### 3.1835 $$\int \frac{a d e+(c d^2+a e^2) x+c d e x^2}{(d+e x)^3} \, dx$$

Optimal. Leaf size=33 $\frac{c d \log (d+e x)}{e^2}-\frac{a-\frac{c d^2}{e^2}}{d+e x}$

[Out]

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

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Rubi [A]  time = 0.0282534, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.061, Rules used = {24, 43} $\frac{c d \log (d+e x)}{e^2}-\frac{a-\frac{c d^2}{e^2}}{d+e x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0127439, size = 36, normalized size = 1.09 $\frac{c d^2-a e^2}{e^2 (d+e x)}+\frac{c d \log (d+e x)}{e^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 39, normalized size = 1.2 \begin{align*}{\frac{cd\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{a}{ex+d}}+{\frac{c{d}^{2}}{{e}^{2} \left ( ex+d \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.1247, size = 53, normalized size = 1.61 \begin{align*} \frac{c d \log \left (e x + d\right )}{e^{2}} + \frac{c d^{2} - a e^{2}}{e^{3} x + d e^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.52604, size = 89, normalized size = 2.7 \begin{align*} \frac{c d^{2} - a e^{2} +{\left (c d e x + c d^{2}\right )} \log \left (e x + d\right )}{e^{3} x + d e^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.538249, size = 32, normalized size = 0.97 \begin{align*} \frac{c d \log{\left (d + e x \right )}}{e^{2}} - \frac{a e^{2} - c d^{2}}{d e^{2} + e^{3} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2081, size = 70, normalized size = 2.12 \begin{align*} c d e^{\left (-2\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{{\left (c d^{3} - a d e^{2} +{\left (c d^{2} e - a e^{3}\right )} x\right )} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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