3.129 \(\int x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx\)
Optimal. Leaf size=101 \[ \frac {256 \tanh ^{-1}(\tanh (a+b x))^{15/2}}{45045 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{13/2}}{3003 b^4}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]
[Out]
2/7*x^4*arctanh(tanh(b*x+a))^(7/2)/b-16/63*x^3*arctanh(tanh(b*x+a))^(9/2)/b^2+32/231*x^2*arctanh(tanh(b*x+a))^
(11/2)/b^3-128/3003*x*arctanh(tanh(b*x+a))^(13/2)/b^4+256/45045*arctanh(tanh(b*x+a))^(15/2)/b^5
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Rubi [A] time = 0.06, antiderivative size = 101, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{2168, 2157, 30} \[ \frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{15/2}}{45045 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{13/2}}{3003 b^4}+\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]
Antiderivative was successfully verified.
[In]
Int[x^4*ArcTanh[Tanh[a + b*x]]^(5/2),x]
[Out]
(2*x^4*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (16*x^3*ArcTanh[Tanh[a + b*x]]^(9/2))/(63*b^2) + (32*x^2*ArcTanh[
Tanh[a + b*x]]^(11/2))/(231*b^3) - (128*x*ArcTanh[Tanh[a + b*x]]^(13/2))/(3003*b^4) + (256*ArcTanh[Tanh[a + b*
x]]^(15/2))/(45045*b^5)
Rule 30
Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]
Rule 2157
Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]
Rule 2168
Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] && !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] && !IntegerQ[m]
) || (ILtQ[m, 0] && !IntegerQ[n]))
Rubi steps
\begin {align*} \int x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 \int x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{7 b}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {16 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2} \, dx}{21 b^2}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {64 \int x \tanh ^{-1}(\tanh (a+b x))^{11/2} \, dx}{231 b^3}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{13/2}}{3003 b^4}+\frac {128 \int \tanh ^{-1}(\tanh (a+b x))^{13/2} \, dx}{3003 b^4}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{13/2}}{3003 b^4}+\frac {128 \operatorname {Subst}\left (\int x^{13/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{3003 b^5}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{13/2}}{3003 b^4}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{15/2}}{45045 b^5}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 83, normalized size = 0.82 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \left (-5720 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+3120 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-960 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+6435 b^4 x^4\right )}{45045 b^5} \]
Antiderivative was successfully verified.
[In]
Integrate[x^4*ArcTanh[Tanh[a + b*x]]^(5/2),x]
[Out]
(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(6435*b^4*x^4 - 5720*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 3120*b^2*x^2*ArcTanh[Tan
h[a + b*x]]^2 - 960*b*x*ArcTanh[Tanh[a + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(45045*b^5)
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fricas [A] time = 0.52, size = 86, normalized size = 0.85 \[ \frac {2 \, {\left (3003 \, b^{7} x^{7} + 7161 \, a b^{6} x^{6} + 4473 \, a^{2} b^{5} x^{5} + 35 \, a^{3} b^{4} x^{4} - 40 \, a^{4} b^{3} x^{3} + 48 \, a^{5} b^{2} x^{2} - 64 \, a^{6} b x + 128 \, a^{7}\right )} \sqrt {b x + a}}{45045 \, b^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")
[Out]
2/45045*(3003*b^7*x^7 + 7161*a*b^6*x^6 + 4473*a^2*b^5*x^5 + 35*a^3*b^4*x^4 - 40*a^4*b^3*x^3 + 48*a^5*b^2*x^2 -
64*a^6*b*x + 128*a^7)*sqrt(b*x + a)/b^5
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giac [B] time = 0.17, size = 344, normalized size = 3.41 \[ \frac {\sqrt {2} {\left (\frac {143 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a^{3}}{b^{4}} + \frac {195 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a^{2}}{b^{4}} + \frac {45 \, \sqrt {2} {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )} a}{b^{4}} + \frac {7 \, \sqrt {2} {\left (429 \, {\left (b x + a\right )}^{\frac {15}{2}} - 3465 \, {\left (b x + a\right )}^{\frac {13}{2}} a + 12285 \, {\left (b x + a\right )}^{\frac {11}{2}} a^{2} - 25025 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{3} + 32175 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{4} - 27027 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} + 15015 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{6} - 6435 \, \sqrt {b x + a} a^{7}\right )}}{b^{4}}\right )}}{45045 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")
[Out]
1/45045*sqrt(2)*(143*sqrt(2)*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x
+ a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*a^3/b^4 + 195*sqrt(2)*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 9
90*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*a^2/b^4
+ 45*sqrt(2)*(231*(b*x + a)^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2)
*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003*sqrt(b*x + a)*a^6)*a/b^4 + 7*sqrt(2)*(429*(b
*x + a)^(15/2) - 3465*(b*x + a)^(13/2)*a + 12285*(b*x + a)^(11/2)*a^2 - 25025*(b*x + a)^(9/2)*a^3 + 32175*(b*x
+ a)^(7/2)*a^4 - 27027*(b*x + a)^(5/2)*a^5 + 15015*(b*x + a)^(3/2)*a^6 - 6435*sqrt(b*x + a)*a^7)/b^4)/b
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maple [A] time = 0.14, size = 154, normalized size = 1.52 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {15}{2}}}{15}+\frac {2 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {13}{2}}}{13}+\frac {2 \left (2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2}+\left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )^{2}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {4 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2} \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}}{b^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^4*arctanh(tanh(b*x+a))^(5/2),x)
