3.34 \(\int (d+c d x)^4 (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=107 \[ \frac{d^4 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}+\frac{b d^4 (c x+1)^4}{20 c}+\frac{2 b d^4 (c x+1)^3}{15 c}+\frac{2 b d^4 (c x+1)^2}{5 c}+\frac{16 b d^4 \log (1-c x)}{5 c}+\frac{8}{5} b d^4 x \]

[Out]

(8*b*d^4*x)/5 + (2*b*d^4*(1 + c*x)^2)/(5*c) + (2*b*d^4*(1 + c*x)^3)/(15*c) + (b*d^4*(1 + c*x)^4)/(20*c) + (d^4
*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/(5*c) + (16*b*d^4*Log[1 - c*x])/(5*c)

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Rubi [A]  time = 0.0548806, antiderivative size = 107, 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} \[ \frac{d^4 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}+\frac{b d^4 (c x+1)^4}{20 c}+\frac{2 b d^4 (c x+1)^3}{15 c}+\frac{2 b d^4 (c x+1)^2}{5 c}+\frac{16 b d^4 \log (1-c x)}{5 c}+\frac{8}{5} b d^4 x \]

Antiderivative was successfully verified.

[In]

Int[(d + c*d*x)^4*(a + b*ArcTanh[c*x]),x]

[Out]

(8*b*d^4*x)/5 + (2*b*d^4*(1 + c*x)^2)/(5*c) + (2*b*d^4*(1 + c*x)^3)/(15*c) + (b*d^4*(1 + c*x)^4)/(20*c) + (d^4
*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/(5*c) + (16*b*d^4*Log[1 - c*x])/(5*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)^4 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}-\frac{b \int \frac{(d+c d x)^5}{1-c^2 x^2} \, dx}{5 d}\\ &=\frac{d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}-\frac{b \int \frac{(d+c d x)^4}{\frac{1}{d}-\frac{c x}{d}} \, dx}{5 d}\\ &=\frac{d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}-\frac{b \int \left (-8 d^5+\frac{16 d^4}{\frac{1}{d}-\frac{c x}{d}}-4 d^4 (d+c d x)-2 d^3 (d+c d x)^2-d^2 (d+c d x)^3\right ) \, dx}{5 d}\\ &=\frac{8}{5} b d^4 x+\frac{2 b d^4 (1+c x)^2}{5 c}+\frac{2 b d^4 (1+c x)^3}{15 c}+\frac{b d^4 (1+c x)^4}{20 c}+\frac{d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}+\frac{16 b d^4 \log (1-c x)}{5 c}\\ \end{align*}

Mathematica [A]  time = 0.178214, size = 146, normalized size = 1.36 \[ \frac{d^4 \left (12 a c^5 x^5+60 a c^4 x^4+120 a c^3 x^3+120 a c^2 x^2+60 a c x+3 b c^4 x^4+20 b c^3 x^3+66 b c^2 x^2+6 b \log \left (1-c^2 x^2\right )+12 b c x \left (c^4 x^4+5 c^3 x^3+10 c^2 x^2+10 c x+5\right ) \tanh ^{-1}(c x)+180 b c x+180 b \log (1-c x)\right )}{60 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + c*d*x)^4*(a + b*ArcTanh[c*x]),x]

[Out]

(d^4*(60*a*c*x + 180*b*c*x + 120*a*c^2*x^2 + 66*b*c^2*x^2 + 120*a*c^3*x^3 + 20*b*c^3*x^3 + 60*a*c^4*x^4 + 3*b*
c^4*x^4 + 12*a*c^5*x^5 + 12*b*c*x*(5 + 10*c*x + 10*c^2*x^2 + 5*c^3*x^3 + c^4*x^4)*ArcTanh[c*x] + 180*b*Log[1 -
c*x] + 6*b*Log[1 - c^2*x^2]))/(60*c)

