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

Optimal. Leaf size=160 \[ -\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^3 d^3 x+3 a c^2 d^3 \log (x)+b c^3 d^3 x \tanh ^{-1}(c x)-\frac {3}{2} b c^2 d^3 \text {Li}_2(-c x)+\frac {3}{2} b c^2 d^3 \text {Li}_2(c x)-b c^2 d^3 \log \left (1-c^2 x^2\right )+3 b c^2 d^3 \log (x)+\frac {1}{2} b c^2 d^3 \tanh ^{-1}(c x)-\frac {b c d^3}{2 x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {5940, 5910, 260, 5916, 325, 206, 266, 36, 29, 31, 5912} \[ -\frac {3}{2} b c^2 d^3 \text {PolyLog}(2,-c x)+\frac {3}{2} b c^2 d^3 \text {PolyLog}(2,c x)-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^3 d^3 x+3 a c^2 d^3 \log (x)-b c^2 d^3 \log \left (1-c^2 x^2\right )+3 b c^2 d^3 \log (x)+\frac {1}{2} b c^2 d^3 \tanh ^{-1}(c x)+b c^3 d^3 x \tanh ^{-1}(c x)-\frac {b c d^3}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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^3} \, dx &=\int \left (c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^3}+\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}+\frac {3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^3 \int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx+\left (3 c d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx+\left (3 c^2 d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx+\left (c^3 d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx\\ &=a c^3 d^3 x-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+3 a c^2 d^3 \log (x)-\frac {3}{2} b c^2 d^3 \text {Li}_2(-c x)+\frac {3}{2} b c^2 d^3 \text {Li}_2(c x)+\frac {1}{2} \left (b c d^3\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (3 b c^2 d^3\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx+\left (b c^3 d^3\right ) \int \tanh ^{-1}(c x) \, dx\\ &=-\frac {b c d^3}{2 x}+a c^3 d^3 x+b c^3 d^3 x \tanh ^{-1}(c x)-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+3 a c^2 d^3 \log (x)-\frac {3}{2} b c^2 d^3 \text {Li}_2(-c x)+\frac {3}{2} b c^2 d^3 \text {Li}_2(c x)+\frac {1}{2} \left (3 b c^2 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )+\frac {1}{2} \left (b c^3 d^3\right ) \int \frac {1}{1-c^2 x^2} \, dx-\left (b c^4 d^3\right ) \int \frac {x}{1-c^2 x^2} \, dx\\ &=-\frac {b c d^3}{2 x}+a c^3 d^3 x+\frac {1}{2} b c^2 d^3 \tanh ^{-1}(c x)+b c^3 d^3 x \tanh ^{-1}(c x)-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+3 a c^2 d^3 \log (x)+\frac {1}{2} b c^2 d^3 \log \left (1-c^2 x^2\right )-\frac {3}{2} b c^2 d^3 \text {Li}_2(-c x)+\frac {3}{2} b c^2 d^3 \text {Li}_2(c x)+\frac {1}{2} \left (3 b c^2 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (3 b c^4 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^3}{2 x}+a c^3 d^3 x+\frac {1}{2} b c^2 d^3 \tanh ^{-1}(c x)+b c^3 d^3 x \tanh ^{-1}(c x)-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+3 a c^2 d^3 \log (x)+3 b c^2 d^3 \log (x)-b c^2 d^3 \log \left (1-c^2 x^2\right )-\frac {3}{2} b c^2 d^3 \text {Li}_2(-c x)+\frac {3}{2} b c^2 d^3 \text {Li}_2(c x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 165, normalized size = 1.03 \[ \frac {d^3 \left (4 a c^3 x^3+12 a c^2 x^2 \log (x)-12 a c x-2 a+4 b c^3 x^3 \tanh ^{-1}(c x)-6 b c^2 x^2 \text {Li}_2(-c x)+6 b c^2 x^2 \text {Li}_2(c x)+12 b c^2 x^2 \log (c x)-b c^2 x^2 \log (1-c x)+b c^2 x^2 \log (c x+1)-4 b c^2 x^2 \log \left (1-c^2 x^2\right )-2 b c x-12 b c x \tanh ^{-1}(c x)-2 b \tanh ^{-1}(c x)\right )}{4 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a c^{3} d^{3} x^{3} + 3 \, a c^{2} d^{3} x^{2} + 3 \, a c d^{3} x + a d^{3} + {\left (b c^{3} d^{3} x^{3} + 3 \, b c^{2} d^{3} x^{2} + 3 \, b c d^{3} x + b d^{3}\right )} \operatorname {artanh}\left (c x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 200, normalized size = 1.25 \[ a \,c^{3} d^{3} x +3 c^{2} d^{3} a \ln \left (c x \right )-\frac {3 c \,d^{3} a}{x}-\frac {d^{3} a}{2 x^{2}}+b \,c^{3} d^{3} x \arctanh \left (c x \right )+3 c^{2} d^{3} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {3 c \,d^{3} b \arctanh \left (c x \right )}{x}-\frac {d^{3} b \arctanh \left (c x \right )}{2 x^{2}}-\frac {3 c^{2} d^{3} b \dilog \left (c x \right )}{2}-\frac {3 c^{2} d^{3} b \dilog \left (c x +1\right )}{2}-\frac {3 c^{2} d^{3} b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {b c \,d^{3}}{2 x}+3 c^{2} d^{3} b \ln \left (c x \right )-\frac {5 c^{2} d^{3} b \ln \left (c x -1\right )}{4}-\frac {3 c^{2} d^{3} b \ln \left (c x +1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a c^{3} d^{3} x + \frac {1}{2} \, {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b c^{2} d^{3} + \frac {3}{2} \, b c^{2} d^{3} \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{x}\,{d x} + 3 \, a c^{2} d^{3} \log \relax (x) - \frac {3}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x}\right )} b c d^{3} + \frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b d^{3} - \frac {3 \, a c d^{3}}{x} - \frac {a d^{3}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________