3.226 \(\int \frac {b c-a d-2 a e x-b e x^2-3 a f x^2-2 b f x^3}{(c+d x+e x^2+f x^3)^2} \, dx\)

Optimal. Leaf size=40 \[ \frac {a}{c+d x+e x^2+f x^3}+\frac {b x}{c+d x+e x^2+f x^3} \]

[Out]

a/(f*x^3+e*x^2+d*x+c)+b*x/(f*x^3+e*x^2+d*x+c)

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Rubi [A]  time = 0.10, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6, 2102, 1588} \[ \frac {a}{c+d x+e x^2+f x^3}+\frac {b x}{c+d x+e x^2+f x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a/(c + d*x + e*x^2 + f*x^3) + (b*x)/(c + d*x + e*x^2 + f*x^3)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 2102

Int[(Pm_)*(Qn_)^(p_.), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x]}, Simp[(Coeff[Pm, x, m]*x^(m - n
+ 1)*Qn^(p + 1))/((m + n*p + 1)*Coeff[Qn, x, n]), x] + Dist[1/((m + n*p + 1)*Coeff[Qn, x, n]), Int[ExpandToSum
[(m + n*p + 1)*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*x^(m - n)*((m - n + 1)*Qn + (p + 1)*x*D[Qn, x]), x]*Qn^p,
x], x] /; LtQ[1, n, m + 1] && m + n*p + 1 < 0] /; FreeQ[p, x] && PolyQ[Pm, x] && PolyQ[Qn, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {b c-a d-2 a e x-b e x^2-3 a f x^2-2 b f x^3}{\left (c+d x+e x^2+f x^3\right )^2} \, dx &=\int \frac {b c-a d-2 a e x+(-b e-3 a f) x^2-2 b f x^3}{\left (c+d x+e x^2+f x^3\right )^2} \, dx\\ &=\frac {b x}{c+d x+e x^2+f x^3}-\frac {\int \frac {2 a d f+4 a e f x+6 a f^2 x^2}{\left (c+d x+e x^2+f x^3\right )^2} \, dx}{2 f}\\ &=\frac {a}{c+d x+e x^2+f x^3}+\frac {b x}{c+d x+e x^2+f x^3}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 23, normalized size = 0.58 \[ \frac {a+b x}{c+d x+e x^2+f x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*x)/(c + d*x + e*x^2 + f*x^3)

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fricas [A]  time = 0.86, size = 23, normalized size = 0.58 \[ \frac {b x + a}{f x^{3} + e x^{2} + d x + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x + a)/(f*x^3 + e*x^2 + d*x + c)

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giac [A]  time = 0.54, size = 24, normalized size = 0.60 \[ \frac {b x + a}{f x^{3} + x^{2} e + d x + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x + a)/(f*x^3 + x^2*e + d*x + c)

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maple [A]  time = 0.01, size = 28, normalized size = 0.70 \[ -\frac {-b x -a}{f \,x^{3}+e \,x^{2}+d x +c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 23, normalized size = 0.58 \[ \frac {b x + a}{f x^{3} + e x^{2} + d x + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x + a)/(f*x^3 + e*x^2 + d*x + c)

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mupad [B]  time = 0.11, size = 23, normalized size = 0.58 \[ \frac {a+b\,x}{f\,x^3+e\,x^2+d\,x+c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a + b*x)/(c + d*x + e*x^2 + f*x^3)

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sympy [A]  time = 32.56, size = 22, normalized size = 0.55 \[ - \frac {- a - b x}{c + d x + e x^{2} + f x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-a - b*x)/(c + d*x + e*x**2 + f*x**3)

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