3.17 \(\int (c e+d e x) (a+b \tanh ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=95 \[ \frac{e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}-\frac{e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+a b e x+\frac{b^2 e \log \left (1-(c+d x)^2\right )}{2 d}+\frac{b^2 e (c+d x) \tanh ^{-1}(c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.135102, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6107, 12, 5916, 5980, 5910, 260, 5948} \[ \frac{e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}-\frac{e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d}+a b e x+\frac{b^2 e \log \left (1-(c+d x)^2\right )}{2 d}+\frac{b^2 e (c+d x) \tanh ^{-1}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)*(a + b*ArcTanh[c + d*x])^2,x]

[Out]

a*b*e*x + (b^2*e*(c + d*x)*ArcTanh[c + d*x])/d - (e*(a + b*ArcTanh[c + d*x])^2)/(2*d) + (e*(c + d*x)^2*(a + b*
ArcTanh[c + d*x])^2)/(2*d) + (b^2*e*Log[1 - (c + d*x)^2])/(2*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 5980

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTanh[c*x])
^p)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.102423, size = 134, normalized size = 1.41 \[ e \left (\frac{a^2 (c+d x)^2}{2 d}+\frac{\left (a b+b^2\right ) \log (-c-d x+1)}{2 d}+\frac{\left (b^2-a b\right ) \log (c+d x+1)}{2 d}+\frac{a b (c+d x)}{d}+\frac{b (c+d x) \tanh ^{-1}(c+d x) (a (c+d x)+b)}{d}+\frac{\left (b^2 (c+d x)^2-b^2\right ) \tanh ^{-1}(c+d x)^2}{2 d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)*(a + b*ArcTanh[c + d*x])^2,x]

[Out]

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

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Maple [B]  time = 0.052, size = 390, normalized size = 4.1 \begin{align*}{\frac{{a}^{2}e{x}^{2}d}{2}}+x{a}^{2}ce+{\frac{{a}^{2}{c}^{2}e}{2\,d}}+{\frac{d \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}{x}^{2}{b}^{2}e}{2}}+ \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}x{b}^{2}ce+{\frac{ \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}{b}^{2}{c}^{2}e}{2\,d}}+{\it Artanh} \left ( dx+c \right ) x{b}^{2}e+{\frac{{\it Artanh} \left ( dx+c \right ){b}^{2}ce}{d}}+{\frac{e{b}^{2}{\it Artanh} \left ( dx+c \right ) \ln \left ( dx+c-1 \right ) }{2\,d}}-{\frac{e{b}^{2}{\it Artanh} \left ( dx+c \right ) \ln \left ( dx+c+1 \right ) }{2\,d}}+{\frac{e{b}^{2} \left ( \ln \left ( dx+c-1 \right ) \right ) ^{2}}{8\,d}}-{\frac{e{b}^{2}\ln \left ( dx+c-1 \right ) }{4\,d}\ln \left ({\frac{1}{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{e{b}^{2}\ln \left ( dx+c-1 \right ) }{2\,d}}+{\frac{e{b}^{2}\ln \left ( dx+c+1 \right ) }{2\,d}}-{\frac{e{b}^{2}\ln \left ( dx+c+1 \right ) }{4\,d}\ln \left ( -{\frac{dx}{2}}-{\frac{c}{2}}+{\frac{1}{2}} \right ) }+{\frac{e{b}^{2}}{4\,d}\ln \left ( -{\frac{dx}{2}}-{\frac{c}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{e{b}^{2} \left ( \ln \left ( dx+c+1 \right ) \right ) ^{2}}{8\,d}}+d{\it Artanh} \left ( dx+c \right ){x}^{2}abe+2\,{\it Artanh} \left ( dx+c \right ) xabce+{\frac{{\it Artanh} \left ( dx+c \right ) ab{c}^{2}e}{d}}+abex+{\frac{abce}{d}}+{\frac{eab\ln \left ( dx+c-1 \right ) }{2\,d}}-{\frac{eab\ln \left ( dx+c+1 \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)*(a+b*arctanh(d*x+c))^2,x)

[Out]

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

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Maxima [B]  time = 1.90262, size = 427, normalized size = 4.49 \begin{align*} \frac{1}{2} \, a^{2} d e x^{2} + \frac{1}{2} \,{\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 )} a b d e + a^{2} c e 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 )} a b c e}{d} + \frac{{\left (b^{2} d^{2} e x^{2} + 2 \, b^{2} c d e x +{\left (c^{2} e - e\right )} b^{2}\right )} \log \left (d x + c + 1\right )^{2} +{\left (b^{2} d^{2} e x^{2} + 2 \, b^{2} c d e x +{\left (c^{2} e - e\right )} b^{2}\right )} \log \left (-d x - c + 1\right )^{2} + 4 \,{\left (b^{2} d e x +{\left (c e + e\right )} b^{2}\right )} \log \left (d x + c + 1\right ) - 2 \,{\left (2 \, b^{2} d e x + 2 \,{\left (c e - e\right )} b^{2} +{\left (b^{2} d^{2} e x^{2} + 2 \, b^{2} c d e x +{\left (c^{2} e - e\right )} b^{2}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)*(a+b*arctanh(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.98767, size = 436, normalized size = 4.59 \begin{align*} \frac{4 \, a^{2} d^{2} e x^{2} + 8 \,{\left (a^{2} c + a b\right )} d e x + 4 \,{\left (a b c^{2} + b^{2} c - a b + b^{2}\right )} e \log \left (d x + c + 1\right ) - 4 \,{\left (a b c^{2} + b^{2} c - a b - b^{2}\right )} e \log \left (d x + c - 1\right ) +{\left (b^{2} d^{2} e x^{2} + 2 \, b^{2} c d e x +{\left (b^{2} c^{2} - b^{2}\right )} e\right )} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 4 \,{\left (a b d^{2} e x^{2} +{\left (2 \, a b c + b^{2}\right )} d e x\right )} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)*(a+b*arctanh(d*x+c))^2,x, algorithm="fricas")

