, a^{2} c - 5 \, {\left (a b c - a^{2} d\right )} x^{2}}{15 \, a^{3} x^{5}} \]

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^6/(b*x^2+a),x, algorithm="maxima")

[Out]

-(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) - 1/15*(15*(b^2*c - a*b*d + a^2*e)*
x^4 + 3*a^2*c - 5*(a*b*c - a^2*d)*x^2)/(a^3*x^5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \left (a+b x^2\right )} \, dx=-\frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {15 \, b^{2} c x^{4} - 15 \, a b d x^{4} + 15 \, a^{2} e x^{4} - 5 \, a b c x^{2} + 5 \, a^{2} d x^{2} + 3 \, a^{2} c}{15 \, a^{3} x^{5}} \]

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^6/(b*x^2+a),x, algorithm="giac")

[Out]

-(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) - 1/15*(15*b^2*c*x^4 - 15*a*b*d*x^4
+ 15*a^2*e*x^4 - 5*a*b*c*x^2 + 5*a^2*d*x^2 + 3*a^2*c)/(a^3*x^5)

Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^6 \left (a+b x^2\right )} \, dx=-\frac {\frac {c}{5\,a}+\frac {x^2\,\left (a\,d-b\,c\right )}{3\,a^2}+\frac {x^4\,\left (e\,a^2-d\,a\,b+c\,b^2\right )}{a^3}}{x^5}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{a^{7/2}\,\sqrt {b}} \]

[In]

int((c + d*x^2 + e*x^4 + f*x^6)/(x^6*(a + b*x^2)),x)

[Out]

- (c/(5*a) + (x^2*(a*d - b*c))/(3*a^2) + (x^4*(b^2*c + a^2*e - a*b*d))/a^3)/x^5 - (atan((b^(1/2)*x)/a^(1/2))*(
b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(a^(7/2)*b^(1/2))