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

Optimal. Leaf size=182 \[ \frac{(d+e x)^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{4 e}+\frac{b \left (c^2 d^4+6 c d^2 e^2+e^4\right ) \log \left (1-c x^2\right )}{8 c^2 e}-\frac{b \left (c^2 d^4-6 c d^2 e^2+e^4\right ) \log \left (c x^2+1\right )}{8 c^2 e}+\frac{b d \left (c d^2-e^2\right ) \tan ^{-1}\left (\sqrt{c} x\right )}{c^{3/2}}-\frac{b d \left (c d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{c} x\right )}{c^{3/2}}+\frac{2 b d e^2 x}{c}+\frac{b e^3 x^2}{4 c} \]

[Out]

(2*b*d*e^2*x)/c + (b*e^3*x^2)/(4*c) + (b*d*(c*d^2 - e^2)*ArcTan[Sqrt[c]*x])/c^(3/2) - (b*d*(c*d^2 + e^2)*ArcTa
nh[Sqrt[c]*x])/c^(3/2) + ((d + e*x)^4*(a + b*ArcTanh[c*x^2]))/(4*e) + (b*(c^2*d^4 + 6*c*d^2*e^2 + e^4)*Log[1 -
c*x^2])/(8*c^2*e) - (b*(c^2*d^4 - 6*c*d^2*e^2 + e^4)*Log[1 + c*x^2])/(8*c^2*e)

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Rubi [A]  time = 0.222603, antiderivative size = 220, normalized size of antiderivative = 1.21, number of steps used = 19, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {6742, 6091, 298, 203, 206, 6097, 260, 321, 212, 275} \[ \frac{a (d+e x)^4}{4 e}+\frac{3 b d^2 e \log \left (1-c^2 x^4\right )}{4 c}-\frac{b d e^2 \tan ^{-1}\left (\sqrt{c} x\right )}{c^{3/2}}-\frac{b d e^2 \tanh ^{-1}\left (\sqrt{c} x\right )}{c^{3/2}}-\frac{b e^3 \tanh ^{-1}\left (c x^2\right )}{4 c^2}+\frac{3}{2} b d^2 e x^2 \tanh ^{-1}\left (c x^2\right )+b d^3 x \tanh ^{-1}\left (c x^2\right )+\frac{b d^3 \tan ^{-1}\left (\sqrt{c} x\right )}{\sqrt{c}}-\frac{b d^3 \tanh ^{-1}\left (\sqrt{c} x\right )}{\sqrt{c}}+b d e^2 x^3 \tanh ^{-1}\left (c x^2\right )+\frac{2 b d e^2 x}{c}+\frac{b e^3 x^2}{4 c}+\frac{1}{4} b e^3 x^4 \tanh ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 6091

