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

Optimal. Leaf size=152 \[ \frac {1}{3} c^3 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {3}{2} c^2 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+3 a c d^3 x+a d^3 \log (x)+\frac {1}{6} b c^2 d^3 x^2+\frac {5}{3} b d^3 \log \left (1-c^2 x^2\right )-\frac {1}{2} b d^3 \text {Li}_2(-c x)+\frac {1}{2} b d^3 \text {Li}_2(c x)+\frac {3}{2} b c d^3 x-\frac {3}{2} b d^3 \tanh ^{-1}(c x)+3 b c d^3 x \tanh ^{-1}(c x) \]

[Out]

3*a*c*d^3*x+3/2*b*c*d^3*x+1/6*b*c^2*d^3*x^2-3/2*b*d^3*arctanh(c*x)+3*b*c*d^3*x*arctanh(c*x)+3/2*c^2*d^3*x^2*(a
+b*arctanh(c*x))+1/3*c^3*d^3*x^3*(a+b*arctanh(c*x))+a*d^3*ln(x)+5/3*b*d^3*ln(-c^2*x^2+1)-1/2*b*d^3*polylog(2,-
c*x)+1/2*b*d^3*polylog(2,c*x)

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Rubi [A]  time = 0.17, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {5940, 5910, 260, 5912, 5916, 321, 206, 266, 43} \[ -\frac {1}{2} b d^3 \text {PolyLog}(2,-c x)+\frac {1}{2} b d^3 \text {PolyLog}(2,c x)+\frac {1}{3} c^3 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {3}{2} c^2 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+3 a c d^3 x+a d^3 \log (x)+\frac {1}{6} b c^2 d^3 x^2+\frac {5}{3} b d^3 \log \left (1-c^2 x^2\right )+\frac {3}{2} b c d^3 x-\frac {3}{2} b d^3 \tanh ^{-1}(c x)+3 b c d^3 x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

3*a*c*d^3*x + (3*b*c*d^3*x)/2 + (b*c^2*d^3*x^2)/6 - (3*b*d^3*ArcTanh[c*x])/2 + 3*b*c*d^3*x*ArcTanh[c*x] + (3*c
^2*d^3*x^2*(a + b*ArcTanh[c*x]))/2 + (c^3*d^3*x^3*(a + b*ArcTanh[c*x]))/3 + a*d^3*Log[x] + (5*b*d^3*Log[1 - c^
2*x^2])/3 - (b*d^3*PolyLog[2, -(c*x)])/2 + (b*d^3*PolyLog[2, c*x])/2

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])

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])

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 321

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 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 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, 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 5940

