3.92 \(\int \frac {1}{1-(c+d x)^2} \, dx\)

Optimal. Leaf size=10 \[ \frac {\tanh ^{-1}(c+d x)}{d} \]

[Out]

arctanh(d*x+c)/d

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Rubi [A]  time = 0.00, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {247, 206} \[ \frac {\tanh ^{-1}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(1 - (c + d*x)^2)^(-1),x]

[Out]

ArcTanh[c + d*x]/d

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps

\begin {align*} \int \frac {1}{1-(c+d x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\tanh ^{-1}(c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 0.01, size = 32, normalized size = 3.20 \[ \frac {\log (c+d x+1)}{2 d}-\frac {\log (-c-d x+1)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - (c + d*x)^2)^(-1),x]

[Out]

-1/2*Log[1 - c - d*x]/d + Log[1 + c + d*x]/(2*d)

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fricas [B]  time = 0.79, size = 22, normalized size = 2.20 \[ \frac {\log \left (d x + c + 1\right ) - \log \left (d x + c - 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.38, size = 27, normalized size = 2.70 \[ \frac {\log \left ({\left | d x + c + 1 \right |}\right )}{2 \, d} - \frac {\log \left ({\left | d x + c - 1 \right |}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-(d*x+c)^2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.01, size = 26, normalized size = 2.60 \[ -\frac {\ln \left (d x +c -1\right )}{2 d}+\frac {\ln \left (d x +c +1\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-(d*x+c)^2),x)

[Out]

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

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maxima [B]  time = 0.49, size = 25, normalized size = 2.50 \[ \frac {\log \left (d x + c + 1\right )}{2 \, d} - \frac {\log \left (d x + c - 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.05, size = 10, normalized size = 1.00 \[ \frac {\mathrm {atanh}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/((c + d*x)^2 - 1),x)

[Out]

atanh(c + d*x)/d

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sympy [B]  time = 0.18, size = 22, normalized size = 2.20 \[ - \frac {\frac {\log {\left (x + \frac {c - 1}{d} \right )}}{2} - \frac {\log {\left (x + \frac {c + 1}{d} \right )}}{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-(d*x+c)**2),x)

[Out]

-(log(x + (c - 1)/d)/2 - log(x + (c + 1)/d)/2)/d

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