3.11.92 \(\int \frac {(d+e x)^m}{(c d^2+2 c d e x+c e^2 x^2)^2} \, dx\) [1092]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {27, 12, 32} \begin {gather*} -\frac {(d+e x)^{m-3}}{c^2 e (3-m)} \end {gather*}

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 12

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

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 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*}

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Mathematica [A]
time = 0.02, size = 21, normalized size = 0.88 \begin {gather*} \frac {(d+e x)^{-3+m}}{c^2 e (-3+m)} \end {gather*}

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 = 0.59, size = 27, normalized size = 1.12

method	result	size
risch	\(\frac {\left (e x +d \right )^{m}}{c^{2} e \left (-3+m \right ) \left (e x +d \right )^{3}}\)	\(27\)
norman	\(\frac {{\mathrm e}^{m \ln \left (e x +d \right )}}{c^{2} e \left (-3+m \right ) \left (e x +d \right )^{3}}\)	\(29\)
gosper	\(\frac {\left (e x +d \right )^{-1+m}}{\left (e^{2} x^{2}+2 d x e +d^{2}\right ) c^{2} e \left (-3+m \right )}\)	\(40\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (21) = 42\).
time = 0.32, size = 64, normalized size = 2.67 \begin {gather*} \frac {{\left (x e + d\right )}^{m}}{c^{2} {\left (m - 3\right )} x^{3} e^{4} + 3 \, c^{2} d {\left (m - 3\right )} x^{2} e^{3} + 3 \, c^{2} d^{2} {\left (m - 3\right )} x e^{2} + c^{2} d^{3} {\left (m - 3\right )} e} \end {gather*}

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]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (21) = 42\).
time = 2.71, size = 91, normalized size = 3.79 \begin {gather*} \frac {{\left (x e + d\right )}^{m}}{{\left (c^{2} m - 3 \, c^{2}\right )} x^{3} e^{4} + 3 \, {\left (c^{2} d m - 3 \, c^{2} d\right )} x^{2} e^{3} + 3 \, {\left (c^{2} d^{2} m - 3 \, c^{2} d^{2}\right )} x e^{2} + {\left (c^{2} d^{3} m - 3 \, c^{2} d^{3}\right )} e} \end {gather*}

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]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (17) = 34\).
time = 0.80, size = 136, normalized size = 5.67 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [B]
time = 0.50, size = 50, normalized size = 2.08 \begin {gather*} \frac {{\left (d+e\,x\right )}^m}{c^2\,e^4\,\left (m-3\right )\,\left (x^3+\frac {d^3}{e^3}+\frac {3\,d\,x^2}{e}+\frac {3\,d^2\,x}{e^2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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