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

Optimal. Leaf size=194 \[ -\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (c x+1)}+\frac {a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (c x+1)^2}+\frac {3 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}+\frac {a x}{c^3 d^3}-\frac {3 b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 c^4 d^3}-\frac {11 b}{8 c^4 d^3 (c x+1)}+\frac {b}{8 c^4 d^3 (c x+1)^2}+\frac {11 b \tanh ^{-1}(c x)}{8 c^4 d^3}+\frac {b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^4 d^3} \]

[Out]

a*x/c^3/d^3+1/8*b/c^4/d^3/(c*x+1)^2-11/8*b/c^4/d^3/(c*x+1)+11/8*b*arctanh(c*x)/c^4/d^3+b*x*arctanh(c*x)/c^3/d^
3+1/2*(a+b*arctanh(c*x))/c^4/d^3/(c*x+1)^2-3*(a+b*arctanh(c*x))/c^4/d^3/(c*x+1)+3*(a+b*arctanh(c*x))*ln(2/(c*x
+1))/c^4/d^3+1/2*b*ln(-c^2*x^2+1)/c^4/d^3-3/2*b*polylog(2,1-2/(c*x+1))/c^4/d^3

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Rubi [A]  time = 0.25, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5940, 5910, 260, 5926, 627, 44, 207, 5918, 2402, 2315} \[ -\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c^4 d^3}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (c x+1)}+\frac {a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (c x+1)^2}+\frac {3 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}+\frac {a x}{c^3 d^3}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac {11 b}{8 c^4 d^3 (c x+1)}+\frac {b}{8 c^4 d^3 (c x+1)^2}+\frac {b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac {11 b \tanh ^{-1}(c x)}{8 c^4 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/(c^3*d^3) + b/(8*c^4*d^3*(1 + c*x)^2) - (11*b)/(8*c^4*d^3*(1 + c*x)) + (11*b*ArcTanh[c*x])/(8*c^4*d^3) +
(b*x*ArcTanh[c*x])/(c^3*d^3) + (a + b*ArcTanh[c*x])/(2*c^4*d^3*(1 + c*x)^2) - (3*(a + b*ArcTanh[c*x]))/(c^4*d
^3*(1 + c*x)) + (3*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/(c^4*d^3) + (b*Log[1 - c^2*x^2])/(2*c^4*d^3) - (3*b*
PolyLog[2, 1 - 2/(1 + c*x)])/(2*c^4*d^3)

