\(\int \frac {a d+a e x+(b d+a f) x^2+b e x^3+(c d+b f) x^4+c e x^5+c f x^6}{a+b x^2+c x^4} \, dx\) [63]

Optimal result

Integrand size = 63, antiderivative size = 20 \[ \int \frac {a d+a e x+(b d+a f) x^2+b e x^3+(c d+b f) x^4+c e x^5+c f x^6}{a+b x^2+c x^4} \, dx=d x+\frac {e x^2}{2}+\frac {f x^3}{3} \]

[Out]

d*x+1/2*e*x^2+1/3*f*x^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {1600} \[ \int \frac {a d+a e x+(b d+a f) x^2+b e x^3+(c d+b f) x^4+c e x^5+c f x^6}{a+b x^2+c x^4} \, dx=d x+\frac {e x^2}{2}+\frac {f x^3}{3} \]

[In]

Int[(a*d + a*e*x + (b*d + a*f)*x^2 + b*e*x^3 + (c*d + b*f)*x^4 + c*e*x^5 + c*f*x^6)/(a + b*x^2 + c*x^4),x]

[Out]

d*x + (e*x^2)/2 + (f*x^3)/3

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (d+e x+f x^2\right ) \, dx \\ & = d x+\frac {e x^2}{2}+\frac {f x^3}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {a d+a e x+(b d+a f) x^2+b e x^3+(c d+b f) x^4+c e x^5+c f x^6}{a+b x^2+c x^4} \, dx=d x+\frac {e x^2}{2}+\frac {f x^3}{3} \]

[In]

Integrate[(a*d + a*e*x + (b*d + a*f)*x^2 + b*e*x^3 + (c*d + b*f)*x^4 + c*e*x^5 + c*f*x^6)/(a + b*x^2 + c*x^4),
x]

[Out]

d*x + (e*x^2)/2 + (f*x^3)/3

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
default	\(d x +\frac {1}{2} e \,x^{2}+\frac {1}{3} f \,x^{3}\)	\(17\)
norman	\(d x +\frac {1}{2} e \,x^{2}+\frac {1}{3} f \,x^{3}\)	\(17\)
risch	\(d x +\frac {1}{2} e \,x^{2}+\frac {1}{3} f \,x^{3}\)	\(17\)
parallelrisch	\(d x +\frac {1}{2} e \,x^{2}+\frac {1}{3} f \,x^{3}\)	\(17\)
parts	\(d x +\frac {1}{2} e \,x^{2}+\frac {1}{3} f \,x^{3}\)	\(17\)
gosper	\(\frac {x \left (2 f \,x^{2}+3 e x +6 d \right )}{6}\)	\(18\)

[In]

int((d*a+a*e*x+(a*f+b*d)*x^2+e*x^3*b+(b*f+c*d)*x^4+c*e*x^5+c*f*x^6)/(c*x^4+b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

d*x+1/2*e*x^2+1/3*f*x^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {a d+a e x+(b d+a f) x^2+b e x^3+(c d+b f) x^4+c e x^5+c f x^6}{a+b x^2+c x^4} \, dx=\frac {1}{3} \, f x^{3} + \frac {1}{2} \, e x^{2} + d x \]

[In]

integrate((a*d+a*e*x+(a*f+b*d)*x^2+b*e*x^3+(b*f+c*d)*x^4+c*e*x^5+c*f*x^6)/(c*x^4+b*x^2+a),x, algorithm="fricas
")

[Out]

1/3*f*x^3 + 1/2*e*x^2 + d*x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {a d+a e x+(b d+a f) x^2+b e x^3+(c d+b f) x^4+c e x^5+c f x^6}{a+b x^2+c x^4} \, dx=d x + \frac {e x^{2}}{2} + \frac {f x^{3}}{3} \]

[In]

integrate((a*d+a*e*x+(a*f+b*d)*x**2+b*e*x**3+(b*f+c*d)*x**4+c*e*x**5+c*f*x**6)/(c*x**4+b*x**2+a),x)

[Out]

d*x + e*x**2/2 + f*x**3/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {a d+a e x+(b d+a f) x^2+b e x^3+(c d+b f) x^4+c e x^5+c f x^6}{a+b x^2+c x^4} \, dx=\frac {1}{3} \, f x^{3} + \frac {1}{2} \, e x^{2} + d x \]

[In]

integrate((a*d+a*e*x+(a*f+b*d)*x^2+b*e*x^3+(b*f+c*d)*x^4+c*e*x^5+c*f*x^6)/(c*x^4+b*x^2+a),x, algorithm="maxima
")

[Out]

1/3*f*x^3 + 1/2*e*x^2 + d*x

Giac [A] (verification not implemented)

none

Time = 0.62 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {a d+a e x+(b d+a f) x^2+b e x^3+(c d+b f) x^4+c e x^5+c f x^6}{a+b x^2+c x^4} \, dx=\frac {1}{3} \, f x^{3} + \frac {1}{2} \, e x^{2} + d x \]

[In]

integrate((a*d+a*e*x+(a*f+b*d)*x^2+b*e*x^3+(b*f+c*d)*x^4+c*e*x^5+c*f*x^6)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

1/3*f*x^3 + 1/2*e*x^2 + d*x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {a d+a e x+(b d+a f) x^2+b e x^3+(c d+b f) x^4+c e x^5+c f x^6}{a+b x^2+c x^4} \, dx=\frac {f\,x^3}{3}+\frac {e\,x^2}{2}+d\,x \]

[In]

int((a*d + x^2*(b*d + a*f) + x^4*(c*d + b*f) + a*e*x + b*e*x^3 + c*e*x^5 + c*f*x^6)/(a + b*x^2 + c*x^4),x)

[Out]

d*x + (e*x^2)/2 + (f*x^3)/3