3.34 \(\int (a+b \tanh ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=40 \[ a x+\frac {b \log \left (1-(c+d x)^2\right )}{2 d}+\frac {b (c+d x) \tanh ^{-1}(c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6103, 5910, 260} \[ a x+\frac {b \log \left (1-(c+d x)^2\right )}{2 d}+\frac {b (c+d x) \tanh ^{-1}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcTanh[c + d*x],x]

[Out]

a*x + (b*(c + d*x)*ArcTanh[c + d*x])/d + (b*Log[1 - (c + d*x)^2])/(2*d)

Rule 260

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

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 6103

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcTanh[x])^p, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 48, normalized size = 1.20 \[ a x+\frac {b ((c+1) \log (c+d x+1)-(c-1) \log (-c-d x+1))}{2 d}+b x \tanh ^{-1}(c+d x) \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcTanh[c + d*x],x]

[Out]

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

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fricas [A]  time = 0.46, size = 61, normalized size = 1.52 \[ \frac {b d x \log \left (-\frac {d x + c + 1}{d x + c - 1}\right ) + 2 \, a d x + {\left (b c + b\right )} \log \left (d x + c + 1\right ) - {\left (b c - b\right )} \log \left (d x + c - 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*arctanh(d*x+c),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.52, size = 200, normalized size = 5.00 \[ \frac {1}{2} \, {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )} b {\left (\frac {\log \left (\frac {{\left | -d x - c - 1 \right |}}{{\left | d x + c - 1 \right |}}\right )}{d^{2}} - \frac {\log \left ({\left | -\frac {d x + c + 1}{d x + c - 1} + 1 \right |}\right )}{d^{2}} + \frac {\log \left (-\frac {c - \frac {{\left (\frac {{\left (d x + c + 1\right )} {\left (c - 1\right )}}{d x + c - 1} - c - 1\right )} d}{\frac {{\left (d x + c + 1\right )} d}{d x + c - 1} - d} + 1}{c - \frac {{\left (\frac {{\left (d x + c + 1\right )} {\left (c - 1\right )}}{d x + c - 1} - c - 1\right )} d}{\frac {{\left (d x + c + 1\right )} d}{d x + c - 1} - d} - 1}\right )}{d^{2} {\left (\frac {d x + c + 1}{d x + c - 1} - 1\right )}}\right )} + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*arctanh(d*x+c),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 44, normalized size = 1.10 \[ a x +b \arctanh \left (d x +c \right ) x +\frac {b \arctanh \left (d x +c \right ) c}{d}+\frac {b \ln \left (1-\left (d x +c \right )^{2}\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a+b*arctanh(d*x+c),x)

[Out]

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

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maxima [A]  time = 0.32, size = 36, normalized size = 0.90 \[ a x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*arctanh(d*x+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.44, size = 48, normalized size = 1.20 \[ a\,x+\frac {\frac {b\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2-1\right )}{2}+b\,c\,\mathrm {atanh}\left (c+d\,x\right )}{d}+b\,x\,\mathrm {atanh}\left (c+d\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*atanh(c + d*x),x)

[Out]

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

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sympy [A]  time = 0.60, size = 46, normalized size = 1.15 \[ a x + b \left (\begin {cases} \frac {c \operatorname {atanh}{\left (c + d x \right )}}{d} + x \operatorname {atanh}{\left (c + d x \right )} + \frac {\log {\left (c + d x + 1 \right )}}{d} - \frac {\operatorname {atanh}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \operatorname {atanh}{\relax (c )} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*atanh(d*x+c),x)

[Out]

a*x + b*Piecewise((c*atanh(c + d*x)/d + x*atanh(c + d*x) + log(c + d*x + 1)/d - atanh(c + d*x)/d, Ne(d, 0)), (
x*atanh(c), True))

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