3.1.14 \(\int \frac {(d+c d x)^2 (a+b \tanh ^{-1}(c x))}{x} \, dx\) [14]

Optimal. Leaf size=114 \[ 2 a c d^2 x+\frac {1}{2} b c d^2 x-\frac {1}{2} b d^2 \tanh ^{-1}(c x)+2 b c d^2 x \tanh ^{-1}(c x)+\frac {1}{2} c^2 d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+a d^2 \log (x)+b d^2 \log \left (1-c^2 x^2\right )-\frac {1}{2} b d^2 \text {PolyLog}(2,-c x)+\frac {1}{2} b d^2 \text {PolyLog}(2,c x) \]

[Out]

2*a*c*d^2*x+1/2*b*c*d^2*x-1/2*b*d^2*arctanh(c*x)+2*b*c*d^2*x*arctanh(c*x)+1/2*c^2*d^2*x^2*(a+b*arctanh(c*x))+a
*d^2*ln(x)+b*d^2*ln(-c^2*x^2+1)-1/2*b*d^2*polylog(2,-c*x)+1/2*b*d^2*polylog(2,c*x)

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6087, 6021, 266, 6031, 6037, 327, 212} \begin {gather*} \frac {1}{2} c^2 d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+2 a c d^2 x+a d^2 \log (x)+b d^2 \log \left (1-c^2 x^2\right )-\frac {1}{2} b d^2 \text {Li}_2(-c x)+\frac {1}{2} b d^2 \text {Li}_2(c x)+\frac {1}{2} b c d^2 x-\frac {1}{2} b d^2 \tanh ^{-1}(c x)+2 b c d^2 x \tanh ^{-1}(c x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

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

Rule 266

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 6021

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

Rule 6031

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b/2)*PolyLog[2, (-c)*x]
, x] + Simp[(b/2)*PolyLog[2, c*x], x]) /; FreeQ[{a, b, c}, x]

Rule 6037

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

Rule 6087

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {(d+c d x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{x} \, dx &=\int \left (2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+c^2 d^2 x \left (a+b \tanh ^{-1}(c x)\right )\right ) \, dx\\ &=d^2 \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx+\left (2 c d^2\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx+\left (c^2 d^2\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx\\ &=2 a c d^2 x+\frac {1}{2} c^2 d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+a d^2 \log (x)-\frac {1}{2} b d^2 \text {Li}_2(-c x)+\frac {1}{2} b d^2 \text {Li}_2(c x)+\left (2 b c d^2\right ) \int \tanh ^{-1}(c x) \, dx-\frac {1}{2} \left (b c^3 d^2\right ) \int \frac {x^2}{1-c^2 x^2} \, dx\\ &=2 a c d^2 x+\frac {1}{2} b c d^2 x+2 b c d^2 x \tanh ^{-1}(c x)+\frac {1}{2} c^2 d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+a d^2 \log (x)-\frac {1}{2} b d^2 \text {Li}_2(-c x)+\frac {1}{2} b d^2 \text {Li}_2(c x)-\frac {1}{2} \left (b c d^2\right ) \int \frac {1}{1-c^2 x^2} \, dx-\left (2 b c^2 d^2\right ) \int \frac {x}{1-c^2 x^2} \, dx\\ &=2 a c d^2 x+\frac {1}{2} b c d^2 x-\frac {1}{2} b d^2 \tanh ^{-1}(c x)+2 b c d^2 x \tanh ^{-1}(c x)+\frac {1}{2} c^2 d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+a d^2 \log (x)+b d^2 \log \left (1-c^2 x^2\right )-\frac {1}{2} b d^2 \text {Li}_2(-c x)+\frac {1}{2} b d^2 \text {Li}_2(c x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.06, size = 103, normalized size = 0.90 \begin {gather*} \frac {1}{4} d^2 \left (8 a c x+2 b c x+2 a c^2 x^2+8 b c x \tanh ^{-1}(c x)+2 b c^2 x^2 \tanh ^{-1}(c x)+4 a \log (x)+b \log (1-c x)-b \log (1+c x)+4 b \log \left (1-c^2 x^2\right )-2 b \text {PolyLog}(2,-c x)+2 b \text {PolyLog}(2,c x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(8*a*c*x + 2*b*c*x + 2*a*c^2*x^2 + 8*b*c*x*ArcTanh[c*x] + 2*b*c^2*x^2*ArcTanh[c*x] + 4*a*Log[x] + b*Log[1
- c*x] - b*Log[1 + c*x] + 4*b*Log[1 - c^2*x^2] - 2*b*PolyLog[2, -(c*x)] + 2*b*PolyLog[2, c*x]))/4

