3.2.99 \(\int \frac {(a+b \tanh ^{-1}(c \sqrt {x}))^2}{x} \, dx\) [199]

Optimal. Leaf size=145 \[ 4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2-2 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right )+2 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {PolyLog}\left (2,-1+\frac {2}{1-c \sqrt {x}}\right )+b^2 \text {PolyLog}\left (3,1-\frac {2}{1-c \sqrt {x}}\right )-b^2 \text {PolyLog}\left (3,-1+\frac {2}{1-c \sqrt {x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6035, 6033, 6199, 6095, 6205, 6745} \begin {gather*} -2 b \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )+2 b \text {Li}_2\left (\frac {2}{1-c \sqrt {x}}-1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )+4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+b^2 \text {Li}_3\left (1-\frac {2}{1-c \sqrt {x}}\right )-b^2 \text {Li}_3\left (\frac {2}{1-c \sqrt {x}}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])^2/x,x]

[Out]

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

Rule 6033

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

Rule 6035

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTanh[c*x])
^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 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 6199

Int[(ArcTanh[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[
Log[1 + u]*((a + b*ArcTanh[c*x])^p/(d + e*x^2)), x], x] - Dist[1/2, Int[Log[1 - u]*((a + b*ArcTanh[c*x])^p/(d
+ e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[u^2 - (1 - 2/(1 - c*x
))^2, 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 \frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{x} \, dx &=2 \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2-(8 b c) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+(4 b c) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )-(4 b c) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2-2 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )+2 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-c \sqrt {x}}\right )+\left (2 b^2 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )-\left (2 b^2 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2-2 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )+2 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-c \sqrt {x}}\right )+b^2 \text {Li}_3\left (1-\frac {2}{1-c \sqrt {x}}\right )-b^2 \text {Li}_3\left (-1+\frac {2}{1-c \sqrt {x}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 164, normalized size = 1.13 \begin {gather*} 4 \tanh ^{-1}\left (1+\frac {2}{-1+c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2-b \left (-2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {PolyLog}\left (2,\frac {1+c \sqrt {x}}{1-c \sqrt {x}}\right )+2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {PolyLog}\left (2,\frac {1+c \sqrt {x}}{-1+c \sqrt {x}}\right )+b \left (\text {PolyLog}\left (3,\frac {1+c \sqrt {x}}{1-c \sqrt {x}}\right )-\text {PolyLog}\left (3,\frac {1+c \sqrt {x}}{-1+c \sqrt {x}}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 2.59, size = 742, normalized size = 5.12

method	result	size
derivativedivides	\(2 a^{2} \ln \left (c \sqrt {x}\right )+2 b^{2} \ln \left (c \sqrt {x}\right ) \arctanh \left (c \sqrt {x}\right )^{2}-2 b^{2} \arctanh \left (c \sqrt {x}\right ) \polylog \left (2, -\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}\right )+b^{2} \polylog \left (3, -\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}\right )-2 b^{2} \arctanh \left (c \sqrt {x}\right )^{2} \ln \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )+2 b^{2} \arctanh \left (c \sqrt {x}\right )^{2} \ln \left (1-\frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )+4 b^{2} \arctanh \left (c \sqrt {x}\right ) \polylog \left (2, \frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )-4 b^{2} \polylog \left (3, \frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )+2 b^{2} \arctanh \left (c \sqrt {x}\right )^{2} \ln \left (1+\frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )+4 b^{2} \arctanh \left (c \sqrt {x}\right ) \polylog \left (2, -\frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )-4 b^{2} \polylog \left (3, -\frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )-i b^{2} \pi \,\mathrm {csgn}\left (\frac {i}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right )^{2} \arctanh \left (c \sqrt {x}\right )^{2}+i b^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right ) \arctanh \left (c \sqrt {x}\right )^{2}+i b^{2} \pi \mathrm {csgn}\left (\frac {i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right )^{3} \arctanh \left (c \sqrt {x}\right )^{2}-i b^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right )^{2} \arctanh \left (c \sqrt {x}\right )^{2}+4 a b \ln \left (c \sqrt {x}\right ) \arctanh \left (c \sqrt {x}\right )-2 a b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )-2 a b \dilog \left (c \sqrt {x}\right )-2 a b \dilog \left (1+c \sqrt {x}\right )\)	\(742\)
default	\(2 a^{2} \ln \left (c \sqrt {x}\right )+2 b^{2} \ln \left (c \sqrt {x}\right ) \arctanh \left (c \sqrt {x}\right )^{2}-2 b^{2} \arctanh \left (c \sqrt {x}\right ) \polylog \left (2, -\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}\right )+b^{2} \polylog \left (3, -\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}\right )-2 b^{2} \arctanh \left (c \sqrt {x}\right )^{2} \ln \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )+2 b^{2} \arctanh \left (c \sqrt {x}\right )^{2} \ln \left (1-\frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )+4 b^{2} \arctanh \left (c \sqrt {x}\right ) \polylog \left (2, \frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )-4 b^{2} \polylog \left (3, \frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )+2 b^{2} \arctanh \left (c \sqrt {x}\right )^{2} \ln \left (1+\frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )+4 b^{2} \arctanh \left (c \sqrt {x}\right ) \polylog \left (2, -\frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )-4 b^{2} \polylog \left (3, -\frac {1+c \sqrt {x}}{\sqrt {-x \,c^{2}+1}}\right )-i b^{2} \pi \,\mathrm {csgn}\left (\frac {i}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right )^{2} \arctanh \left (c \sqrt {x}\right )^{2}+i b^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right ) \arctanh \left (c \sqrt {x}\right )^{2}+i b^{2} \pi \mathrm {csgn}\left (\frac {i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right )^{3} \arctanh \left (c \sqrt {x}\right )^{2}-i b^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}-1\right )}{1+\frac {\left (1+c \sqrt {x}\right )^{2}}{-x \,c^{2}+1}}\right )^{2} \arctanh \left (c \sqrt {x}\right )^{2}+4 a b \ln \left (c \sqrt {x}\right ) \arctanh \left (c \sqrt {x}\right )-2 a b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )-2 a b \dilog \left (c \sqrt {x}\right )-2 a b \dilog \left (1+c \sqrt {x}\right )\)	\(742\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))^2/x,x,method=_RETURNVERBOSE)

