3.64 \(\int \frac {a+b \tanh ^{-1}(c x)}{x^2 (d+c d x)^3} \, dx\)

Optimal. Leaf size=218 \[ -\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (c x+1)}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (c x+1)^2}-\frac {a+b \tanh ^{-1}(c x)}{d^3 x}-\frac {3 c \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac {3 a c \log (x)}{d^3}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d^3}+\frac {3 b c \text {Li}_2(-c x)}{2 d^3}-\frac {3 b c \text {Li}_2(c x)}{2 d^3}+\frac {3 b c \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 d^3}-\frac {9 b c}{8 d^3 (c x+1)}-\frac {b c}{8 d^3 (c x+1)^2}+\frac {b c \log (x)}{d^3}+\frac {9 b c \tanh ^{-1}(c x)}{8 d^3} \]

[Out]

-1/8*b*c/d^3/(c*x+1)^2-9/8*b*c/d^3/(c*x+1)+9/8*b*c*arctanh(c*x)/d^3+(-a-b*arctanh(c*x))/d^3/x-1/2*c*(a+b*arcta
nh(c*x))/d^3/(c*x+1)^2-2*c*(a+b*arctanh(c*x))/d^3/(c*x+1)-3*a*c*ln(x)/d^3+b*c*ln(x)/d^3-3*c*(a+b*arctanh(c*x))
*ln(2/(c*x+1))/d^3-1/2*b*c*ln(-c^2*x^2+1)/d^3+3/2*b*c*polylog(2,-c*x)/d^3-3/2*b*c*polylog(2,c*x)/d^3+3/2*b*c*p
olylog(2,1-2/(c*x+1))/d^3

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Rubi [A]  time = 0.27, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 14, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5940, 5916, 266, 36, 29, 31, 5912, 5926, 627, 44, 207, 5918, 2402, 2315} \[ \frac {3 b c \text {PolyLog}(2,-c x)}{2 d^3}-\frac {3 b c \text {PolyLog}(2,c x)}{2 d^3}+\frac {3 b c \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 d^3}-\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (c x+1)}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (c x+1)^2}-\frac {a+b \tanh ^{-1}(c x)}{d^3 x}-\frac {3 c \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac {3 a c \log (x)}{d^3}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d^3}-\frac {9 b c}{8 d^3 (c x+1)}-\frac {b c}{8 d^3 (c x+1)^2}+\frac {b c \log (x)}{d^3}+\frac {9 b c \tanh ^{-1}(c x)}{8 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(x^2*(d + c*d*x)^3),x]

[Out]

-(b*c)/(8*d^3*(1 + c*x)^2) - (9*b*c)/(8*d^3*(1 + c*x)) + (9*b*c*ArcTanh[c*x])/(8*d^3) - (a + b*ArcTanh[c*x])/(
d^3*x) - (c*(a + b*ArcTanh[c*x]))/(2*d^3*(1 + c*x)^2) - (2*c*(a + b*ArcTanh[c*x]))/(d^3*(1 + c*x)) - (3*a*c*Lo
g[x])/d^3 + (b*c*Log[x])/d^3 - (3*c*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/d^3 - (b*c*Log[1 - c^2*x^2])/(2*d^3
) + (3*b*c*PolyLog[2, -(c*x)])/(2*d^3) - (3*b*c*PolyLog[2, c*x])/(2*d^3) + (3*b*c*PolyLog[2, 1 - 2/(1 + c*x)])
/(2*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 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 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^2 (d+c d x)^3} \, dx &=\int \left (\frac {a+b \tanh ^{-1}(c x)}{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 )}{d^3 (1+c x)^3}+\frac {2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)^2}+\frac {3 c^2 \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^2} \, dx}{d^3}-\frac {(3 c) \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx}{d^3}+\frac {c^2 \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{d^3}+\frac {\left (2 c^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{d^3}+\frac {\left (3 c^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1+c x} \, dx}{d^3}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d^3 x}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}-\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}-\frac {3 a c \log (x)}{d^3}-\frac {3 c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}+\frac {3 b c \text {Li}_2(-c x)}{2 d^3}-\frac {3 b c \text {Li}_2(c x)}{2 d^3}+\frac {(b c) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx}{d^3}+\frac {\left (b c^2\right ) \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{2 d^3}+\frac {\left (2 b c^2\right ) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{d^3}+\frac {\left (3 b c^2\right ) \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d^3}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d^3 x}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}-\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}-\frac {3 a c \log (x)}{d^3}-\frac {3 c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}+\frac {3 b c \text {Li}_2(-c x)}{2 d^3}-\frac {3 b c \text {Li}_2(c x)}{2 d^3}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )}{2 d^3}+\frac {(3 b c) \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^2\right ) \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{2 d^3}+\frac {\left (2 b c^2\right ) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{d^3}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d^3 x}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}-\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}-\frac {3 a c \log (x)}{d^3}-\frac {3 c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}+\frac {3 b c \text {Li}_2(-c x)}{2 d^3}-\frac {3 b c \text {Li}_2(c x)}{2 d^3}+\frac {3 b c \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^3}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d^3}+\frac {\left (b c^2\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 (2 b c^2\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 (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )}{2 d^3}\\ &=-\frac {b c}{8 d^3 (1+c x)^2}-\frac {9 b c}{8 d^3 (1+c x)}-\frac {a+b \tanh ^{-1}(c x)}{d^3 x}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}-\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}-\frac {3 a c \log (x)}{d^3}+\frac {b c \log (x)}{d^3}-\frac {3 c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d^3}+\frac {3 b c \text {Li}_2(-c x)}{2 d^3}-\frac {3 b c \text {Li}_2(c x)}{2 d^3}+\frac {3 b c \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^3}-\frac {\left (b c^2\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{8 d^3}-\frac {\left (b c^2\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{d^3}\\ &=-\frac {b c}{8 d^3 (1+c x)^2}-\frac {9 b c}{8 d^3 (1+c x)}+\frac {9 b c \tanh ^{-1}(c x)}{8 d^3}-\frac {a+b \tanh ^{-1}(c x)}{d^3 x}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}-\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}-\frac {3 a c \log (x)}{d^3}+\frac {b c \log (x)}{d^3}-\frac {3 c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d^3}+\frac {3 b c \text {Li}_2(-c x)}{2 d^3}-\frac {3 b c \text {Li}_2(c x)}{2 d^3}+\frac {3 b c \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 1.24, size = 186, normalized size = 0.85 \[ \frac {-\frac {64 a c}{c x+1}-\frac {16 a c}{(c x+1)^2}-96 a c \log (x)+96 a c \log (c x+1)-\frac {32 a}{x}+b c \left (32 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+48 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+20 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )-20 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (-\frac {8}{c x}-24 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+10 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )-10 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )\right )\right )}{32 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])/(x^2*(d + c*d*x)^3),x]

