3.1104 \(\int \frac{(c d^2+2 c d e x+c e^2 x^2)^p}{d+e x} \, dx\)

Optimal. Leaf size=32 \[ \frac{\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{2 e p} \]

[Out]

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

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Rubi [A]  time = 0.0212296, antiderivative size = 32, 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} \[ \frac{\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{2 e p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0064535, size = 21, normalized size = 0.66 \[ \frac{\left (c (d+e x)^2\right )^p}{2 e p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(d + e*x)^2)^p/(2*e*p)

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Maple [A]  time = 0.039, size = 31, normalized size = 1. \begin{align*}{\frac{ \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{p}}{2\,ep}} \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),x)

[Out]

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

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Maxima [A]  time = 1.15541, size = 27, normalized size = 0.84 \begin{align*} \frac{{\left (e x + d\right )}^{2 \, p} c^{p}}{2 \, e p} \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),x, algorithm="maxima")

[Out]

1/2*(e*x + d)^(2*p)*c^p/(e*p)

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Fricas [A]  time = 2.51845, size = 61, normalized size = 1.91 \begin{align*} \frac{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \, e p} \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),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.348116, size = 48, normalized size = 1.5 \begin{align*} \begin{cases} \frac{x}{d} & \text{for}\: e = 0 \wedge p = 0 \\\frac{\log{\left (\frac{d}{e} + x \right )}}{e} & \text{for}\: p = 0 \\\frac{x \left (c d^{2}\right )^{p}}{d} & \text{for}\: e = 0 \\\frac{\left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p} & \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),x)

[Out]

Piecewise((x/d, Eq(e, 0) & Eq(p, 0)), (log(d/e + x)/e, Eq(p, 0)), (x*(c*d**2)**p/d, Eq(e, 0)), ((c*d**2 + 2*c*
d*e*x + c*e**2*x**2)**p/(2*e*p), 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}}{e x + d}\,{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),x, algorithm="giac")

[Out]

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