3.2.97 \(\int x (a+b \tanh ^{-1}(c \sqrt {x}))^2 \, dx\) [197]

Optimal. Leaf size=129 \[ \frac {a b \sqrt {x}}{c^3}+\frac {b^2 x}{6 c^2}+\frac {b^2 \sqrt {x} \tanh ^{-1}\left (c \sqrt {x}\right )}{c^3}+\frac {b x^{3/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{3 c}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{2 c^4}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+\frac {2 b^2 \log \left (1-c^2 x\right )}{3 c^4} \]

[Out]

1/6*b^2*x/c^2+1/3*b*x^(3/2)*(a+b*arctanh(c*x^(1/2)))/c-1/2*(a+b*arctanh(c*x^(1/2)))^2/c^4+1/2*x^2*(a+b*arctanh
(c*x^(1/2)))^2+2/3*b^2*ln(-c^2*x+1)/c^4+a*b*x^(1/2)/c^3+b^2*arctanh(c*x^(1/2))*x^(1/2)/c^3

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Rubi [A]
time = 0.18, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6039, 6037, 6127, 272, 45, 6021, 266, 6095} \begin {gather*} -\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{2 c^4}+\frac {a b \sqrt {x}}{c^3}+\frac {b x^{3/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{3 c}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+\frac {b^2 \sqrt {x} \tanh ^{-1}\left (c \sqrt {x}\right )}{c^3}+\frac {b^2 x}{6 c^2}+\frac {2 b^2 \log \left (1-c^2 x\right )}{3 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c*Sqrt[x]])^2,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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 272

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

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 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 6039

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

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]

Rubi steps

\begin {align*} \int x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \, dx &=\int x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \, dx\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 160, normalized size = 1.24 \begin {gather*} \frac {6 a b c \sqrt {x}+b^2 c^2 x+2 a b c^3 x^{3/2}+3 a^2 c^4 x^2+2 b c \sqrt {x} \left (3 a c^3 x^{3/2}+b \left (3+c^2 x\right )\right ) \tanh ^{-1}\left (c \sqrt {x}\right )+3 b^2 \left (-1+c^4 x^2\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^2+b (3 a+4 b) \log \left (1-c \sqrt {x}\right )-3 a b \log \left (1+c \sqrt {x}\right )+4 b^2 \log \left (1+c \sqrt {x}\right )}{6 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(296\) vs. \(2(105)=210\).
time = 0.18, size = 297, normalized size = 2.30

