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

Optimal. Leaf size=46 \[ \frac{\left (\frac{c d}{a}-e\right ) \log (a+c x)}{2 c^2}-\frac{\left (\frac{c d}{a}+e\right ) \log (a-c x)}{2 c^2} \]

[Out]

-(((c*d)/a + e)*Log[a - c*x])/(2*c^2) + (((c*d)/a - e)*Log[a + c*x])/(2*c^2)

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Rubi [A]  time = 0.0231256, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {633, 31} \[ \frac{\left (\frac{c d}{a}-e\right ) \log (a+c x)}{2 c^2}-\frac{\left (\frac{c d}{a}+e\right ) \log (a-c x)}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(a^2 - c^2*x^2),x]

[Out]

-(((c*d)/a + e)*Log[a - c*x])/(2*c^2) + (((c*d)/a - e)*Log[a + c*x])/(2*c^2)

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

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

Mathematica [A]  time = 0.0069652, size = 37, normalized size = 0.8 \[ \frac{d \tanh ^{-1}\left (\frac{c x}{a}\right )}{a c}-\frac{e \log \left (a^2-c^2 x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*ArcTanh[(c*x)/a])/(a*c) - (e*Log[a^2 - c^2*x^2])/(2*c^2)

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Maple [A]  time = 0.006, size = 60, normalized size = 1.3 \begin{align*} -{\frac{\ln \left ( cx+a \right ) e}{2\,{c}^{2}}}+{\frac{\ln \left ( cx+a \right ) d}{2\,ac}}-{\frac{\ln \left ( cx-a \right ) e}{2\,{c}^{2}}}-{\frac{\ln \left ( cx-a \right ) d}{2\,ac}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.997951, size = 62, normalized size = 1.35 \begin{align*} \frac{{\left (c d - a e\right )} \log \left (c x + a\right )}{2 \, a c^{2}} - \frac{{\left (c d + a e\right )} \log \left (c x - a\right )}{2 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.47897, size = 90, normalized size = 1.96 \begin{align*} \frac{{\left (c d - a e\right )} \log \left (c x + a\right ) -{\left (c d + a e\right )} \log \left (c x - a\right )}{2 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.31694, size = 71, normalized size = 1.54 \begin{align*} - \frac{\left (a e - c d\right ) \log{\left (x + \frac{a^{2} e - a \left (a e - c d\right )}{c^{2} d} \right )}}{2 a c^{2}} - \frac{\left (a e + c d\right ) \log{\left (x + \frac{a^{2} e - a \left (a e + c d\right )}{c^{2} d} \right )}}{2 a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14586, size = 68, normalized size = 1.48 \begin{align*} \frac{{\left (c d - a e\right )} \log \left ({\left | c x + a \right |}\right )}{2 \, a c^{2}} - \frac{{\left (c d + a e\right )} \log \left ({\left | c x - a \right |}\right )}{2 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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