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 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 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 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 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 {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{(d+c d x)^3} \, dx &=\int \left (\frac {a+b \tanh ^{-1}(c x)}{c^3 d^3}-\frac {a+b \tanh ^{-1}(c x)}{c^3 d^3 (1+c x)^3}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^3 (1+c x)}\right ) \, dx\\ &=\frac {\int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^3 d^3}-\frac {\int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{c^3 d^3}+\frac {3 \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{c^3 d^3}-\frac {3 \int \frac {a+b \tanh ^{-1}(c x)}{1+c x} \, dx}{c^3 d^3}\\ &=\frac {a x}{c^3 d^3}+\frac {a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {b \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{2 c^3 d^3}+\frac {b \int \tanh ^{-1}(c x) \, dx}{c^3 d^3}+\frac {(3 b) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{c^3 d^3}-\frac {(3 b) \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^3 d^3}\\ &=\frac {a x}{c^3 d^3}+\frac {b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac {a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{c^4 d^3}-\frac {b \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{2 c^3 d^3}+\frac {(3 b) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{c^3 d^3}-\frac {b \int \frac {x}{1-c^2 x^2} \, dx}{c^2 d^3}\\ &=\frac {a x}{c^3 d^3}+\frac {b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac {a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac {3 b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 c^4 d^3}-\frac {b \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 c^3 d^3}+\frac {(3 b) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^3 d^3}\\ &=\frac {a x}{c^3 d^3}+\frac {b}{8 c^4 d^3 (1+c x)^2}-\frac {11 b}{8 c^4 d^3 (1+c x)}+\frac {b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac {a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac {3 b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 c^4 d^3}+\frac {b \int \frac {1}{-1+c^2 x^2} \, dx}{8 c^3 d^3}-\frac {(3 b) \int \frac {1}{-1+c^2 x^2} \, dx}{2 c^3 d^3}\\ &=\frac {a x}{c^3 d^3}+\frac {b}{8 c^4 d^3 (1+c x)^2}-\frac {11 b}{8 c^4 d^3 (1+c x)}+\frac {11 b \tanh ^{-1}(c x)}{8 c^4 d^3}+\frac {b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac {a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac {3 b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 c^4 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.70, size = 167, normalized size = 0.86 \[ \frac {32 a c x-\frac {96 a}{c x+1}+\frac {16 a}{(c x+1)^2}-96 a \log (c x+1)+b \left (16 \log \left (1-c^2 x^2\right )-48 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+20 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )-20 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (8 c x+24 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+10 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )-10 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )\right )\right )}{32 c^4 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(32*a*c*x + (16*a)/(1 + c*x)^2 - (96*a)/(1 + c*x) - 96*a*Log[1 + c*x] + b*(-20*Cosh[2*ArcTanh[c*x]] + Cosh[4*A
rcTanh[c*x]] + 16*Log[1 - c^2*x^2] - 48*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 20*Sinh[2*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*Ar
cTanh[c*x]] - Sinh[4*ArcTanh[c*x]]) - Sinh[4*ArcTanh[c*x]]))/(32*c^4*d^3)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \operatorname {artanh}\left (c x\right ) + a x^{3}}{c^{3} d^{3} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c d^{3} x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 270, normalized size = 1.39 \[ \frac {a x}{c^{3} d^{3}}+\frac {a}{2 c^{4} d^{3} \left (c x +1\right )^{2}}-\frac {3 a}{c^{4} d^{3} \left (c x +1\right )}-\frac {3 a \ln \left (c x +1\right )}{c^{4} d^{3}}+\frac {b x \arctanh \left (c x \right )}{c^{3} d^{3}}+\frac {b \arctanh \left (c x \right )}{2 c^{4} d^{3} \left (c x +1\right )^{2}}-\frac {3 b \arctanh \left (c x \right )}{c^{4} d^{3} \left (c x +1\right )}-\frac {3 b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{c^{4} d^{3}}+\frac {3 b \ln \left (c x +1\right )^{2}}{4 c^{4} d^{3}}-\frac {3 b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 c^{4} d^{3}}+\frac {3 b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{4} d^{3}}+\frac {3 b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{4} d^{3}}+\frac {b}{8 c^{4} d^{3} \left (c x +1\right )^{2}}-\frac {11 b}{8 c^{4} d^{3} \left (c x +1\right )}+\frac {19 b \ln \left (c x +1\right )}{16 c^{4} d^{3}}-\frac {3 b \ln \left (c x -1\right )}{16 c^{4} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x/c^3/d^3+1/2/c^4*a/d^3/(c*x+1)^2-3/c^4*a/d^3/(c*x+1)-3/c^4*a/d^3*ln(c*x+1)+b*x*arctanh(c*x)/c^3/d^3+1/2/c^4