method	result	size
derivativedivides	\(\frac {\frac {c^{4} x^{2} a^{2}}{2}+\frac {b^{2} c^{4} x^{2} \arctanh \left (c \sqrt {x}\right )^{2}}{2}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{3}+b^{2} \arctanh \left (c \sqrt {x}\right ) c \sqrt {x}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}-\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b^{2} \ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {b^{2} \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}-\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{4}+\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {b^{2} \ln \left (1+c \sqrt {x}\right )^{2}}{8}+\frac {b^{2} c^{2} x}{6}+\frac {2 b^{2} \ln \left (c \sqrt {x}-1\right )}{3}+\frac {2 b^{2} \ln \left (1+c \sqrt {x}\right )}{3}+a b \,c^{4} x^{2} \arctanh \left (c \sqrt {x}\right )+\frac {a b \,c^{3} x^{\frac {3}{2}}}{3}+a b c \sqrt {x}+\frac {a b \ln \left (c \sqrt {x}-1\right )}{2}-\frac {a b \ln \left (1+c \sqrt {x}\right )}{2}}{c^{4}}\)	\(297\)
default	\(\frac {\frac {c^{4} x^{2} a^{2}}{2}+\frac {b^{2} c^{4} x^{2} \arctanh \left (c \sqrt {x}\right )^{2}}{2}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{3}+b^{2} \arctanh \left (c \sqrt {x}\right ) c \sqrt {x}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}-\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b^{2} \ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {b^{2} \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}-\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{4}+\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {b^{2} \ln \left (1+c \sqrt {x}\right )^{2}}{8}+\frac {b^{2} c^{2} x}{6}+\frac {2 b^{2} \ln \left (c \sqrt {x}-1\right )}{3}+\frac {2 b^{2} \ln \left (1+c \sqrt {x}\right )}{3}+a b \,c^{4} x^{2} \arctanh \left (c \sqrt {x}\right )+\frac {a b \,c^{3} x^{\frac {3}{2}}}{3}+a b c \sqrt {x}+\frac {a b \ln \left (c \sqrt {x}-1\right )}{2}-\frac {a b \ln \left (1+c \sqrt {x}\right )}{2}}{c^{4}}\)	\(297\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/c^4*(1/4*c^4*x^2*a^2+1/4*b^2*c^4*x^2*arctanh(c*x^(1/2))^2+1/6*b^2*arctanh(c*x^(1/2))*c^3*x^(3/2)+1/2*b^2*arc
tanh(c*x^(1/2))*c*x^(1/2)+1/4*b^2*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1)-1/4*b^2*arctanh(c*x^(1/2))*ln(1+c*x^(1/2)
)+1/16*b^2*ln(c*x^(1/2)-1)^2-1/8*b^2*ln(c*x^(1/2)-1)*ln(1/2*c*x^(1/2)+1/2)-1/8*b^2*ln(-1/2*c*x^(1/2)+1/2)*ln(1
+c*x^(1/2))+1/8*b^2*ln(-1/2*c*x^(1/2)+1/2)*ln(1/2*c*x^(1/2)+1/2)+1/16*b^2*ln(1+c*x^(1/2))^2+1/12*b^2*c^2*x+1/3
*b^2*ln(c*x^(1/2)-1)+1/3*b^2*ln(1+c*x^(1/2))+1/2*a*b*c^4*x^2*arctanh(c*x^(1/2))+1/6*a*b*c^3*x^(3/2)+1/2*a*b*c*
x^(1/2)+1/4*a*b*ln(c*x^(1/2)-1)-1/4*a*b*ln(1+c*x^(1/2)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (105) = 210\).
time = 0.27, size = 215, normalized size = 1.67 \begin {gather*} \frac {1}{2} \, b^{2} x^{2} \operatorname {artanh}\left (c \sqrt {x}\right )^{2} + \frac {1}{2} \, a^{2} x^{2} + \frac {1}{6} \, {\left (6 \, x^{2} \operatorname {artanh}\left (c \sqrt {x}\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{\frac {3}{2}} + 3 \, \sqrt {x}\right )}}{c^{4}} - \frac {3 \, \log \left (c \sqrt {x} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c \sqrt {x} - 1\right )}{c^{5}}\right )}\right )} a b + \frac {1}{24} \, {\left (4 \, c {\left (\frac {2 \, {\left (c^{2} x^{\frac {3}{2}} + 3 \, \sqrt {x}\right )}}{c^{4}} - \frac {3 \, \log \left (c \sqrt {x} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c \sqrt {x} - 1\right )}{c^{5}}\right )} \operatorname {artanh}\left (c \sqrt {x}\right ) + \frac {4 \, c^{2} x - 2 \, {\left (3 \, \log \left (c \sqrt {x} - 1\right ) - 8\right )} \log \left (c \sqrt {x} + 1\right ) + 3 \, \log \left (c \sqrt {x} + 1\right )^{2} + 3 \, \log \left (c \sqrt {x} - 1\right )^{2} + 16 \, \log \left (c \sqrt {x} - 1\right )}{c^{4}}\right )} b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^2*arctanh(c*sqrt(x))^2 + 1/2*a^2*x^2 + 1/6*(6*x^2*arctanh(c*sqrt(x)) + c*(2*(c^2*x^(3/2) + 3*sqrt(x)
)/c^4 - 3*log(c*sqrt(x) + 1)/c^5 + 3*log(c*sqrt(x) - 1)/c^5))*a*b + 1/24*(4*c*(2*(c^2*x^(3/2) + 3*sqrt(x))/c^4
- 3*log(c*sqrt(x) + 1)/c^5 + 3*log(c*sqrt(x) - 1)/c^5)*arctanh(c*sqrt(x)) + (4*c^2*x - 2*(3*log(c*sqrt(x) - 1
) - 8)*log(c*sqrt(x) + 1) + 3*log(c*sqrt(x) + 1)^2 + 3*log(c*sqrt(x) - 1)^2 + 16*log(c*sqrt(x) - 1))/c^4)*b^2

