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

Optimal. Leaf size=84 \[ \frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c}+\frac{b d^3 (c x+1)^3}{12 c}+\frac{b d^3 (c x+1)^2}{4 c}+\frac{2 b d^3 \log (1-c x)}{c}+b d^3 x \]

[Out]

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

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Rubi [A]  time = 0.051169, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {5926, 627, 43} \[ \frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c}+\frac{b d^3 (c x+1)^3}{12 c}+\frac{b d^3 (c x+1)^2}{4 c}+\frac{2 b d^3 \log (1-c x)}{c}+b d^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.144953, size = 115, normalized size = 1.37 \[ \frac{d^3 \left (6 a c^4 x^4+24 a c^3 x^3+36 a c^2 x^2+24 a c x+2 b c^3 x^3+12 b c^2 x^2+6 b c x \left (c^3 x^3+4 c^2 x^2+6 c x+4\right ) \tanh ^{-1}(c x)+42 b c x+45 b \log (1-c x)+3 b \log (c x+1)\right )}{24 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(24*a*c*x + 42*b*c*x + 36*a*c^2*x^2 + 12*b*c^2*x^2 + 24*a*c^3*x^3 + 2*b*c^3*x^3 + 6*a*c^4*x^4 + 6*b*c*x*(
4 + 6*c*x + 4*c^2*x^2 + c^3*x^3)*ArcTanh[c*x] + 45*b*Log[1 - c*x] + 3*b*Log[1 + c*x]))/(24*c)

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Maple [B]  time = 0.029, size = 162, normalized size = 1.9 \begin{align*}{\frac{{c}^{3}{x}^{4}a{d}^{3}}{4}}+{c}^{2}{x}^{3}a{d}^{3}+{\frac{3\,c{x}^{2}a{d}^{3}}{2}}+ax{d}^{3}+{\frac{{d}^{3}a}{4\,c}}+{\frac{{c}^{3}{d}^{3}b{\it Artanh} \left ( cx \right ){x}^{4}}{4}}+{c}^{2}{d}^{3}b{\it Artanh} \left ( cx \right ){x}^{3}+{\frac{3\,c{d}^{3}b{\it Artanh} \left ( cx \right ){x}^{2}}{2}}+{d}^{3}b{\it Artanh} \left ( cx \right ) x+{\frac{{d}^{3}b{\it Artanh} \left ( cx \right ) }{4\,c}}+{\frac{{c}^{2}{d}^{3}b{x}^{3}}{12}}+{\frac{c{d}^{3}b{x}^{2}}{2}}+{\frac{7\,b{d}^{3}x}{4}}+2\,{\frac{{d}^{3}b\ln \left ( cx-1 \right ) }{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c^3*x^4*a*d^3+c^2*x^3*a*d^3+3/2*c*x^2*a*d^3+a*x*d^3+1/4/c*d^3*a+1/4*c^3*d^3*b*arctanh(c*x)*x^4+c^2*d^3*b*a
rctanh(c*x)*x^3+3/2*c*d^3*b*arctanh(c*x)*x^2+d^3*b*arctanh(c*x)*x+1/4/c*d^3*b*arctanh(c*x)+1/12*c^2*d^3*b*x^3+
1/2*c*d^3*b*x^2+7/4*b*d^3*x+2/c*d^3*b*ln(c*x-1)