Int[ArcTanh[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcTanh[c*x^n], x] - Dist[c*n, Int[x^n/(1 - c^2*x^(2*n)), x]
, x] /; FreeQ[{c, n}, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin{align*} \int (d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \, dx &=\int \left (a (d+e x)^3+b (d+e x)^3 \tanh ^{-1}\left (c x^2\right )\right ) \, dx\\ &=\frac{a (d+e x)^4}{4 e}+b \int (d+e x)^3 \tanh ^{-1}\left (c x^2\right ) \, dx\\ &=\frac{a (d+e x)^4}{4 e}+b \int \left (d^3 \tanh ^{-1}\left (c x^2\right )+3 d^2 e x \tanh ^{-1}\left (c x^2\right )+3 d e^2 x^2 \tanh ^{-1}\left (c x^2\right )+e^3 x^3 \tanh ^{-1}\left (c x^2\right )\right ) \, dx\\ &=\frac{a (d+e x)^4}{4 e}+\left (b d^3\right ) \int \tanh ^{-1}\left (c x^2\right ) \, dx+\left (3 b d^2 e\right ) \int x \tanh ^{-1}\left (c x^2\right ) \, dx+\left (3 b d e^2\right ) \int x^2 \tanh ^{-1}\left (c x^2\right ) \, dx+\left (b e^3\right ) \int x^3 \tanh ^{-1}\left (c x^2\right ) \, dx\\ &=\frac{a (d+e x)^4}{4 e}+b d^3 x \tanh ^{-1}\left (c x^2\right )+\frac{3}{2} b d^2 e x^2 \tanh ^{-1}\left (c x^2\right )+b d e^2 x^3 \tanh ^{-1}\left (c x^2\right )+\frac{1}{4} b e^3 x^4 \tanh ^{-1}\left (c x^2\right )-\left (2 b c d^3\right ) \int \frac{x^2}{1-c^2 x^4} \, dx-\left (3 b c d^2 e\right ) \int \frac{x^3}{1-c^2 x^4} \, dx-\left (2 b c d e^2\right ) \int \frac{x^4}{1-c^2 x^4} \, dx-\frac{1}{2} \left (b c e^3\right ) \int \frac{x^5}{1-c^2 x^4} \, dx\\ &=\frac{2 b d e^2 x}{c}+\frac{a (d+e x)^4}{4 e}+b d^3 x \tanh ^{-1}\left (c x^2\right )+\frac{3}{2} b d^2 e x^2 \tanh ^{-1}\left (c x^2\right )+b d e^2 x^3 \tanh ^{-1}\left (c x^2\right )+\frac{1}{4} b e^3 x^4 \tanh ^{-1}\left (c x^2\right )+\frac{3 b d^2 e \log \left (1-c^2 x^4\right )}{4 c}-\left (b d^3\right ) \int \frac{1}{1-c x^2} \, dx+\left (b d^3\right ) \int \frac{1}{1+c x^2} \, dx-\frac{\left (2 b d e^2\right ) \int \frac{1}{1-c^2 x^4} \, dx}{c}-\frac{1}{4} \left (b c e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\frac{2 b d e^2 x}{c}+\frac{b e^3 x^2}{4 c}+\frac{a (d+e x)^4}{4 e}+\frac{b d^3 \tan ^{-1}\left (\sqrt{c} x\right )}{\sqrt{c}}-\frac{b d^3 \tanh ^{-1}\left (\sqrt{c} x\right )}{\sqrt{c}}+b d^3 x \tanh ^{-1}\left (c x^2\right )+\frac{3}{2} b d^2 e x^2 \tanh ^{-1}\left (c x^2\right )+b d e^2 x^3 \tanh ^{-1}\left (c x^2\right )+\frac{1}{4} b e^3 x^4 \tanh ^{-1}\left (c x^2\right )+\frac{3 b d^2 e \log \left (1-c^2 x^4\right )}{4 c}-\frac{\left (b d e^2\right ) \int \frac{1}{1-c x^2} \, dx}{c}-\frac{\left (b d e^2\right ) \int \frac{1}{1+c x^2} \, dx}{c}-\frac{\left (b e^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,x^2\right )}{4 c}\\ &=\frac{2 b d e^2 x}{c}+\frac{b e^3 x^2}{4 c}+\frac{a (d+e x)^4}{4 e}+\frac{b d^3 \tan ^{-1}\left (\sqrt{c} x\right )}{\sqrt{c}}-\frac{b d e^2 \tan ^{-1}\left (\sqrt{c} x\right )}{c^{3/2}}-\frac{b d^3 \tanh ^{-1}\left (\sqrt{c} x\right )}{\sqrt{c}}-\frac{b d e^2 \tanh ^{-1}\left (\sqrt{c} x\right )}{c^{3/2}}-\frac{b e^3 \tanh ^{-1}\left (c x^2\right )}{4 c^2}+b d^3 x \tanh ^{-1}\left (c x^2\right )+\frac{3}{2} b d^2 e x^2 \tanh ^{-1}\left (c x^2\right )+b d e^2 x^3 \tanh ^{-1}\left (c x^2\right )+\frac{1}{4} b e^3 x^4 \tanh ^{-1}\left (c x^2\right )+\frac{3 b d^2 e \log \left (1-c^2 x^4\right )}{4 c}\\ \end{align*}

