3.3.5 \(\int (a+b \tanh ^{-1}(c \sqrt {x}))^3 \, dx\) [205]

Optimal. Leaf size=142 \[ \frac {3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{c^2}+\frac {3 b \sqrt {x} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{c}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{c^2}+x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-\frac {6 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1-c \sqrt {x}}\right )}{c^2}-\frac {3 b^3 \text {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right )}{c^2} \]

[Out]

3*b*(a+b*arctanh(c*x^(1/2)))^2/c^2-(a+b*arctanh(c*x^(1/2)))^3/c^2+x*(a+b*arctanh(c*x^(1/2)))^3-6*b^2*(a+b*arct
anh(c*x^(1/2)))*ln(2/(1-c*x^(1/2)))/c^2-3*b^3*polylog(2,1-2/(1-c*x^(1/2)))/c^2+3*b*(a+b*arctanh(c*x^(1/2)))^2*
x^(1/2)/c

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Rubi [A]
time = 0.20, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6027, 6037, 6127, 6021, 6131, 6055, 2449, 2352, 6095} \begin {gather*} -\frac {6 b^2 \log \left (\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^2}+\frac {3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{c^2}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{c^2}+\frac {3 b \sqrt {x} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{c}+x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )}{c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])^3,x]

[Out]

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

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 6027

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x
^(k - 1)*(a + b*ArcTanh[c*x^(k*n)])^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1] && FractionQ[
n]

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 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 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]

Rubi steps

\begin {align*} \int \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3 \, dx &=\int \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3 \, dx\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 201, normalized size = 1.42 \begin {gather*} \frac {6 b^2 \left (-1+c \sqrt {x}\right ) \left (a+b+a c \sqrt {x}\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^2+2 b^3 \left (-1+c^2 x\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^3+6 b \tanh ^{-1}\left (c \sqrt {x}\right ) \left (2 a b c \sqrt {x}+a^2 c^2 x-2 b^2 \log \left (1+e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )\right )+a \left (6 a b c \sqrt {x}+2 a^2 c^2 x+3 a b \log \left (1-c \sqrt {x}\right )-3 a b \log \left (1+c \sqrt {x}\right )+6 b^2 \log \left (1-c^2 x\right )\right )+6 b^3 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b^2*(-1 + c*Sqrt[x])*(a + b + a*c*Sqrt[x])*ArcTanh[c*Sqrt[x]]^2 + 2*b^3*(-1 + c^2*x)*ArcTanh[c*Sqrt[x]]^3 +
6*b*ArcTanh[c*Sqrt[x]]*(2*a*b*c*Sqrt[x] + a^2*c^2*x - 2*b^2*Log[1 + E^(-2*ArcTanh[c*Sqrt[x]])]) + a*(6*a*b*c*
Sqrt[x] + 2*a^2*c^2*x + 3*a*b*Log[1 - c*Sqrt[x]] - 3*a*b*Log[1 + c*Sqrt[x]] + 6*b^2*Log[1 - c^2*x]) + 6*b^3*Po
lyLog[2, -E^(-2*ArcTanh[c*Sqrt[x]])])/(2*c^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 6.80, size = 5972, normalized size = 42.06

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(5972\)
default	\(\text {Expression too large to display}\)	\(5972\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))^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((a+b*arctanh(c*x^(1/2)))^3,x, algorithm="maxima")

[Out]

3/2*(c*(2*sqrt(x)/c^2 - log(c*sqrt(x) + 1)/c^3 + log(c*sqrt(x) - 1)/c^3) + 2*x*arctanh(c*sqrt(x)))*a^2*b + 3/4
*(4*c*(2*sqrt(x)/c^2 - log(c*sqrt(x) + 1)/c^3 + log(c*sqrt(x) - 1)/c^3)*arctanh(c*sqrt(x)) + 4*x*arctanh(c*sqr
t(x))^2 - (2*(log(c*sqrt(x) - 1) - 2)*log(c*sqrt(x) + 1) - log(c*sqrt(x) + 1)^2 - log(c*sqrt(x) - 1)^2 - 4*log
(c*sqrt(x) - 1))/c^2)*a*b^2 + a^3*x - 1/32*b^3*(((4*log(-c*sqrt(x) + 1)^3 - 6*log(-c*sqrt(x) + 1)^2 + 6*log(-c
*sqrt(x) + 1) - 3)*(c*sqrt(x) - 1)^2 + 8*(log(-c*sqrt(x) + 1)^3 - 3*log(-c*sqrt(x) + 1)^2 + 6*log(-c*sqrt(x) +
1) - 6)*(c*sqrt(x) - 1))/c^2 - 4*integrate(log(c*sqrt(x) + 1)^3 - 3*log(c*sqrt(x) + 1)^2*log(-c*sqrt(x) + 1)
+ 3*log(c*sqrt(x) + 1)*log(-c*sqrt(x) + 1)^2, 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((a+b*arctanh(c*x^(1/2)))^3,x, algorithm="fricas")

[Out]

integral(b^3*arctanh(c*sqrt(x))^3 + 3*a*b^2*arctanh(c*sqrt(x))^2 + 3*a^2*b*arctanh(c*sqrt(x)) + a^3, 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 \sqrt {x} \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**(1/2)))**3,x)

[Out]

Integral((a + b*atanh(c*sqrt(x)))**3, 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((a+b*arctanh(c*x^(1/2)))^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*sqrt(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\,\sqrt {x}\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^(1/2)))^3,x)

[Out]

int((a + b*atanh(c*x^(1/2)))^3, x)

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