3.2 \(\int (d+e x)^3 (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=125 \[ \frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 e}+\frac{b e x \left (6 c^2 d^2+e^2\right )}{4 c^3}-\frac{b (c d-e)^4 \log (c x+1)}{8 c^4 e}+\frac{b (c d+e)^4 \log (1-c x)}{8 c^4 e}+\frac{b d e^2 x^2}{2 c}+\frac{b e^3 x^3}{12 c} \]

[Out]

(b*e*(6*c^2*d^2 + e^2)*x)/(4*c^3) + (b*d*e^2*x^2)/(2*c) + (b*e^3*x^3)/(12*c) + ((d + e*x)^4*(a + b*ArcTanh[c*x
]))/(4*e) + (b*(c*d + e)^4*Log[1 - c*x])/(8*c^4*e) - (b*(c*d - e)^4*Log[1 + c*x])/(8*c^4*e)

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Rubi [A]  time = 0.141063, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5926, 702, 633, 31} \[ \frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 e}+\frac{b e x \left (6 c^2 d^2+e^2\right )}{4 c^3}-\frac{b (c d-e)^4 \log (c x+1)}{8 c^4 e}+\frac{b (c d+e)^4 \log (1-c x)}{8 c^4 e}+\frac{b d e^2 x^2}{2 c}+\frac{b e^3 x^3}{12 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(a + b*ArcTanh[c*x]),x]

[Out]

(b*e*(6*c^2*d^2 + e^2)*x)/(4*c^3) + (b*d*e^2*x^2)/(2*c) + (b*e^3*x^3)/(12*c) + ((d + e*x)^4*(a + b*ArcTanh[c*x
]))/(4*e) + (b*(c*d + e)^4*Log[1 - c*x])/(8*c^4*e) - (b*(c*d - e)^4*Log[1 + c*x])/(8*c^4*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)^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 e}-\frac{(b c) \int \frac{(d+e x)^4}{1-c^2 x^2} \, dx}{4 e}\\ &=\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 e}-\frac{(b c) \int \left (-\frac{e^2 \left (6 c^2 d^2+e^2\right )}{c^4}-\frac{4 d e^3 x}{c^2}-\frac{e^4 x^2}{c^2}+\frac{c^4 d^4+6 c^2 d^2 e^2+e^4+4 c^2 d e \left (c^2 d^2+e^2\right ) x}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx}{4 e}\\ &=\frac{b e \left (6 c^2 d^2+e^2\right ) x}{4 c^3}+\frac{b d e^2 x^2}{2 c}+\frac{b e^3 x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 e}-\frac{b \int \frac{c^4 d^4+6 c^2 d^2 e^2+e^4+4 c^2 d e \left (c^2 d^2+e^2\right ) x}{1-c^2 x^2} \, dx}{4 c^3 e}\\ &=\frac{b e \left (6 c^2 d^2+e^2\right ) x}{4 c^3}+\frac{b d e^2 x^2}{2 c}+\frac{b e^3 x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 e}+\frac{\left (b (c d-e)^4\right ) \int \frac{1}{-c-c^2 x} \, dx}{8 c^2 e}-\frac{\left (b (c d+e)^4\right ) \int \frac{1}{c-c^2 x} \, dx}{8 c^2 e}\\ &=\frac{b e \left (6 c^2 d^2+e^2\right ) x}{4 c^3}+\frac{b d e^2 x^2}{2 c}+\frac{b e^3 x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 e}+\frac{b (c d+e)^4 \log (1-c x)}{8 c^4 e}-\frac{b (c d-e)^4 \log (1+c x)}{8 c^4 e}\\ \end{align*}

Mathematica [A]  time = 0.131718, size = 205, normalized size = 1.64 \[ \frac{6 c x \left (4 a c^3 d^3+b e \left (6 c^2 d^2+e^2\right )\right )+2 c^3 e^2 x^3 (12 a c d+b e)+12 c^3 d e x^2 (3 a c d+b e)+6 a c^4 e^3 x^4+6 b c^4 x \tanh ^{-1}(c x) \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+3 b \left (6 c^2 d^2 e+4 c^3 d^3+4 c d e^2+e^3\right ) \log (1-c x)+3 b \left (-6 c^2 d^2 e+4 c^3 d^3+4 c d e^2-e^3\right ) \log (c x+1)}{24 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*(a + b*ArcTanh[c*x]),x]