Mathematica [A]  time = 0.283916, size = 254, normalized size = 1.4 \[ \frac{1}{8} \left (\frac{2 e x^2 \left (6 a c d^2+b e^2\right )}{c}+\frac{8 d x \left (a c d^2+2 b e^2\right )}{c}+8 a d e^2 x^3+2 a e^3 x^4+\frac{b \left (4 c^{3/2} d^3+4 \sqrt{c} d e^2+e^3\right ) \log \left (1-\sqrt{c} x\right )}{c^2}+\frac{b \left (-4 c^2 d^3-4 c d e^2+\sqrt{c} e^3\right ) \log \left (\sqrt{c} x+1\right )}{c^{5/2}}+\frac{8 b d \left (c d^2-e^2\right ) \tan ^{-1}\left (\sqrt{c} x\right )}{c^{3/2}}+\frac{6 b d^2 e \log \left (1-c^2 x^4\right )}{c}-\frac{b e^3 \log \left (c x^2+1\right )}{c^2}+2 b x \tanh ^{-1}\left (c x^2\right ) \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*d*(a*c*d^2 + 2*b*e^2)*x)/c + (2*e*(6*a*c*d^2 + b*e^2)*x^2)/c + 8*a*d*e^2*x^3 + 2*a*e^3*x^4 + (8*b*d*(c*d^2
- e^2)*ArcTan[Sqrt[c]*x])/c^(3/2) + 2*b*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3)*ArcTanh[c*x^2] + (b*(4*
c^(3/2)*d^3 + 4*Sqrt[c]*d*e^2 + e^3)*Log[1 - Sqrt[c]*x])/c^2 + (b*(-4*c^2*d^3 - 4*c*d*e^2 + Sqrt[c]*e^3)*Log[1
+ Sqrt[c]*x])/c^(5/2) - (b*e^3*Log[1 + c*x^2])/c^2 + (6*b*d^2*e*Log[1 - c^2*x^4])/c)/8

