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

Optimal. Leaf size=175 \[ \frac {1}{3} b c d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^2 (c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c}-\frac {8 b d^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 c}+2 a b d^2 x+\frac {b^2 d^2 \log \left (1-c^2 x^2\right )}{c}-\frac {4 b^2 d^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{3 c}-\frac {b^2 d^2 \tanh ^{-1}(c x)}{3 c}+2 b^2 d^2 x \tanh ^{-1}(c x)+\frac {1}{3} b^2 d^2 x \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {5928, 5910, 260, 5916, 321, 206, 1586, 5918, 2402, 2315} \[ -\frac {4 b^2 d^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{3 c}+\frac {1}{3} b c d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^2 (c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 c}-\frac {8 b d^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 c}+2 a b d^2 x+\frac {b^2 d^2 \log \left (1-c^2 x^2\right )}{c}-\frac {b^2 d^2 \tanh ^{-1}(c x)}{3 c}+2 b^2 d^2 x \tanh ^{-1}(c x)+\frac {1}{3} b^2 d^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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 5928

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

Rubi steps

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

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Mathematica [A]  time = 0.66, size = 227, normalized size = 1.30 \[ \frac {d^2 \left (a^2 c^3 x^3+3 a^2 c^2 x^2+3 a^2 c x+a b c^2 x^2+3 a b \log \left (1-c^2 x^2\right )+a b \log \left (c^2 x^2-1\right )+b \tanh ^{-1}(c x) \left (2 a c x \left (c^2 x^2+3 c x+3\right )+b \left (c^2 x^2+6 c x-1\right )-8 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+6 a b c x+3 a b \log (1-c x)-3 a b \log (c x+1)+3 b^2 \log \left (1-c^2 x^2\right )+b^2 \left (c^3 x^3+3 c^2 x^2+3 c x-7\right ) \tanh ^{-1}(c x)^2+4 b^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+b^2 c x\right )}{3 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^2*(3*a^2*c*x + 6*a*b*c*x + b^2*c*x + 3*a^2*c^2*x^2 + a*b*c^2*x^2 + a^2*c^3*x^3 + b^2*(-7 + 3*c*x + 3*c^2*x^
2 + c^3*x^3)*ArcTanh[c*x]^2 + b*ArcTanh[c*x]*(2*a*c*x*(3 + 3*c*x + c^2*x^2) + b*(-1 + 6*c*x + c^2*x^2) - 8*b*L
og[1 + E^(-2*ArcTanh[c*x])]) + 3*a*b*Log[1 - c*x] - 3*a*b*Log[1 + c*x] + 3*a*b*Log[1 - c^2*x^2] + 3*b^2*Log[1
- c^2*x^2] + a*b*Log[-1 + c^2*x^2] + 4*b^2*PolyLog[2, -E^(-2*ArcTanh[c*x])]))/(3*c)

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} c^{2} d^{2} x^{2} + 2 \, a^{2} c d^{2} x + a^{2} d^{2} + {\left (b^{2} c^{2} d^{2} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} d^{2}\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c^{2} d^{2} x^{2} + 2 \, a b c d^{2} x + a b d^{2}\right )} \operatorname {artanh}\left (c x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 372, normalized size = 2.13 \[ \frac {b^{2} d^{2} x}{3}+2 c \,d^{2} a b \arctanh \left (c x \right ) x^{2}+\frac {2 c^{2} d^{2} a b \arctanh \left (c x \right ) x^{3}}{3}-\frac {4 d^{2} b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{3 c}+2 b^{2} d^{2} x \arctanh \left (c x \right )+d^{2} b^{2} \arctanh \left (c x \right )^{2} x +\frac {d^{2} b^{2} \arctanh \left (c x \right )^{2}}{3 c}+c \,d^{2} a^{2} x^{2}+\frac {c^{2} d^{2} a^{2} x^{3}}{3}+\frac {5 d^{2} b^{2} \ln \left (c x +1\right )}{6 c}+\frac {7 d^{2} b^{2} \ln \left (c x -1\right )}{6 c}+\frac {2 d^{2} b^{2} \ln \left (c x -1\right )^{2}}{3 c}-\frac {d^{2} b^{2}}{3 c}+\frac {d^{2} a^{2}}{3 c}+x \,a^{2} d^{2}+2 a b \,d^{2} x +\frac {c \,d^{2} b^{2} \arctanh \left (c x \right ) x^{2}}{3}+2 d^{2} a b \arctanh \left (c x \right ) x +\frac {8 d^{2} b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{3 c}+\frac {2 d^{2} a b \arctanh \left (c x \right )}{3 c}+\frac {c^{2} d^{2} b^{2} \arctanh \left (c x \right )^{2} x^{3}}{3}+c \,d^{2} b^{2} \arctanh \left (c x \right )^{2} x^{2}+\frac {8 d^{2} a b \ln \left (c x -1\right )}{3 c}-\frac {4 d^{2} b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{3 c}+\frac {c \,d^{2} a b \,x^{2}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*d^2*x+2/3*c^2*d^2*a*b*arctanh(c*x)*x^3+2*c*d^2*a*b*arctanh(c*x)*x^2+c*d^2*a^2*x^2+1/3*c^2*d^2*a^2*x^3+
5/6/c*d^2*b^2*ln(c*x+1)+d^2*b^2*arctanh(c*x)^2*x+7/6/c*d^2*b^2*ln(c*x-1)+1/3/c*d^2*b^2*arctanh(c*x)^2-4/3/c*d^
2*b^2*dilog(1/2+1/2*c*x)+2/3/c*d^2*b^2*ln(c*x-1)^2-1/3/c*d^2*b^2+1/3/c*d^2*a^2+x*a^2*d^2+2*a*b*d^2*x+2*b^2*d^2
*x*arctanh(c*x)+2*d^2*a*b*arctanh(c*x)*x+8/3/c*d^2*a*b*ln(c*x-1)+8/3/c*d^2*b^2*arctanh(c*x)*ln(c*x-1)-4/3/c*d^
2*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+2/3/c*d^2*a*b*arctanh(c*x)+1/3*c^2*d^2*b^2*arctanh(c*x)^2*x^3+c*d^2*b^2*arctan
h(c*x)^2*x^2+1/3*c*d^2*b^2*arctanh(c*x)*x^2+1/3*c*d^2*a*b*x^2