[Out]

1/8*(4*a^2*d^2*e*x^2 + 8*(a^2*c + a*b)*d*e*x + 4*(a*b*c^2 + b^2*c - a*b + b^2)*e*log(d*x + c + 1) - 4*(a*b*c^2
+ b^2*c - a*b - b^2)*e*log(d*x + c - 1) + (b^2*d^2*e*x^2 + 2*b^2*c*d*e*x + (b^2*c^2 - b^2)*e)*log(-(d*x + c +
1)/(d*x + c - 1))^2 + 4*(a*b*d^2*e*x^2 + (2*a*b*c + b^2)*d*e*x)*log(-(d*x + c + 1)/(d*x + c - 1)))/d

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Sympy [A]  time = 6.19507, size = 238, normalized size = 2.51 \begin{align*} \begin{cases} a^{2} c e x + \frac{a^{2} d e x^{2}}{2} + \frac{a b c^{2} e \operatorname{atanh}{\left (c + d x \right )}}{d} + 2 a b c e x \operatorname{atanh}{\left (c + d x \right )} + a b d e x^{2} \operatorname{atanh}{\left (c + d x \right )} + a b e x - \frac{a b e \operatorname{atanh}{\left (c + d x \right )}}{d} + \frac{b^{2} c^{2} e \operatorname{atanh}^{2}{\left (c + d x \right )}}{2 d} + b^{2} c e x \operatorname{atanh}^{2}{\left (c + d x \right )} + \frac{b^{2} c e \operatorname{atanh}{\left (c + d x \right )}}{d} + \frac{b^{2} d e x^{2} \operatorname{atanh}^{2}{\left (c + d x \right )}}{2} + b^{2} e x \operatorname{atanh}{\left (c + d x \right )} + \frac{b^{2} e \log{\left (\frac{c}{d} + x + \frac{1}{d} \right )}}{d} - \frac{b^{2} e \operatorname{atanh}^{2}{\left (c + d x \right )}}{2 d} - \frac{b^{2} e \operatorname{atanh}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\c e x \left (a + b \operatorname{atanh}{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)*(a+b*atanh(d*x+c))**2,x)

[Out]

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

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Giac [B]  time = 1.34392, size = 470, normalized size = 4.95 \begin{align*} \frac{b^{2} d^{2} x^{2} e \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 4 \, a b d^{2} x^{2} e \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 2 \, b^{2} c d x e \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 4 \, a^{2} d^{2} x^{2} e + 8 \, a b c d x e \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + b^{2} c^{2} e \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 8 \, a^{2} c d x e + 4 \, a b c^{2} e \log \left (d x + c + 1\right ) - 4 \, a b c^{2} e \log \left (d x + c - 1\right ) + 4 \, b^{2} d x e \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 8 \, a b d x e + 4 \, b^{2} c e \log \left (d x + c + 1\right ) - 4 \, b^{2} c e \log \left (d x + c - 1\right ) - b^{2} e \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} - 4 \, a b e \log \left (d x + c + 1\right ) + 4 \, b^{2} e \log \left (d x + c + 1\right ) + 4 \, a b e \log \left (d x + c - 1\right ) + 4 \, b^{2} e \log \left (d x + c - 1\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)*(a+b*arctanh(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(b^2*d^2*x^2*e*log(-(d*x + c + 1)/(d*x + c - 1))^2 + 4*a*b*d^2*x^2*e*log(-(d*x + c + 1)/(d*x + c - 1)) + 2
*b^2*c*d*x*e*log(-(d*x + c + 1)/(d*x + c - 1))^2 + 4*a^2*d^2*x^2*e + 8*a*b*c*d*x*e*log(-(d*x + c + 1)/(d*x + c
- 1)) + b^2*c^2*e*log(-(d*x + c + 1)/(d*x + c - 1))^2 + 8*a^2*c*d*x*e + 4*a*b*c^2*e*log(d*x + c + 1) - 4*a*b*
c^2*e*log(d*x + c - 1) + 4*b^2*d*x*e*log(-(d*x + c + 1)/(d*x + c - 1)) + 8*a*b*d*x*e + 4*b^2*c*e*log(d*x + c +
1) - 4*b^2*c*e*log(d*x + c - 1) - b^2*e*log(-(d*x + c + 1)/(d*x + c - 1))^2 - 4*a*b*e*log(d*x + c + 1) + 4*b^
2*e*log(d*x + c + 1) + 4*a*b*e*log(d*x + c - 1) + 4*b^2*e*log(d*x + c - 1))/d