### 3.1105 $$\int \frac{(c d^2+2 c d e x+c e^2 x^2)^p}{(d+e x)^2} \, dx$$

Optimal. Leaf size=42 $-\frac{c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{p-1}}{e (1-2 p)}$

[Out]

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

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Rubi [A]  time = 0.0256373, antiderivative size = 42, 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 = {642, 609} $-\frac{c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{p-1}}{e (1-2 p)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 642

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^m/c^(m/2), Int[(a +
b*x + c*x^2)^(p + m/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/2]

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-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)^2} \, dx &=c \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{-1+p} \, dx\\ &=-\frac{c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{-1+p}}{e (1-2 p)}\\ \end{align*}

Mathematica [A]  time = 0.0166086, size = 30, normalized size = 0.71 $\frac{c (d+e x) \left (c (d+e x)^2\right )^{p-1}}{e (2 p-1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 41, normalized size = 1. \begin{align*}{\frac{ \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{p}}{ \left ( ex+d \right ) \left ( -1+2\,p \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.31708, size = 46, normalized size = 1.1 \begin{align*} \frac{{\left (e x + d\right )}^{2 \, p} c^{p}}{e^{2}{\left (2 \, p - 1\right )} x + d e{\left (2 \, p - 1\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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.44252, size = 96, normalized size = 2.29 \begin{align*} \frac{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \, d e p - d e +{\left (2 \, e^{2} p - e^{2}\right )} 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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)**2,x)

[Out]

Exception raised: TypeError

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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 )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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