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

Optimal. Leaf size=149 \[ \frac{(d+e x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 e}+\frac{b e^2 x^2 \left (10 c^2 d^2+e^2\right )}{10 c^3}+\frac{b d e x \left (2 c^2 d^2+e^2\right )}{c^3}-\frac{b (c d-e)^5 \log (c x+1)}{10 c^5 e}+\frac{b (c d+e)^5 \log (1-c x)}{10 c^5 e}+\frac{b d e^3 x^3}{3 c}+\frac{b e^4 x^4}{20 c} \]

[Out]

(b*d*e*(2*c^2*d^2 + e^2)*x)/c^3 + (b*e^2*(10*c^2*d^2 + e^2)*x^2)/(10*c^3) + (b*d*e^3*x^3)/(3*c) + (b*e^4*x^4)/
(20*c) + ((d + e*x)^5*(a + b*ArcTanh[c*x]))/(5*e) + (b*(c*d + e)^5*Log[1 - c*x])/(10*c^5*e) - (b*(c*d - e)^5*L
og[1 + c*x])/(10*c^5*e)

________________________________________________________________________________________

Rubi [A]  time = 0.140575, antiderivative size = 149, 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} \[ \frac{(d+e x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 e}+\frac{b e^2 x^2 \left (10 c^2 d^2+e^2\right )}{10 c^3}+\frac{b d e x \left (2 c^2 d^2+e^2\right )}{c^3}-\frac{b (c d-e)^5 \log (c x+1)}{10 c^5 e}+\frac{b (c d+e)^5 \log (1-c x)}{10 c^5 e}+\frac{b d e^3 x^3}{3 c}+\frac{b e^4 x^4}{20 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*e*(2*c^2*d^2 + e^2)*x)/c^3 + (b*e^2*(10*c^2*d^2 + e^2)*x^2)/(10*c^3) + (b*d*e^3*x^3)/(3*c) + (b*e^4*x^4)/
(20*c) + ((d + e*x)^5*(a + b*ArcTanh[c*x]))/(5*e) + (b*(c*d + e)^5*Log[1 - c*x])/(10*c^5*e) - (b*(c*d - e)^5*L
og[1 + c*x])/(10*c^5*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)^4 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac{(d+e x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 e}-\frac{(b c) \int \frac{(d+e x)^5}{1-c^2 x^2} \, dx}{5 e}\\ &=\frac{(d+e x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 e}-\frac{(b c) \int \left (-\frac{5 d e^2 \left (2 c^2 d^2+e^2\right )}{c^4}-\frac{e^3 \left (10 c^2 d^2+e^2\right ) x}{c^4}-\frac{5 d e^4 x^2}{c^2}-\frac{e^5 x^3}{c^2}+\frac{c^4 d^5+10 c^2 d^3 e^2+5 d e^4+e \left (5 c^4 d^4+10 c^2 d^2 e^2+e^4\right ) x}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx}{5 e}\\ &=\frac{b d e \left (2 c^2 d^2+e^2\right ) x}{c^3}+\frac{b e^2 \left (10 c^2 d^2+e^2\right ) x^2}{10 c^3}+\frac{b d e^3 x^3}{3 c}+\frac{b e^4 x^4}{20 c}+\frac{(d+e x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 e}-\frac{b \int \frac{c^4 d^5+10 c^2 d^3 e^2+5 d e^4+e \left (5 c^4 d^4+10 c^2 d^2 e^2+e^4\right ) x}{1-c^2 x^2} \, dx}{5 c^3 e}\\ &=\frac{b d e \left (2 c^2 d^2+e^2\right ) x}{c^3}+\frac{b e^2 \left (10 c^2 d^2+e^2\right ) x^2}{10 c^3}+\frac{b d e^3 x^3}{3 c}+\frac{b e^4 x^4}{20 c}+\frac{(d+e x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 e}+\frac{\left (b (c d-e)^5\right ) \int \frac{1}{-c-c^2 x} \, dx}{10 c^3 e}-\frac{\left (b (c d+e)^5\right ) \int \frac{1}{c-c^2 x} \, dx}{10 c^3 e}\\ &=\frac{b d e \left (2 c^2 d^2+e^2\right ) x}{c^3}+\frac{b e^2 \left (10 c^2 d^2+e^2\right ) x^2}{10 c^3}+\frac{b d e^3 x^3}{3 c}+\frac{b e^4 x^4}{20 c}+\frac{(d+e x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 e}+\frac{b (c d+e)^5 \log (1-c x)}{10 c^5 e}-\frac{b (c d-e)^5 \log (1+c x)}{10 c^5 e}\\ \end{align*}