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Maple [A]  time = 0.031, size = 306, normalized size = 1.7 \begin{align*}{\frac{a{e}^{3}{x}^{4}}{4}}+a{e}^{2}{x}^{3}d+{\frac{3\,ae{x}^{2}{d}^{2}}{2}}+ax{d}^{3}+{\frac{a{d}^{4}}{4\,e}}+{\frac{b{e}^{3}{\it Artanh} \left ( c{x}^{2} \right ){x}^{4}}{4}}+b{e}^{2}{\it Artanh} \left ( c{x}^{2} \right ){x}^{3}d+{\frac{3\,be{\it Artanh} \left ( c{x}^{2} \right ){x}^{2}{d}^{2}}{2}}+b{\it Artanh} \left ( c{x}^{2} \right ) x{d}^{3}+{\frac{b{\it Artanh} \left ( c{x}^{2} \right ){d}^{4}}{4\,e}}+{\frac{b{e}^{3}{x}^{2}}{4\,c}}+2\,{\frac{b{e}^{2}dx}{c}}-{\frac{b\ln \left ( c{x}^{2}+1 \right ){d}^{4}}{8\,e}}+{\frac{3\,be\ln \left ( c{x}^{2}+1 \right ){d}^{2}}{4\,c}}-{\frac{b{e}^{3}\ln \left ( c{x}^{2}+1 \right ) }{8\,{c}^{2}}}+{b{d}^{3}\arctan \left ( x\sqrt{c} \right ){\frac{1}{\sqrt{c}}}}-{b{e}^{2}d\arctan \left ( x\sqrt{c} \right ){c}^{-{\frac{3}{2}}}}+{\frac{b\ln \left ( c{x}^{2}-1 \right ){d}^{4}}{8\,e}}+{\frac{3\,be\ln \left ( c{x}^{2}-1 \right ){d}^{2}}{4\,c}}+{\frac{b{e}^{3}\ln \left ( c{x}^{2}-1 \right ) }{8\,{c}^{2}}}-{b{d}^{3}{\it Artanh} \left ( x\sqrt{c} \right ){\frac{1}{\sqrt{c}}}}-{b{e}^{2}d{\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)^3*(a+b*arctanh(c*x^2)),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.33936, size = 1145, normalized size = 6.29 \begin{align*} \left [\frac{2 \, a c^{2} e^{3} x^{4} + 8 \, a c^{2} d e^{2} x^{3} + 2 \,{\left (6 \, a c^{2} d^{2} e + b c e^{3}\right )} x^{2} + 8 \,{\left (b c d^{3} - b d e^{2}\right )} \sqrt{c} \arctan \left (\sqrt{c} x\right ) + 4 \,{\left (b c d^{3} + b d e^{2}\right )} \sqrt{c} \log \left (\frac{c x^{2} - 2 \, \sqrt{c} x + 1}{c x^{2} - 1}\right ) + 8 \,{\left (a c^{2} d^{3} + 2 \, b c d e^{2}\right )} x +{\left (6 \, b c d^{2} e - b e^{3}\right )} \log \left (c x^{2} + 1\right ) +{\left (6 \, b c d^{2} e + b e^{3}\right )} \log \left (c x^{2} - 1\right ) +{\left (b c^{2} e^{3} x^{4} + 4 \, b c^{2} d e^{2} x^{3} + 6 \, b c^{2} d^{2} e x^{2} + 4 \, b c^{2} d^{3} x\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{8 \, c^{2}}, \frac{2 \, a c^{2} e^{3} x^{4} + 8 \, a c^{2} d e^{2} x^{3} + 2 \,{\left (6 \, a c^{2} d^{2} e + b c e^{3}\right )} x^{2} + 8 \,{\left (b c d^{3} + b d e^{2}\right )} \sqrt{-c} \arctan \left (\sqrt{-c} x\right ) + 4 \,{\left (b c d^{3} - b d e^{2}\right )} \sqrt{-c} \log \left (\frac{c x^{2} + 2 \, \sqrt{-c} x - 1}{c x^{2} + 1}\right ) + 8 \,{\left (a c^{2} d^{3} + 2 \, b c d e^{2}\right )} x +{\left (6 \, b c d^{2} e - b e^{3}\right )} \log \left (c x^{2} + 1\right ) +{\left (6 \, b c d^{2} e + b e^{3}\right )} \log \left (c x^{2} - 1\right ) +{\left (b c^{2} e^{3} x^{4} + 4 \, b c^{2} d e^{2} x^{3} + 6 \, b c^{2} d^{2} e x^{2} + 4 \, b c^{2} d^{3} x\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{8 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(2*a*c^2*e^3*x^4 + 8*a*c^2*d*e^2*x^3 + 2*(6*a*c^2*d^2*e + b*c*e^3)*x^2 + 8*(b*c*d^3 - b*d*e^2)*sqrt(c)*ar
ctan(sqrt(c)*x) + 4*(b*c*d^3 + b*d*e^2)*sqrt(c)*log((c*x^2 - 2*sqrt(c)*x + 1)/(c*x^2 - 1)) + 8*(a*c^2*d^3 + 2*
b*c*d*e^2)*x + (6*b*c*d^2*e - b*e^3)*log(c*x^2 + 1) + (6*b*c*d^2*e + b*e^3)*log(c*x^2 - 1) + (b*c^2*e^3*x^4 +
4*b*c^2*d*e^2*x^3 + 6*b*c^2*d^2*e*x^2 + 4*b*c^2*d^3*x)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^2, 1/8*(2*a*c^2*e^3*x^
4 + 8*a*c^2*d*e^2*x^3 + 2*(6*a*c^2*d^2*e + b*c*e^3)*x^2 + 8*(b*c*d^3 + b*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x) +
4*(b*c*d^3 - b*d*e^2)*sqrt(-c)*log((c*x^2 + 2*sqrt(-c)*x - 1)/(c*x^2 + 1)) + 8*(a*c^2*d^3 + 2*b*c*d*e^2)*x + (
6*b*c*d^2*e - b*e^3)*log(c*x^2 + 1) + (6*b*c*d^2*e + b*e^3)*log(c*x^2 - 1) + (b*c^2*e^3*x^4 + 4*b*c^2*d*e^2*x^
3 + 6*b*c^2*d^2*e*x^2 + 4*b*c^2*d^3*x)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^2]

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Sympy [A]  time = 45.2278, size = 1083, normalized size = 5.95 \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)**3*(a+b*atanh(c*x**2)),x)

[Out]