[Out]

(6*c*(4*a*c^3*d^3 + b*e*(6*c^2*d^2 + e^2))*x + 12*c^3*d*e*(3*a*c*d + b*e)*x^2 + 2*c^3*e^2*(12*a*c*d + b*e)*x^3
+ 6*a*c^4*e^3*x^4 + 6*b*c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3)*ArcTanh[c*x] + 3*b*(4*c^3*d^3 + 6*c
^2*d^2*e + 4*c*d*e^2 + e^3)*Log[1 - c*x] + 3*b*(4*c^3*d^3 - 6*c^2*d^2*e + 4*c*d*e^2 - e^3)*Log[1 + c*x])/(24*c
^4)

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Maple [B]  time = 0.03, size = 308, normalized size = 2.5 \begin{align*}{\frac{a{e}^{3}{x}^{4}}{4}}+a{e}^{2}{x}^{3}d+{\frac{3\,ae{x}^{2}{d}^{2}}{2}}+ax{d}^{3}+{\frac{a{d}^{4}}{4\,e}}+{\frac{b{e}^{3}{\it Artanh} \left ( cx \right ){x}^{4}}{4}}+b{e}^{2}{\it Artanh} \left ( cx \right ){x}^{3}d+{\frac{3\,be{\it Artanh} \left ( cx \right ){x}^{2}{d}^{2}}{2}}+b{\it Artanh} \left ( cx \right ) x{d}^{3}+{\frac{b{\it Artanh} \left ( cx \right ){d}^{4}}{4\,e}}+{\frac{b{e}^{3}{x}^{3}}{12\,c}}+{\frac{b{e}^{2}d{x}^{2}}{2\,c}}+{\frac{3\,bex{d}^{2}}{2\,c}}+{\frac{bx{e}^{3}}{4\,{c}^{3}}}+{\frac{b\ln \left ( cx-1 \right ){d}^{4}}{8\,e}}+{\frac{b\ln \left ( cx-1 \right ){d}^{3}}{2\,c}}+{\frac{3\,be\ln \left ( cx-1 \right ){d}^{2}}{4\,{c}^{2}}}+{\frac{b{e}^{2}\ln \left ( cx-1 \right ) d}{2\,{c}^{3}}}+{\frac{b{e}^{3}\ln \left ( cx-1 \right ) }{8\,{c}^{4}}}-{\frac{b\ln \left ( cx+1 \right ){d}^{4}}{8\,e}}+{\frac{b\ln \left ( cx+1 \right ){d}^{3}}{2\,c}}-{\frac{3\,be\ln \left ( cx+1 \right ){d}^{2}}{4\,{c}^{2}}}+{\frac{b{e}^{2}\ln \left ( cx+1 \right ) d}{2\,{c}^{3}}}-{\frac{b{e}^{3}\ln \left ( cx+1 \right ) }{8\,{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(a+b*arctanh(c*x)),x)

[Out]

1/4*a*e^3*x^4+a*e^2*x^3*d+3/2*a*e*x^2*d^2+a*x*d^3+1/4*a/e*d^4+1/4*b*e^3*arctanh(c*x)*x^4+b*e^2*arctanh(c*x)*x^
3*d+3/2*b*e*arctanh(c*x)*x^2*d^2+b*arctanh(c*x)*x*d^3+1/4*b/e*arctanh(c*x)*d^4+1/12*b*e^3*x^3/c+1/2*b*d*e^2*x^
2/c+3/2*b/c*e*x*d^2+1/4*b/c^3*x*e^3+1/8*b/e*ln(c*x-1)*d^4+1/2/c*b*ln(c*x-1)*d^3+3/4/c^2*b*e*ln(c*x-1)*d^2+1/2/
c^3*b*e^2*ln(c*x-1)*d+1/8/c^4*b*e^3*ln(c*x-1)-1/8*b/e*ln(c*x+1)*d^4+1/2/c*b*ln(c*x+1)*d^3-3/4/c^2*b*e*ln(c*x+1
)*d^2+1/2/c^3*b*e^2*ln(c*x+1)*d-1/8/c^4*b*e^3*ln(c*x+1)

