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

Optimal. Leaf size=181 \[ \frac {a+b \tanh ^{-1}(c x)}{c^4 d^2 (c x+1)}-\frac {3 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2 d^2}-\frac {2 a x}{c^3 d^2}+\frac {3 b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 c^4 d^2}+\frac {b}{2 c^4 d^2 (c x+1)}-\frac {b \tanh ^{-1}(c x)}{c^4 d^2}+\frac {b x}{2 c^3 d^2}-\frac {2 b x \tanh ^{-1}(c x)}{c^3 d^2}-\frac {b \log \left (1-c^2 x^2\right )}{c^4 d^2} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {5940, 5910, 260, 5916, 321, 206, 5926, 627, 44, 207, 5918, 2402, 2315} \[ \frac {3 b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c^4 d^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2 d^2}+\frac {a+b \tanh ^{-1}(c x)}{c^4 d^2 (c x+1)}-\frac {3 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^2}-\frac {2 a x}{c^3 d^2}-\frac {b \log \left (1-c^2 x^2\right )}{c^4 d^2}+\frac {b x}{2 c^3 d^2}+\frac {b}{2 c^4 d^2 (c x+1)}-\frac {2 b x \tanh ^{-1}(c x)}{c^3 d^2}-\frac {b \tanh ^{-1}(c x)}{c^4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.70, size = 142, normalized size = 0.78 \[ \frac {2 a c^2 x^2-8 a c x+\frac {4 a}{c x+1}+12 a \log (c x+1)+b \left (-4 \log \left (1-c^2 x^2\right )+2 \tanh ^{-1}(c x) \left (c^2 x^2-4 c x-6 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )-1\right )+6 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+2 c x-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )\right )}{4 c^4 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-8*a*c*x + 2*a*c^2*x^2 + (4*a)/(1 + c*x) + 12*a*Log[1 + c*x] + b*(2*c*x + Cosh[2*ArcTanh[c*x]] - 4*Log[1 - c^
2*x^2] + 6*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 2*ArcTanh[c*x]*(-1 - 4*c*x + c^2*x^2 + Cosh[2*ArcTanh[c*x]] - 6*
Log[1 + E^(-2*ArcTanh[c*x])] - Sinh[2*ArcTanh[c*x]]) - Sinh[2*ArcTanh[c*x]]))/(4*c^4*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 265, normalized size = 1.46 \[ \frac {a \,x^{2}}{2 c^{2} d^{2}}-\frac {2 a x}{c^{3} d^{2}}+\frac {a}{c^{4} d^{2} \left (c x +1\right )}+\frac {3 a \ln \left (c x +1\right )}{c^{4} d^{2}}+\frac {b \arctanh \left (c x \right ) x^{2}}{2 c^{2} d^{2}}-\frac {2 b x \arctanh \left (c x \right )}{c^{3} d^{2}}+\frac {b \arctanh \left (c x \right )}{c^{4} d^{2} \left (c x +1\right )}+\frac {3 b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{c^{4} d^{2}}-\frac {3 b \ln \left (c x +1\right )^{2}}{4 c^{4} d^{2}}+\frac {3 b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 c^{4} d^{2}}-\frac {3 b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{4} d^{2}}-\frac {3 b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{4} d^{2}}+\frac {b x}{2 c^{3} d^{2}}+\frac {b}{2 c^{4} d^{2}}+\frac {b}{2 c^{4} d^{2} \left (c x +1\right )}-\frac {3 b \ln \left (c x +1\right )}{2 c^{4} d^{2}}-\frac {b \ln \left (c x -1\right )}{2 c^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{16} \, {\left (c^{4} {\left (\frac {2}{c^{9} d^{2} x + c^{8} d^{2}} + \frac {2 \, {\left (c x^{2} - 2 \, x\right )}}{c^{7} d^{2}} + \frac {7 \, \log \left (c x + 1\right )}{c^{8} d^{2}} + \frac {\log \left (c x - 1\right )}{c^{8} d^{2}}\right )} + 16 \, c^{4} \int \frac {x^{4} \log \left (c x + 1\right )}{2 \, {\left (c^{6} d^{2} x^{3} + c^{5} d^{2} x^{2} - c^{4} d^{2} x - c^{3} d^{2}\right )}}\,{d x} + 2 \, c^{3} {\left (\frac {2}{c^{8} d^{2} x + c^{7} d^{2}} - \frac {4 \, x}{c^{6} d^{2}} + \frac {5 \, \log \left (c x + 1\right )}{c^{7} d^{2}} - \frac {\log \left (c x - 1\right )}{c^{7} d^{2}}\right )} - 16 \, c^{3} \int \frac {x^{3} \log \left (c x + 1\right )}{2 \, {\left (c^{6} d^{2} x^{3} + c^{5} d^{2} x^{2} - c^{4} d^{2} x - c^{3} d^{2}\right )}}\,{d x} - 7 \, c^{2} {\left (\frac {2}{c^{7} d^{2} x + c^{6} d^{2}} + \frac {3 \, \log \left (c x + 1\right )}{c^{6} d^{2}} + \frac {\log \left (c x - 1\right )}{c^{6} d^{2}}\right )} + 48 \, c^{2} \int \frac {x^{2} \log \left (c x + 1\right )}{2 \, {\left (c^{6} d^{2} x^{3} + c^{5} d^{2} x^{2} - c^{4} d^{2} x - c^{3} d^{2}\right )}}\,{d x} + 2 \, c {\left (\frac {2}{c^{6} d^{2} x + c^{5} d^{2}} + \frac {\log \left (c x + 1\right )}{c^{5} d^{2}} - \frac {\log \left (c x - 1\right )}{c^{5} d^{2}}\right )} + 96 \, c \int \frac {x \log \left (c x + 1\right )}{2 \, {\left (c^{6} d^{2} x^{3} + c^{5} d^{2} x^{2} - c^{4} d^{2} x - c^{3} d^{2}\right )}}\,{d x} - \frac {4 \, {\left (c^{3} x^{3} - 3 \, c^{2} x^{2} - 4 \, c x + 6 \, {\left (c x + 1\right )} \log \left (c x + 1\right ) + 2\right )} \log \left (-c x + 1\right )}{c^{5} d^{2} x + c^{4} d^{2}} + \frac {4}{c^{5} d^{2} x + c^{4} d^{2}} - \frac {2 \, \log \left (c x + 1\right )}{c^{4} d^{2}} + \frac {2 \, \log \left (c x - 1\right )}{c^{4} d^{2}} + 48 \, \int \frac {\log \left (c x + 1\right )}{2 \, {\left (c^{6} d^{2} x^{3} + c^{5} d^{2} x^{2} - c^{4} d^{2} x - c^{3} d^{2}\right )}}\,{d x}\right )} b + \frac {1}{2} \, a {\left (\frac {2}{c^{5} d^{2} x + c^{4} d^{2}} + \frac {c x^{2} - 4 \, x}{c^{3} d^{2}} + \frac {6 \, \log \left (c x + 1\right )}{c^{4} d^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(c^4*(2/(c^9*d^2*x + c^8*d^2) + 2*(c*x^2 - 2*x)/(c^7*d^2) + 7*log(c*x + 1)/(c^8*d^2) + log(c*x - 1)/(c^8*
d^2)) + 16*c^4*integrate(1/2*x^4*log(c*x + 1)/(c^6*d^2*x^3 + c^5*d^2*x^2 - c^4*d^2*x - c^3*d^2), x) + 2*c^3*(2
/(c^8*d^2*x + c^7*d^2) - 4*x/(c^6*d^2) + 5*log(c*x + 1)/(c^7*d^2) - log(c*x - 1)/(c^7*d^2)) - 16*c^3*integrate
(1/2*x^3*log(c*x + 1)/(c^6*d^2*x^3 + c^5*d^2*x^2 - c^4*d^2*x - c^3*d^2), x) - 7*c^2*(2/(c^7*d^2*x + c^6*d^2) +
3*log(c*x + 1)/(c^6*d^2) + log(c*x - 1)/(c^6*d^2)) + 48*c^2*integrate(1/2*x^2*log(c*x + 1)/(c^6*d^2*x^3 + c^5
*d^2*x^2 - c^4*d^2*x - c^3*d^2), x) + 2*c*(2/(c^6*d^2*x + c^5*d^2) + log(c*x + 1)/(c^5*d^2) - log(c*x - 1)/(c^
5*d^2)) + 96*c*integrate(1/2*x*log(c*x + 1)/(c^6*d^2*x^3 + c^5*d^2*x^2 - c^4*d^2*x - c^3*d^2), x) - 4*(c^3*x^3
- 3*c^2*x^2 - 4*c*x + 6*(c*x + 1)*log(c*x + 1) + 2)*log(-c*x + 1)/(c^5*d^2*x + c^4*d^2) + 4/(c^5*d^2*x + c^4*
d^2) - 2*log(c*x + 1)/(c^4*d^2) + 2*log(c*x - 1)/(c^4*d^2) + 48*integrate(1/2*log(c*x + 1)/(c^6*d^2*x^3 + c^5*
d^2*x^2 - c^4*d^2*x - c^3*d^2), x))*b + 1/2*a*(2/(c^5*d^2*x + c^4*d^2) + (c*x^2 - 4*x)/(c^3*d^2) + 6*log(c*x +
1)/(c^4*d^2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}{{\left (d+c\,d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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