[Out]

((-32*a)/x - (16*a*c)/(1 + c*x)^2 - (64*a*c)/(1 + c*x) - 96*a*c*Log[x] + 96*a*c*Log[1 + c*x] + b*c*(-20*Cosh[2
*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] + 32*Log[(c*x)/Sqrt[1 - c^2*x^2]] + 48*PolyLog[2, E^(-2*ArcTanh[c*x])] +
20*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(-8/(c*x) - 10*Cosh[2*ArcTanh[c*x]] - Cosh[4*
ArcTanh[c*x]] - 24*Log[1 - E^(-2*ArcTanh[c*x])] + 10*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]])))/(32*d^3)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{c^{3} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{4} + 3 \, c d^{3} x^{3} + d^{3} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^2/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^2/(c*d*x+d)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 319, normalized size = 1.46 \[ -\frac {a}{d^{3} x}-\frac {3 c a \ln \left (c x \right )}{d^{3}}-\frac {c a}{2 d^{3} \left (c x +1\right )^{2}}-\frac {2 c a}{d^{3} \left (c x +1\right )}+\frac {3 c a \ln \left (c x +1\right )}{d^{3}}-\frac {b \arctanh \left (c x \right )}{d^{3} x}-\frac {3 c b \arctanh \left (c x \right ) \ln \left (c x \right )}{d^{3}}-\frac {c b \arctanh \left (c x \right )}{2 d^{3} \left (c x +1\right )^{2}}-\frac {2 c b \arctanh \left (c x \right )}{d^{3} \left (c x +1\right )}+\frac {3 c b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d^{3}}+\frac {c b \ln \left (c x \right )}{d^{3}}-\frac {17 c b \ln \left (c x -1\right )}{16 d^{3}}-\frac {b c}{8 d^{3} \left (c x +1\right )^{2}}-\frac {9 b c}{8 d^{3} \left (c x +1\right )}+\frac {c b \ln \left (c x +1\right )}{16 d^{3}}+\frac {3 c b \dilog \left (c x \right )}{2 d^{3}}+\frac {3 c b \dilog \left (c x +1\right )}{2 d^{3}}+\frac {3 c b \ln \left (c x \right ) \ln \left (c x +1\right )}{2 d^{3}}-\frac {3 c b \ln \left (c x +1\right )^{2}}{4 d^{3}}+\frac {3 c b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 d^{3}}-\frac {3 c b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 d^{3}}-\frac {3 c b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/d^3/x-3*c*a/d^3*ln(c*x)-1/2*c*a/d^3/(c*x+1)^2-2*c*a/d^3/(c*x+1)+3*c*a/d^3*ln(c*x+1)-b/d^3*arctanh(c*x)/x-3*
c*b/d^3*arctanh(c*x)*ln(c*x)-1/2*c*b/d^3*arctanh(c*x)/(c*x+1)^2-2*c*b/d^3*arctanh(c*x)/(c*x+1)+3*c*b/d^3*arcta
nh(c*x)*ln(c*x+1)+c*b/d^3*ln(c*x)-17/16*c*b/d^3*ln(c*x-1)-1/8*b*c/d^3/(c*x+1)^2-9/8*b*c/d^3/(c*x+1)+1/16*c*b/d
^3*ln(c*x+1)+3/2*c*b/d^3*dilog(c*x)+3/2*c*b/d^3*dilog(c*x+1)+3/2*c*b/d^3*ln(c*x)*ln(c*x+1)-3/4*c*b/d^3*ln(c*x+
1)^2+3/2*c*b/d^3*ln(-1/2*c*x+1/2)*ln(c*x+1)-3/2*c*b/d^3*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-3/2*c*b/d^3*dilog(1/2
+1/2*c*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {6 \, c^{2} x^{2} + 9 \, c x + 2}{c^{2} d^{3} x^{3} + 2 \, c d^{3} x^{2} + d^{3} x} - \frac {6 \, c \log \left (c x + 1\right )}{d^{3}} + \frac {6 \, c \log \relax (x)}{d^{3}}\right )} + \frac {1}{2} \, b \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{c^{3} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{4} + 3 \, c d^{3} x^{3} + d^{3} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^2/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

-1/2*a*((6*c^2*x^2 + 9*c*x + 2)/(c^2*d^3*x^3 + 2*c*d^3*x^2 + d^3*x) - 6*c*log(c*x + 1)/d^3 + 6*c*log(x)/d^3) +
1/2*b*integrate((log(c*x + 1) - log(-c*x + 1))/(c^3*d^3*x^5 + 3*c^2*d^3*x^4 + 3*c*d^3*x^3 + d^3*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{x^2\,{\left (d+c\,d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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