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Maxima [A]  time = 0.966265, size = 282, normalized size = 2.26 \begin{align*} \frac{1}{4} \, a e^{3} x^{4} + a d e^{2} x^{3} + \frac{3}{2} \, a d^{2} e x^{2} + \frac{3}{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 d^{2} e + \frac{1}{2} \,{\left (2 \, x^{3} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{x^{2}}{c^{2}} + \frac{\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d e^{2} + \frac{1}{24} \,{\left (6 \, x^{4} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \,{\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac{3 \, \log \left (c x + 1\right )}{c^{5}} + \frac{3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b e^{3} + a d^{3} x + \frac{{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{3}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/4*a*e^3*x^4 + a*d*e^2*x^3 + 3/2*a*d^2*e*x^2 + 3/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(
c*x - 1)/c^3))*b*d^2*e + 1/2*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*d*e^2 + 1/24*(6*x^4*a
rctanh(c*x) + c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*b*e^3 + a*d^3*x + 1/2*(2*c*
x*arctanh(c*x) + log(-c^2*x^2 + 1))*b*d^3/c

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Fricas [B]  time = 1.79442, size = 527, normalized size = 4.22 \begin{align*} \frac{6 \, a c^{4} e^{3} x^{4} + 2 \,{\left (12 \, a c^{4} d e^{2} + b c^{3} e^{3}\right )} x^{3} + 12 \,{\left (3 \, a c^{4} d^{2} e + b c^{3} d e^{2}\right )} x^{2} + 6 \,{\left (4 \, a c^{4} d^{3} + 6 \, b c^{3} d^{2} e + b c e^{3}\right )} x + 3 \,{\left (4 \, b c^{3} d^{3} - 6 \, b c^{2} d^{2} e + 4 \, b c d e^{2} - b e^{3}\right )} \log \left (c x + 1\right ) + 3 \,{\left (4 \, b c^{3} d^{3} + 6 \, b c^{2} d^{2} e + 4 \, b c d e^{2} + b e^{3}\right )} \log \left (c x - 1\right ) + 3 \,{\left (b c^{4} e^{3} x^{4} + 4 \, b c^{4} d e^{2} x^{3} + 6 \, b c^{4} d^{2} e x^{2} + 4 \, b c^{4} d^{3} x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{24 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/24*(6*a*c^4*e^3*x^4 + 2*(12*a*c^4*d*e^2 + b*c^3*e^3)*x^3 + 12*(3*a*c^4*d^2*e + b*c^3*d*e^2)*x^2 + 6*(4*a*c^4
*d^3 + 6*b*c^3*d^2*e + b*c*e^3)*x + 3*(4*b*c^3*d^3 - 6*b*c^2*d^2*e + 4*b*c*d*e^2 - b*e^3)*log(c*x + 1) + 3*(4*
b*c^3*d^3 + 6*b*c^2*d^2*e + 4*b*c*d*e^2 + b*e^3)*log(c*x - 1) + 3*(b*c^4*e^3*x^4 + 4*b*c^4*d*e^2*x^3 + 6*b*c^4
*d^2*e*x^2 + 4*b*c^4*d^3*x)*log(-(c*x + 1)/(c*x - 1)))/c^4

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Sympy [A]  time = 3.34995, size = 279, normalized size = 2.23 \begin{align*} \begin{cases} a d^{3} x + \frac{3 a d^{2} e x^{2}}{2} + a d e^{2} x^{3} + \frac{a e^{3} x^{4}}{4} + b d^{3} x \operatorname{atanh}{\left (c x \right )} + \frac{3 b d^{2} e x^{2} \operatorname{atanh}{\left (c x \right )}}{2} + b d e^{2} x^{3} \operatorname{atanh}{\left (c x \right )} + \frac{b e^{3} x^{4} \operatorname{atanh}{\left (c x \right )}}{4} + \frac{b d^{3} \log{\left (x - \frac{1}{c} \right )}}{c} + \frac{b d^{3} \operatorname{atanh}{\left (c x \right )}}{c} + \frac{3 b d^{2} e x}{2 c} + \frac{b d e^{2} x^{2}}{2 c} + \frac{b e^{3} x^{3}}{12 c} - \frac{3 b d^{2} e \operatorname{atanh}{\left (c x \right )}}{2 c^{2}} + \frac{b d e^{2} \log{\left (x - \frac{1}{c} \right )}}{c^{3}} + \frac{b d e^{2} \operatorname{atanh}{\left (c x \right )}}{c^{3}} + \frac{b e^{3} x}{4 c^{3}} - \frac{b e^{3} \operatorname{atanh}{\left (c x \right )}}{4 c^{4}} & \text{for}\: c \neq 0 \\a \left (d^{3} x + \frac{3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac{e^{3} x^{4}}{4}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(a+b*atanh(c*x)),x)

