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

Optimal. Leaf size=158 \[ \frac{(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 e}+\frac{b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (\sqrt{c} x\right )}{3 c^{3/2}}-\frac{b \left (3 c d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{c} x\right )}{3 c^{3/2}}+\frac{b d \left (c d^2+3 e^2\right ) \log \left (1-c x^2\right )}{6 c e}-\frac{b d \left (c d^2-3 e^2\right ) \log \left (c x^2+1\right )}{6 c e}+\frac{2 b e^2 x}{3 c} \]

[Out]

(2*b*e^2*x)/(3*c) + (b*(3*c*d^2 - e^2)*ArcTan[Sqrt[c]*x])/(3*c^(3/2)) - (b*(3*c*d^2 + e^2)*ArcTanh[Sqrt[c]*x])
/(3*c^(3/2)) + ((d + e*x)^3*(a + b*ArcTanh[c*x^2]))/(3*e) + (b*d*(c*d^2 + 3*e^2)*Log[1 - c*x^2])/(6*c*e) - (b*
d*(c*d^2 - 3*e^2)*Log[1 + c*x^2])/(6*c*e)

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Rubi [A]  time = 0.210324, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {6273, 12, 1831, 1248, 633, 31, 1280, 1167, 205, 208} \[ \frac{(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 e}+\frac{b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (\sqrt{c} x\right )}{3 c^{3/2}}-\frac{b \left (3 c d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{c} x\right )}{3 c^{3/2}}+\frac{b d \left (c d^2+3 e^2\right ) \log \left (1-c x^2\right )}{6 c e}-\frac{b d \left (c d^2-3 e^2\right ) \log \left (c x^2+1\right )}{6 c e}+\frac{2 b e^2 x}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*e^2*x)/(3*c) + (b*(3*c*d^2 - e^2)*ArcTan[Sqrt[c]*x])/(3*c^(3/2)) - (b*(3*c*d^2 + e^2)*ArcTanh[Sqrt[c]*x])
/(3*c^(3/2)) + ((d + e*x)^3*(a + b*ArcTanh[c*x^2]))/(3*e) + (b*d*(c*d^2 + 3*e^2)*Log[1 - c*x^2])/(6*c*e) - (b*
d*(c*d^2 - 3*e^2)*Log[1 + c*x^2])/(6*c*e)

Rule 6273

Int[((a_.) + ArcTanh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 - u^2), x],
x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m
+ 1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]

Rule 12

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

Rule 1831

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[((c*x)^(m + ii)*(Coeff[Pq,
x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2)))/(c^ii*(a + b*x^n)), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; Fr
eeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 1280

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f*(f*x)^(m - 1)*
(a + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m - 2)*(a + c*x^4)^p*(a*e*
(m - 1) - c*d*(m + 4*p + 3)*x^2), x], x] /; FreeQ[{a, c, d, e, f, p}, x] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] &
& IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1167

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q)
, Int[1/(-q + c*x^2), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x^2), x], x]] /; FreeQ[{a, c, d, e}, x] &&
NeQ[c*d^2 - a*e^2, 0] && PosQ[-(a*c)]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

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

Rubi steps

\begin{align*} \int (d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 e}-\frac{b \int \frac{2 c x (d+e x)^3}{1-c^2 x^4} \, dx}{3 e}\\ &=\frac{(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 e}-\frac{(2 b c) \int \frac{x (d+e x)^3}{1-c^2 x^4} \, dx}{3 e}\\ &=\frac{(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 e}-\frac{(2 b c) \int \left (\frac{x \left (d^3+3 d e^2 x^2\right )}{1-c^2 x^4}+\frac{x^2 \left (3 d^2 e+e^3 x^2\right )}{1-c^2 x^4}\right ) \, dx}{3 e}\\ &=\frac{(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 e}-\frac{(2 b c) \int \frac{x \left (d^3+3 d e^2 x^2\right )}{1-c^2 x^4} \, dx}{3 e}-\frac{(2 b c) \int \frac{x^2 \left (3 d^2 e+e^3 x^2\right )}{1-c^2 x^4} \, dx}{3 e}\\ &=\frac{2 b e^2 x}{3 c}+\frac{(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 e}-\frac{(2 b) \int \frac{e^3+3 c^2 d^2 e x^2}{1-c^2 x^4} \, dx}{3 c e}-\frac{(b c) \operatorname{Subst}\left (\int \frac{d^3+3 d e^2 x}{1-c^2 x^2} \, dx,x,x^2\right )}{3 e}\\ &=\frac{2 b e^2 x}{3 c}+\frac{(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 e}+\frac{\left (b c d \left (c d^2-3 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-c-c^2 x} \, dx,x,x^2\right )}{6 e}-\frac{1}{3} \left (b \left (3 c d^2-e^2\right )\right ) \int \frac{1}{-c-c^2 x^2} \, dx-\frac{1}{3} \left (b \left (3 c d^2+e^2\right )\right ) \int \frac{1}{c-c^2 x^2} \, dx-\frac{\left (b c d \left (c d^2+3 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c-c^2 x} \, dx,x,x^2\right )}{6 e}\\ &=\frac{2 b e^2 x}{3 c}+\frac{b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (\sqrt{c} x\right )}{3 c^{3/2}}-\frac{b \left (3 c d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{c} x\right )}{3 c^{3/2}}+\frac{(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{3 e}+\frac{b d \left (c d^2+3 e^2\right ) \log \left (1-c x^2\right )}{6 c e}-\frac{b d \left (c d^2-3 e^2\right ) \log \left (1+c x^2\right )}{6 c e}\\ \end{align*}

