3.1092 \(\int \frac{(d+e x)^m}{(c d^2+2 c d e x+c e^2 x^2)^2} \, dx\)

Optimal. Leaf size=24 \[ -\frac{(d+e x)^{m-3}}{c^2 e (3-m)} \]

[Out]

-((d + e*x)^(-3 + m)/(c^2*e*(3 - m)))

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Rubi [A]  time = 0.0101768, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {27, 12, 32} \[ -\frac{(d+e x)^{m-3}}{c^2 e (3-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d + e*x)^(-3 + m)/(c^2*e*(3 - m)))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0148665, size = 21, normalized size = 0.88 \[ \frac{(d+e x)^{m-3}}{c^2 e (m-3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x)^(-3 + m)/(c^2*e*(-3 + m))

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Maple [A]  time = 0.04, size = 40, normalized size = 1.7 \begin{align*}{\frac{ \left ( ex+d \right ) ^{-1+m}}{ \left ({e}^{2}{x}^{2}+2\,dex+{d}^{2} \right ){c}^{2}e \left ( -3+m \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.37361, size = 88, normalized size = 3.67 \begin{align*} \frac{{\left (e x + d\right )}^{m}}{c^{2} e^{4}{\left (m - 3\right )} x^{3} + 3 \, c^{2} d e^{3}{\left (m - 3\right )} x^{2} + 3 \, c^{2} d^{2} e^{2}{\left (m - 3\right )} x + c^{2} d^{3} e{\left (m - 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x + d)^m/(c^2*e^4*(m - 3)*x^3 + 3*c^2*d*e^3*(m - 3)*x^2 + 3*c^2*d^2*e^2*(m - 3)*x + c^2*d^3*e*(m - 3))

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Fricas [B]  time = 2.47713, size = 196, normalized size = 8.17 \begin{align*} \frac{{\left (e x + d\right )}^{m}}{c^{2} d^{3} e m - 3 \, c^{2} d^{3} e +{\left (c^{2} e^{4} m - 3 \, c^{2} e^{4}\right )} x^{3} + 3 \,{\left (c^{2} d e^{3} m - 3 \, c^{2} d e^{3}\right )} x^{2} + 3 \,{\left (c^{2} d^{2} e^{2} m - 3 \, c^{2} d^{2} e^{2}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x + d)^m/(c^2*d^3*e*m - 3*c^2*d^3*e + (c^2*e^4*m - 3*c^2*e^4)*x^3 + 3*(c^2*d*e^3*m - 3*c^2*d*e^3)*x^2 + 3*(
c^2*d^2*e^2*m - 3*c^2*d^2*e^2)*x)

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Sympy [A]  time = 2.14732, size = 136, normalized size = 5.67 \begin{align*} \begin{cases} \frac{x}{c^{2} d} & \text{for}\: e = 0 \wedge m = 3 \\\frac{d^{m} x}{c^{2} d^{4}} & \text{for}\: e = 0 \\\frac{\log{\left (\frac{d}{e} + x \right )}}{c^{2} e} & \text{for}\: m = 3 \\\frac{\left (d + e x\right )^{m}}{c^{2} d^{3} e m - 3 c^{2} d^{3} e + 3 c^{2} d^{2} e^{2} m x - 9 c^{2} d^{2} e^{2} x + 3 c^{2} d e^{3} m x^{2} - 9 c^{2} d e^{3} x^{2} + c^{2} e^{4} m x^{3} - 3 c^{2} e^{4} x^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/(c**2*d), Eq(e, 0) & Eq(m, 3)), (d**m*x/(c**2*d**4), Eq(e, 0)), (log(d/e + x)/(c**2*e), Eq(m, 3))
, ((d + e*x)**m/(c**2*d**3*e*m - 3*c**2*d**3*e + 3*c**2*d**2*e**2*m*x - 9*c**2*d**2*e**2*x + 3*c**2*d*e**3*m*x
**2 - 9*c**2*d*e**3*x**2 + c**2*e**4*m*x**3 - 3*c**2*e**4*x**3), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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