3.2.95 \(\int x^3 (a+b \tanh ^{-1}(c \sqrt {x}))^2 \, dx\) [195]

Optimal. Leaf size=211 \[ \frac {a b \sqrt {x}}{2 c^7}+\frac {71 b^2 x}{420 c^6}+\frac {3 b^2 x^2}{70 c^4}+\frac {b^2 x^3}{84 c^2}+\frac {b^2 \sqrt {x} \tanh ^{-1}\left (c \sqrt {x}\right )}{2 c^7}+\frac {b x^{3/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{6 c^5}+\frac {b x^{5/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{10 c^3}+\frac {b x^{7/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{14 c}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{4 c^8}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+\frac {44 b^2 \log \left (1-c^2 x\right )}{105 c^8} \]

[Out]

71/420*b^2*x/c^6+3/70*b^2*x^2/c^4+1/84*b^2*x^3/c^2+1/6*b*x^(3/2)*(a+b*arctanh(c*x^(1/2)))/c^5+1/10*b*x^(5/2)*(
a+b*arctanh(c*x^(1/2)))/c^3+1/14*b*x^(7/2)*(a+b*arctanh(c*x^(1/2)))/c-1/4*(a+b*arctanh(c*x^(1/2)))^2/c^8+1/4*x
^4*(a+b*arctanh(c*x^(1/2)))^2+44/105*b^2*ln(-c^2*x+1)/c^8+1/2*a*b*x^(1/2)/c^7+1/2*b^2*arctanh(c*x^(1/2))*x^(1/
2)/c^7

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Rubi [A]
time = 0.37, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, 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}{4 c^8}+\frac {a b \sqrt {x}}{2 c^7}+\frac {b x^{3/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{6 c^5}+\frac {b x^{5/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{10 c^3}+\frac {b x^{7/2} \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{14 c}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+\frac {b^2 \sqrt {x} \tanh ^{-1}\left (c \sqrt {x}\right )}{2 c^7}+\frac {71 b^2 x}{420 c^6}+\frac {3 b^2 x^2}{70 c^4}+\frac {b^2 x^3}{84 c^2}+\frac {44 b^2 \log \left (1-c^2 x\right )}{105 c^8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*Sqrt[x])/(2*c^7) + (71*b^2*x)/(420*c^6) + (3*b^2*x^2)/(70*c^4) + (b^2*x^3)/(84*c^2) + (b^2*Sqrt[x]*ArcTan
h[c*Sqrt[x]])/(2*c^7) + (b*x^(3/2)*(a + b*ArcTanh[c*Sqrt[x]]))/(6*c^5) + (b*x^(5/2)*(a + b*ArcTanh[c*Sqrt[x]])
)/(10*c^3) + (b*x^(7/2)*(a + b*ArcTanh[c*Sqrt[x]]))/(14*c) - (a + b*ArcTanh[c*Sqrt[x]])^2/(4*c^8) + (x^4*(a +
b*ArcTanh[c*Sqrt[x]])^2)/4 + (44*b^2*Log[1 - c^2*x])/(105*c^8)

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^3 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \, dx &=\int x^3 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \, dx\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 224, normalized size = 1.06 \begin {gather*} \frac {210 a b c \sqrt {x}+71 b^2 c^2 x+70 a b c^3 x^{3/2}+18 b^2 c^4 x^2+42 a b c^5 x^{5/2}+5 b^2 c^6 x^3+30 a b c^7 x^{7/2}+105 a^2 c^8 x^4+2 b c \sqrt {x} \left (105 a c^7 x^{7/2}+b \left (105+35 c^2 x+21 c^4 x^2+15 c^6 x^3\right )\right ) \tanh ^{-1}\left (c \sqrt {x}\right )+105 b^2 \left (-1+c^8 x^4\right ) \tanh ^{-1}\left (c \sqrt {x}\right )^2+b (105 a+176 b) \log \left (1-c \sqrt {x}\right )-105 a b \log \left (1+c \sqrt {x}\right )+176 b^2 \log \left (1+c \sqrt {x}\right )}{420 c^8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(210*a*b*c*Sqrt[x] + 71*b^2*c^2*x + 70*a*b*c^3*x^(3/2) + 18*b^2*c^4*x^2 + 42*a*b*c^5*x^(5/2) + 5*b^2*c^6*x^3 +
30*a*b*c^7*x^(7/2) + 105*a^2*c^8*x^4 + 2*b*c*Sqrt[x]*(105*a*c^7*x^(7/2) + b*(105 + 35*c^2*x + 21*c^4*x^2 + 15
*c^6*x^3))*ArcTanh[c*Sqrt[x]] + 105*b^2*(-1 + c^8*x^4)*ArcTanh[c*Sqrt[x]]^2 + b*(105*a + 176*b)*Log[1 - c*Sqrt
[x]] - 105*a*b*Log[1 + c*Sqrt[x]] + 176*b^2*Log[1 + c*Sqrt[x]])/(420*c^8)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(372\) vs. \(2(167)=334\).
time = 0.18, size = 373, normalized size = 1.77