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maxima [B]  time = 0.55, size = 464, normalized size = 2.65 \[ \frac {1}{3} \, a^{2} c^{2} d^{2} x^{3} + \frac {1}{3} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} a b c^{2} d^{2} + a^{2} c d^{2} x^{2} + {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} a b c d^{2} + a^{2} d^{2} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} a b d^{2}}{c} + \frac {4 \, {\left (\log \left (c x + 1\right ) \log \left (-\frac {1}{2} \, c x + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c x + \frac {1}{2}\right )\right )} b^{2} d^{2}}{3 \, c} + \frac {5 \, b^{2} d^{2} \log \left (c x + 1\right )}{6 \, c} + \frac {7 \, b^{2} d^{2} \log \left (c x - 1\right )}{6 \, c} + \frac {4 \, b^{2} c d^{2} x + {\left (b^{2} c^{3} d^{2} x^{3} + 3 \, b^{2} c^{2} d^{2} x^{2} + 3 \, b^{2} c d^{2} x + b^{2} d^{2}\right )} \log \left (c x + 1\right )^{2} + {\left (b^{2} c^{3} d^{2} x^{3} + 3 \, b^{2} c^{2} d^{2} x^{2} + 3 \, b^{2} c d^{2} x - 7 \, b^{2} d^{2}\right )} \log \left (-c x + 1\right )^{2} + 2 \, {\left (b^{2} c^{2} d^{2} x^{2} + 6 \, b^{2} c d^{2} x\right )} \log \left (c x + 1\right ) - 2 \, {\left (b^{2} c^{2} d^{2} x^{2} + 6 \, b^{2} c d^{2} x + {\left (b^{2} c^{3} d^{2} x^{3} + 3 \, b^{2} c^{2} d^{2} x^{2} + 3 \, b^{2} c d^{2} x + b^{2} d^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{12 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*c^2*d^2*x^3 + 1/3*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a*b*c^2*d^2 + a^2*c*d^2*x^
2 + (2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*a*b*c*d^2 + a^2*d^2*x + (2*c*x*ar
ctanh(c*x) + log(-c^2*x^2 + 1))*a*b*d^2/c + 4/3*(log(c*x + 1)*log(-1/2*c*x + 1/2) + dilog(1/2*c*x + 1/2))*b^2*
d^2/c + 5/6*b^2*d^2*log(c*x + 1)/c + 7/6*b^2*d^2*log(c*x - 1)/c + 1/12*(4*b^2*c*d^2*x + (b^2*c^3*d^2*x^3 + 3*b
^2*c^2*d^2*x^2 + 3*b^2*c*d^2*x + b^2*d^2)*log(c*x + 1)^2 + (b^2*c^3*d^2*x^3 + 3*b^2*c^2*d^2*x^2 + 3*b^2*c*d^2*
x - 7*b^2*d^2)*log(-c*x + 1)^2 + 2*(b^2*c^2*d^2*x^2 + 6*b^2*c*d^2*x)*log(c*x + 1) - 2*(b^2*c^2*d^2*x^2 + 6*b^2
*c*d^2*x + (b^2*c^3*d^2*x^3 + 3*b^2*c^2*d^2*x^2 + 3*b^2*c*d^2*x + b^2*d^2)*log(c*x + 1))*log(-c*x + 1))/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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