3.11.16 \(\int \frac {1}{(d+e x)^2 (c d^2+2 c d e x+c e^2 x^2)^2} \, dx\) [1016]

Optimal. Leaf size=17 \[ -\frac {1}{5 c^2 e (d+e x)^5} \]

[Out]

-1/5/c^2/e/(e*x+d)^5

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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}{5 c^2 e (d+e x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*1/(c^2*e*(d + e*x)^5)

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)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )^2} \, dx &=\int \frac {1}{c^2 (d+e x)^6} \, dx\\ &=\frac {\int \frac {1}{(d+e x)^6} \, dx}{c^2}\\ &=-\frac {1}{5 c^2 e (d+e x)^5}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 17, normalized size = 1.00 \begin {gather*} -\frac {1}{5 c^2 e (d+e x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*1/(c^2*e*(d + e*x)^5)

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Maple [A]
time = 0.68, size = 16, normalized size = 0.94

method	result	size
default	\(-\frac {1}{5 c^{2} e \left (e x +d \right )^{5}}\)	\(16\)
norman	\(-\frac {1}{5 c^{2} e \left (e x +d \right )^{5}}\)	\(16\)
gosper	\(-\frac {1}{5 \left (e x +d \right ) e \,c^{2} \left (e^{2} x^{2}+2 d x e +d^{2}\right )^{2}}\)	\(34\)
risch	\(-\frac {1}{5 e \left (e x +d \right )^{3} c^{2} \left (e^{2} x^{2}+2 d x e +d^{2}\right )}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/c^2/e/(e*x+d)^5

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (15) = 30\).
time = 0.30, size = 71, normalized size = 4.18 \begin {gather*} -\frac {1}{5 \, {\left (c^{2} x^{5} e^{6} + 5 \, c^{2} d x^{4} e^{5} + 10 \, c^{2} d^{2} x^{3} e^{4} + 10 \, c^{2} d^{3} x^{2} e^{3} + 5 \, c^{2} d^{4} x e^{2} + c^{2} d^{5} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/(c^2*x^5*e^6 + 5*c^2*d*x^4*e^5 + 10*c^2*d^2*x^3*e^4 + 10*c^2*d^3*x^2*e^3 + 5*c^2*d^4*x*e^2 + c^2*d^5*e)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (15) = 30\).
time = 2.77, size = 71, normalized size = 4.18 \begin {gather*} -\frac {1}{5 \, {\left (c^{2} x^{5} e^{6} + 5 \, c^{2} d x^{4} e^{5} + 10 \, c^{2} d^{2} x^{3} e^{4} + 10 \, c^{2} d^{3} x^{2} e^{3} + 5 \, c^{2} d^{4} x e^{2} + c^{2} d^{5} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/(c^2*x^5*e^6 + 5*c^2*d*x^4*e^5 + 10*c^2*d^2*x^3*e^4 + 10*c^2*d^3*x^2*e^3 + 5*c^2*d^4*x*e^2 + c^2*d^5*e)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (15) = 30\).
time = 0.18, size = 82, normalized size = 4.82 \begin {gather*} - \frac {1}{5 c^{2} d^{5} e + 25 c^{2} d^{4} e^{2} x + 50 c^{2} d^{3} e^{3} x^{2} + 50 c^{2} d^{2} e^{4} x^{3} + 25 c^{2} d e^{5} x^{4} + 5 c^{2} e^{6} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(5*c**2*d**5*e + 25*c**2*d**4*e**2*x + 50*c**2*d**3*e**3*x**2 + 50*c**2*d**2*e**4*x**3 + 25*c**2*d*e**5*x**
4 + 5*c**2*e**6*x**5)

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Giac [A]
time = 1.19, size = 15, normalized size = 0.88 \begin {gather*} -\frac {e^{\left (-1\right )}}{5 \, {\left (x e + d\right )}^{5} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*e^(-1)/((x*e + d)^5*c^2)

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Mupad [B]
time = 0.07, size = 77, normalized size = 4.53 \begin {gather*} -\frac {1}{5\,c^2\,d^5\,e+25\,c^2\,d^4\,e^2\,x+50\,c^2\,d^3\,e^3\,x^2+50\,c^2\,d^2\,e^4\,x^3+25\,c^2\,d\,e^5\,x^4+5\,c^2\,e^6\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(5*c^2*d^5*e + 5*c^2*e^6*x^5 + 25*c^2*d^4*e^2*x + 25*c^2*d*e^5*x^4 + 50*c^2*d^3*e^3*x^2 + 50*c^2*d^2*e^4*x^
3)

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