method	result	size
derivativedivides	\(\frac {\frac {71 b^{2} c^{2} x}{420}+\frac {c^{8} x^{4} a^{2}}{4}+\frac {a b \,c^{8} x^{4} \arctanh \left (c \sqrt {x}\right )}{2}-\frac {b^{2} \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{8}+\frac {a b \ln \left (c \sqrt {x}-1\right )}{4}-\frac {a b \ln \left (1+c \sqrt {x}\right )}{4}+\frac {b^{2} c^{6} x^{3}}{84}+\frac {3 b^{2} c^{4} x^{2}}{70}+\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{8}-\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{8}-\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{4}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{4}+\frac {b^{2} \ln \left (1+c \sqrt {x}\right )^{2}}{16}+\frac {44 b^{2} \ln \left (c \sqrt {x}-1\right )}{105}+\frac {44 b^{2} \ln \left (1+c \sqrt {x}\right )}{105}+\frac {b^{2} \ln \left (c \sqrt {x}-1\right )^{2}}{16}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c \sqrt {x}}{2}+\frac {b^{2} c^{8} x^{4} \arctanh \left (c \sqrt {x}\right )^{2}}{4}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c^{7} x^{\frac {7}{2}}}{14}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{10}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{6}+\frac {c^{7} x^{\frac {7}{2}} a b}{14}+\frac {a b \,c^{3} x^{\frac {3}{2}}}{6}+\frac {a b c \sqrt {x}}{2}+\frac {a b \,c^{5} x^{\frac {5}{2}}}{10}}{c^{8}}\)	\(373\)
default	\(\frac {\frac {71 b^{2} c^{2} x}{420}+\frac {c^{8} x^{4} a^{2}}{4}+\frac {a b \,c^{8} x^{4} \arctanh \left (c \sqrt {x}\right )}{2}-\frac {b^{2} \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{8}+\frac {a b \ln \left (c \sqrt {x}-1\right )}{4}-\frac {a b \ln \left (1+c \sqrt {x}\right )}{4}+\frac {b^{2} c^{6} x^{3}}{84}+\frac {3 b^{2} c^{4} x^{2}}{70}+\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{8}-\frac {b^{2} \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{8}-\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{4}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{4}+\frac {b^{2} \ln \left (1+c \sqrt {x}\right )^{2}}{16}+\frac {44 b^{2} \ln \left (c \sqrt {x}-1\right )}{105}+\frac {44 b^{2} \ln \left (1+c \sqrt {x}\right )}{105}+\frac {b^{2} \ln \left (c \sqrt {x}-1\right )^{2}}{16}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c \sqrt {x}}{2}+\frac {b^{2} c^{8} x^{4} \arctanh \left (c \sqrt {x}\right )^{2}}{4}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c^{7} x^{\frac {7}{2}}}{14}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{10}+\frac {b^{2} \arctanh \left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{6}+\frac {c^{7} x^{\frac {7}{2}} a b}{14}+\frac {a b \,c^{3} x^{\frac {3}{2}}}{6}+\frac {a b c \sqrt {x}}{2}+\frac {a b \,c^{5} x^{\frac {5}{2}}}{10}}{c^{8}}\)	\(373\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x^(1/2)))^2,x,method=_RETURNVERBOSE)

