3.2.22 \(\int (1+c x) (a+b \tanh ^{-1}(c x))^3 \, dx\) [122]

Optimal. Leaf size=191 \[ \frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^3 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c}+\frac {3 b^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 c} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6065, 6021, 6131, 6055, 2449, 2352, 1600, 6095, 6205, 6745} \begin {gather*} -\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac {3 b^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {(c x+1)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac {3 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c}-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{2 c}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + c*x)*(a + b*ArcTanh[c*x])^3,x]

[Out]

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

Rule 1600

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 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 6021

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 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 6065

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]

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -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 6205

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

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int (1+c x) \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac {(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac {1}{2} (3 b) \int \left (-\left (a+b \tanh ^{-1}(c x)\right )^2+\frac {2 (1+c x) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2}\right ) \, dx\\ &=\frac {(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}+\frac {1}{2} (3 b) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx-(3 b) \int \frac {(1+c x) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac {3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-(3 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx-\left (3 b^2 c\right ) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\left (3 b^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx+\left (6 b^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\left (3 b^3\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx+\left (3 b^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c}\\ &=\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {3}{2} b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c}\\ \end {align*}

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Mathematica [A]
time = 0.48, size = 334, normalized size = 1.75 \begin {gather*} \frac {4 a^3 c x+6 a^2 b c x+2 a^3 c^2 x^2+12 a^2 b c x \tanh ^{-1}(c x)+12 a b^2 c x \tanh ^{-1}(c x)+6 a^2 b c^2 x^2 \tanh ^{-1}(c x)-18 a b^2 \tanh ^{-1}(c x)^2-6 b^3 \tanh ^{-1}(c x)^2+12 a b^2 c x \tanh ^{-1}(c x)^2+6 b^3 c x \tanh ^{-1}(c x)^2+6 a b^2 c^2 x^2 \tanh ^{-1}(c x)^2-6 b^3 \tanh ^{-1}(c x)^3+4 b^3 c x \tanh ^{-1}(c x)^3+2 b^3 c^2 x^2 \tanh ^{-1}(c x)^3-24 a b^2 \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-12 b^3 \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-12 b^3 \tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+9 a^2 b \log (1-c x)+3 a^2 b \log (1+c x)+6 a b^2 \log \left (1-c^2 x^2\right )+6 b^2 \left (2 a+b+2 b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+6 b^3 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + c*x)*(a + b*ArcTanh[c*x])^3,x]

[Out]

(4*a^3*c*x + 6*a^2*b*c*x + 2*a^3*c^2*x^2 + 12*a^2*b*c*x*ArcTanh[c*x] + 12*a*b^2*c*x*ArcTanh[c*x] + 6*a^2*b*c^2
*x^2*ArcTanh[c*x] - 18*a*b^2*ArcTanh[c*x]^2 - 6*b^3*ArcTanh[c*x]^2 + 12*a*b^2*c*x*ArcTanh[c*x]^2 + 6*b^3*c*x*A
rcTanh[c*x]^2 + 6*a*b^2*c^2*x^2*ArcTanh[c*x]^2 - 6*b^3*ArcTanh[c*x]^3 + 4*b^3*c*x*ArcTanh[c*x]^3 + 2*b^3*c^2*x
^2*ArcTanh[c*x]^3 - 24*a*b^2*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - 12*b^3*ArcTanh[c*x]*Log[1 + E^(-2*Arc
Tanh[c*x])] - 12*b^3*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 9*a^2*b*Log[1 - c*x] + 3*a^2*b*Log[1 + c*x]
+ 6*a*b^2*Log[1 - c^2*x^2] + 6*b^2*(2*a + b + 2*b*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 6*b^3*Poly
Log[3, -E^(-2*ArcTanh[c*x])])/(4*c)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 10.72, size = 6152, normalized size = 32.21

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(6152\)
default	\(\text {Expression too large to display}\)	\(6152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x+1)*(a+b*arctanh(c*x))^3,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)*(a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

1/2*a^3*c*x^2 + 3/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*a^2*b*c + a^3*x +
3/2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a^2*b/c - 1/16*((b^3*c^2*x^2 + 2*b^3*c*x - 3*b^3)*log(-c*x + 1)^
3 - 3*(2*a*b^2*c^2*x^2 + 2*(2*a*b^2*c + b^3*c)*x + (b^3*c^2*x^2 + 2*b^3*c*x + b^3)*log(c*x + 1))*log(-c*x + 1)
^2)/c - integrate(-1/8*((b^3*c^2*x^2 - b^3)*log(c*x + 1)^3 + 6*(a*b^2*c^2*x^2 - a*b^2)*log(c*x + 1)^2 - 3*(2*a
*b^2*c^2*x^2 + (b^3*c^2*x^2 - b^3)*log(c*x + 1)^2 + 2*(2*a*b^2*c + b^3*c)*x + (2*b^3*c*x - 4*a*b^2 + b^3 + (4*
a*b^2*c^2 + b^3*c^2)*x^2)*log(c*x + 1))*log(-c*x + 1))/(c*x - 1), x)

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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((c*x+1)*(a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

integral(a^3*c*x + (b^3*c*x + b^3)*arctanh(c*x)^3 + a^3 + 3*(a*b^2*c*x + a*b^2)*arctanh(c*x)^2 + 3*(a^2*b*c*x
+ a^2*b)*arctanh(c*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3} \left (c x + 1\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)*(a+b*atanh(c*x))**3,x)

[Out]

Integral((a + b*atanh(c*x))**3*(c*x + 1), 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((c*x+1)*(a+b*arctanh(c*x))^3,x, algorithm="giac")

[Out]

integrate((c*x + 1)*(b*arctanh(c*x) + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3\,\left (c\,x+1\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^3*(c*x + 1),x)

[Out]

int((a + b*atanh(c*x))^3*(c*x + 1), x)

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