3.1.16 \(\int (c e+d e x)^2 (a+b \tanh ^{-1}(c+d x))^2 \, dx\) [16]

Optimal. Leaf size=179 \[ \frac {1}{3} b^2 e^2 x-\frac {b^2 e^2 \tanh ^{-1}(c+d x)}{3 d}+\frac {b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d}+\frac {e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d}+\frac {e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d}-\frac {2 b e^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{3 d}-\frac {b^2 e^2 \text {PolyLog}\left (2,-\frac {1+c+d x}{1-c-d x}\right )}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {6242, 12, 6037, 6127, 327, 212, 6131, 6055, 2449, 2352} \begin {gather*} \frac {e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d}+\frac {b e^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d}+\frac {e^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{3 d}-\frac {2 b e^2 \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d}-\frac {b^2 e^2 \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{3 d}-\frac {b^2 e^2 \tanh ^{-1}(c+d x)}{3 d}+\frac {1}{3} b^2 e^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 212

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

Rule 327

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 2352

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

Rule 2449

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 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rule 6055

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 6127

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

Rule 6131

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

Rule 6242

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

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Mathematica [A]
time = 0.29, size = 150, normalized size = 0.84 \begin {gather*} \frac {e^2 \left (a^2 (c+d x)^3+a b \left ((c+d x)^2+2 (c+d x)^3 \tanh ^{-1}(c+d x)+\log \left (-1+(c+d x)^2\right )\right )+b^2 \left (c+d x-\tanh ^{-1}(c+d x)+(c+d x)^2 \tanh ^{-1}(c+d x)-\tanh ^{-1}(c+d x)^2+(c+d x)^3 \tanh ^{-1}(c+d x)^2-2 \tanh ^{-1}(c+d x) \log \left (1+e^{-2 \tanh ^{-1}(c+d x)}\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(341\) vs. \(2(167)=334\).
time = 4.12, size = 342, normalized size = 1.91 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^2*(a+b*arctanh(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 572 vs. \(2 (150) = 300\).
time = 0.45, size = 572, normalized size = 3.20 \begin {gather*} \frac {1}{3} \, a^{2} d^{2} x^{3} e^{2} + a^{2} c d x^{2} e^{2} + {\left (2 \, x^{2} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {2 \, x}{d^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} a b c d e^{2} + \frac {1}{3} \, {\left (2 \, x^{3} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} + \frac {{\left (c^{3} + 3 \, c^{2} + 3 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{4}} - \frac {{\left (c^{3} - 3 \, c^{2} + 3 \, c - 1\right )} \log \left (d x + c - 1\right )}{d^{4}}\right )}\right )} a b d^{2} e^{2} + a^{2} c^{2} x e^{2} + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} a b c^{2} e^{2}}{d} + \frac {{\left (c^{2} - 1\right )} b^{2} e^{2} \log \left (d x + c + 1\right )}{6 \, d} - \frac {{\left (c^{2} - 1\right )} b^{2} e^{2} \log \left (d x + c - 1\right )}{6 \, d} + \frac {{\left (\log \left (d x + c + 1\right ) \log \left (-\frac {1}{2} \, d x - \frac {1}{2} \, c + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, d x + \frac {1}{2} \, c + \frac {1}{2}\right )\right )} b^{2} e^{2}}{3 \, d} + \frac {4 \, b^{2} d x e^{2} + {\left (b^{2} d^{3} x^{3} e^{2} + 3 \, b^{2} c d^{2} x^{2} e^{2} + 3 \, b^{2} c^{2} d x e^{2} + {\left (c^{3} + 1\right )} b^{2} e^{2}\right )} \log \left (d x + c + 1\right )^{2} + {\left (b^{2} d^{3} x^{3} e^{2} + 3 \, b^{2} c d^{2} x^{2} e^{2} + 3 \, b^{2} c^{2} d x e^{2} + {\left (c^{3} - 1\right )} b^{2} e^{2}\right )} \log \left (-d x - c + 1\right )^{2} + 2 \, {\left (b^{2} d^{2} x^{2} e^{2} + 2 \, b^{2} c d x e^{2}\right )} \log \left (d x + c + 1\right ) - 2 \, {\left (b^{2} d^{2} x^{2} e^{2} + 2 \, b^{2} c d x e^{2} + {\left (b^{2} d^{3} x^{3} e^{2} + 3 \, b^{2} c d^{2} x^{2} e^{2} + 3 \, b^{2} c^{2} d x e^{2} + {\left (c^{3} + 1\right )} b^{2} e^{2}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{2} \left (\int a^{2} c^{2}\, dx + \int a^{2} d^{2} x^{2}\, dx + \int b^{2} c^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}\, dx + \int 2 a b c^{2} \operatorname {atanh}{\left (c + d x \right )}\, dx + \int 2 a^{2} c d x\, dx + \int b^{2} d^{2} x^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}\, dx + \int 2 a b d^{2} x^{2} \operatorname {atanh}{\left (c + d x \right )}\, dx + \int 2 b^{2} c d x \operatorname {atanh}^{2}{\left (c + d x \right )}\, dx + \int 4 a b c d x \operatorname {atanh}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e**2*(Integral(a**2*c**2, x) + Integral(a**2*d**2*x**2, x) + Integral(b**2*c**2*atanh(c + d*x)**2, x) + Integr
al(2*a*b*c**2*atanh(c + d*x), x) + Integral(2*a**2*c*d*x, x) + Integral(b**2*d**2*x**2*atanh(c + d*x)**2, x) +
Integral(2*a*b*d**2*x**2*atanh(c + d*x), x) + Integral(2*b**2*c*d*x*atanh(c + d*x)**2, x) + Integral(4*a*b*c*
d*x*atanh(c + d*x), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,e+d\,e\,x\right )}^2\,{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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