[Out]

2/c^8*(71/840*b^2*c^2*x+1/8*c^8*x^4*a^2-1/8*b^2*arctanh(c*x^(1/2))*ln(1+c*x^(1/2))-1/16*b^2*ln(c*x^(1/2)-1)*ln
(1/2*c*x^(1/2)+1/2)+1/8*b^2*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1)+1/8*a*b*ln(c*x^(1/2)-1)-1/8*a*b*ln(1+c*x^(1/2))
+1/168*b^2*c^6*x^3+3/140*b^2*c^4*x^2+1/16*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(-1/2*c*
x^(1/2)+1/2)*ln(1+c*x^(1/2))+1/32*b^2*ln(1+c*x^(1/2))^2+22/105*b^2*ln(c*x^(1/2)-1)+22/105*b^2*ln(1+c*x^(1/2))+
1/32*b^2*ln(c*x^(1/2)-1)^2+1/8*b^2*c^8*x^4*arctanh(c*x^(1/2))^2+1/28*b^2*arctanh(c*x^(1/2))*c^7*x^(7/2)+1/20*b
^2*arctanh(c*x^(1/2))*c^5*x^(5/2)+1/12*b^2*arctanh(c*x^(1/2))*c^3*x^(3/2)+1/4*b^2*arctanh(c*x^(1/2))*c*x^(1/2)
+1/28*c^7*x^(7/2)*a*b+1/12*a*b*c^3*x^(3/2)+1/4*a*b*c*x^(1/2)+1/20*a*b*c^5*x^(5/2)+1/4*a*b*c^8*x^4*arctanh(c*x^
(1/2)))

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Maxima [A]
time = 0.27, size = 265, normalized size = 1.26 \begin {gather*} \frac {1}{4} \, b^{2} x^{4} \operatorname {artanh}\left (c \sqrt {x}\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{420} \, {\left (210 \, x^{4} \operatorname {artanh}\left (c \sqrt {x}\right ) + c {\left (\frac {2 \, {\left (15 \, c^{6} x^{\frac {7}{2}} + 21 \, c^{4} x^{\frac {5}{2}} + 35 \, c^{2} x^{\frac {3}{2}} + 105 \, \sqrt {x}\right )}}{c^{8}} - \frac {105 \, \log \left (c \sqrt {x} + 1\right )}{c^{9}} + \frac {105 \, \log \left (c \sqrt {x} - 1\right )}{c^{9}}\right )}\right )} a b + \frac {1}{1680} \, {\left (4 \, c {\left (\frac {2 \, {\left (15 \, c^{6} x^{\frac {7}{2}} + 21 \, c^{4} x^{\frac {5}{2}} + 35 \, c^{2} x^{\frac {3}{2}} + 105 \, \sqrt {x}\right )}}{c^{8}} - \frac {105 \, \log \left (c \sqrt {x} + 1\right )}{c^{9}} + \frac {105 \, \log \left (c \sqrt {x} - 1\right )}{c^{9}}\right )} \operatorname {artanh}\left (c \sqrt {x}\right ) + \frac {20 \, c^{6} x^{3} + 72 \, c^{4} x^{2} + 284 \, c^{2} x - 2 \, {\left (105 \, \log \left (c \sqrt {x} - 1\right ) - 352\right )} \log \left (c \sqrt {x} + 1\right ) + 105 \, \log \left (c \sqrt {x} + 1\right )^{2} + 105 \, \log \left (c \sqrt {x} - 1\right )^{2} + 704 \, \log \left (c \sqrt {x} - 1\right )}{c^{8}}\right )} b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4*arctanh(c*sqrt(x))^2 + 1/4*a^2*x^4 + 1/420*(210*x^4*arctanh(c*sqrt(x)) + c*(2*(15*c^6*x^(7/2) + 21
*c^4*x^(5/2) + 35*c^2*x^(3/2) + 105*sqrt(x))/c^8 - 105*log(c*sqrt(x) + 1)/c^9 + 105*log(c*sqrt(x) - 1)/c^9))*a
*b + 1/1680*(4*c*(2*(15*c^6*x^(7/2) + 21*c^4*x^(5/2) + 35*c^2*x^(3/2) + 105*sqrt(x))/c^8 - 105*log(c*sqrt(x) +
1)/c^9 + 105*log(c*sqrt(x) - 1)/c^9)*arctanh(c*sqrt(x)) + (20*c^6*x^3 + 72*c^4*x^2 + 284*c^2*x - 2*(105*log(c
*sqrt(x) - 1) - 352)*log(c*sqrt(x) + 1) + 105*log(c*sqrt(x) + 1)^2 + 105*log(c*sqrt(x) - 1)^2 + 704*log(c*sqrt
(x) - 1))/c^8)*b^2