[Out]

2*a^2*ln(c*x^(1/2))+2*b^2*ln(c*x^(1/2))*arctanh(c*x^(1/2))^2-2*b^2*arctanh(c*x^(1/2))*polylog(2,-(1+c*x^(1/2))
^2/(-c^2*x+1))+b^2*polylog(3,-(1+c*x^(1/2))^2/(-c^2*x+1))-2*b^2*arctanh(c*x^(1/2))^2*ln((1+c*x^(1/2))^2/(-c^2*
x+1)-1)+2*b^2*arctanh(c*x^(1/2))^2*ln(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+4*b^2*arctanh(c*x^(1/2))*polylog(2,(1+
c*x^(1/2))/(-c^2*x+1)^(1/2))-4*b^2*polylog(3,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+2*b^2*arctanh(c*x^(1/2))^2*ln(1+(
1+c*x^(1/2))/(-c^2*x+1)^(1/2))+4*b^2*arctanh(c*x^(1/2))*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-4*b^2*polyl
og(3,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+I*b^2*Pi*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/(1+(1+c*x^(1/2))^2/(-c^2*
x+1)))^3*arctanh(c*x^(1/2))^2-I*b^2*Pi*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1))*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+
1)-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*arctanh(c*x^(1/2))^2-I*b^2*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*c
sgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*arctanh(c*x^(1/2))^2+I*b^2*Pi*csgn(I*((
1+c*x^(1/2))^2/(-c^2*x+1)-1))*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/(1+
(1+c*x^(1/2))^2/(-c^2*x+1)))*arctanh(c*x^(1/2))^2+4*a*b*ln(c*x^(1/2))*arctanh(c*x^(1/2))-2*a*b*ln(c*x^(1/2))*l
n(1+c*x^(1/2))-2*a*b*dilog(c*x^(1/2))-2*a*b*dilog(1+c*x^(1/2))

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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)))^2/x,x, algorithm="maxima")

[Out]

1/4*b^2*integrate(log(c*sqrt(x) + 1)^2/x, x) - 1/2*b^2*integrate(log(c*sqrt(x) + 1)*log(-c*sqrt(x) + 1)/x, x)
+ 1/4*b^2*integrate(log(-c*sqrt(x) + 1)^2/x, x) + a*b*integrate(log(c*sqrt(x) + 1)/x, x) - a*b*integrate(log(-
c*sqrt(x) + 1)/x, x) + a^2*log(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)))^2/x,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )^{2}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^(1/2)))^2/x,x)

[Out]

int((a + b*atanh(c*x^(1/2)))^2/x, x)

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