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

Optimal. Leaf size=268 \[ \frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (c x+1)}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (c x+1)^2}+\frac {6 c^2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 d^3 x^2}+\frac {3 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 x}+\frac {6 a c^2 \log (x)}{d^3}-\frac {3 b c^2 \text {Li}_2(-c x)}{d^3}+\frac {3 b c^2 \text {Li}_2(c x)}{d^3}-\frac {3 b c^2 \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{d^3}+\frac {3 b c^2 \log \left (1-c^2 x^2\right )}{2 d^3}+\frac {13 b c^2}{8 d^3 (c x+1)}+\frac {b c^2}{8 d^3 (c x+1)^2}-\frac {3 b c^2 \log (x)}{d^3}-\frac {9 b c^2 \tanh ^{-1}(c x)}{8 d^3}-\frac {b c}{2 d^3 x} \]

[Out]

-1/2*b*c/d^3/x+1/8*b*c^2/d^3/(c*x+1)^2+13/8*b*c^2/d^3/(c*x+1)-9/8*b*c^2*arctanh(c*x)/d^3+1/2*(-a-b*arctanh(c*x
))/d^3/x^2+3*c*(a+b*arctanh(c*x))/d^3/x+1/2*c^2*(a+b*arctanh(c*x))/d^3/(c*x+1)^2+3*c^2*(a+b*arctanh(c*x))/d^3/
(c*x+1)+6*a*c^2*ln(x)/d^3-3*b*c^2*ln(x)/d^3+6*c^2*(a+b*arctanh(c*x))*ln(2/(c*x+1))/d^3+3/2*b*c^2*ln(-c^2*x^2+1
)/d^3-3*b*c^2*polylog(2,-c*x)/d^3+3*b*c^2*polylog(2,c*x)/d^3-3*b*c^2*polylog(2,1-2/(c*x+1))/d^3

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Rubi [A]  time = 0.32, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 16, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5940, 5916, 325, 206, 266, 36, 29, 31, 5912, 5926, 627, 44, 207, 5918, 2402, 2315} \[ -\frac {3 b c^2 \text {PolyLog}(2,-c x)}{d^3}+\frac {3 b c^2 \text {PolyLog}(2,c x)}{d^3}-\frac {3 b c^2 \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{d^3}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (c x+1)}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (c x+1)^2}+\frac {6 c^2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 d^3 x^2}+\frac {3 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 x}+\frac {6 a c^2 \log (x)}{d^3}+\frac {3 b c^2 \log \left (1-c^2 x^2\right )}{2 d^3}+\frac {13 b c^2}{8 d^3 (c x+1)}+\frac {b c^2}{8 d^3 (c x+1)^2}-\frac {3 b c^2 \log (x)}{d^3}-\frac {9 b c^2 \tanh ^{-1}(c x)}{8 d^3}-\frac {b c}{2 d^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(2*d^3*x) + (b*c^2)/(8*d^3*(1 + c*x)^2) + (13*b*c^2)/(8*d^3*(1 + c*x)) - (9*b*c^2*ArcTanh[c*x])/(8*d^3)
- (a + b*ArcTanh[c*x])/(2*d^3*x^2) + (3*c*(a + b*ArcTanh[c*x]))/(d^3*x) + (c^2*(a + b*ArcTanh[c*x]))/(2*d^3*(
1 + c*x)^2) + (3*c^2*(a + b*ArcTanh[c*x]))/(d^3*(1 + c*x)) + (6*a*c^2*Log[x])/d^3 - (3*b*c^2*Log[x])/d^3 + (6*
c^2*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/d^3 + (3*b*c^2*Log[1 - c^2*x^2])/(2*d^3) - (3*b*c^2*PolyLog[2, -(c*
x)])/d^3 + (3*b*c^2*PolyLog[2, c*x])/d^3 - (3*b*c^2*PolyLog[2, 1 - 2/(1 + c*x)])/d^3

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 207

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

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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 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 5918

