3.131 \(\int x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx\)
Optimal. Leaf size=59 \[ \frac {16 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{693 b^3}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]
[Out]
2/7*x^2*arctanh(tanh(b*x+a))^(7/2)/b-8/63*x*arctanh(tanh(b*x+a))^(9/2)/b^2+16/693*arctanh(tanh(b*x+a))^(11/2)/
b^3
________________________________________________________________________________________
Rubi [A] time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{2168, 2157, 30} \[ \frac {16 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{693 b^3}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]
Antiderivative was successfully verified.
[In]
Int[x^2*ArcTanh[Tanh[a + b*x]]^(5/2),x]
[Out]
(2*x^2*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (8*x*ArcTanh[Tanh[a + b*x]]^(9/2))/(63*b^2) + (16*ArcTanh[Tanh[a
+ b*x]]^(11/2))/(693*b^3)
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^2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx &=\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {4 \int x \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{7 b}\\ &=\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {8 \int \tanh ^{-1}(\tanh (a+b x))^{9/2} \, dx}{63 b^2}\\ &=\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {8 \operatorname {Subst}\left (\int x^{9/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{63 b^3}\\ &=\frac {2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {16 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{693 b^3}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.04, size = 49, normalized size = 0.83 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \left (-44 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+99 b^2 x^2\right )}{693 b^3} \]
Antiderivative was successfully verified.
[In]
Integrate[x^2*ArcTanh[Tanh[a + b*x]]^(5/2),x]
[Out]
(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(99*b^2*x^2 - 44*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh[a + b*x]]^2))/(69
3*b^3)
________________________________________________________________________________________
fricas [A] time = 0.53, size = 64, normalized size = 1.08 \[ \frac {2 \, {\left (63 \, b^{5} x^{5} + 161 \, a b^{4} x^{4} + 113 \, a^{2} b^{3} x^{3} + 3 \, a^{3} b^{2} x^{2} - 4 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt {b x + a}}{693 \, b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")
[Out]
2/693*(63*b^5*x^5 + 161*a*b^4*x^4 + 113*a^2*b^3*x^3 + 3*a^3*b^2*x^2 - 4*a^4*b*x + 8*a^5)*sqrt(b*x + a)/b^3
________________________________________________________________________________________
giac [B] time = 1.14, size = 248, normalized size = 4.20 \[ \frac {\sqrt {2} {\left (\frac {231 \, \sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{3}}{b^{2}} + \frac {297 \, \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{2}}{b^{2}} + \frac {33 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a}{b^{2}} + \frac {5 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )}}{b^{2}}\right )}}{3465 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")
[Out]
1/3465*sqrt(2)*(231*sqrt(2)*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a^3/b^2 + 297*sq
rt(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^2/b^2 + 33*
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/b^2 + 5*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)/b^2)/b
________________________________________________________________________________________
maple [A] time = 0.14, size = 69, normalized size = 1.17 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}}{b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*arctanh(tanh(b*x+a))^(5/2),x)
[Out]
2/b^3*(1/11*arctanh(tanh(b*x+a))^(11/2)+1/9*(-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)))^2*arctanh(tanh(b*x+a))^(7/2))
________________________________________________________________________________________
maxima [A] time = 0.54, size = 42, normalized size = 0.71 \[ \frac {2 \, {\left (63 \, b^{3} x^{3} + 35 \, a b^{2} x^{2} - 20 \, a^{2} b x + 8 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{693 \, b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")
[Out]
2/693*(63*b^3*x^3 + 35*a*b^2*x^2 - 20*a^2*b*x + 8*a^3)*(b*x + a)^(5/2)/b^3
________________________________________________________________________________________
mupad [B] time = 1.14, size = 1789, normalized size = 30.32 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*atanh(tanh(a + b*x))^(5/2),x)
[Out]
(2*b^2*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))/11 + (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)*((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 - (8*(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) - (20*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))/11)*(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) - (x^4*(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) - (20*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))/11)*(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))/(9*b) - (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 - (6*((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 - (8*(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) - (20*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))/11)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(e
xp(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))/(e
xp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(7*b)))/(5*b) - (8*(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 - (6*((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 - (8*(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) - (20*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))/11)*(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))/(7*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)/(15*b^3) - (4*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 - (6*((3*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/2 - (8*(3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - lo
g((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x) - (20*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))/11)*(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))/(7*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))/(15*b^2)
________________________________________________________________________________________
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**2*atanh(tanh(b*x+a))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________