3.11.4 \(\int \frac {1}{(d+e x) (c d^2+2 c d e x+c e^2 x^2)} \, dx\) [1004]

Optimal. Leaf size=17 \[ -\frac {1}{2 c e (d+e x)^2} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 17, 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 {1}{2 c e (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)),x]

[Out]

-1/2*1/(c*e*(d + e*x)^2)

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

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Mathematica [A]
time = 0.00, size = 17, normalized size = 1.00 \begin {gather*} -\frac {1}{2 c e (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)),x]

[Out]

-1/2*1/(c*e*(d + e*x)^2)

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Maple [A]
time = 0.61, size = 16, normalized size = 0.94

method	result	size
default	\(-\frac {1}{2 c e \left (e x +d \right )^{2}}\)	\(16\)
norman	\(-\frac {1}{2 c e \left (e x +d \right )^{2}}\)	\(16\)
risch	\(-\frac {1}{2 c e \left (e x +d \right )^{2}}\)	\(16\)
gosper	\(-\frac {1}{2 e c \left (e^{2} x^{2}+2 d x e +d^{2}\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 26, normalized size = 1.53 \begin {gather*} -\frac {1}{2 \, {\left (c x^{2} e^{3} + 2 \, c d x e^{2} + c d^{2} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.73, size = 26, normalized size = 1.53 \begin {gather*} -\frac {1}{2 \, {\left (c x^{2} e^{3} + 2 \, c d x e^{2} + c d^{2} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (14) = 28\).
time = 0.08, size = 31, normalized size = 1.82 \begin {gather*} - \frac {1}{2 c d^{2} e + 4 c d e^{2} x + 2 c e^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e**2*x**2+2*c*d*e*x+c*d**2),x)

[Out]

-1/(2*c*d**2*e + 4*c*d*e**2*x + 2*c*e**3*x**2)

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Giac [A]
time = 0.93, size = 15, normalized size = 0.88 \begin {gather*} -\frac {e^{\left (-1\right )}}{2 \, {\left (x e + d\right )}^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2),x, algorithm="giac")

[Out]

-1/2*e^(-1)/((x*e + d)^2*c)

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Mupad [B]
time = 0.03, size = 29, normalized size = 1.71 \begin {gather*} -\frac {1}{2\,c\,d^2\,e+4\,c\,d\,e^2\,x+2\,c\,e^3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*c*e^3*x^2 + 2*c*d^2*e + 4*c*d*e^2*x)

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