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Fricas [A]
time = 0.39, size = 273, normalized size = 1.29 \begin {gather*} \frac {420 \, a^{2} c^{8} x^{4} + 20 \, b^{2} c^{6} x^{3} + 72 \, b^{2} c^{4} x^{2} + 284 \, b^{2} c^{2} x + 105 \, {\left (b^{2} c^{8} x^{4} - b^{2}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right )^{2} + 4 \, {\left (105 \, a b c^{8} - 105 \, a b + 176 \, b^{2}\right )} \log \left (c \sqrt {x} + 1\right ) - 4 \, {\left (105 \, a b c^{8} - 105 \, a b - 176 \, b^{2}\right )} \log \left (c \sqrt {x} - 1\right ) + 4 \, {\left (105 \, a b c^{8} x^{4} - 105 \, a b c^{8} + {\left (15 \, b^{2} c^{7} x^{3} + 21 \, b^{2} c^{5} x^{2} + 35 \, b^{2} c^{3} x + 105 \, 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 (15 \, a b c^{7} x^{3} + 21 \, a b c^{5} x^{2} + 35 \, a b c^{3} x + 105 \, a b c\right )} \sqrt {x}}{1680 \, c^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1680*(420*a^2*c^8*x^4 + 20*b^2*c^6*x^3 + 72*b^2*c^4*x^2 + 284*b^2*c^2*x + 105*(b^2*c^8*x^4 - b^2)*log(-(c^2*
x + 2*c*sqrt(x) + 1)/(c^2*x - 1))^2 + 4*(105*a*b*c^8 - 105*a*b + 176*b^2)*log(c*sqrt(x) + 1) - 4*(105*a*b*c^8
- 105*a*b - 176*b^2)*log(c*sqrt(x) - 1) + 4*(105*a*b*c^8*x^4 - 105*a*b*c^8 + (15*b^2*c^7*x^3 + 21*b^2*c^5*x^2
+ 35*b^2*c^3*x + 105*b^2*c)*sqrt(x))*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) + 8*(15*a*b*c^7*x^3 + 21*a*b*
c^5*x^2 + 35*a*b*c^3*x + 105*a*b*c)*sqrt(x))/c^8

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \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**3*(a+b*atanh(c*x**(1/2)))**2,x)

[Out]

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

[Out]

