\(\int \frac {1}{d f+(e f+d g) x+e g x^2} \, dx\) [547]

Optimal result

Integrand size = 21, antiderivative size = 36 \[ \int \frac {1}{d f+(e f+d g) x+e g x^2} \, dx=\frac {\log (d+e x)}{e f-d g}-\frac {\log (f+g x)}{e f-d g} \]

[Out]

ln(e*x+d)/(-d*g+e*f)-ln(g*x+f)/(-d*g+e*f)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {630, 31} \[ \int \frac {1}{d f+(e f+d g) x+e g x^2} \, dx=\frac {\log (d+e x)}{e f-d g}-\frac {\log (f+g x)}{e f-d g} \]

[In]

Int[(d*f + (e*f + d*g)*x + e*g*x^2)^(-1),x]

[Out]

Log[d + e*x]/(e*f - d*g) - Log[f + g*x]/(e*f - d*g)

Rule 31

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

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = -\frac {(e g) \int \frac {1}{e f+e g x} \, dx}{e f-d g}+\frac {(e g) \int \frac {1}{d g+e g x} \, dx}{e f-d g} \\ & = \frac {\log (d+e x)}{e f-d g}-\frac {\log (f+g x)}{e f-d g} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.72 \[ \int \frac {1}{d f+(e f+d g) x+e g x^2} \, dx=\frac {\log (d+e x)-\log (f+g x)}{e f-d g} \]

[In]

Integrate[(d*f + (e*f + d*g)*x + e*g*x^2)^(-1),x]

[Out]

(Log[d + e*x] - Log[f + g*x])/(e*f - d*g)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78

method	result	size
parallelrisch	\(-\frac {\ln \left (e x +d \right )-\ln \left (g x +f \right )}{d g -e f}\)	\(28\)
default	\(-\frac {\ln \left (e x +d \right )}{d g -e f}+\frac {\ln \left (g x +f \right )}{d g -e f}\)	\(37\)
norman	\(-\frac {\ln \left (e x +d \right )}{d g -e f}+\frac {\ln \left (g x +f \right )}{d g -e f}\)	\(37\)
risch	\(\frac {\ln \left (-g x -f \right )}{d g -e f}-\frac {\ln \left (e x +d \right )}{d g -e f}\)	\(40\)

[In]

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

[Out]

-(ln(e*x+d)-ln(g*x+f))/(d*g-e*f)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.72 \[ \int \frac {1}{d f+(e f+d g) x+e g x^2} \, dx=\frac {\log \left (e x + d\right ) - \log \left (g x + f\right )}{e f - d g} \]

[In]

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

[Out]

(log(e*x + d) - log(g*x + f))/(e*f - d*g)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (26) = 52\).

Time = 0.18 (sec) , antiderivative size = 128, normalized size of antiderivative = 3.56 \[ \int \frac {1}{d f+(e f+d g) x+e g x^2} \, dx=\frac {\log {\left (x + \frac {- \frac {d^{2} g^{2}}{d g - e f} + \frac {2 d e f g}{d g - e f} + d g - \frac {e^{2} f^{2}}{d g - e f} + e f}{2 e g} \right )}}{d g - e f} - \frac {\log {\left (x + \frac {\frac {d^{2} g^{2}}{d g - e f} - \frac {2 d e f g}{d g - e f} + d g + \frac {e^{2} f^{2}}{d g - e f} + e f}{2 e g} \right )}}{d g - e f} \]

[In]

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

[Out]

log(x + (-d**2*g**2/(d*g - e*f) + 2*d*e*f*g/(d*g - e*f) + d*g - e**2*f**2/(d*g - e*f) + e*f)/(2*e*g))/(d*g - e
*f) - log(x + (d**2*g**2/(d*g - e*f) - 2*d*e*f*g/(d*g - e*f) + d*g + e**2*f**2/(d*g - e*f) + e*f)/(2*e*g))/(d*
g - e*f)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {1}{d f+(e f+d g) x+e g x^2} \, dx=\frac {\log \left (e x + d\right )}{e f - d g} - \frac {\log \left (g x + f\right )}{e f - d g} \]

[In]

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

[Out]

log(e*x + d)/(e*f - d*g) - log(g*x + f)/(e*f - d*g)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {1}{d f+(e f+d g) x+e g x^2} \, dx=\frac {e \log \left ({\left | e x + d \right |}\right )}{e^{2} f - d e g} - \frac {g \log \left ({\left | g x + f \right |}\right )}{e f g - d g^{2}} \]

[In]

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

[Out]

e*log(abs(e*x + d))/(e^2*f - d*e*g) - g*log(abs(g*x + f))/(e*f*g - d*g^2)

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {1}{d f+(e f+d g) x+e g x^2} \, dx=\frac {\mathrm {atan}\left (\frac {e\,f\,2{}\mathrm {i}+e\,g\,x\,2{}\mathrm {i}}{d\,g-e\,f}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d\,g-e\,f} \]

[In]

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

[Out]

(atan((e*f*2i + e*g*x*2i)/(d*g - e*f) + 1i)*2i)/(d*g - e*f)