3.2358 \(\int \frac{x (a+b x+c x^2)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=84 \[ \frac{d \left (a e^2-b d e+c d^2\right )}{e^4 (d+e x)}+\frac{\log (d+e x) \left (3 c d^2-e (2 b d-a e)\right )}{e^4}-\frac{x (2 c d-b e)}{e^3}+\frac{c x^2}{2 e^2} \]

[Out]

-(((2*c*d - b*e)*x)/e^3) + (c*x^2)/(2*e^2) + (d*(c*d^2 - b*d*e + a*e^2))/(e^4*(d + e*x)) + ((3*c*d^2 - e*(2*b*
d - a*e))*Log[d + e*x])/e^4

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Rubi [A]  time = 0.0904978, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {771} \[ \frac{d \left (a e^2-b d e+c d^2\right )}{e^4 (d+e x)}+\frac{\log (d+e x) \left (3 c d^2-e (2 b d-a e)\right )}{e^4}-\frac{x (2 c d-b e)}{e^3}+\frac{c x^2}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*c*d - b*e)*x)/e^3) + (c*x^2)/(2*e^2) + (d*(c*d^2 - b*d*e + a*e^2))/(e^4*(d + e*x)) + ((3*c*d^2 - e*(2*b*
d - a*e))*Log[d + e*x])/e^4

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [A]  time = 0.0657236, size = 79, normalized size = 0.94 \[ \frac{\frac{2 \left (d e (a e-b d)+c d^3\right )}{d+e x}+2 \log (d+e x) \left (e (a e-2 b d)+3 c d^2\right )+2 e x (b e-2 c d)+c e^2 x^2}{2 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(-2*c*d + b*e)*x + c*e^2*x^2 + (2*(c*d^3 + d*e*(-(b*d) + a*e)))/(d + e*x) + 2*(3*c*d^2 + e*(-2*b*d + a*e)
)*Log[d + e*x])/(2*e^4)

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Maple [A]  time = 0.007, size = 108, normalized size = 1.3 \begin{align*}{\frac{c{x}^{2}}{2\,{e}^{2}}}+{\frac{bx}{{e}^{2}}}-2\,{\frac{cdx}{{e}^{3}}}+{\frac{\ln \left ( ex+d \right ) a}{{e}^{2}}}-2\,{\frac{\ln \left ( ex+d \right ) bd}{{e}^{3}}}+3\,{\frac{\ln \left ( ex+d \right ) c{d}^{2}}{{e}^{4}}}+{\frac{ad}{{e}^{2} \left ( ex+d \right ) }}-{\frac{b{d}^{2}}{{e}^{3} \left ( ex+d \right ) }}+{\frac{{d}^{3}c}{{e}^{4} \left ( ex+d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c*x^2/e^2+1/e^2*b*x-2/e^3*c*d*x+1/e^2*ln(e*x+d)*a-2/e^3*ln(e*x+d)*b*d+3/e^4*ln(e*x+d)*c*d^2+d/e^2/(e*x+d)*
a-d^2/e^3/(e*x+d)*b+d^3/e^4/(e*x+d)*c

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Maxima [A]  time = 1.00651, size = 115, normalized size = 1.37 \begin{align*} \frac{c d^{3} - b d^{2} e + a d e^{2}}{e^{5} x + d e^{4}} + \frac{c e x^{2} - 2 \,{\left (2 \, c d - b e\right )} x}{2 \, e^{3}} + \frac{{\left (3 \, c d^{2} - 2 \, b d e + a e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.31729, size = 281, normalized size = 3.35 \begin{align*} \frac{c e^{3} x^{3} + 2 \, c d^{3} - 2 \, b d^{2} e + 2 \, a d e^{2} -{\left (3 \, c d e^{2} - 2 \, b e^{3}\right )} x^{2} - 2 \,{\left (2 \, c d^{2} e - b d e^{2}\right )} x + 2 \,{\left (3 \, c d^{3} - 2 \, b d^{2} e + a d e^{2} +{\left (3 \, c d^{2} e - 2 \, b d e^{2} + a e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (e^{5} x + d e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.627243, size = 80, normalized size = 0.95 \begin{align*} \frac{c x^{2}}{2 e^{2}} + \frac{a d e^{2} - b d^{2} e + c d^{3}}{d e^{4} + e^{5} x} + \frac{x \left (b e - 2 c d\right )}{e^{3}} + \frac{\left (a e^{2} - 2 b d e + 3 c d^{2}\right ) \log{\left (d + e x \right )}}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x**2/(2*e**2) + (a*d*e**2 - b*d**2*e + c*d**3)/(d*e**4 + e**5*x) + x*(b*e - 2*c*d)/e**3 + (a*e**2 - 2*b*d*e
+ 3*c*d**2)*log(d + e*x)/e**4

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Giac [A]  time = 1.10659, size = 177, normalized size = 2.11 \begin{align*} \frac{1}{2} \,{\left ({\left (x e + d\right )}^{2}{\left (c - \frac{2 \,{\left (3 \, c d e - b e^{2}\right )} e^{\left (-1\right )}}{x e + d}\right )} e^{\left (-3\right )} - 2 \,{\left (3 \, c d^{2} - 2 \, b d e + a e^{2}\right )} e^{\left (-3\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + 2 \,{\left (\frac{c d^{3} e^{2}}{x e + d} - \frac{b d^{2} e^{3}}{x e + d} + \frac{a d e^{4}}{x e + d}\right )} e^{\left (-5\right )}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((x*e + d)^2*(c - 2*(3*c*d*e - b*e^2)*e^(-1)/(x*e + d))*e^(-3) - 2*(3*c*d^2 - 2*b*d*e + a*e^2)*e^(-3)*log(
abs(x*e + d)*e^(-1)/(x*e + d)^2) + 2*(c*d^3*e^2/(x*e + d) - b*d^2*e^3/(x*e + d) + a*d*e^4/(x*e + d))*e^(-5))*e
^(-1)