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

Optimal. Leaf size=89 \[ -\frac {3 c d^2-e (2 b d-a e)}{e^4 (d+e x)}+\frac {d \left (a e^2-b d e+c d^2\right )}{2 e^4 (d+e x)^2}-\frac {(3 c d-b e) \log (d+e x)}{e^4}+\frac {c x}{e^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 89, 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 )}{2 e^4 (d+e x)^2}-\frac {3 c d^2-e (2 b d-a e)}{e^4 (d+e x)}-\frac {(3 c d-b e) \log (d+e x)}{e^4}+\frac {c x}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.05, size = 80, normalized size = 0.90 \begin {gather*} \frac {\frac {d e (a e-b d)+c d^3}{(d+e x)^2}-\frac {2 \left (e (a e-2 b d)+3 c d^2\right )}{d+e x}+2 (b e-3 c d) \log (d+e x)+2 c e x}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*e*x + (c*d^3 + d*e*(-(b*d) + a*e))/(d + e*x)^2 - (2*(3*c*d^2 + e*(-2*b*d + a*e)))/(d + e*x) + 2*(-3*c*d +
b*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)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.38, size = 147, normalized size = 1.65 \begin {gather*} \frac {2 \, c e^{3} x^{3} + 4 \, c d e^{2} x^{2} - 5 \, c d^{3} + 3 \, b d^{2} e - a d e^{2} - 2 \, {\left (2 \, c d^{2} e - 2 \, b d e^{2} + a e^{3}\right )} x - 2 \, {\left (3 \, c d^{3} - b d^{2} e + {\left (3 \, c d e^{2} - b e^{3}\right )} x^{2} + 2 \, {\left (3 \, c d^{2} e - b d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} 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)^3,x, algorithm="fricas")

[Out]

1/2*(2*c*e^3*x^3 + 4*c*d*e^2*x^2 - 5*c*d^3 + 3*b*d^2*e - a*d*e^2 - 2*(2*c*d^2*e - 2*b*d*e^2 + a*e^3)*x - 2*(3*
c*d^3 - b*d^2*e + (3*c*d*e^2 - b*e^3)*x^2 + 2*(3*c*d^2*e - b*d*e^2)*x)*log(e*x + d))/(e^6*x^2 + 2*d*e^5*x + d^
2*e^4)

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giac [A]  time = 0.15, size = 82, normalized size = 0.92 \begin {gather*} c x e^{\left (-3\right )} - {\left (3 \, c d - b e\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) - \frac {{\left (5 \, c d^{3} - 3 \, b d^{2} e + a d e^{2} + 2 \, {\left (3 \, c d^{2} e - 2 \, b d e^{2} + a e^{3}\right )} x\right )} e^{\left (-4\right )}}{2 \, {\left (x e + d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 121, normalized size = 1.36 \begin {gather*} \frac {a d}{2 \left (e x +d \right )^{2} e^{2}}-\frac {b \,d^{2}}{2 \left (e x +d \right )^{2} e^{3}}+\frac {c \,d^{3}}{2 \left (e x +d \right )^{2} e^{4}}-\frac {a}{\left (e x +d \right ) e^{2}}+\frac {2 b d}{\left (e x +d \right ) e^{3}}+\frac {b \ln \left (e x +d \right )}{e^{3}}-\frac {3 c \,d^{2}}{\left (e x +d \right ) e^{4}}-\frac {3 c d \ln \left (e x +d \right )}{e^{4}}+\frac {c 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)^3,x)

[Out]

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

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maxima [A]  time = 0.63, size = 96, normalized size = 1.08 \begin {gather*} -\frac {5 \, c d^{3} - 3 \, b d^{2} e + a d e^{2} + 2 \, {\left (3 \, c d^{2} e - 2 \, b d e^{2} + a e^{3}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {c x}{e^{3}} - \frac {{\left (3 \, c d - b e\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)^3,x, algorithm="maxima")

[Out]

-1/2*(5*c*d^3 - 3*b*d^2*e + a*d*e^2 + 2*(3*c*d^2*e - 2*b*d*e^2 + a*e^3)*x)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) + c
*x/e^3 - (3*c*d - b*e)*log(e*x + d)/e^4

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mupad [B]  time = 2.35, size = 96, normalized size = 1.08 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (b\,e-3\,c\,d\right )}{e^4}-\frac {x\,\left (3\,c\,d^2-2\,b\,d\,e+a\,e^2\right )+\frac {5\,c\,d^3-3\,b\,d^2\,e+a\,d\,e^2}{2\,e}}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}+\frac {c\,x}{e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.76, size = 97, normalized size = 1.09 \begin {gather*} \frac {c x}{e^{3}} + \frac {- a d e^{2} + 3 b d^{2} e - 5 c d^{3} + x \left (- 2 a e^{3} + 4 b d e^{2} - 6 c d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} + \frac {\left (b e - 3 c d\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)**3,x)

[Out]

c*x/e**3 + (-a*d*e**2 + 3*b*d**2*e - 5*c*d**3 + x*(-2*a*e**3 + 4*b*d*e**2 - 6*c*d**2*e))/(2*d**2*e**4 + 4*d*e*
*5*x + 2*e**6*x**2) + (b*e - 3*c*d)*log(d + e*x)/e**4

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