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)^3 \left (a+b \tanh ^{-1}(c x)\right )}{x} \, dx &=\int \left (3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )+c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )\right ) \, dx\\ &=d^3 \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx+\left (3 c d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx+\left (3 c^2 d^3\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx+\left (c^3 d^3\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx\\ &=3 a c d^3 x+\frac {3}{2} c^2 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{3} c^3 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+a d^3 \log (x)-\frac {1}{2} b d^3 \text {Li}_2(-c x)+\frac {1}{2} b d^3 \text {Li}_2(c x)+\left (3 b c d^3\right ) \int \tanh ^{-1}(c x) \, dx-\frac {1}{2} \left (3 b c^3 d^3\right ) \int \frac {x^2}{1-c^2 x^2} \, dx-\frac {1}{3} \left (b c^4 d^3\right ) \int \frac {x^3}{1-c^2 x^2} \, dx\\ &=3 a c d^3 x+\frac {3}{2} b c d^3 x+3 b c d^3 x \tanh ^{-1}(c x)+\frac {3}{2} c^2 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{3} c^3 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+a d^3 \log (x)-\frac {1}{2} b d^3 \text {Li}_2(-c x)+\frac {1}{2} b d^3 \text {Li}_2(c x)-\frac {1}{2} \left (3 b c d^3\right ) \int \frac {1}{1-c^2 x^2} \, dx-\left (3 b c^2 d^3\right ) \int \frac {x}{1-c^2 x^2} \, dx-\frac {1}{6} \left (b c^4 d^3\right ) \operatorname {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )\\ &=3 a c d^3 x+\frac {3}{2} b c d^3 x-\frac {3}{2} b d^3 \tanh ^{-1}(c x)+3 b c d^3 x \tanh ^{-1}(c x)+\frac {3}{2} c^2 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{3} c^3 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+a d^3 \log (x)+\frac {3}{2} b d^3 \log \left (1-c^2 x^2\right )-\frac {1}{2} b d^3 \text {Li}_2(-c x)+\frac {1}{2} b d^3 \text {Li}_2(c x)-\frac {1}{6} \left (b c^4 d^3\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=3 a c d^3 x+\frac {3}{2} b c d^3 x+\frac {1}{6} b c^2 d^3 x^2-\frac {3}{2} b d^3 \tanh ^{-1}(c x)+3 b c d^3 x \tanh ^{-1}(c x)+\frac {3}{2} c^2 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{3} c^3 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+a d^3 \log (x)+\frac {5}{3} b d^3 \log \left (1-c^2 x^2\right )-\frac {1}{2} b d^3 \text {Li}_2(-c x)+\frac {1}{2} b d^3 \text {Li}_2(c x)\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 148, normalized size = 0.97 \[ \frac {1}{12} d^3 \left (4 a c^3 x^3+18 a c^2 x^2+36 a c x+12 a \log (x)+4 b c^3 x^3 \tanh ^{-1}(c x)+2 b c^2 x^2+18 b \log \left (1-c^2 x^2\right )+2 b \log \left (c^2 x^2-1\right )+18 b c^2 x^2 \tanh ^{-1}(c x)-6 b \text {Li}_2(-c x)+6 b \text {Li}_2(c x)+18 b c x+9 b \log (1-c x)-9 b \log (c x+1)+36 b c x \tanh ^{-1}(c x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^3*(36*a*c*x + 18*b*c*x + 18*a*c^2*x^2 + 2*b*c^2*x^2 + 4*a*c^3*x^3 + 36*b*c*x*ArcTanh[c*x] + 18*b*c^2*x^2*Ar
cTanh[c*x] + 4*b*c^3*x^3*ArcTanh[c*x] + 12*a*Log[x] + 9*b*Log[1 - c*x] - 9*b*Log[1 + c*x] + 18*b*Log[1 - c^2*x
^2] + 2*b*Log[-1 + c^2*x^2] - 6*b*PolyLog[2, -(c*x)] + 6*b*PolyLog[2, c*x]))/12

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a c^{3} d^{3} x^{3} + 3 \, a c^{2} d^{3} x^{2} + 3 \, a c d^{3} x + a d^{3} + {\left (b c^{3} d^{3} x^{3} + 3 \, b c^{2} d^{3} x^{2} + 3 \, b c d^{3} x + b d^{3}\right )} \operatorname {artanh}\left (c x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*c^3*d^3*x^3 + 3*a*c^2*d^3*x^2 + 3*a*c*d^3*x + a*d^3 + (b*c^3*d^3*x^3 + 3*b*c^2*d^3*x^2 + 3*b*c*d^3
*x + b*d^3)*arctanh(c*x))/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d x + d\right )}^{3} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*d*x + d)^3*(b*arctanh(c*x) + a)/x, x)