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Fricas [A]
time = 0.36, size = 207, normalized size = 1.60 \begin {gather*} \frac {12 \, a^{2} c^{4} x^{2} + 4 \, b^{2} c^{2} x + 3 \, {\left (b^{2} c^{4} x^{2} - b^{2}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right )^{2} + 4 \, {\left (3 \, a b c^{4} - 3 \, a b + 4 \, b^{2}\right )} \log \left (c \sqrt {x} + 1\right ) - 4 \, {\left (3 \, a b c^{4} - 3 \, a b - 4 \, b^{2}\right )} \log \left (c \sqrt {x} - 1\right ) + 4 \, {\left (3 \, a b c^{4} x^{2} - 3 \, a b c^{4} + {\left (b^{2} c^{3} x + 3 \, b^{2} c\right )} \sqrt {x}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) + 8 \, {\left (a b c^{3} x + 3 \, a b c\right )} \sqrt {x}}{24 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x^(1/2)))^2,x, algorithm="fricas")

[Out]

1/24*(12*a^2*c^4*x^2 + 4*b^2*c^2*x + 3*(b^2*c^4*x^2 - b^2)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1))^2 + 4*(
3*a*b*c^4 - 3*a*b + 4*b^2)*log(c*sqrt(x) + 1) - 4*(3*a*b*c^4 - 3*a*b - 4*b^2)*log(c*sqrt(x) - 1) + 4*(3*a*b*c^
4*x^2 - 3*a*b*c^4 + (b^2*c^3*x + 3*b^2*c)*sqrt(x))*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) + 8*(a*b*c^3*x
+ 3*a*b*c)*sqrt(x))/c^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 1.28, size = 143, normalized size = 1.11 \begin {gather*} \frac {a^2\,x^2}{2}-\frac {b^2\,{\mathrm {atanh}\left (c\,\sqrt {x}\right )}^2}{2\,c^4}+\frac {2\,b^2\,\ln \left (c^2\,x-1\right )}{3\,c^4}+\frac {b^2\,x^2\,{\mathrm {atanh}\left (c\,\sqrt {x}\right )}^2}{2}+\frac {b^2\,x}{6\,c^2}+\frac {b^2\,x^{3/2}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{3\,c}+\frac {b^2\,\sqrt {x}\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{c^3}+\frac {a\,b\,x^{3/2}}{3\,c}+\frac {a\,b\,\sqrt {x}}{c^3}-\frac {a\,b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{c^4}+a\,b\,x^2\,\mathrm {atanh}\left (c\,\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atanh(c*x^(1/2)))^2,x)

[Out]

(a^2*x^2)/2 - (b^2*atanh(c*x^(1/2))^2)/(2*c^4) + (2*b^2*log(c^2*x - 1))/(3*c^4) + (b^2*x^2*atanh(c*x^(1/2))^2)
/2 + (b^2*x)/(6*c^2) + (b^2*x^(3/2)*atanh(c*x^(1/2)))/(3*c) + (b^2*x^(1/2)*atanh(c*x^(1/2)))/c^3 + (a*b*x^(3/2
))/(3*c) + (a*b*x^(1/2))/c^3 - (a*b*atanh(c*x^(1/2)))/c^4 + a*b*x^2*atanh(c*x^(1/2))

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