Mathematica [A]  time = 0.167183, size = 170, normalized size = 1.08 \[ \frac{1}{6} \left (6 a d^2 x+6 a d e x^2+2 a e^2 x^3+\frac{b \left (3 c d^2+e^2\right ) \log \left (1-\sqrt{c} x\right )}{c^{3/2}}-\frac{b \left (3 c d^2+e^2\right ) \log \left (\sqrt{c} x+1\right )}{c^{3/2}}+\frac{2 b \left (3 c d^2-e^2\right ) \tan ^{-1}\left (\sqrt{c} x\right )}{c^{3/2}}+\frac{3 b d e \log \left (1-c^2 x^4\right )}{c}+2 b x \tanh ^{-1}\left (c x^2\right ) \left (3 d^2+3 d e x+e^2 x^2\right )+\frac{4 b e^2 x}{c}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*d^2*x + (4*b*e^2*x)/c + 6*a*d*e*x^2 + 2*a*e^2*x^3 + (2*b*(3*c*d^2 - e^2)*ArcTan[Sqrt[c]*x])/c^(3/2) + 2*b
*x*(3*d^2 + 3*d*e*x + e^2*x^2)*ArcTanh[c*x^2] + (b*(3*c*d^2 + e^2)*Log[1 - Sqrt[c]*x])/c^(3/2) - (b*(3*c*d^2 +
e^2)*Log[1 + Sqrt[c]*x])/c^(3/2) + (3*b*d*e*Log[1 - c^2*x^4])/c)/6