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maple [A]  time = 0.05, size = 182, normalized size = 1.20 \[ \frac {d^{3} a \,c^{3} x^{3}}{3}+\frac {3 d^{3} a \,c^{2} x^{2}}{2}+3 a c \,d^{3} x +d^{3} a \ln \left (c x \right )+\frac {d^{3} b \arctanh \left (c x \right ) c^{3} x^{3}}{3}+\frac {3 d^{3} b \arctanh \left (c x \right ) c^{2} x^{2}}{2}+3 b c \,d^{3} x \arctanh \left (c x \right )+d^{3} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {d^{3} b \dilog \left (c x \right )}{2}-\frac {d^{3} b \dilog \left (c x +1\right )}{2}-\frac {d^{3} b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {b \,c^{2} d^{3} x^{2}}{6}+\frac {3 b c \,d^{3} x}{2}+\frac {29 d^{3} b \ln \left (c x -1\right )}{12}+\frac {11 d^{3} b \ln \left (c x +1\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*d^3*a*c^3*x^3+3/2*d^3*a*c^2*x^2+3*a*c*d^3*x+d^3*a*ln(c*x)+1/3*d^3*b*arctanh(c*x)*c^3*x^3+3/2*d^3*b*arctanh
(c*x)*c^2*x^2+3*b*c*d^3*x*arctanh(c*x)+d^3*b*arctanh(c*x)*ln(c*x)-1/2*d^3*b*dilog(c*x)-1/2*d^3*b*dilog(c*x+1)-
1/2*d^3*b*ln(c*x)*ln(c*x+1)+1/6*b*c^2*d^3*x^2+3/2*b*c*d^3*x+29/12*d^3*b*ln(c*x-1)+11/12*d^3*b*ln(c*x+1)

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maxima [A]  time = 0.46, size = 228, normalized size = 1.50 \[ \frac {1}{3} \, a c^{3} d^{3} x^{3} + \frac {3}{2} \, a c^{2} d^{3} x^{2} + \frac {1}{6} \, b c^{2} d^{3} x^{2} + 3 \, a c d^{3} x + \frac {3}{2} \, b c d^{3} x + \frac {3}{2} \, {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{3} - \frac {1}{2} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} b d^{3} + \frac {1}{2} \, {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} b d^{3} - \frac {7}{12} \, b d^{3} \log \left (c x + 1\right ) + \frac {11}{12} \, b d^{3} \log \left (c x - 1\right ) + a d^{3} \log \relax (x) + \frac {1}{12} \, {\left (2 \, b c^{3} d^{3} x^{3} + 9 \, b c^{2} d^{3} x^{2}\right )} \log \left (c x + 1\right ) - \frac {1}{12} \, {\left (2 \, b c^{3} d^{3} x^{3} + 9 \, b c^{2} d^{3} x^{2}\right )} \log \left (-c x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*c^3*d^3*x^3 + 3/2*a*c^2*d^3*x^2 + 1/6*b*c^2*d^3*x^2 + 3*a*c*d^3*x + 3/2*b*c*d^3*x + 3/2*(2*c*x*arctanh(c
*x) + log(-c^2*x^2 + 1))*b*d^3 - 1/2*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1))*b*d^3 + 1/2*(log(c*x + 1)*log(
-c*x) + dilog(c*x + 1))*b*d^3 - 7/12*b*d^3*log(c*x + 1) + 11/12*b*d^3*log(c*x - 1) + a*d^3*log(x) + 1/12*(2*b*
c^3*d^3*x^3 + 9*b*c^2*d^3*x^2)*log(c*x + 1) - 1/12*(2*b*c^3*d^3*x^3 + 9*b*c^2*d^3*x^2)*log(-c*x + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^3}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + c*d*x)^3)/x,x)

[Out]

int(((a + b*atanh(c*x))*(d + c*d*x)^3)/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{3} \left (\int 3 a c\, dx + \int \frac {a}{x}\, dx + \int 3 a c^{2} x\, dx + \int a c^{3} x^{2}\, dx + \int 3 b c \operatorname {atanh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x}\, dx + \int 3 b c^{2} x \operatorname {atanh}{\left (c x \right )}\, dx + \int b c^{3} x^{2} \operatorname {atanh}{\left (c x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*(Integral(3*a*c, x) + Integral(a/x, x) + Integral(3*a*c**2*x, x) + Integral(a*c**3*x**2, x) + Integral(3*
b*c*atanh(c*x), x) + Integral(b*atanh(c*x)/x, x) + Integral(3*b*c**2*x*atanh(c*x), x) + Integral(b*c**3*x**2*a
tanh(c*x), x))

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