Mathematica [A]  time = 0.179837, size = 274, normalized size = 1.84 \[ \frac{6 c^2 e x^2 \left (20 a c^3 d^3+b e \left (10 c^2 d^2+e^2\right )\right )+60 c^2 d x \left (a c^3 d^3+b e \left (2 c^2 d^2+e^2\right )\right )+3 c^4 e^3 x^4 (20 a c d+b e)+20 c^4 d e^2 x^3 (6 a c d+b e)+12 a c^5 e^4 x^5+12 b c^5 x \tanh ^{-1}(c x) \left (10 d^2 e^2 x^2+10 d^3 e x+5 d^4+5 d e^3 x^3+e^4 x^4\right )+6 b \left (10 c^2 d^2 e^2+10 c^3 d^3 e+5 c^4 d^4+5 c d e^3+e^4\right ) \log (1-c x)+6 b \left (10 c^2 d^2 e^2-10 c^3 d^3 e+5 c^4 d^4-5 c d e^3+e^4\right ) \log (c x+1)}{60 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.031, size = 395, normalized size = 2.7 \begin{align*} 2\,b{e}^{2}{\it Artanh} \left ( cx \right ){x}^{3}{d}^{2}+{\frac{a{d}^{5}}{5\,e}}+{\frac{a{e}^{4}{x}^{5}}{5}}+ax{d}^{4}+{\frac{b{e}^{3}dx}{{c}^{3}}}+{\frac{b{e}^{4}{x}^{2}}{10\,{c}^{3}}}+{\frac{b{e}^{4}\ln \left ( cx-1 \right ) }{10\,{c}^{5}}}+{\frac{b{e}^{4}\ln \left ( cx+1 \right ) }{10\,{c}^{5}}}-{\frac{b\ln \left ( cx+1 \right ){d}^{5}}{10\,e}}+{\frac{b\ln \left ( cx-1 \right ){d}^{5}}{10\,e}}+{\frac{b\ln \left ( cx+1 \right ){d}^{4}}{2\,c}}+{\frac{b\ln \left ( cx-1 \right ){d}^{4}}{2\,c}}+a{e}^{3}{x}^{4}d+2\,a{e}^{2}{x}^{3}{d}^{2}+2\,ae{x}^{2}{d}^{3}+{\frac{b{\it Artanh} \left ( cx \right ){d}^{5}}{5\,e}}+{\frac{b{e}^{4}{\it Artanh} \left ( cx \right ){x}^{5}}{5}}+b{\it Artanh} \left ( cx \right ) x{d}^{4}+2\,be{\it Artanh} \left ( cx \right ){x}^{2}{d}^{3}+b{e}^{3}{\it Artanh} \left ( cx \right ){x}^{4}d+2\,{\frac{be{d}^{3}x}{c}}+{\frac{b{e}^{2}{x}^{2}{d}^{2}}{c}}-{\frac{be\ln \left ( cx+1 \right ){d}^{3}}{{c}^{2}}}+{\frac{b{e}^{2}\ln \left ( cx-1 \right ){d}^{2}}{{c}^{3}}}+{\frac{b{e}^{2}\ln \left ( cx+1 \right ){d}^{2}}{{c}^{3}}}+{\frac{be\ln \left ( cx-1 \right ){d}^{3}}{{c}^{2}}}+{\frac{b{e}^{3}\ln \left ( cx-1 \right ) d}{2\,{c}^{4}}}-{\frac{b{e}^{3}\ln \left ( cx+1 \right ) d}{2\,{c}^{4}}}+{\frac{b{e}^{4}{x}^{4}}{20\,c}}+{\frac{b{e}^{3}d{x}^{3}}{3\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a/e*d^5+1/5*a*e^4*x^5+a*x*d^4+b/c^3*e^3*d*x+1/10/c^3*b*e^4*x^2+1/10/c^5*b*e^4*ln(c*x-1)+1/10/c^5*b*e^4*ln(
c*x+1)-1/10*b/e*ln(c*x+1)*d^5+1/10*b/e*ln(c*x-1)*d^5+1/5*b/e*arctanh(c*x)*d^5+1/5*b*e^4*arctanh(c*x)*x^5+b*arc
tanh(c*x)*x*d^4+1/2/c*b*ln(c*x+1)*d^4+1/2/c*b*ln(c*x-1)*d^4+a*e^3*x^4*d+2*a*e^2*x^3*d^2+2*a*e*x^2*d^3+2*b*e^2*
arctanh(c*x)*x^3*d^2+2*b/c*e*d^3*x+1/c*b*e^2*x^2*d^2+2*b*e*arctanh(c*x)*x^2*d^3+b*e^3*arctanh(c*x)*x^4*d-1/c^2
*b*e*ln(c*x+1)*d^3+1/c^3*b*e^2*ln(c*x-1)*d^2+1/c^3*b*e^2*ln(c*x+1)*d^2+1/c^2*b*e*ln(c*x-1)*d^3+1/2/c^4*b*e^3*l
n(c*x-1)*d-1/2/c^4*b*e^3*ln(c*x+1)*d+1/20*b*e^4*x^4/c+1/3*b*d*e^3*x^3/c