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Maple [A]  time = 0.032, size = 223, normalized size = 1.4 \begin{align*}{\frac{a{x}^{3}{e}^{2}}{3}}+a{x}^{2}de+ax{d}^{2}+{\frac{a{d}^{3}}{3\,e}}+{\frac{b{e}^{2}{\it Artanh} \left ( c{x}^{2} \right ){x}^{3}}{3}}+be{\it Artanh} \left ( c{x}^{2} \right ){x}^{2}d+b{\it Artanh} \left ( c{x}^{2} \right ) x{d}^{2}+{\frac{b{\it Artanh} \left ( c{x}^{2} \right ){d}^{3}}{3\,e}}+{\frac{2\,b{e}^{2}x}{3\,c}}-{\frac{b\ln \left ( c{x}^{2}+1 \right ){d}^{3}}{6\,e}}+{\frac{be\ln \left ( c{x}^{2}+1 \right ) d}{2\,c}}+{b{d}^{2}\arctan \left ( x\sqrt{c} \right ){\frac{1}{\sqrt{c}}}}-{\frac{b{e}^{2}}{3}\arctan \left ( x\sqrt{c} \right ){c}^{-{\frac{3}{2}}}}+{\frac{b\ln \left ( c{x}^{2}-1 \right ){d}^{3}}{6\,e}}+{\frac{be\ln \left ( c{x}^{2}-1 \right ) d}{2\,c}}-{b{d}^{2}{\it Artanh} \left ( x\sqrt{c} \right ){\frac{1}{\sqrt{c}}}}-{\frac{b{e}^{2}}{3}{\it Artanh} \left ( x\sqrt{c} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80859, size = 918, normalized size = 5.81 \begin{align*} \left [\frac{2 \, a c^{2} e^{2} x^{3} + 6 \, a c^{2} d e x^{2} + 3 \, b c d e \log \left (c x^{2} + 1\right ) + 3 \, b c d e \log \left (c x^{2} - 1\right ) + 2 \,{\left (3 \, b c d^{2} - b e^{2}\right )} \sqrt{c} \arctan \left (\sqrt{c} x\right ) +{\left (3 \, b c d^{2} + b e^{2}\right )} \sqrt{c} \log \left (\frac{c x^{2} - 2 \, \sqrt{c} x + 1}{c x^{2} - 1}\right ) + 2 \,{\left (3 \, a c^{2} d^{2} + 2 \, b c e^{2}\right )} x +{\left (b c^{2} e^{2} x^{3} + 3 \, b c^{2} d e x^{2} + 3 \, b c^{2} d^{2} x\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{6 \, c^{2}}, \frac{2 \, a c^{2} e^{2} x^{3} + 6 \, a c^{2} d e x^{2} + 3 \, b c d e \log \left (c x^{2} + 1\right ) + 3 \, b c d e \log \left (c x^{2} - 1\right ) + 2 \,{\left (3 \, b c d^{2} + b e^{2}\right )} \sqrt{-c} \arctan \left (\sqrt{-c} x\right ) +{\left (3 \, b c d^{2} - b e^{2}\right )} \sqrt{-c} \log \left (\frac{c x^{2} + 2 \, \sqrt{-c} x - 1}{c x^{2} + 1}\right ) + 2 \,{\left (3 \, a c^{2} d^{2} + 2 \, b c e^{2}\right )} x +{\left (b c^{2} e^{2} x^{3} + 3 \, b c^{2} d e x^{2} + 3 \, b c^{2} d^{2} x\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{6 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(2*a*c^2*e^2*x^3 + 6*a*c^2*d*e*x^2 + 3*b*c*d*e*log(c*x^2 + 1) + 3*b*c*d*e*log(c*x^2 - 1) + 2*(3*b*c*d^2 -
b*e^2)*sqrt(c)*arctan(sqrt(c)*x) + (3*b*c*d^2 + b*e^2)*sqrt(c)*log((c*x^2 - 2*sqrt(c)*x + 1)/(c*x^2 - 1)) + 2
*(3*a*c^2*d^2 + 2*b*c*e^2)*x + (b*c^2*e^2*x^3 + 3*b*c^2*d*e*x^2 + 3*b*c^2*d^2*x)*log(-(c*x^2 + 1)/(c*x^2 - 1))
)/c^2, 1/6*(2*a*c^2*e^2*x^3 + 6*a*c^2*d*e*x^2 + 3*b*c*d*e*log(c*x^2 + 1) + 3*b*c*d*e*log(c*x^2 - 1) + 2*(3*b*c
*d^2 + b*e^2)*sqrt(-c)*arctan(sqrt(-c)*x) + (3*b*c*d^2 - b*e^2)*sqrt(-c)*log((c*x^2 + 2*sqrt(-c)*x - 1)/(c*x^2
+ 1)) + 2*(3*a*c^2*d^2 + 2*b*c*e^2)*x + (b*c^2*e^2*x^3 + 3*b*c^2*d*e*x^2 + 3*b*c^2*d^2*x)*log(-(c*x^2 + 1)/(c
*x^2 - 1)))/c^2]

