3.21 \(\int \frac {(a+b \tanh ^{-1}(c+d x))^2}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=180 \[ -\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4}+\frac {2 b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4}-\frac {b^2 \text {Li}_2\left (\frac {2}{c+d x+1}-1\right )}{3 d e^4}-\frac {b^2}{3 d e^4 (c+d x)}+\frac {b^2 \tanh ^{-1}(c+d x)}{3 d e^4} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {6107, 12, 5916, 5982, 325, 206, 5988, 5932, 2447} \[ -\frac {b^2 \text {PolyLog}\left (2,\frac {2}{c+d x+1}-1\right )}{3 d e^4}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d e^4}+\frac {2 b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d e^4}-\frac {b^2}{3 d e^4 (c+d x)}+\frac {b^2 \tanh ^{-1}(c+d x)}{3 d e^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, 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 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 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 5982

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTanh[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 5988

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

Rule 6107

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

Rubi steps

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

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Mathematica [A]  time = 0.54, size = 218, normalized size = 1.21 \[ -\frac {a^2-a b \left ((c+d x) \left (c^2+2 c d x+2 (c+d x)^2 \log \left (\frac {c+d x}{\sqrt {1-(c+d x)^2}}\right )+d^2 x^2-1\right )-2 \tanh ^{-1}(c+d x)\right )+b^2 \left ((c+d x)^3 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c+d x)}\right )+(c+d x)^2+(c+d x)^2 \tanh ^{-1}(c+d x)^2+\left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^2+(c+d x) \tanh ^{-1}(c+d x) \left (-(c+d x)^2+(c+d x)^2 \left (-\tanh ^{-1}(c+d x)\right )-2 (c+d x)^2 \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )+1\right )\right )}{3 d e^4 (c+d x)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {artanh}\left (d x + c\right ) + a^{2}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

integrate((b*arctanh(d*x + c) + a)^2/(d*e*x + c*e)^4, x)

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maple [B]  time = 0.08, size = 492, normalized size = 2.73 \[ -\frac {a^{2}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{2} \arctanh \left (d x +c \right )^{2}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{2} \arctanh \left (d x +c \right )}{3 d \,e^{4} \left (d x +c \right )^{2}}+\frac {2 b^{2} \ln \left (d x +c \right ) \arctanh \left (d x +c \right )}{3 d \,e^{4}}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{3 d \,e^{4}}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{3 d \,e^{4}}-\frac {b^{2}}{3 d \,e^{4} \left (d x +c \right )}-\frac {b^{2} \ln \left (d x +c -1\right )}{6 d \,e^{4}}+\frac {b^{2} \ln \left (d x +c +1\right )}{6 d \,e^{4}}-\frac {b^{2} \ln \left (d x +c -1\right )^{2}}{12 d \,e^{4}}+\frac {b^{2} \dilog \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{3 d \,e^{4}}+\frac {b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{6 d \,e^{4}}+\frac {b^{2} \ln \left (d x +c +1\right )^{2}}{12 d \,e^{4}}-\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{6 d \,e^{4}}+\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{6 d \,e^{4}}-\frac {b^{2} \dilog \left (d x +c \right )}{3 d \,e^{4}}-\frac {b^{2} \dilog \left (d x +c +1\right )}{3 d \,e^{4}}-\frac {b^{2} \ln \left (d x +c \right ) \ln \left (d x +c +1\right )}{3 d \,e^{4}}-\frac {2 a b \arctanh \left (d x +c \right )}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {a b}{3 d \,e^{4} \left (d x +c \right )^{2}}+\frac {2 a b \ln \left (d x +c \right )}{3 d \,e^{4}}-\frac {a b \ln \left (d x +c -1\right )}{3 d \,e^{4}}-\frac {a b \ln \left (d x +c +1\right )}{3 d \,e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(d*x+c))^2/(d*e*x+c*e)^4,x)

[Out]

-1/3/d*a^2/e^4/(d*x+c)^3-1/3/d*b^2/e^4/(d*x+c)^3*arctanh(d*x+c)^2-1/3/d*b^2/e^4*arctanh(d*x+c)/(d*x+c)^2+2/3/d
*b^2/e^4*ln(d*x+c)*arctanh(d*x+c)-1/3/d*b^2/e^4*arctanh(d*x+c)*ln(d*x+c-1)-1/3/d*b^2/e^4*arctanh(d*x+c)*ln(d*x
+c+1)-1/3*b^2/d/e^4/(d*x+c)-1/6/d*b^2/e^4*ln(d*x+c-1)+1/6/d*b^2/e^4*ln(d*x+c+1)-1/12/d*b^2/e^4*ln(d*x+c-1)^2+1
/3/d*b^2/e^4*dilog(1/2+1/2*d*x+1/2*c)+1/6/d*b^2/e^4*ln(d*x+c-1)*ln(1/2+1/2*d*x+1/2*c)+1/12/d*b^2/e^4*ln(d*x+c+
1)^2-1/6/d*b^2/e^4*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)+1/6/d*b^2/e^4*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2+1/2*d*x+1/2*
c)-1/3/d*b^2/e^4*dilog(d*x+c)-1/3/d*b^2/e^4*dilog(d*x+c+1)-1/3/d*b^2/e^4*ln(d*x+c)*ln(d*x+c+1)-2/3/d*a*b/e^4/(
d*x+c)^3*arctanh(d*x+c)-1/3/d*a*b/e^4/(d*x+c)^2+2/3/d*a*b/e^4*ln(d*x+c)-1/3/d*a*b/e^4*ln(d*x+c-1)-1/3/d*a*b/e^
4*ln(d*x+c+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

-1/3*(d*(1/(d^4*e^4*x^2 + 2*c*d^3*e^4*x + c^2*d^2*e^4) + log(d*x + c + 1)/(d^2*e^4) - 2*log(d*x + c)/(d^2*e^4)
+ log(d*x + c - 1)/(d^2*e^4)) + 2*arctanh(d*x + c)/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e
^4))*a*b - 1/12*b^2*(log(-d*x - c + 1)^2/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) + 3*int
egrate(-1/3*(3*(d*x + c - 1)*log(d*x + c + 1)^2 + 2*(d*x - 3*(d*x + c - 1)*log(d*x + c + 1) + c)*log(-d*x - c
+ 1))/(d^5*e^4*x^5 + c^5*e^4 - c^4*e^4 + (5*c*d^4*e^4 - d^4*e^4)*x^4 + 2*(5*c^2*d^3*e^4 - 2*c*d^3*e^4)*x^3 + 2
*(5*c^3*d^2*e^4 - 3*c^2*d^2*e^4)*x^2 + (5*c^4*d*e^4 - 4*c^3*d*e^4)*x), x)) - 1/3*a^2/(d^4*e^4*x^3 + 3*c*d^3*e^
4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c + d*x))^2/(c*e + d*e*x)^4,x)

[Out]

int((a + b*atanh(c + d*x))^2/(c*e + d*e*x)^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))**2/(d*e*x+c*e)**4,x)

[Out]

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

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