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

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

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Rubi [A]  time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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*}

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Mathematica [A]  time = 0.06, size = 79, normalized size = 0.94 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (a+b x+c x^2\right )}{(d+e x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 131, normalized size = 1.56 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.20, size = 131, normalized size = 1.56 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.05, size = 108, normalized size = 1.29 \begin {gather*} \frac {c \,x^{2}}{2 e^{2}}+\frac {a d}{\left (e x +d \right ) e^{2}}+\frac {a \ln \left (e x +d \right )}{e^{2}}-\frac {b \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {2 b d \ln \left (e x +d \right )}{e^{3}}+\frac {b x}{e^{2}}+\frac {c \,d^{3}}{\left (e x +d \right ) e^{4}}+\frac {3 c \,d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {2 c d x}{e^{3}} \end {gather*}

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

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maxima [A]  time = 0.51, size = 85, normalized size = 1.01 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 2.34, size = 88, normalized size = 1.05 \begin {gather*} x\,\left (\frac {b}{e^2}-\frac {2\,c\,d}{e^3}\right )+\frac {c\,x^2}{2\,e^2}+\frac {\ln \left (d+e\,x\right )\,\left (3\,c\,d^2-2\,b\,d\,e+a\,e^2\right )}{e^4}+\frac {c\,d^3-b\,d^2\,e+a\,d\,e^2}{e\,\left (x\,e^4+d\,e^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.42, size = 82, normalized size = 0.98 \begin {gather*} \frac {c x^{2}}{2 e^{2}} + x \left (\frac {b}{e^{2}} - \frac {2 c d}{e^{3}}\right ) + \frac {a d e^{2} - b d^{2} e + c d^{3}}{d e^{4} + e^{5} x} + \frac {\left (a e^{2} - 2 b d e + 3 c d^{2}\right ) \log {\left (d + e x \right )}}{e^{4}} \end {gather*}

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

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