3.1002 \(\int \frac {d+e x}{c d^2+2 c d e x+c e^2 x^2} \, dx\)

Optimal. Leaf size=13 \[ \frac {\log (d+e x)}{c e} \]

[Out]

ln(e*x+d)/c/e

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Rubi [A]  time = 0.00, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {27, 12, 31} \[ \frac {\log (d+e x)}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[d + e*x]/(c*e)

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin {align*} \int \frac {d+e x}{c d^2+2 c d e x+c e^2 x^2} \, dx &=\int \frac {1}{c (d+e x)} \, dx\\ &=\frac {\int \frac {1}{d+e x} \, dx}{c}\\ &=\frac {\log (d+e x)}{c e}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.23 \[ \frac {\log (c d+c e x)}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[c*d + c*e*x]/(c*e)

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fricas [A]  time = 1.00, size = 13, normalized size = 1.00 \[ \frac {\log \left (e x + d\right )}{c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e*x + d)/(c*e)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: exp(1)*1/2/c/exp(2)*ln(x^2*exp(2)+2*x*d*
exp(1)+d^2)+(2*exp(2)*d-2*d*exp(1)^2)/c/exp(2)*1/2/d/sqrt(-exp(1)^2+exp(2))*atan((d*exp(1)+x*exp(2))/d/sqrt(-e
xp(1)^2+exp(2)))

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maple [A]  time = 0.04, size = 14, normalized size = 1.08 \[ \frac {\ln \left (e x +d \right )}{c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(e*x+d)/c/e

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maxima [B]  time = 1.33, size = 29, normalized size = 2.23 \[ \frac {\log \left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}{2 \, c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 13, normalized size = 1.00 \[ \frac {\ln \left (d+e\,x\right )}{c\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(d + e*x)/(c*e)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.92 \[ \frac {\log {\left (c d + c e x \right )}}{c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*d + c*e*x)/(c*e)

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