[Out]

Piecewise((a*d**3*x + 3*a*d**2*e*x**2/2 + a*d*e**2*x**3 + a*e**3*x**4/4 + b*d**3*x*atanh(c*x) + 3*b*d**2*e*x**
2*atanh(c*x)/2 + b*d*e**2*x**3*atanh(c*x) + b*e**3*x**4*atanh(c*x)/4 + b*d**3*log(x - 1/c)/c + b*d**3*atanh(c*
x)/c + 3*b*d**2*e*x/(2*c) + b*d*e**2*x**2/(2*c) + b*e**3*x**3/(12*c) - 3*b*d**2*e*atanh(c*x)/(2*c**2) + b*d*e*
*2*log(x - 1/c)/c**3 + b*d*e**2*atanh(c*x)/c**3 + b*e**3*x/(4*c**3) - b*e**3*atanh(c*x)/(4*c**4), Ne(c, 0)), (
a*(d**3*x + 3*d**2*e*x**2/2 + d*e**2*x**3 + e**3*x**4/4), True))

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Giac [B]  time = 1.33147, size = 397, normalized size = 3.18 \begin{align*} \frac{3 \, b c^{4} x^{4} e^{3} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 12 \, b c^{4} d x^{3} e^{2} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 18 \, b c^{4} d^{2} x^{2} e \log \left (-\frac{c x + 1}{c x - 1}\right ) + 6 \, a c^{4} x^{4} e^{3} + 24 \, a c^{4} d x^{3} e^{2} + 36 \, a c^{4} d^{2} x^{2} e + 12 \, b c^{4} d^{3} x \log \left (-\frac{c x + 1}{c x - 1}\right ) + 24 \, a c^{4} d^{3} x + 2 \, b c^{3} x^{3} e^{3} + 12 \, b c^{3} d x^{2} e^{2} + 36 \, b c^{3} d^{2} x e + 12 \, b c^{3} d^{3} \log \left (c^{2} x^{2} - 1\right ) - 18 \, b c^{2} d^{2} e \log \left (c x + 1\right ) + 18 \, b c^{2} d^{2} e \log \left (c x - 1\right ) + 12 \, b c d e^{2} \log \left (c^{2} x^{2} - 1\right ) + 6 \, b c x e^{3} - 3 \, b e^{3} \log \left (c x + 1\right ) + 3 \, b e^{3} \log \left (c x - 1\right )}{24 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

1/24*(3*b*c^4*x^4*e^3*log(-(c*x + 1)/(c*x - 1)) + 12*b*c^4*d*x^3*e^2*log(-(c*x + 1)/(c*x - 1)) + 18*b*c^4*d^2*
x^2*e*log(-(c*x + 1)/(c*x - 1)) + 6*a*c^4*x^4*e^3 + 24*a*c^4*d*x^3*e^2 + 36*a*c^4*d^2*x^2*e + 12*b*c^4*d^3*x*l
og(-(c*x + 1)/(c*x - 1)) + 24*a*c^4*d^3*x + 2*b*c^3*x^3*e^3 + 12*b*c^3*d*x^2*e^2 + 36*b*c^3*d^2*x*e + 12*b*c^3
*d^3*log(c^2*x^2 - 1) - 18*b*c^2*d^2*e*log(c*x + 1) + 18*b*c^2*d^2*e*log(c*x - 1) + 12*b*c*d*e^2*log(c^2*x^2 -
1) + 6*b*c*x*e^3 - 3*b*e^3*log(c*x + 1) + 3*b*e^3*log(c*x - 1))/c^4