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Maple [B]  time = 0.026, size = 202, normalized size = 1.9 \begin{align*}{\frac{{c}^{4}{x}^{5}a{d}^{4}}{5}}+{c}^{3}{x}^{4}a{d}^{4}+2\,{c}^{2}{x}^{3}a{d}^{4}+2\,c{x}^{2}a{d}^{4}+xa{d}^{4}+{\frac{{d}^{4}a}{5\,c}}+{\frac{{c}^{4}{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{5}}{5}}+{c}^{3}{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{4}+2\,{c}^{2}{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{3}+2\,c{d}^{4}b{\it Artanh} \left ( cx \right ){x}^{2}+{d}^{4}b{\it Artanh} \left ( cx \right ) x+{\frac{{d}^{4}b{\it Artanh} \left ( cx \right ) }{5\,c}}+{\frac{{c}^{3}{d}^{4}b{x}^{4}}{20}}+{\frac{{c}^{2}{d}^{4}b{x}^{3}}{3}}+{\frac{11\,c{d}^{4}b{x}^{2}}{10}}+3\,b{d}^{4}x+{\frac{16\,{d}^{4}b\ln \left ( cx-1 \right ) }{5\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^4*(a+b*arctanh(c*x)),x)

[Out]

1/5*c^4*x^5*a*d^4+c^3*x^4*a*d^4+2*c^2*x^3*a*d^4+2*c*x^2*a*d^4+x*a*d^4+1/5/c*d^4*a+1/5*c^4*d^4*b*arctanh(c*x)*x
^5+c^3*d^4*b*arctanh(c*x)*x^4+2*c^2*d^4*b*arctanh(c*x)*x^3+2*c*d^4*b*arctanh(c*x)*x^2+d^4*b*arctanh(c*x)*x+1/5
/c*d^4*b*arctanh(c*x)+1/20*c^3*d^4*b*x^4+1/3*c^2*d^4*b*x^3+11/10*c*d^4*b*x^2+3*b*d^4*x+16/5/c*d^4*b*ln(c*x-1)

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Maxima [B]  time = 0.949735, size = 382, normalized size = 3.57 \begin{align*} \frac{1}{5} \, a c^{4} d^{4} x^{5} + a c^{3} d^{4} x^{4} + \frac{1}{20} \,{\left (4 \, x^{5} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{4} d^{4} + 2 \, a c^{2} d^{4} x^{3} + \frac{1}{6} \,{\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^{4} +{\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^{4} + 2 \, a c d^{4} x^{2} +{\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^{4} + a d^{4} x + \frac{{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/5*a*c^4*d^4*x^5 + a*c^3*d^4*x^4 + 1/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c
^6))*b*c^4*d^4 + 2*a*c^2*d^4*x^3 + 1/6*(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^4 + (2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*c^2*d^4 + 2*a*c*d^
4*x^2 + (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^4 + a*d^4*x + 1/2*(2*c*
x*arctanh(c*x) + log(-c^2*x^2 + 1))*b*d^4/c

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Fricas [A]  time = 2.04621, size = 402, normalized size = 3.76 \begin{align*} \frac{12 \, a c^{5} d^{4} x^{5} + 3 \,{\left (20 \, a + b\right )} c^{4} d^{4} x^{4} + 20 \,{\left (6 \, a + b\right )} c^{3} d^{4} x^{3} + 6 \,{\left (20 \, a + 11 \, b\right )} c^{2} d^{4} x^{2} + 60 \,{\left (a + 3 \, b\right )} c d^{4} x + 6 \, b d^{4} \log \left (c x + 1\right ) + 186 \, b d^{4} \log \left (c x - 1\right ) + 6 \,{\left (b c^{5} d^{4} x^{5} + 5 \, b c^{4} d^{4} x^{4} + 10 \, b c^{3} d^{4} x^{3} + 10 \, b c^{2} d^{4} x^{2} + 5 \, b c d^{4} x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{60 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/60*(12*a*c^5*d^4*x^5 + 3*(20*a + b)*c^4*d^4*x^4 + 20*(6*a + b)*c^3*d^4*x^3 + 6*(20*a + 11*b)*c^2*d^4*x^2 + 6
0*(a + 3*b)*c*d^4*x + 6*b*d^4*log(c*x + 1) + 186*b*d^4*log(c*x - 1) + 6*(b*c^5*d^4*x^5 + 5*b*c^4*d^4*x^4 + 10*
b*c^3*d^4*x^3 + 10*b*c^2*d^4*x^2 + 5*b*c*d^4*x)*log(-(c*x + 1)/(c*x - 1)))/c