[Out]
2/b^5*(1/15*arctanh(tanh(b*x+a))^(15/2)+1/13*(-4*arctanh(tanh(b*x+a))+4*b*x)*arctanh(tanh(b*x+a))^(13/2)+1/11*
(2*(b*x-arctanh(tanh(b*x+a)))^2+(-2*arctanh(tanh(b*x+a))+2*b*x)^2)*arctanh(tanh(b*x+a))^(11/2)+2/9*(b*x-arctan
h(tanh(b*x+a)))^2*(-2*arctanh(tanh(b*x+a))+2*b*x)*arctanh(tanh(b*x+a))^(9/2)+1/7*(b*x-arctanh(tanh(b*x+a)))^4*
arctanh(tanh(b*x+a))^(7/2))
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maxima [A] time = 0.53, size = 64, normalized size = 0.63 \[ \frac {2 \, {\left (3003 \, b^{5} x^{5} + 1155 \, a b^{4} x^{4} - 840 \, a^{2} b^{3} x^{3} + 560 \, a^{3} b^{2} x^{2} - 320 \, a^{4} b x + 128 \, a^{5}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{45045 \, b^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")
[Out]
2/45045*(3003*b^5*x^5 + 1155*a*b^4*x^4 - 840*a^2*b^3*x^3 + 560*a^3*b^2*x^2 - 320*a^4*b*x + 128*a^5)*(b*x + a)^
(5/2)/b^5
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mupad [B] time = 1.28, size = 2681, normalized size = 26.54 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^4*atanh(tanh(a + b*x))^(5/2),x)
[Out]
(2*b^2*x^7*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/
2))/15 + (x^5*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^
(1/2)*((3*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x
)^2)/2 - (12*(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))
+ 2*b*x) - (28*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1
))/2 + b*x))/15)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/
2 + b*x))/(13*b)))/(11*b) - (x^6*(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2
*a)*exp(2*b*x) + 1)) + 2*b*x) - (28*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp
(2*a)*exp(2*b*x) + 1))/2 + b*x))/15)*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*
a)*exp(2*b*x) + 1))/2)^(1/2))/(13*b) - (x^4*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/
(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*((log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)
*exp(2*b*x) + 1)) + 2*b*x)^3/4 - (10*((3*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/2 - (12*(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b
*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (28*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2
*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/15)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*
x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(13*b))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b
*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(11*b)))/(9*b) - (128*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b
*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*((log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*e
xp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3/4 - (10*((3*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(
2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/2 - (12*(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log
((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (28*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - l
og((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/15)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log(
(2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(13*b))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log
((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(11*b))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - lo
g((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)^4)/(315*b^5) - (8*x^3*(log((2*exp(2*a)*exp(2*b*x
))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*((log(2/(exp(2*a)*exp(2*b*x) + 1))
- log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3/4 - (10*((3*b*(log(2/(exp(2*a)*exp(2*b*x)
+ 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/2 - (12*(3*b^2*(log(2/(exp(2*a)*ex
p(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (28*b^2*(log(2/(exp(2*a)*ex
p(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/15)*(log(2/(exp(2*a)*exp(2
*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(13*b))*(log(2/(exp(2*a)*exp(
2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(11*b))*(log(2/(exp(2*a)*exp
(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(63*b^2) - (64*x*(log((2*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*((log(2/(exp(2*a)*
exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3/4 - (10*((3*b*(log(2/(exp
(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/2 - (12*(3*b^2*(lo
g(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (28*b^2*(lo
g(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/15)*(log(2
/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(13*b))*(log(
2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(11*b))*(log
(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)^3)/(315*b^4)
- (16*x^2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/
2)*((log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3/4 -
(10*((3*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^
2)/2 - (12*(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) +
2*b*x) - (28*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))
/2 + b*x))/15)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2
+ b*x))/(13*b))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2
+ b*x))/(11*b))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/
2 + b*x)^2)/(105*b^3)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**4*atanh(tanh(b*x+a))**(5/2),x)
[Out]
Timed out
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