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

Optimal. Leaf size=72 \[ \frac{e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )}{4 d}+\frac{b e^3 (c+d x)^3}{12 d}-\frac{b e^3 \tanh ^{-1}(c+d x)}{4 d}+\frac{1}{4} b e^3 x \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0666851, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {6107, 12, 5916, 302, 206} \[ \frac{e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )}{4 d}+\frac{b e^3 (c+d x)^3}{12 d}-\frac{b e^3 \tanh ^{-1}(c+d x)}{4 d}+\frac{1}{4} b e^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6107

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((f*x)/d)^m*(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f,
0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.10449, size = 78, normalized size = 1.08 \[ \frac{e^3 \left (6 a (c+d x)^4+2 b (c+d x)^3+6 b (c+d x)+3 b \log (-c-d x+1)-3 b \log (c+d x+1)+6 b (c+d x)^4 \tanh ^{-1}(c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.035, size = 242, normalized size = 3.4 \begin{align*}{\frac{{d}^{3}{x}^{4}a{e}^{3}}{4}}+{d}^{2}{x}^{3}ac{e}^{3}+{\frac{3\,d{x}^{2}a{c}^{2}{e}^{3}}{2}}+xa{c}^{3}{e}^{3}+{\frac{a{c}^{4}{e}^{3}}{4\,d}}+{\frac{{d}^{3}{\it Artanh} \left ( dx+c \right ){x}^{4}b{e}^{3}}{4}}+{d}^{2}{\it Artanh} \left ( dx+c \right ){x}^{3}bc{e}^{3}+{\frac{3\,d{\it Artanh} \left ( dx+c \right ){x}^{2}b{c}^{2}{e}^{3}}{2}}+{\it Artanh} \left ( dx+c \right ) xb{c}^{3}{e}^{3}+{\frac{{\it Artanh} \left ( dx+c \right ) b{c}^{4}{e}^{3}}{4\,d}}+{\frac{{d}^{2}{x}^{3}b{e}^{3}}{12}}+{\frac{d{x}^{2}bc{e}^{3}}{4}}+{\frac{xb{c}^{2}{e}^{3}}{4}}+{\frac{b{c}^{3}{e}^{3}}{12\,d}}+{\frac{b{e}^{3}x}{4}}+{\frac{bc{e}^{3}}{4\,d}}+{\frac{{e}^{3}b\ln \left ( dx+c-1 \right ) }{8\,d}}-{\frac{{e}^{3}b\ln \left ( dx+c+1 \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 0.995547, size = 482, normalized size = 6.69 \begin{align*} \frac{1}{4} \, a d^{3} e^{3} x^{4} + a c d^{2} e^{3} x^{3} + \frac{3}{2} \, a c^{2} d e^{3} x^{2} + \frac{3}{4} \,{\left (2 \, x^{2} \operatorname{artanh}\left (d x + c\right ) + d{\left (\frac{2 \, x}{d^{2}} - \frac{{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac{{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} b c^{2} d e^{3} + \frac{1}{2} \,{\left (2 \, x^{3} \operatorname{artanh}\left (d x + c\right ) + d{\left (\frac{d x^{2} - 4 \, c x}{d^{3}} + \frac{{\left (c^{3} + 3 \, c^{2} + 3 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{4}} - \frac{{\left (c^{3} - 3 \, c^{2} + 3 \, c - 1\right )} \log \left (d x + c - 1\right )}{d^{4}}\right )}\right )} b c d^{2} e^{3} + \frac{1}{24} \,{\left (6 \, x^{4} \operatorname{artanh}\left (d x + c\right ) + d{\left (\frac{2 \,{\left (d^{2} x^{3} - 3 \, c d x^{2} + 3 \,{\left (3 \, c^{2} + 1\right )} x\right )}}{d^{4}} - \frac{3 \,{\left (c^{4} + 4 \, c^{3} + 6 \, c^{2} + 4 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{5}} + \frac{3 \,{\left (c^{4} - 4 \, c^{3} + 6 \, c^{2} - 4 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{5}}\right )}\right )} b d^{3} e^{3} + a c^{3} e^{3} x + \frac{{\left (2 \,{\left (d x + c\right )} \operatorname{artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b c^{3} e^{3}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.23478, size = 342, normalized size = 4.75 \begin{align*} \frac{6 \, a d^{4} e^{3} x^{4} + 2 \,{\left (12 \, a c + b\right )} d^{3} e^{3} x^{3} + 6 \,{\left (6 \, a c^{2} + b c\right )} d^{2} e^{3} x^{2} + 6 \,{\left (4 \, a c^{3} + b c^{2} + b\right )} d e^{3} x + 3 \,{\left (b d^{4} e^{3} x^{4} + 4 \, b c d^{3} e^{3} x^{3} + 6 \, b c^{2} d^{2} e^{3} x^{2} + 4 \, b c^{3} d e^{3} x +{\left (b c^{4} - b\right )} e^{3}\right )} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 26.819, size = 231, normalized size = 3.21 \begin{align*} \begin{cases} a c^{3} e^{3} x + \frac{3 a c^{2} d e^{3} x^{2}}{2} + a c d^{2} e^{3} x^{3} + \frac{a d^{3} e^{3} x^{4}}{4} + \frac{b c^{4} e^{3} \operatorname{atanh}{\left (c + d x \right )}}{4 d} + b c^{3} e^{3} x \operatorname{atanh}{\left (c + d x \right )} + \frac{3 b c^{2} d e^{3} x^{2} \operatorname{atanh}{\left (c + d x \right )}}{2} + \frac{b c^{2} e^{3} x}{4} + b c d^{2} e^{3} x^{3} \operatorname{atanh}{\left (c + d x \right )} + \frac{b c d e^{3} x^{2}}{4} + \frac{b d^{3} e^{3} x^{4} \operatorname{atanh}{\left (c + d x \right )}}{4} + \frac{b d^{2} e^{3} x^{3}}{12} + \frac{b e^{3} x}{4} - \frac{b e^{3} \operatorname{atanh}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname{atanh}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.29449, size = 362, normalized size = 5.03 \begin{align*} \frac{3 \, b d^{4} x^{4} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 6 \, a d^{4} x^{4} e^{3} + 12 \, b c d^{3} x^{3} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 24 \, a c d^{3} x^{3} e^{3} + 18 \, b c^{2} d^{2} x^{2} e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 36 \, a c^{2} d^{2} x^{2} e^{3} + 2 \, b d^{3} x^{3} e^{3} + 12 \, b c^{3} d x e^{3} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 24 \, a c^{3} d x e^{3} + 6 \, b c d^{2} x^{2} e^{3} + 3 \, b c^{4} e^{3} \log \left (d x + c + 1\right ) - 3 \, b c^{4} e^{3} \log \left (-d x - c + 1\right ) + 6 \, b c^{2} d x e^{3} + 6 \, b d x e^{3} - 3 \, b e^{3} \log \left (d x + c + 1\right ) + 3 \, b e^{3} \log \left (-d x - c + 1\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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