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Sympy [A]  time = 5.00716, size = 226, normalized size = 2.11 \begin{align*} \begin{cases} \frac{a c^{4} d^{4} x^{5}}{5} + a c^{3} d^{4} x^{4} + 2 a c^{2} d^{4} x^{3} + 2 a c d^{4} x^{2} + a d^{4} x + \frac{b c^{4} d^{4} x^{5} \operatorname{atanh}{\left (c x \right )}}{5} + b c^{3} d^{4} x^{4} \operatorname{atanh}{\left (c x \right )} + \frac{b c^{3} d^{4} x^{4}}{20} + 2 b c^{2} d^{4} x^{3} \operatorname{atanh}{\left (c x \right )} + \frac{b c^{2} d^{4} x^{3}}{3} + 2 b c d^{4} x^{2} \operatorname{atanh}{\left (c x \right )} + \frac{11 b c d^{4} x^{2}}{10} + b d^{4} x \operatorname{atanh}{\left (c x \right )} + 3 b d^{4} x + \frac{16 b d^{4} \log{\left (x - \frac{1}{c} \right )}}{5 c} + \frac{b d^{4} \operatorname{atanh}{\left (c x \right )}}{5 c} & \text{for}\: c \neq 0 \\a d^{4} x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)**4*(a+b*atanh(c*x)),x)

[Out]

Piecewise((a*c**4*d**4*x**5/5 + a*c**3*d**4*x**4 + 2*a*c**2*d**4*x**3 + 2*a*c*d**4*x**2 + a*d**4*x + b*c**4*d*
*4*x**5*atanh(c*x)/5 + b*c**3*d**4*x**4*atanh(c*x) + b*c**3*d**4*x**4/20 + 2*b*c**2*d**4*x**3*atanh(c*x) + b*c
**2*d**4*x**3/3 + 2*b*c*d**4*x**2*atanh(c*x) + 11*b*c*d**4*x**2/10 + b*d**4*x*atanh(c*x) + 3*b*d**4*x + 16*b*d
**4*log(x - 1/c)/(5*c) + b*d**4*atanh(c*x)/(5*c), Ne(c, 0)), (a*d**4*x, True))

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Giac [B]  time = 1.28042, size = 261, normalized size = 2.44 \begin{align*} \frac{1}{5} \, a c^{4} d^{4} x^{5} + \frac{1}{20} \,{\left (20 \, a c^{3} d^{4} + b c^{3} d^{4}\right )} x^{4} + \frac{b d^{4} \log \left (c x + 1\right )}{10 \, c} + \frac{31 \, b d^{4} \log \left (c x - 1\right )}{10 \, c} + \frac{1}{3} \,{\left (6 \, a c^{2} d^{4} + b c^{2} d^{4}\right )} x^{3} + \frac{1}{10} \,{\left (20 \, a c d^{4} + 11 \, b c d^{4}\right )} x^{2} +{\left (a d^{4} + 3 \, b d^{4}\right )} x + \frac{1}{10} \,{\left (b c^{4} d^{4} x^{5} + 5 \, b c^{3} d^{4} x^{4} + 10 \, b c^{2} d^{4} x^{3} + 10 \, b c d^{4} x^{2} + 5 \, b d^{4} 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)^4*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

1/5*a*c^4*d^4*x^5 + 1/20*(20*a*c^3*d^4 + b*c^3*d^4)*x^4 + 1/10*b*d^4*log(c*x + 1)/c + 31/10*b*d^4*log(c*x - 1)
/c + 1/3*(6*a*c^2*d^4 + b*c^2*d^4)*x^3 + 1/10*(20*a*c*d^4 + 11*b*c*d^4)*x^2 + (a*d^4 + 3*b*d^4)*x + 1/10*(b*c^
4*d^4*x^5 + 5*b*c^3*d^4*x^4 + 10*b*c^2*d^4*x^3 + 10*b*c*d^4*x^2 + 5*b*d^4*x)*log(-(c*x + 1)/(c*x - 1))