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

Optimal. Leaf size=44 \[ \frac {d (c x+1)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c}+\frac {b d \log (1-c x)}{c}+\frac {b d x}{2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5926, 627, 43} \[ \frac {d (c x+1)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c}+\frac {b d \log (1-c x)}{c}+\frac {b d x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(d + c*d*x)*(a + b*ArcTanh[c*x]),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 5926

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcTanh[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rubi steps

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

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Mathematica [B]  time = 0.01, size = 95, normalized size = 2.16 \[ \frac {1}{2} a c d x^2+a d x+\frac {b d \log \left (1-c^2 x^2\right )}{2 c}+\frac {1}{2} b c d x^2 \tanh ^{-1}(c x)+\frac {b d \log (1-c x)}{4 c}-\frac {b d \log (c x+1)}{4 c}+b d x \tanh ^{-1}(c x)+\frac {b d x}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + c*d*x)*(a + b*ArcTanh[c*x]),x]

[Out]

a*d*x + (b*d*x)/2 + (a*c*d*x^2)/2 + b*d*x*ArcTanh[c*x] + (b*c*d*x^2*ArcTanh[c*x])/2 + (b*d*Log[1 - c*x])/(4*c)
- (b*d*Log[1 + c*x])/(4*c) + (b*d*Log[1 - c^2*x^2])/(2*c)

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fricas [A]  time = 0.59, size = 77, normalized size = 1.75 \[ \frac {2 \, a c^{2} d x^{2} + 2 \, {\left (2 \, a + b\right )} c d x + b d \log \left (c x + 1\right ) + 3 \, b d \log \left (c x - 1\right ) + {\left (b c^{2} d x^{2} + 2 \, b c d x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*c^2*d*x^2 + 2*(2*a + b)*c*d*x + b*d*log(c*x + 1) + 3*b*d*log(c*x - 1) + (b*c^2*d*x^2 + 2*b*c*d*x)*log
(-(c*x + 1)/(c*x - 1)))/c

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giac [B]  time = 0.18, size = 211, normalized size = 4.80 \[ -c {\left (\frac {b d \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{2}} - \frac {{\left (\frac {2 \, {\left (c x + 1\right )} b d}{c x - 1} - b d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} - \frac {2 \, {\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}} - \frac {b d \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{2}} - \frac {\frac {4 \, {\left (c x + 1\right )} a d}{c x - 1} - 2 \, a d + \frac {{\left (c x + 1\right )} b d}{c x - 1} - b d}{\frac {{\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} - \frac {2 \, {\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*(b*d*log(-(c*x + 1)/(c*x - 1) + 1)/c^2 - (2*(c*x + 1)*b*d/(c*x - 1) - b*d)*log(-(c*x + 1)/(c*x - 1))/((c*x
+ 1)^2*c^2/(c*x - 1)^2 - 2*(c*x + 1)*c^2/(c*x - 1) + c^2) - b*d*log(-(c*x + 1)/(c*x - 1))/c^2 - (4*(c*x + 1)*a
*d/(c*x - 1) - 2*a*d + (c*x + 1)*b*d/(c*x - 1) - b*d)/((c*x + 1)^2*c^2/(c*x - 1)^2 - 2*(c*x + 1)*c^2/(c*x - 1)
+ c^2))

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maple [A]  time = 0.03, size = 65, normalized size = 1.48 \[ \frac {c d a \,x^{2}}{2}+a d x +\frac {c d b \arctanh \left (c x \right ) x^{2}}{2}+d b \arctanh \left (c x \right ) x +\frac {b d x}{2}+\frac {3 d b \ln \left (c x -1\right )}{4 c}+\frac {d b \ln \left (c x +1\right )}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c*d*a*x^2+a*d*x+1/2*c*d*b*arctanh(c*x)*x^2+d*b*arctanh(c*x)*x+1/2*b*d*x+3/4/c*d*b*ln(c*x-1)+1/4/c*d*b*ln(c
*x+1)

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maxima [B]  time = 0.31, size = 85, normalized size = 1.93 \[ \frac {1}{2} \, a c d x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b c d + a d x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*c*d*x^2 + 1/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*b*c*d + a*d*x + 1
/2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*b*d/c

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mupad [B]  time = 0.86, size = 65, normalized size = 1.48 \[ \frac {d\,\left (2\,a\,x+b\,x+2\,b\,x\,\mathrm {atanh}\left (c\,x\right )\right )}{2}+\frac {c\,d\,\left (a\,x^2+b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{2}-\frac {d\,\left (b\,\mathrm {atanh}\left (c\,x\right )-b\,\ln \left (c^2\,x^2-1\right )\right )}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))*(d + c*d*x),x)

[Out]

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

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sympy [A]  time = 0.58, size = 75, normalized size = 1.70 \[ \begin {cases} \frac {a c d x^{2}}{2} + a d x + \frac {b c d x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + b d x \operatorname {atanh}{\left (c x \right )} + \frac {b d x}{2} + \frac {b d \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b d \operatorname {atanh}{\left (c x \right )}}{2 c} & \text {for}\: c \neq 0 \\a d x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c*d*x**2/2 + a*d*x + b*c*d*x**2*atanh(c*x)/2 + b*d*x*atanh(c*x) + b*d*x/2 + b*d*log(x - 1/c)/c +
b*d*atanh(c*x)/(2*c), Ne(c, 0)), (a*d*x, True))

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