### 3.1106 $$\int \frac{(c d^2+2 c d e x+c e^2 x^2)^p}{(d+e x)^3} \, dx$$

Optimal. Leaf size=39 $-\frac{c \left (c d^2+2 c d e x+c e^2 x^2\right )^{p-1}}{2 e (1-p)}$

[Out]

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

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Rubi [A]  time = 0.0256884, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.067, Rules used = {643, 629} $-\frac{c \left (c d^2+2 c d e x+c e^2 x^2\right )^{p-1}}{2 e (1-p)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 643

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c,
0] &&  !IntegerQ[p] && EqQ[2*c*d - b*e, 0] && IntegerQ[(m - 1)/2]

Rule 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^3} \, dx &=c^2 \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{-2+p} \, dx\\ &=-\frac{c \left (c d^2+2 c d e x+c e^2 x^2\right )^{-1+p}}{2 e (1-p)}\\ \end{align*}

Mathematica [A]  time = 0.0124948, size = 26, normalized size = 0.67 $\frac{c \left (c (d+e x)^2\right )^{p-1}}{2 e (p-1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 40, normalized size = 1. \begin{align*}{\frac{ \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{p}}{2\, \left ( ex+d \right ) ^{2} \left ( p-1 \right ) e}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.27352, size = 61, normalized size = 1.56 \begin{align*} \frac{{\left (e x + d\right )}^{2 \, p} c^{p}}{2 \,{\left (e^{3}{\left (p - 1\right )} x^{2} + 2 \, d e^{2}{\left (p - 1\right )} x + d^{2} e{\left (p - 1\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e*x + d)^(2*p)*c^p/(e^3*(p - 1)*x^2 + 2*d*e^2*(p - 1)*x + d^2*e*(p - 1))

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Fricas [A]  time = 2.48723, size = 136, normalized size = 3.49 \begin{align*} \frac{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \,{\left (d^{2} e p - d^{2} e +{\left (e^{3} p - e^{3}\right )} x^{2} + 2 \,{\left (d e^{2} p - d e^{2}\right )} x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)^p/(d^2*e*p - d^2*e + (e^3*p - e^3)*x^2 + 2*(d*e^2*p - d*e^2)*x)

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Sympy [A]  time = 1.16808, size = 100, normalized size = 2.56 \begin{align*} \begin{cases} \frac{c x}{d} & \text{for}\: e = 0 \wedge p = 1 \\\frac{x \left (c d^{2}\right )^{p}}{d^{3}} & \text{for}\: e = 0 \\\frac{c \log{\left (\frac{d}{e} + x \right )}}{e} & \text{for}\: p = 1 \\\frac{\left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 d^{2} e p - 2 d^{2} e + 4 d e^{2} p x - 4 d e^{2} x + 2 e^{3} p x^{2} - 2 e^{3} x^{2}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c*x/d, Eq(e, 0) & Eq(p, 1)), (x*(c*d**2)**p/d**3, Eq(e, 0)), (c*log(d/e + x)/e, Eq(p, 1)), ((c*d**2
+ 2*c*d*e*x + c*e**2*x**2)**p/(2*d**2*e*p - 2*d**2*e + 4*d*e**2*p*x - 4*d*e**2*x + 2*e**3*p*x**2 - 2*e**3*x**
2), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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