________________________________________________________________________________________

Maple [A]
time = 0.18, size = 142, normalized size = 1.25

method	result	size
derivativedivides	\(\frac {d^{2} a \,c^{2} x^{2}}{2}+2 d^{2} a c x +d^{2} a \ln \left (c x \right )+\frac {d^{2} b \arctanh \left (c x \right ) c^{2} x^{2}}{2}+2 b c \,d^{2} x \arctanh \left (c x \right )+d^{2} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {d^{2} b \dilog \left (c x \right )}{2}-\frac {d^{2} b \dilog \left (c x +1\right )}{2}-\frac {d^{2} b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {b c \,d^{2} x}{2}+\frac {5 d^{2} b \ln \left (c x -1\right )}{4}+\frac {3 d^{2} b \ln \left (c x +1\right )}{4}\)	\(142\)
default	\(\frac {d^{2} a \,c^{2} x^{2}}{2}+2 d^{2} a c x +d^{2} a \ln \left (c x \right )+\frac {d^{2} b \arctanh \left (c x \right ) c^{2} x^{2}}{2}+2 b c \,d^{2} x \arctanh \left (c x \right )+d^{2} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {d^{2} b \dilog \left (c x \right )}{2}-\frac {d^{2} b \dilog \left (c x +1\right )}{2}-\frac {d^{2} b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {b c \,d^{2} x}{2}+\frac {5 d^{2} b \ln \left (c x -1\right )}{4}+\frac {3 d^{2} b \ln \left (c x +1\right )}{4}\)	\(142\)
risch	\(\frac {d^{2} a \,c^{2} x^{2}}{2}+2 d^{2} a c x -\frac {5 d^{2} a}{2}+d^{2} a \ln \left (-c x \right )-\frac {d^{2} \ln \left (-c x +1\right ) x^{2} b \,c^{2}}{4}-d^{2} b \ln \left (-c x +1\right ) c x +\frac {5 d^{2} b \ln \left (-c x +1\right )}{4}+\frac {b c \,d^{2} x}{2}-2 d^{2} b +\frac {d^{2} \dilog \left (-c x +1\right ) b}{2}+d^{2} b \ln \left (c x +1\right ) c x +\frac {3 d^{2} b \ln \left (c x +1\right )}{4}+\frac {d^{2} b \ln \left (c x +1\right ) x^{2} c^{2}}{4}-\frac {d^{2} b \dilog \left (c x +1\right )}{2}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^2*(a+b*arctanh(c*x))/x,x,method=_RETURNVERBOSE)

[Out]

1/2*d^2*a*c^2*x^2+2*d^2*a*c*x+d^2*a*ln(c*x)+1/2*d^2*b*arctanh(c*x)*c^2*x^2+2*b*c*d^2*x*arctanh(c*x)+d^2*b*arct
anh(c*x)*ln(c*x)-1/2*d^2*b*dilog(c*x)-1/2*d^2*b*dilog(c*x+1)-1/2*d^2*b*ln(c*x)*ln(c*x+1)+1/2*b*c*d^2*x+5/4*d^2
*b*ln(c*x-1)+3/4*d^2*b*ln(c*x+1)

________________________________________________________________________________________

Maxima [A]
time = 0.35, size = 173, normalized size = 1.52 \begin {gather*} \frac {1}{4} \, b c^{2} d^{2} x^{2} \log \left (c x + 1\right ) - \frac {1}{4} \, b c^{2} d^{2} x^{2} \log \left (-c x + 1\right ) + \frac {1}{2} \, a c^{2} d^{2} x^{2} + 2 \, a c d^{2} x + \frac {1}{2} \, b c d^{2} x + {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{2} - \frac {1}{2} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} b d^{2} + \frac {1}{2} \, {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} b d^{2} - \frac {1}{4} \, b d^{2} \log \left (c x + 1\right ) + \frac {1}{4} \, b d^{2} \log \left (c x - 1\right ) + a d^{2} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*c^2*d^2*x^2 + 2*a*c*d^2*x + a*d^2 + (b*c^2*d^2*x^2 + 2*b*c*d^2*x + b*d^2)*arctanh(c*x))/x, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{2} \left (\int 2 a c\, dx + \int \frac {a}{x}\, dx + \int a c^{2} x\, dx + \int 2 b c \operatorname {atanh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x}\, dx + \int b c^{2} x \operatorname {atanh}{\left (c x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(2*a*c, x) + Integral(a/x, x) + Integral(a*c**2*x, x) + Integral(2*b*c*atanh(c*x), x) + Integral
(b*atanh(c*x)/x, x) + Integral(b*c**2*x*atanh(c*x), x))

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*d*x + d)^2*(b*arctanh(c*x) + a)/x, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + c*d*x)^2)/x,x)

[Out]

int(((a + b*atanh(c*x))*(d + c*d*x)^2)/x, x)

________________________________________________________________________________________