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

Rule 5926

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcTanh[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -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 {a+b \tanh ^{-1}(c x)}{x^3 (d+c d x)^3} \, dx &=\int \left (\frac {a+b \tanh ^{-1}(c x)}{d^3 x^3}-\frac {3 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 x^2}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 x}-\frac {c^3 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)^3}-\frac {3 c^3 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)^2}-\frac {6 c^3 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx}{d^3}-\frac {(3 c) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx}{d^3}+\frac {\left (6 c^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx}{d^3}-\frac {c^3 \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{d^3}-\frac {\left (3 c^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{d^3}-\frac {\left (6 c^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1+c x} \, dx}{d^3}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{2 d^3 x^2}+\frac {3 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}+\frac {6 a c^2 \log (x)}{d^3}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {3 b c^2 \text {Li}_2(-c x)}{d^3}+\frac {3 b c^2 \text {Li}_2(c x)}{d^3}+\frac {(b c) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx}{2 d^3}-\frac {\left (3 b c^2\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx}{d^3}-\frac {\left (b c^3\right ) \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{2 d^3}-\frac {\left (3 b c^3\right ) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{d^3}-\frac {\left (6 b c^3\right ) \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d^3}\\ &=-\frac {b c}{2 d^3 x}-\frac {a+b \tanh ^{-1}(c x)}{2 d^3 x^2}+\frac {3 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}+\frac {6 a c^2 \log (x)}{d^3}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {3 b c^2 \text {Li}_2(-c x)}{d^3}+\frac {3 b c^2 \text {Li}_2(c x)}{d^3}-\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )}{2 d^3}-\frac {\left (6 b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{d^3}-\frac {\left (b c^3\right ) \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{2 d^3}+\frac {\left (b c^3\right ) \int \frac {1}{1-c^2 x^2} \, dx}{2 d^3}-\frac {\left (3 b c^3\right ) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{d^3}\\ &=-\frac {b c}{2 d^3 x}+\frac {b c^2 \tanh ^{-1}(c x)}{2 d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 d^3 x^2}+\frac {3 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}+\frac {6 a c^2 \log (x)}{d^3}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {3 b c^2 \text {Li}_2(-c x)}{d^3}+\frac {3 b c^2 \text {Li}_2(c x)}{d^3}-\frac {3 b c^2 \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d^3}-\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d^3}-\frac {\left (b c^3\right ) \int \left (\frac {1}{2 (1+c x)^3}+\frac {1}{4 (1+c x)^2}-\frac {1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{2 d^3}-\frac {\left (3 b c^3\right ) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{d^3}-\frac {\left (3 b c^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )}{2 d^3}\\ &=-\frac {b c}{2 d^3 x}+\frac {b c^2}{8 d^3 (1+c x)^2}+\frac {13 b c^2}{8 d^3 (1+c x)}+\frac {b c^2 \tanh ^{-1}(c x)}{2 d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 d^3 x^2}+\frac {3 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}+\frac {6 a c^2 \log (x)}{d^3}-\frac {3 b c^2 \log (x)}{d^3}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}+\frac {3 b c^2 \log \left (1-c^2 x^2\right )}{2 d^3}-\frac {3 b c^2 \text {Li}_2(-c x)}{d^3}+\frac {3 b c^2 \text {Li}_2(c x)}{d^3}-\frac {3 b c^2 \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d^3}+\frac {\left (b c^3\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{8 d^3}+\frac {\left (3 b c^3\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{2 d^3}\\ &=-\frac {b c}{2 d^3 x}+\frac {b c^2}{8 d^3 (1+c x)^2}+\frac {13 b c^2}{8 d^3 (1+c x)}-\frac {9 b c^2 \tanh ^{-1}(c x)}{8 d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 d^3 x^2}+\frac {3 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}+\frac {6 a c^2 \log (x)}{d^3}-\frac {3 b c^2 \log (x)}{d^3}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}+\frac {3 b c^2 \log \left (1-c^2 x^2\right )}{2 d^3}-\frac {3 b c^2 \text {Li}_2(-c x)}{d^3}+\frac {3 b c^2 \text {Li}_2(c x)}{d^3}-\frac {3 b c^2 \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d^3}\\ \end {align*}