________________________________________________________________________________________

Maxima [A]  time = 0.977438, size = 369, normalized size = 2.48 \begin{align*} \frac{1}{5} \, a e^{4} x^{5} + a d e^{3} x^{4} + 2 \, a d^{2} e^{2} x^{3} + 2 \, a d^{3} e 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 d^{3} e +{\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^{2} e^{2} + \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 d e^{3} + \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 e^{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((e*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.7327, size = 697, normalized size = 4.68 \begin{align*} \frac{12 \, a c^{5} e^{4} x^{5} + 3 \,{\left (20 \, a c^{5} d e^{3} + b c^{4} e^{4}\right )} x^{4} + 20 \,{\left (6 \, a c^{5} d^{2} e^{2} + b c^{4} d e^{3}\right )} x^{3} + 6 \,{\left (20 \, a c^{5} d^{3} e + 10 \, b c^{4} d^{2} e^{2} + b c^{2} e^{4}\right )} x^{2} + 60 \,{\left (a c^{5} d^{4} + 2 \, b c^{4} d^{3} e + b c^{2} d e^{3}\right )} x + 6 \,{\left (5 \, b c^{4} d^{4} - 10 \, b c^{3} d^{3} e + 10 \, b c^{2} d^{2} e^{2} - 5 \, b c d e^{3} + b e^{4}\right )} \log \left (c x + 1\right ) + 6 \,{\left (5 \, b c^{4} d^{4} + 10 \, b c^{3} d^{3} e + 10 \, b c^{2} d^{2} e^{2} + 5 \, b c d e^{3} + b e^{4}\right )} \log \left (c x - 1\right ) + 6 \,{\left (b c^{5} e^{4} x^{5} + 5 \, b c^{5} d e^{3} x^{4} + 10 \, b c^{5} d^{2} e^{2} x^{3} + 10 \, b c^{5} d^{3} e x^{2} + 5 \, b c^{5} d^{4} x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{60 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 5.24591, size = 381, normalized size = 2.56 \begin{align*} \begin{cases} a d^{4} x + 2 a d^{3} e x^{2} + 2 a d^{2} e^{2} x^{3} + a d e^{3} x^{4} + \frac{a e^{4} x^{5}}{5} + b d^{4} x \operatorname{atanh}{\left (c x \right )} + 2 b d^{3} e x^{2} \operatorname{atanh}{\left (c x \right )} + 2 b d^{2} e^{2} x^{3} \operatorname{atanh}{\left (c x \right )} + b d e^{3} x^{4} \operatorname{atanh}{\left (c x \right )} + \frac{b e^{4} x^{5} \operatorname{atanh}{\left (c x \right )}}{5} + \frac{b d^{4} \log{\left (x - \frac{1}{c} \right )}}{c} + \frac{b d^{4} \operatorname{atanh}{\left (c x \right )}}{c} + \frac{2 b d^{3} e x}{c} + \frac{b d^{2} e^{2} x^{2}}{c} + \frac{b d e^{3} x^{3}}{3 c} + \frac{b e^{4} x^{4}}{20 c} - \frac{2 b d^{3} e \operatorname{atanh}{\left (c x \right )}}{c^{2}} + \frac{2 b d^{2} e^{2} \log{\left (x - \frac{1}{c} \right )}}{c^{3}} + \frac{2 b d^{2} e^{2} \operatorname{atanh}{\left (c x \right )}}{c^{3}} + \frac{b d e^{3} x}{c^{3}} + \frac{b e^{4} x^{2}}{10 c^{3}} - \frac{b d e^{3} \operatorname{atanh}{\left (c x \right )}}{c^{4}} + \frac{b e^{4} \log{\left (x - \frac{1}{c} \right )}}{5 c^{5}} + \frac{b e^{4} \operatorname{atanh}{\left (c x \right )}}{5 c^{5}} & \text{for}\: c \neq 0 \\a \left (d^{4} x + 2 d^{3} e x^{2} + 2 d^{2} e^{2} x^{3} + d e^{3} x^{4} + \frac{e^{4} x^{5}}{5}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.33342, size = 524, normalized size = 3.52 \begin{align*} \frac{6 \, b c^{5} x^{5} e^{4} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 30 \, b c^{5} d x^{4} e^{3} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 60 \, b c^{5} d^{2} x^{3} e^{2} \log \left (-\frac{c x + 1}{c x - 1}\right ) + 60 \, b c^{5} d^{3} x^{2} e \log \left (-\frac{c x + 1}{c x - 1}\right ) + 12 \, a c^{5} x^{5} e^{4} + 60 \, a c^{5} d x^{4} e^{3} + 120 \, a c^{5} d^{2} x^{3} e^{2} + 120 \, a c^{5} d^{3} x^{2} e + 30 \, b c^{5} d^{4} x \log \left (-\frac{c x + 1}{c x - 1}\right ) + 60 \, a c^{5} d^{4} x + 3 \, b c^{4} x^{4} e^{4} + 20 \, b c^{4} d x^{3} e^{3} + 60 \, b c^{4} d^{2} x^{2} e^{2} + 120 \, b c^{4} d^{3} x e + 30 \, b c^{4} d^{4} \log \left (c^{2} x^{2} - 1\right ) - 60 \, b c^{3} d^{3} e \log \left (c x + 1\right ) + 60 \, b c^{3} d^{3} e \log \left (c x - 1\right ) + 60 \, b c^{2} d^{2} e^{2} \log \left (c^{2} x^{2} - 1\right ) + 6 \, b c^{2} x^{2} e^{4} + 60 \, b c^{2} d x e^{3} - 30 \, b c d e^{3} \log \left (c x + 1\right ) + 30 \, b c d e^{3} \log \left (c x - 1\right ) + 6 \, b e^{4} \log \left (c^{2} x^{2} - 1\right )}{60 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(6*b*c^5*x^5*e^4*log(-(c*x + 1)/(c*x - 1)) + 30*b*c^5*d*x^4*e^3*log(-(c*x + 1)/(c*x - 1)) + 60*b*c^5*d^2*
x^3*e^2*log(-(c*x + 1)/(c*x - 1)) + 60*b*c^5*d^3*x^2*e*log(-(c*x + 1)/(c*x - 1)) + 12*a*c^5*x^5*e^4 + 60*a*c^5
*d*x^4*e^3 + 120*a*c^5*d^2*x^3*e^2 + 120*a*c^5*d^3*x^2*e + 30*b*c^5*d^4*x*log(-(c*x + 1)/(c*x - 1)) + 60*a*c^5
*d^4*x + 3*b*c^4*x^4*e^4 + 20*b*c^4*d*x^3*e^3 + 60*b*c^4*d^2*x^2*e^2 + 120*b*c^4*d^3*x*e + 30*b*c^4*d^4*log(c^
2*x^2 - 1) - 60*b*c^3*d^3*e*log(c*x + 1) + 60*b*c^3*d^3*e*log(c*x - 1) + 60*b*c^2*d^2*e^2*log(c^2*x^2 - 1) + 6
*b*c^2*x^2*e^4 + 60*b*c^2*d*x*e^3 - 30*b*c*d*e^3*log(c*x + 1) + 30*b*c*d*e^3*log(c*x - 1) + 6*b*e^4*log(c^2*x^
2 - 1))/c^5