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Sympy [A]  time = 34.1347, size = 1652, normalized size = 10.46 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-12*a*c**2*d**2*x*sqrt(1/c)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*I*a*c**2*d**2*x*sqrt(1/
c)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*a*c**2*d*e*x**2*sqrt(1/c)/(-12*c**2*sqrt(1/c) - 12*I*c**2*s
qrt(1/c)) - 12*I*a*c**2*d*e*x**2*sqrt(1/c)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 4*a*c**2*e**2*x**3*sqr
t(1/c)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 4*I*a*c**2*e**2*x**3*sqrt(1/c)/(-12*c**2*sqrt(1/c) - 12*I*
c**2*sqrt(1/c)) - 12*b*c**2*d**2*x*sqrt(1/c)*atanh(c*x**2)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*I*b
*c**2*d**2*x*sqrt(1/c)*atanh(c*x**2)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) + 12*I*b*c**2*d**2*log(x - I*s
qrt(1/c))/(-12*c**3*sqrt(1/c) - 12*I*c**3*sqrt(1/c)) - 12*I*b*c**2*d**2*log(x - sqrt(1/c))/(-12*c**3*sqrt(1/c)
- 12*I*c**3*sqrt(1/c)) - 12*I*b*c**2*d**2*atanh(c*x**2)/(-12*c**3*sqrt(1/c) - 12*I*c**3*sqrt(1/c)) - 12*b*c**
2*d*e*x**2*sqrt(1/c)*atanh(c*x**2)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*I*b*c**2*d*e*x**2*sqrt(1/c)
*atanh(c*x**2)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 4*b*c**2*e**2*x**3*sqrt(1/c)*atanh(c*x**2)/(-12*c*
*2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 4*I*b*c**2*e**2*x**3*sqrt(1/c)*atanh(c*x**2)/(-12*c**2*sqrt(1/c) - 12*I*
c**2*sqrt(1/c)) - 2*I*b*c**2*e**2*log(x + I*sqrt(1/c))/(-12*c**4*sqrt(1/c) - 12*I*c**4*sqrt(1/c)) + 12*b*c*d**
2*log(x + I*sqrt(1/c))/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*b*c*d**2*log(x - sqrt(1/c))/(-12*c**2*s
qrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*b*c*d**2*atanh(c*x**2)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*b*
c*d*e*sqrt(1/c)*log(x - I*sqrt(1/c))/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*I*b*c*d*e*sqrt(1/c)*log(x
- I*sqrt(1/c))/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*b*c*d*e*sqrt(1/c)*log(x + I*sqrt(1/c))/(-12*c*
*2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 12*I*b*c*d*e*sqrt(1/c)*log(x + I*sqrt(1/c))/(-12*c**2*sqrt(1/c) - 12*I*c
**2*sqrt(1/c)) + 12*b*c*d*e*sqrt(1/c)*atanh(c*x**2)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) + 12*I*b*c*d*e*
sqrt(1/c)*atanh(c*x**2)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 8*b*c*e**2*x*sqrt(1/c)/(-12*c**2*sqrt(1/c
) - 12*I*c**2*sqrt(1/c)) - 8*I*b*c*e**2*x*sqrt(1/c)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) + 6*I*b*c*e**2*
log(x + I*sqrt(1/c))/(-12*c**3*sqrt(1/c) - 12*I*c**3*sqrt(1/c)) - 4*I*b*c*e**2*log(x - sqrt(1/c))/(-12*c**3*sq
rt(1/c) - 12*I*c**3*sqrt(1/c)) - 4*I*b*c*e**2*atanh(c*x**2)/(-12*c**3*sqrt(1/c) - 12*I*c**3*sqrt(1/c)) + 4*b*e
**2*log(x - I*sqrt(1/c))/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)) - 4*b*e**2*log(x - sqrt(1/c))/(-12*c**2*sq
rt(1/c) - 12*I*c**2*sqrt(1/c)) - 4*b*e**2*atanh(c*x**2)/(-12*c**2*sqrt(1/c) - 12*I*c**2*sqrt(1/c)), Ne(c, 0)),
(a*(d**2*x + d*e*x**2 + e**2*x**3/3), True))

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Giac [A]  time = 1.91503, size = 289, normalized size = 1.83 \begin{align*} -\frac{1}{3} \, b c^{5}{\left (\frac{\arctan \left (\sqrt{c} x\right )}{c^{\frac{13}{2}}} - \frac{\arctan \left (\frac{c x}{\sqrt{-c}}\right )}{\sqrt{-c} c^{6}}\right )} e^{2} + b c^{3} d^{2}{\left (\frac{\arctan \left (\sqrt{c} x\right )}{c^{\frac{7}{2}}} + \frac{\arctan \left (\frac{c x}{\sqrt{-c}}\right )}{\sqrt{-c} c^{3}}\right )} + \frac{b c x^{3} e^{2} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 3 \, b c d x^{2} e \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a c x^{3} e^{2} + 6 \, a c d x^{2} e + 3 \, b c d^{2} x \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a c d^{2} x + 3 \, b d e \log \left (c^{2} x^{4} - 1\right ) + 4 \, b x e^{2}}{6 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b*c^5*(arctan(sqrt(c)*x)/c^(13/2) - arctan(c*x/sqrt(-c))/(sqrt(-c)*c^6))*e^2 + b*c^3*d^2*(arctan(sqrt(c)*
x)/c^(7/2) + arctan(c*x/sqrt(-c))/(sqrt(-c)*c^3)) + 1/6*(b*c*x^3*e^2*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 3*b*c*d*x
^2*e*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 2*a*c*x^3*e^2 + 6*a*c*d*x^2*e + 3*b*c*d^2*x*log(-(c*x^2 + 1)/(c*x^2 - 1))
+ 6*a*c*d^2*x + 3*b*d*e*log(c^2*x^4 - 1) + 4*b*x*e^2)/c