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Mathematica [A]  time = 1.47, size = 220, normalized size = 0.82 \[ \frac {\frac {96 a c^2}{c x+1}+\frac {16 a c^2}{(c x+1)^2}+192 a c^2 \log (x)-192 a c^2 \log (c x+1)+\frac {96 a c}{x}-\frac {16 a}{x^2}+b c^2 \left (-96 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+4 \tanh ^{-1}(c x) \left (-\frac {4}{c^2 x^2}+\frac {24}{c x}+48 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-14 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )+14 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )+4\right )-96 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )-\frac {16}{c x}-28 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )+28 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )\right )}{32 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-16*a)/x^2 + (96*a*c)/x + (16*a*c^2)/(1 + c*x)^2 + (96*a*c^2)/(1 + c*x) + 192*a*c^2*Log[x] - 192*a*c^2*Log[1
+ c*x] + b*c^2*(-16/(c*x) + 28*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] - 96*Log[(c*x)/Sqrt[1 - c^2*x^2]]
- 96*PolyLog[2, E^(-2*ArcTanh[c*x])] - 28*Sinh[2*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(4 - 4/(c^2*x^2) + 24/(c*x) +
14*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] + 48*Log[1 - E^(-2*ArcTanh[c*x])] - 14*Sinh[2*ArcTanh[c*x]] - S
inh[4*ArcTanh[c*x]]) - Sinh[4*ArcTanh[c*x]]))/(32*d^3)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{c^{3} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{5} + 3 \, c d^{3} x^{4} + d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctanh(c*x) + a)/(c^3*d^3*x^6 + 3*c^2*d^3*x^5 + 3*c*d^3*x^4 + d^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 394, normalized size = 1.47 \[ \frac {3 c^{2} b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{d^{3}}-\frac {b \arctanh \left (c x \right )}{2 d^{3} x^{2}}-\frac {3 c^{2} b \dilog \left (c x \right )}{d^{3}}-\frac {3 c^{2} b \dilog \left (c x +1\right )}{d^{3}}+\frac {3 c a}{d^{3} x}+\frac {c^{2} a}{2 d^{3} \left (c x +1\right )^{2}}+\frac {3 c^{2} a}{d^{3} \left (c x +1\right )}+\frac {3 c^{2} b \ln \left (c x +1\right )^{2}}{2 d^{3}}-\frac {6 c^{2} a \ln \left (c x +1\right )}{d^{3}}+\frac {6 c^{2} a \ln \left (c x \right )}{d^{3}}+\frac {15 c^{2} b \ln \left (c x +1\right )}{16 d^{3}}+\frac {33 c^{2} b \ln \left (c x -1\right )}{16 d^{3}}-\frac {3 c^{2} b \ln \left (c x \right )}{d^{3}}-\frac {b c}{2 d^{3} x}+\frac {b \,c^{2}}{8 d^{3} \left (c x +1\right )^{2}}+\frac {13 b \,c^{2}}{8 d^{3} \left (c x +1\right )}-\frac {a}{2 d^{3} x^{2}}-\frac {3 c^{2} b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{d^{3}}-\frac {3 c^{2} b \ln \left (c x \right ) \ln \left (c x +1\right )}{d^{3}}+\frac {3 c^{2} b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{d^{3}}-\frac {6 c^{2} b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d^{3}}+\frac {6 c^{2} b \arctanh \left (c x \right ) \ln \left (c x \right )}{d^{3}}+\frac {3 c b \arctanh \left (c x \right )}{d^{3} x}+\frac {3 c^{2} b \arctanh \left (c x \right )}{d^{3} \left (c x +1\right )}+\frac {c^{2} b \arctanh \left (c x \right )}{2 d^{3} \left (c x +1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*c*a/d^3/x+1/2*c^2*a/d^3/(c*x+1)^2+3*c^2*a/d^3/(c*x+1)+3/2*c^2*b/d^3*ln(c*x+1)^2+3*c^2*b/d^3*dilog(1/2+1/2*c*
x)-6*c^2*a/d^3*ln(c*x+1)+6*c^2*a/d^3*ln(c*x)-1/2*b/d^3*arctanh(c*x)/x^2+15/16*c^2*b/d^3*ln(c*x+1)+33/16*c^2*b/
d^3*ln(c*x-1)-3*c^2*b/d^3*ln(c*x)-3*c^2*b/d^3*dilog(c*x)-3*c^2*b/d^3*dilog(c*x+1)-1/2*b*c/d^3/x+1/8*b*c^2/d^3/
(c*x+1)^2+13/8*b*c^2/d^3/(c*x+1)-1/2*a/d^3/x^2-6*c^2*b/d^3*arctanh(c*x)*ln(c*x+1)-3*c^2*b/d^3*ln(-1/2*c*x+1/2)
*ln(c*x+1)-3*c^2*b/d^3*ln(c*x)*ln(c*x+1)+6*c^2*b/d^3*arctanh(c*x)*ln(c*x)+3*c^2*b/d^3*ln(-1/2*c*x+1/2)*ln(1/2+
1/2*c*x)+3*c*b/d^3*arctanh(c*x)/x+3*c^2*b/d^3*arctanh(c*x)/(c*x+1)+1/2*c^2*b/d^3*arctanh(c*x)/(c*x+1)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a {\left (\frac {12 \, c^{3} x^{3} + 18 \, c^{2} x^{2} + 4 \, c x - 1}{c^{2} d^{3} x^{4} + 2 \, c d^{3} x^{3} + d^{3} x^{2}} - \frac {12 \, c^{2} \log \left (c x + 1\right )}{d^{3}} + \frac {12 \, c^{2} \log \relax (x)}{d^{3}}\right )} + \frac {1}{2} \, b \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{c^{3} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{5} + 3 \, c d^{3} x^{4} + d^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*((12*c^3*x^3 + 18*c^2*x^2 + 4*c*x - 1)/(c^2*d^3*x^4 + 2*c*d^3*x^3 + d^3*x^2) - 12*c^2*log(c*x + 1)/d^3 +
12*c^2*log(x)/d^3) + 1/2*b*integrate((log(c*x + 1) - log(-c*x + 1))/(c^3*d^3*x^6 + 3*c^2*d^3*x^5 + 3*c*d^3*x^
4 + d^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{3} x^{6} + 3 c^{2} x^{5} + 3 c x^{4} + x^{3}}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{6} + 3 c^{2} x^{5} + 3 c x^{4} + x^{3}}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**3*x**6 + 3*c**2*x**5 + 3*c*x**4 + x**3), x) + Integral(b*atanh(c*x)/(c**3*x**6 + 3*c**2*x**5 +
3*c*x**4 + x**3), x))/d**3

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