3.19.69 \(\int \frac {d+e x}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [1869]

Optimal. Leaf size=16 \[ \frac {\log (a e+c d x)}{c d} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {640, 31} \begin {gather*} \frac {\log (a e+c d x)}{c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[a*e + c*d*x]/(c*d)

Rule 31

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

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 16, normalized size = 1.00 \begin {gather*} \frac {\log (a e+c d x)}{c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[a*e + c*d*x]/(c*d)

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Maple [A]
time = 0.63, size = 17, normalized size = 1.06

method	result	size
default	\(\frac {\ln \left (c d x +a e \right )}{c d}\)	\(17\)
norman	\(\frac {\ln \left (c d x +a e \right )}{c d}\)	\(17\)
risch	\(\frac {\ln \left (c d x +a e \right )}{c d}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 17, normalized size = 1.06 \begin {gather*} \frac {\log \left (c d x + a e\right )}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*d*x + a*e)/(c*d)

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Fricas [A]
time = 2.52, size = 17, normalized size = 1.06 \begin {gather*} \frac {\log \left (c d x + a e\right )}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*d*x + a*e)/(c*d)

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Sympy [A]
time = 0.02, size = 12, normalized size = 0.75 \begin {gather*} \frac {\log {\left (a e + c d x \right )}}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*e + c*d*x)/(c*d)

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Giac [A]
time = 1.48, size = 18, normalized size = 1.12 \begin {gather*} \frac {\log \left ({\left | c d x + a e \right |}\right )}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(c*d*x + a*e))/(c*d)

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Mupad [B]
time = 0.56, size = 16, normalized size = 1.00 \begin {gather*} \frac {\ln \left (a\,e+c\,d\,x\right )}{c\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*e + c*d*x)/(c*d)

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