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

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Mupad [B]
time = 4.87, size = 453, normalized size = 2.15 \begin {gather*} \frac {a^2\,x^4}{4}+\frac {44\,b^2\,\ln \left (c\,\sqrt {x}-1\right )}{105\,c^8}+\frac {44\,b^2\,\ln \left (c\,\sqrt {x}+1\right )}{105\,c^8}+\frac {71\,b^2\,x}{420\,c^6}-\frac {b^2\,{\ln \left (c\,\sqrt {x}+1\right )}^2}{16\,c^8}-\frac {b^2\,{\ln \left (1-c\,\sqrt {x}\right )}^2}{16\,c^8}+\frac {b^2\,x^3}{84\,c^2}+\frac {3\,b^2\,x^2}{70\,c^4}+\frac {b^2\,x^4\,{\ln \left (c\,\sqrt {x}+1\right )}^2}{16}+\frac {b^2\,x^4\,{\ln \left (1-c\,\sqrt {x}\right )}^2}{16}+\frac {b^2\,x^{7/2}\,\ln \left (c\,\sqrt {x}+1\right )}{28\,c}+\frac {b^2\,x^{5/2}\,\ln \left (c\,\sqrt {x}+1\right )}{20\,c^3}+\frac {b^2\,x^{3/2}\,\ln \left (c\,\sqrt {x}+1\right )}{12\,c^5}+\frac {b^2\,\sqrt {x}\,\ln \left (c\,\sqrt {x}+1\right )}{4\,c^7}-\frac {b^2\,x^{7/2}\,\ln \left (1-c\,\sqrt {x}\right )}{28\,c}-\frac {b^2\,x^{5/2}\,\ln \left (1-c\,\sqrt {x}\right )}{20\,c^3}-\frac {b^2\,x^{3/2}\,\ln \left (1-c\,\sqrt {x}\right )}{12\,c^5}-\frac {b^2\,\sqrt {x}\,\ln \left (1-c\,\sqrt {x}\right )}{4\,c^7}+\frac {a\,b\,\ln \left (c\,\sqrt {x}-1\right )}{4\,c^8}-\frac {a\,b\,\ln \left (c\,\sqrt {x}+1\right )}{4\,c^8}+\frac {a\,b\,x^4\,\ln \left (c\,\sqrt {x}+1\right )}{4}-\frac {a\,b\,x^4\,\ln \left (1-c\,\sqrt {x}\right )}{4}+\frac {b^2\,\ln \left (c\,\sqrt {x}+1\right )\,\ln \left (1-c\,\sqrt {x}\right )}{8\,c^8}+\frac {a\,b\,x^{7/2}}{14\,c}+\frac {a\,b\,x^{5/2}}{10\,c^3}+\frac {a\,b\,x^{3/2}}{6\,c^5}+\frac {a\,b\,\sqrt {x}}{2\,c^7}-\frac {b^2\,x^4\,\ln \left (c\,\sqrt {x}+1\right )\,\ln \left (1-c\,\sqrt {x}\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c*x^(1/2)))^2,x)

[Out]

(a^2*x^4)/4 + (44*b^2*log(c*x^(1/2) - 1))/(105*c^8) + (44*b^2*log(c*x^(1/2) + 1))/(105*c^8) + (71*b^2*x)/(420*
c^6) - (b^2*log(c*x^(1/2) + 1)^2)/(16*c^8) - (b^2*log(1 - c*x^(1/2))^2)/(16*c^8) + (b^2*x^3)/(84*c^2) + (3*b^2
*x^2)/(70*c^4) + (b^2*x^4*log(c*x^(1/2) + 1)^2)/16 + (b^2*x^4*log(1 - c*x^(1/2))^2)/16 + (b^2*x^(7/2)*log(c*x^
(1/2) + 1))/(28*c) + (b^2*x^(5/2)*log(c*x^(1/2) + 1))/(20*c^3) + (b^2*x^(3/2)*log(c*x^(1/2) + 1))/(12*c^5) + (
b^2*x^(1/2)*log(c*x^(1/2) + 1))/(4*c^7) - (b^2*x^(7/2)*log(1 - c*x^(1/2)))/(28*c) - (b^2*x^(5/2)*log(1 - c*x^(
1/2)))/(20*c^3) - (b^2*x^(3/2)*log(1 - c*x^(1/2)))/(12*c^5) - (b^2*x^(1/2)*log(1 - c*x^(1/2)))/(4*c^7) + (a*b*
log(c*x^(1/2) - 1))/(4*c^8) - (a*b*log(c*x^(1/2) + 1))/(4*c^8) + (a*b*x^4*log(c*x^(1/2) + 1))/4 - (a*b*x^4*log
(1 - c*x^(1/2)))/4 + (b^2*log(c*x^(1/2) + 1)*log(1 - c*x^(1/2)))/(8*c^8) + (a*b*x^(7/2))/(14*c) + (a*b*x^(5/2)
)/(10*c^3) + (a*b*x^(3/2))/(6*c^5) + (a*b*x^(1/2))/(2*c^7) - (b^2*x^4*log(c*x^(1/2) + 1)*log(1 - c*x^(1/2)))/8

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