Piecewise((a*d**3*x + 3*a*d**2*e*x**2/2 + a*d*e**2*x**3 + a*e**3*x**4/4 - b*c**2*d**3*(1/c)**(3/2)*log(x + I*s
qrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) + I*b*c**2*d**3*(1/c)**(3/2)*log(x + I*sqrt(1/c))/(-2*I*c**3/c**2 + 1
0*I*c**2/c) - 2*I*b*c**2*d**3*(1/c)**(3/2)*log(x - sqrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) + 4*b*c*d**3*sqrt
(1/c)*log(x - I*sqrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) - 4*I*b*c*d**3*sqrt(1/c)*log(x - I*sqrt(1/c))/(-2*I*
c**3/c**2 + 10*I*c**2/c) - 3*b*c*d**3*sqrt(1/c)*log(x + I*sqrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) - 5*I*b*c*
d**3*sqrt(1/c)*log(x + I*sqrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) + 10*I*b*c*d**3*sqrt(1/c)*log(x - sqrt(1/c)
)/(-2*I*c**3/c**2 + 10*I*c**2/c) + 8*I*b*c*d**3*sqrt(1/c)*atanh(c*x**2)/(-2*I*c**3/c**2 + 10*I*c**2/c) - b*c*d
*e**2*(1/c)**(3/2)*log(x + I*sqrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) + I*b*c*d*e**2*(1/c)**(3/2)*log(x + I*s
qrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) - 2*I*b*c*d*e**2*(1/c)**(3/2)*log(x - sqrt(1/c))/(-2*I*c**3/c**2 + 10
*I*c**2/c) + b*d**3*x*atanh(c*x**2) + 3*b*d**2*e*x**2*atanh(c*x**2)/2 + b*d*e**2*x**3*atanh(c*x**2) - 4*b*d*e*
*2*sqrt(1/c)*log(x - I*sqrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) - 4*I*b*d*e**2*sqrt(1/c)*log(x - I*sqrt(1/c))
/(-2*I*c**3/c**2 + 10*I*c**2/c) + 5*b*d*e**2*sqrt(1/c)*log(x + I*sqrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) - 5
*I*b*d*e**2*sqrt(1/c)*log(x + I*sqrt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) + 10*I*b*d*e**2*sqrt(1/c)*log(x - sq
rt(1/c))/(-2*I*c**3/c**2 + 10*I*c**2/c) + 8*I*b*d*e**2*sqrt(1/c)*atanh(c*x**2)/(-2*I*c**3/c**2 + 10*I*c**2/c)
+ b*e**3*x**4*atanh(c*x**2)/4 + 3*b*d**2*e*log(x - I*sqrt(1/c))/(2*c) + 3*b*d**2*e*log(x + I*sqrt(1/c))/(2*c)
- 3*b*d**2*e*atanh(c*x**2)/(2*c) + 2*b*d*e**2*x/c + b*e**3*x**2/(4*c) - b*e**3*atanh(c*x**2)/(4*c**2), Ne(c, 0
)), (a*(d**3*x + 3*d**2*e*x**2/2 + d*e**2*x**3 + e**3*x**4/4), True))

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Giac [A]  time = 2.39938, size = 423, normalized size = 2.32 \begin{align*} -b c^{5} d{\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^{3}{\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^{2} x^{4} e^{3} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 4 \, b c^{2} d x^{3} e^{2} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, b c^{2} d^{2} x^{2} e \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a c^{2} x^{4} e^{3} + 8 \, a c^{2} d x^{3} e^{2} + 12 \, a c^{2} d^{2} x^{2} e + 4 \, b c^{2} d^{3} x \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 8 \, a c^{2} d^{3} x + 6 \, b c d^{2} e \log \left (c^{2} x^{4} - 1\right ) + 2 \, b c x^{2} e^{3} + 16 \, b c d x e^{2} - b e^{3} \log \left (c x^{2} + 1\right ) + b e^{3} \log \left (c x^{2} - 1\right )}{8 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*c^5*d*(arctan(sqrt(c)*x)/c^(13/2) - arctan(c*x/sqrt(-c))/(sqrt(-c)*c^6))*e^2 + b*c^3*d^3*(arctan(sqrt(c)*x)
/c^(7/2) + arctan(c*x/sqrt(-c))/(sqrt(-c)*c^3)) + 1/8*(b*c^2*x^4*e^3*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 4*b*c^2*d
*x^3*e^2*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 6*b*c^2*d^2*x^2*e*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 2*a*c^2*x^4*e^3 + 8
*a*c^2*d*x^3*e^2 + 12*a*c^2*d^2*x^2*e + 4*b*c^2*d^3*x*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 8*a*c^2*d^3*x + 6*b*c*d^
2*e*log(c^2*x^4 - 1) + 2*b*c*x^2*e^3 + 16*b*c*d*x*e^2 - b*e^3*log(c*x^2 + 1) + b*e^3*log(c*x^2 - 1))/c^2