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

Optimal. Leaf size=84 \[ \frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac{b (c d-e)^2 \log (c x+1)}{4 c^2 e}+\frac{b (c d+e)^2 \log (1-c x)}{4 c^2 e}+\frac{b e x}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.0762702, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5926, 702, 633, 31} \[ \frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac{b (c d-e)^2 \log (c x+1)}{4 c^2 e}+\frac{b (c d+e)^2 \log (1-c x)}{4 c^2 e}+\frac{b e x}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcTanh[c*x]),x]

[Out]

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

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]

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 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 (d+e x) \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac{(b c) \int \frac{(d+e x)^2}{1-c^2 x^2} \, dx}{2 e}\\ &=\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac{(b c) \int \left (-\frac{e^2}{c^2}+\frac{c^2 d^2+e^2+2 c^2 d e x}{c^2 \left (1-c^2 x^2\right )}\right ) \, dx}{2 e}\\ &=\frac{b e x}{2 c}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac{b \int \frac{c^2 d^2+e^2+2 c^2 d e x}{1-c^2 x^2} \, dx}{2 c e}\\ &=\frac{b e x}{2 c}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}+\frac{\left (b (c d-e)^2\right ) \int \frac{1}{-c-c^2 x} \, dx}{4 e}-\frac{\left (b (c d+e)^2\right ) \int \frac{1}{c-c^2 x} \, dx}{4 e}\\ &=\frac{b e x}{2 c}+\frac{(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}+\frac{b (c d+e)^2 \log (1-c x)}{4 c^2 e}-\frac{b (c d-e)^2 \log (1+c x)}{4 c^2 e}\\ \end{align*}

Mathematica [A]  time = 0.0096586, size = 96, normalized size = 1.14 \[ a d x+\frac{1}{2} a e x^2+\frac{b d \log \left (1-c^2 x^2\right )}{2 c}+\frac{b e \log (1-c x)}{4 c^2}-\frac{b e \log (c x+1)}{4 c^2}+b d x \tanh ^{-1}(c x)+\frac{1}{2} b e x^2 \tanh ^{-1}(c x)+\frac{b e x}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.029, size = 92, normalized size = 1.1 \begin{align*}{\frac{a{x}^{2}e}{2}}+adx+{\frac{b{\it Artanh} \left ( cx \right ){x}^{2}e}{2}}+b{\it Artanh} \left ( cx \right ) xd+{\frac{bex}{2\,c}}+{\frac{b\ln \left ( cx-1 \right ) d}{2\,c}}+{\frac{b\ln \left ( cx-1 \right ) e}{4\,{c}^{2}}}+{\frac{b\ln \left ( cx+1 \right ) d}{2\,c}}-{\frac{b\ln \left ( cx+1 \right ) e}{4\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.972936, size = 112, normalized size = 1.33 \begin{align*} \frac{1}{2} \, a e 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 e + 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.64868, size = 228, normalized size = 2.71 \begin{align*} \frac{2 \, a c^{2} e x^{2} + 2 \,{\left (2 \, a c^{2} d + b c e\right )} x +{\left (2 \, b c d - b e\right )} \log \left (c x + 1\right ) +{\left (2 \, b c d + b e\right )} \log \left (c x - 1\right ) +{\left (b c^{2} e x^{2} + 2 \, b c^{2} d x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.18385, size = 92, normalized size = 1.1 \begin{align*} \begin{cases} a d x + \frac{a e x^{2}}{2} + b d x \operatorname{atanh}{\left (c x \right )} + \frac{b e x^{2} \operatorname{atanh}{\left (c x \right )}}{2} + \frac{b d \log{\left (x - \frac{1}{c} \right )}}{c} + \frac{b d \operatorname{atanh}{\left (c x \right )}}{c} + \frac{b e x}{2 c} - \frac{b e \operatorname{atanh}{\left (c x \right )}}{2 c^{2}} & \text{for}\: c \neq 0 \\a \left (d x + \frac{e x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16253, size = 157, normalized size = 1.87 \begin{align*} \frac{b c^{2} x^{2} e \log \left (-\frac{c x + 1}{c x - 1}\right ) + 2 \, a c^{2} x^{2} e + 2 \, b c^{2} d x \log \left (-\frac{c x + 1}{c x - 1}\right ) + 4 \, a c^{2} d x + 2 \, b c x e + 2 \, b c d \log \left (c^{2} x^{2} - 1\right ) - b e \log \left (c x + 1\right ) + b e \log \left (c x - 1\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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