3.130 \(\int x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx\)
Optimal. Leaf size=80 \[ -\frac {32 \tanh ^{-1}(\tanh (a+b x))^{13/2}}{3003 b^4}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{21 b^2}+\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]
[Out]
2/7*x^3*arctanh(tanh(b*x+a))^(7/2)/b-4/21*x^2*arctanh(tanh(b*x+a))^(9/2)/b^2+16/231*x*arctanh(tanh(b*x+a))^(11
/2)/b^3-32/3003*arctanh(tanh(b*x+a))^(13/2)/b^4
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Rubi [A] time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{2168, 2157, 30} \[ -\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{21 b^2}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{13/2}}{3003 b^4}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}+\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]
Antiderivative was successfully verified.
[In]
Int[x^3*ArcTanh[Tanh[a + b*x]]^(5/2),x]
[Out]
(2*x^3*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (4*x^2*ArcTanh[Tanh[a + b*x]]^(9/2))/(21*b^2) + (16*x*ArcTanh[Tan
h[a + b*x]]^(11/2))/(231*b^3) - (32*ArcTanh[Tanh[a + b*x]]^(13/2))/(3003*b^4)
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^3 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {6 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{7 b}\\ &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{21 b^2}+\frac {8 \int x \tanh ^{-1}(\tanh (a+b x))^{9/2} \, dx}{21 b^2}\\ &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{21 b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {16 \int \tanh ^{-1}(\tanh (a+b x))^{11/2} \, dx}{231 b^3}\\ &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{21 b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {16 \operatorname {Subst}\left (\int x^{11/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{231 b^4}\\ &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{21 b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{13/2}}{3003 b^4}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 66, normalized size = 0.82 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \left (-286 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+104 b x \tanh ^{-1}(\tanh (a+b x))^2-16 \tanh ^{-1}(\tanh (a+b x))^3+429 b^3 x^3\right )}{3003 b^4} \]
Antiderivative was successfully verified.
[In]
Integrate[x^3*ArcTanh[Tanh[a + b*x]]^(5/2),x]
[Out]
(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(429*b^3*x^3 - 286*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 104*b*x*ArcTanh[Tanh[a + b
*x]]^2 - 16*ArcTanh[Tanh[a + b*x]]^3))/(3003*b^4)
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fricas [A] time = 0.45, size = 75, normalized size = 0.94 \[ \frac {2 \, {\left (231 \, b^{6} x^{6} + 567 \, a b^{5} x^{5} + 371 \, a^{2} b^{4} x^{4} + 5 \, a^{3} b^{3} x^{3} - 6 \, a^{4} b^{2} x^{2} + 8 \, a^{5} b x - 16 \, a^{6}\right )} \sqrt {b x + a}}{3003 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")
[Out]
2/3003*(231*b^6*x^6 + 567*a*b^5*x^5 + 371*a^2*b^4*x^4 + 5*a^3*b^3*x^3 - 6*a^4*b^2*x^2 + 8*a^5*b*x - 16*a^6)*sq
rt(b*x + a)/b^4
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giac [B] time = 0.13, size = 296, normalized size = 3.70 \[ \frac {\sqrt {2} {\left (\frac {429 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{3}}{b^{3}} + \frac {143 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a^{2}}{b^{3}} + \frac {65 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a}{b^{3}} + \frac {5 \, \sqrt {2} {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )}}{b^{3}}\right )}}{15015 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")
[Out]
1/15015*sqrt(2)*(429*sqrt(2)*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x
+ a)*a^3)*a^3/b^3 + 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^2/b^3 + 65*sqrt(2)*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a +
990*(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/b^3
+ 5*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)/b^3)/b
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maple [A] time = 0.14, size = 124, normalized size = 1.55 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {13}{2}}}{13}+\frac {2 \left (-3 \arctanh \left (\tanh \left (b x +a \right )\right )+3 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )+\left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*arctanh(tanh(b*x+a))^(5/2),x)
[Out]
2/b^4*(1/13*arctanh(tanh(b*x+a))^(13/2)+1/11*(-3*arctanh(tanh(b*x+a))+3*b*x)*arctanh(tanh(b*x+a))^(11/2)+1/9*(
(b*x-arctanh(tanh(b*x+a)))*(-2*arctanh(tanh(b*x+a))+2*b*x)+(b*x-arctanh(tanh(b*x+a)))^2)*arctanh(tanh(b*x+a))^
(9/2)+1/7*(b*x-arctanh(tanh(b*x+a)))^3*arctanh(tanh(b*x+a))^(7/2))
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maxima [A] time = 0.54, size = 53, normalized size = 0.66 \[ \frac {2 \, {\left (231 \, b^{4} x^{4} + 105 \, a b^{3} x^{3} - 70 \, a^{2} b^{2} x^{2} + 40 \, a^{3} b x - 16 \, a^{4}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{3003 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")
[Out]
2/3003*(231*b^4*x^4 + 105*a*b^3*x^3 - 70*a^2*b^2*x^2 + 40*a^3*b*x - 16*a^4)*(b*x + a)^(5/2)/b^4
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mupad [B] time = 1.15, size = 2235, normalized size = 27.94 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*atanh(tanh(a + b*x))^(5/2),x)
[Out]
(2*b^2*x^6*(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 + (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)*((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 - (10*(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) - (24*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))/13)*(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) - (x^5*(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) - (24*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))/13)*(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))/(11*b) - (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 - (8*((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 - (10*(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) - (24*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))/13)*(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))/(9*b)))/(7*b) - (16*(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 - (8*((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 - (10*(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*e
xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (24*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))/13)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*ex
p(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*e
xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(9*b))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*e
xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)^3)/(35*b^4) - (6*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 - (8*((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 - (10*(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) - (24*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))/13)*(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))/(9*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))/(35*b^2) - (8*x*(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 - (8*((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 - (10*(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) - (24*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))/13)*(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))/(9*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)/(35*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**3*atanh(tanh(b*x+a))**(5/2),x)
[Out]
Timed out
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