*b/d^3*arctanh(c*x)/(c*x+1)^2-3/c^4*b/d^3*arctanh(c*x)/(c*x+1)-3/c^4*b/d^3*arctanh(c*x)*ln(c*x+1)+3/4/c^4*b/d^
3*ln(c*x+1)^2-3/2/c^4*b/d^3*ln(-1/2*c*x+1/2)*ln(c*x+1)+3/2/c^4*b/d^3*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+3/2/c^4*
b/d^3*dilog(1/2+1/2*c*x)+1/8*b/c^4/d^3/(c*x+1)^2-11/8*b/c^4/d^3/(c*x+1)+19/16/c^4*b/d^3*ln(c*x+1)-3/16/c^4*b/d
^3*ln(c*x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{32} \, {\left (2 \, c^{4} {\left (\frac {2 \, {\left (7 \, c x + 6\right )}}{c^{10} d^{3} x^{2} + 2 \, c^{9} d^{3} x + c^{8} d^{3}} - \frac {8 \, x}{c^{7} d^{3}} + \frac {17 \, \log \left (c x + 1\right )}{c^{8} d^{3}} - \frac {\log \left (c x - 1\right )}{c^{8} d^{3}}\right )} - 32 \, c^{4} \int \frac {x^{4} \log \left (c x + 1\right )}{2 \, {\left (c^{7} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{3} - 2 \, c^{4} d^{3} x - c^{3} d^{3}\right )}}\,{d x} - 6 \, c^{3} {\left (\frac {2 \, {\left (5 \, c x + 4\right )}}{c^{9} d^{3} x^{2} + 2 \, c^{8} d^{3} x + c^{7} d^{3}} + \frac {7 \, \log \left (c x + 1\right )}{c^{7} d^{3}} + \frac {\log \left (c x - 1\right )}{c^{7} d^{3}}\right )} + 128 \, c^{3} \int \frac {x^{3} \log \left (c x + 1\right )}{2 \, {\left (c^{7} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{3} - 2 \, c^{4} d^{3} x - c^{3} d^{3}\right )}}\,{d x} + 288 \, c^{2} \int \frac {x^{2} \log \left (c x + 1\right )}{2 \, {\left (c^{7} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{3} - 2 \, c^{4} d^{3} x - c^{3} d^{3}\right )}}\,{d x} + 9 \, c {\left (\frac {2 \, x}{c^{6} d^{3} x^{2} + 2 \, c^{5} d^{3} x + c^{4} d^{3}} - \frac {\log \left (c x + 1\right )}{c^{5} d^{3}} + \frac {\log \left (c x - 1\right )}{c^{5} d^{3}}\right )} + 288 \, c \int \frac {x \log \left (c x + 1\right )}{2 \, {\left (c^{7} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{3} - 2 \, c^{4} d^{3} x - c^{3} d^{3}\right )}}\,{d x} + \frac {8 \, {\left (2 \, c^{3} x^{3} + 4 \, c^{2} x^{2} - 4 \, c x - 6 \, {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x + 1\right ) - 5\right )} \log \left (-c x + 1\right )}{c^{6} d^{3} x^{2} + 2 \, c^{5} d^{3} x + c^{4} d^{3}} + \frac {10 \, {\left (c x + 2\right )}}{c^{6} d^{3} x^{2} + 2 \, c^{5} d^{3} x + c^{4} d^{3}} - \frac {5 \, \log \left (c x + 1\right )}{c^{4} d^{3}} + \frac {5 \, \log \left (c x - 1\right )}{c^{4} d^{3}} + 96 \, \int \frac {\log \left (c x + 1\right )}{2 \, {\left (c^{7} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{3} - 2 \, c^{4} d^{3} x - c^{3} d^{3}\right )}}\,{d x}\right )} b - \frac {1}{2} \, a {\left (\frac {6 \, c x + 5}{c^{6} d^{3} x^{2} + 2 \, c^{5} d^{3} x + c^{4} d^{3}} - \frac {2 \, x}{c^{3} d^{3}} + \frac {6 \, \log \left (c x + 1\right )}{c^{4} d^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(2*c^4*(2*(7*c*x + 6)/(c^10*d^3*x^2 + 2*c^9*d^3*x + c^8*d^3) - 8*x/(c^7*d^3) + 17*log(c*x + 1)/(c^8*d^3)
- log(c*x - 1)/(c^8*d^3)) - 32*c^4*integrate(1/2*x^4*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x
- c^3*d^3), x) - 6*c^3*(2*(5*c*x + 4)/(c^9*d^3*x^2 + 2*c^8*d^3*x + c^7*d^3) + 7*log(c*x + 1)/(c^7*d^3) + log(c
*x - 1)/(c^7*d^3)) + 128*c^3*integrate(1/2*x^3*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d
^3), x) + 288*c^2*integrate(1/2*x^2*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d^3), x) + 9
*c*(2*x/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - log(c*x + 1)/(c^5*d^3) + log(c*x - 1)/(c^5*d^3)) + 288*c*integ
rate(1/2*x*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d^3), x) + 8*(2*c^3*x^3 + 4*c^2*x^2 -
4*c*x - 6*(c^2*x^2 + 2*c*x + 1)*log(c*x + 1) - 5)*log(-c*x + 1)/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) + 10*(c
*x + 2)/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - 5*log(c*x + 1)/(c^4*d^3) + 5*log(c*x - 1)/(c^4*d^3) + 96*integ
rate(1/2*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d^3), x))*b - 1/2*a*((6*c*x + 5)/(c^6*d
^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - 2*x/(c^3*d^3) + 6*log(c*x + 1)/(c^4*d^3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x^{3}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b x^{3} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x**3/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b*x**3*atanh(c*x)/(c**3*x**3 + 3*c**2*x*
*2 + 3*c*x + 1), x))/d**3

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