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Maxima [B]  time = 0.970664, size = 296, normalized size = 3.52 \begin{align*} \frac{1}{4} \, a c^{3} d^{3} x^{4} + a c^{2} d^{3} x^{3} + \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 c^{3} d^{3} + \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 c^{2} d^{3} + \frac{3}{2} \, a c d^{3} 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 c d^{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((c*d*x+d)^3*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/4*a*c^3*d^3*x^4 + a*c^2*d^3*x^3 + 1/24*(6*x^4*arctanh(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*c^3*d^3 + 1/2*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*c^2*d^3 + 3/
2*a*c*d^3*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*c*d^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 [A]  time = 2.01528, size = 333, normalized size = 3.96 \begin{align*} \frac{6 \, a c^{4} d^{3} x^{4} + 2 \,{\left (12 \, a + b\right )} c^{3} d^{3} x^{3} + 12 \,{\left (3 \, a + b\right )} c^{2} d^{3} x^{2} + 6 \,{\left (4 \, a + 7 \, b\right )} c d^{3} x + 3 \, b d^{3} \log \left (c x + 1\right ) + 45 \, b d^{3} \log \left (c x - 1\right ) + 3 \,{\left (b c^{4} d^{3} x^{4} + 4 \, b c^{3} d^{3} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c d^{3} x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{24 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*a*c^4*d^3*x^4 + 2*(12*a + b)*c^3*d^3*x^3 + 12*(3*a + b)*c^2*d^3*x^2 + 6*(4*a + 7*b)*c*d^3*x + 3*b*d^3*
log(c*x + 1) + 45*b*d^3*log(c*x - 1) + 3*(b*c^4*d^3*x^4 + 4*b*c^3*d^3*x^3 + 6*b*c^2*d^3*x^2 + 4*b*c*d^3*x)*log
(-(c*x + 1)/(c*x - 1)))/c

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Sympy [A]  time = 2.59779, size = 182, normalized size = 2.17 \begin{align*} \begin{cases} \frac{a c^{3} d^{3} x^{4}}{4} + a c^{2} d^{3} x^{3} + \frac{3 a c d^{3} x^{2}}{2} + a d^{3} x + \frac{b c^{3} d^{3} x^{4} \operatorname{atanh}{\left (c x \right )}}{4} + b c^{2} d^{3} x^{3} \operatorname{atanh}{\left (c x \right )} + \frac{b c^{2} d^{3} x^{3}}{12} + \frac{3 b c d^{3} x^{2} \operatorname{atanh}{\left (c x \right )}}{2} + \frac{b c d^{3} x^{2}}{2} + b d^{3} x \operatorname{atanh}{\left (c x \right )} + \frac{7 b d^{3} x}{4} + \frac{2 b d^{3} \log{\left (x - \frac{1}{c} \right )}}{c} + \frac{b d^{3} \operatorname{atanh}{\left (c x \right )}}{4 c} & \text{for}\: c \neq 0 \\a d^{3} x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**3*d**3*x**4/4 + a*c**2*d**3*x**3 + 3*a*c*d**3*x**2/2 + a*d**3*x + b*c**3*d**3*x**4*atanh(c*x)/
4 + b*c**2*d**3*x**3*atanh(c*x) + b*c**2*d**3*x**3/12 + 3*b*c*d**3*x**2*atanh(c*x)/2 + b*c*d**3*x**2/2 + b*d**
3*x*atanh(c*x) + 7*b*d**3*x/4 + 2*b*d**3*log(x - 1/c)/c + b*d**3*atanh(c*x)/(4*c), Ne(c, 0)), (a*d**3*x, True)
)

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Giac [B]  time = 1.2448, size = 215, normalized size = 2.56 \begin{align*} \frac{1}{4} \, a c^{3} d^{3} x^{4} + \frac{1}{12} \,{\left (12 \, a c^{2} d^{3} + b c^{2} d^{3}\right )} x^{3} + \frac{b d^{3} \log \left (c x + 1\right )}{8 \, c} + \frac{15 \, b d^{3} \log \left (c x - 1\right )}{8 \, c} + \frac{1}{2} \,{\left (3 \, a c d^{3} + b c d^{3}\right )} x^{2} + \frac{1}{4} \,{\left (4 \, a d^{3} + 7 \, b d^{3}\right )} x + \frac{1}{8} \,{\left (b c^{3} d^{3} x^{4} + 4 \, b c^{2} d^{3} x^{3} + 6 \, b c d^{3} x^{2} + 4 \, b d^{3} x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*c^3*d^3*x^4 + 1/12*(12*a*c^2*d^3 + b*c^2*d^3)*x^3 + 1/8*b*d^3*log(c*x + 1)/c + 15/8*b*d^3*log(c*x - 1)/c
+ 1/2*(3*a*c*d^3 + b*c*d^3)*x^2 + 1/4*(4*a*d^3 + 7*b*d^3)*x + 1/8*(b*c^3*d^3*x^4 + 4*b*c^2*d^3*x^3 + 6*b*c*d^
3*x^2 + 4*b*d^3*x)*log(-(c*x + 1)/(c*x - 1))