3.18.7 \(\int \frac {a+b x}{(d+e x) (a^2+2 a b x+b^2 x^2)} \, dx\)

Optimal. Leaf size=36 \[ \frac {\log (a+b x)}{b d-a e}-\frac {\log (d+e x)}{b d-a e} \]

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Rubi [A]  time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {27, 36, 31} \begin {gather*} \frac {\log (a+b x)}{b d-a e}-\frac {\log (d+e x)}{b d-a e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[a + b*x]/(b*d - a*e) - Log[d + e*x]/(b*d - a*e)

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]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.72 \begin {gather*} \frac {\log (a+b x)-\log (d+e x)}{b d-a e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[a + b*x] - Log[d + e*x])/(b*d - a*e)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b x}{(d+e x) \left (a^2+2 a b x+b^2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)/((d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

IntegrateAlgebraic[(a + b*x)/((d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)), x]

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fricas [A]  time = 0.40, size = 26, normalized size = 0.72 \begin {gather*} \frac {\log \left (b x + a\right ) - \log \left (e x + d\right )}{b d - a e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(b*x + a) - log(e*x + d))/(b*d - a*e)

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giac [A]  time = 0.16, size = 49, normalized size = 1.36 \begin {gather*} \frac {b \log \left ({\left | b x + a \right |}\right )}{b^{2} d - a b e} - \frac {e \log \left ({\left | x e + d \right |}\right )}{b d e - a e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(abs(b*x + a))/(b^2*d - a*b*e) - e*log(abs(x*e + d))/(b*d*e - a*e^2)

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maple [A]  time = 0.05, size = 37, normalized size = 1.03 \begin {gather*} -\frac {\ln \left (b x +a \right )}{a e -b d}+\frac {\ln \left (e x +d \right )}{a e -b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*e-b*d)*ln(b*x+a)+1/(a*e-b*d)*ln(e*x+d)

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maxima [A]  time = 0.62, size = 36, normalized size = 1.00 \begin {gather*} \frac {\log \left (b x + a\right )}{b d - a e} - \frac {\log \left (e x + d\right )}{b d - a e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b*x + a)/(b*d - a*e) - log(e*x + d)/(b*d - a*e)

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mupad [B]  time = 0.08, size = 40, normalized size = 1.11 \begin {gather*} \frac {\mathrm {atan}\left (\frac {b\,d\,2{}\mathrm {i}+b\,e\,x\,2{}\mathrm {i}}{a\,e-b\,d}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{a\,e-b\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((b*d*2i + b*e*x*2i)/(a*e - b*d) + 1i)*2i)/(a*e - b*d)

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sympy [B]  time = 0.38, size = 128, normalized size = 3.56 \begin {gather*} \frac {\log {\left (x + \frac {- \frac {a^{2} e^{2}}{a e - b d} + \frac {2 a b d e}{a e - b d} + a e - \frac {b^{2} d^{2}}{a e - b d} + b d}{2 b e} \right )}}{a e - b d} - \frac {\log {\left (x + \frac {\frac {a^{2} e^{2}}{a e - b d} - \frac {2 a b d e}{a e - b d} + a e + \frac {b^{2} d^{2}}{a e - b d} + b d}{2 b e} \right )}}{a e - b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + (-a**2*e**2/(a*e - b*d) + 2*a*b*d*e/(a*e - b*d) + a*e - b**2*d**2/(a*e - b*d) + b*d)/(2*b*e))/(a*e - b
*d) - log(x + (a**2*e**2/(a*e - b*d) - 2*a*b*d*e/(a*e - b*d) + a*e + b**2*d**2/(a*e - b*d) + b*d)/(2*b*e))/(a*
e - b*d)

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